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Maharashtra Board Class 12 Chemistry Important Questions Chapter 6 Chemical Kinetics

July 23, 2024July 22, 2024 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 6 Chemical Kinetics Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 6 Chemical Kinetics

Question 1.
What is chemical kinetics?
Answer:
Chemical kinetics is a branch of physical chemistry which involves the study of the rates and mechanisms of chemical reactions and the influence of various factors like temperature, pressure, catalyst, etc., on the rates of reactions.

Question 2.
What is the importance of chemical kinetics?
Answer:

It deals with the study of the rates and mechanism of reactions.
The effect of temperature on the reaction rates can be studied.
The influence of catalysts can be studied.
The conditions for altering the rates and mechanisms of chemical reactions can be predicted.
Thermodynamic parameters like energy, enthalpy changes, Δ5, ΔG of the reactions can be calculated.

Question 3.
How are reactions classified according to their rates? Give one example of each.
Answer:
According to the rates of the reactions, they can be classified as :
(1) Fast reactions,
(2) Very slow reactions,
(3) Moderately slow reactions.

(1) Fitst actions : In this, reactants react almost instantaneously, e.g., neutralisation reaction between H+ and OH-, forming water.
\(\mathrm{H}_{(\mathrm{xa})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{0 \mathrm{D}}\)

(2) Very slow reactions : In this, the reactants react extremely slow, so that there is no appreciable change in the concentrations of the reactants over a long period of time. E.g., reaction of silica with mineral acids, rusting of iron, etc.

(3) Moderately slow reactions : In this, the reactants react moderately slow with a measurable velocity, e.g., the hydrolysis of the esters.
\(\begin{aligned}
\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{COOH} \\
&+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}
\end{aligned}\)

Question 4.
Define rate of a reaction.
Answer:
Definition : The rate of a chemical reaction is defined as the change in the concentration of the reactants or products per unit time.

It is often expressed in mol dm^-3s^-1.

Question 5.
Explain the following :
(A) Rate of the reaction in terms of the concentration of the reactants.
(B) Rate of reaction in terms of the concentration of the products.
Answer:
(A) Rate of the reaction in terms of the concentration of the reactants :
If c₁ and c₂ are the concentrations of the reactant A at time t₁ and t₂ respectively, then, the change in concentration, Δc = c₂ – c₁
Since c₂ < c₁, the term Δc is negative often written as – Δc.
The time interval is, Δt – t₂ – t₁
If Δ [A] is the change in concentration of A, then A[A] = C₂– C₁
∴ Rate of the reaction = \(\mathrm{A}=\frac{-\Delta[\mathrm{A}]}{\Delta t}\)
∴ Rate of the reaction = \(\frac{-\Delta c}{\Delta t}\)

(B) Rate of the reaction in terms of the concentration of the products :
If x₁ and x₂ are the concentrations of the product B at time t₁ and t₂ respectively, then the change in concentration, Δx = x₂ – x₁.

∴ x₂ > x₁, the term Δx is positive.
The time interval is, Δt = t₂ – t₁

If Δ B is the change in concentration of product B, then Δ[B] = x₂ – x₁ = Δx
∴ Rate of formation of \(\mathrm{B}=+\frac{\Delta[\mathrm{B}]}{\Delta t}\)
∴ Rate of the reaction \(=\frac{\Delta x}{\Delta t}\)

Question 6.
What are the units of rate of a chemical reaction?
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 3
∴ The unit of the rate of a chemical reaction : mol L^-1 3t^-1 or mol dm^-3s^-1 (According to IUPAC, the rate of a chemical reaction should be expressed in mol m^-3s^-1 [SI unit]).

Question 7.
Mention the factors that affect the rate of a chemical reaction.
Answer:
The rate of a chemical reaction depends on the following factors :

Nature of the reactants.
The concentration of the reactants. In case of a gaseous reaction the rate depends on the pressures of the reactants.
Temperature of the reaction.
The presence of a catalyst and its nature.

Question 8.
Explain the term Average rate of a reaction.
Answer:
In chemical kinetics the rate of a reaction is measured in terms of the changes in the concentrations of the reactants or the products per unit time. Average rate of a chemical reaction : It is expressed as a finite change in concentration (- Δc) of the reactant divided by the time interval (Δt) for the change in concentration.

Consider a reaction,
A → B
The rate of a reaction, \(R=\frac{-\Delta[\mathrm{A}]}{\Delta t}=\frac{-\Delta c}{\Delta t}=\frac{c_{2}-c_{1}}{t_{2}-t_{1}}\)

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 5
∴ Average rate \(=\frac{-\Delta[\mathrm{A}]}{\Delta t}\) (in mol dm^-3s^-1)

Δc is negative, since the concentrartion of the reactant decreases with the time.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 6
The rate of a reaction is also measured in terms of a finite change in the concentration (Δx) of the product divided by the time interval (Δt), for the change.

For the reaction,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 7

Question 9.
Explain the term Instantaneous rate of a reaction.
Answer:
Instantaneous rate of a reaction : It is defined as a rate of a reaction at a specific instant during a course of the reaction.

If the average reaction rate is calculated over shorter and shorter intervals (making Δt very small) then instantaneous rate is obtained.

In case of reactant A, the instantaneous rate is represented as, \(R=\frac{-d[\mathrm{~A}]}{d t}\) and in case of product B, it is represented as \(R=\frac{+d[B]}{d t}\)

Question 10.
Define :
(a) Average rate of reaction.
(b) Instantaneous rate of reaction.
Answer:
(a) Average rate of a chemical reaction : It is expressed as a finite change in concentration (- Δc) of the reactant divided by the time interval (Δt) for the change in concentration.

∴ Average rate, \(R=\frac{-\Delta c}{\Delta t}\)

(b) Instantaneous rate of reaction : It is defined as a rate of a reaction at a specific instant during a course of the reaction.

Instantaneous rate \(=\frac{-d c}{d t}\)

Question 11.
Represent the average rates of the following reaction. N_2(g) + 3H_2(g) → 2NH_3(g).
Answer:
For the reation,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 12
This is because the rate of consumption of H₂ is thrice the rate of consumption of N₂ while the rate of formation of NH₃ will be twice the rate of consumption of N₂.

Question 12.
Express the rate of a reaction in terms of change in concentration of each constituent in the following reaction : aA+bB → cC+ dD
Answer:
The rate of a reaction may be expressed in terms of decrease in the concentration of the reactants or in-crease in the concentration of the product per unit time,

∴ For the given reaction, aA T bB → cC +dD
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 13

Question 13.
For a hypothetical reaction, A + 2B → products, the concentration of A and B at different intervals of time are given in the following table. Find the rates of the reaction in terms of concentration changes in A and B.

The equilibrium concentration of A and B at different time intervals :

Time t/minute	[A]/mol L^-1	[B]/ml L^-1
0	1.000	2.000
10	0.534	1.068
20	0.342	0.360
30	0.180	0.360

Answer:
Rate of a reaction = \(\frac{-\Delta[\mathrm{A}]}{\Delta t}=-\frac{1}{2} \frac{\Delta[\mathrm{B}]}{\Delta t}\)
(1) Over time interval from O to 10 minutes
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 14
(Note that the rate of a reaction in terms of changes in concentration of any reactant or product at the given time remains the same.)

(2) Over the time interval from 10 to 20 minutes,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 15

Question 14.
Show that the rate of reaction is the same whether expressed in terms of the rate of consumption of any reactant or of the formation of any product.
2N₂O_5(g) → 4NO_2(g)+ O_2(g)
The concentrations of reactants and products at different time intervals are given in the following table :
Concentrations of various species at different times for the reaction N₂O_5(g) → 4NO_2(g) + O_2(g):

Time/s	[N₂O₅]/M	[NO₂]/M	[O₂]/M
0	0.0300	0	0
200	0.0213	0.0174	0.00435
400	0.0152	0.0296	0.00740
600	0.0108	0.0384	0.00960

Answer:
The rate of the reaction can be expressed in terms of rate of consumption of reactants or rate of formation of products.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 16
Consider concentrations at time t₁ = 200 seconds and t₂ = 400 seconds

The constant values of rate of reaction proves that the rate of the reaction may be measured in terms of concentration changes of reactants or products per unit time.

Question 15.
Define Rate law (or differential rate law).
Answer:
Rate law (or differential rate law) : It is defined as an experimentally determined mathematical equation which expresses the rate of a chemical reaction in terms of molar concentrations of the reactants which influence the rate of the reaction. For example, for a reaction, A + B → Products By rate law, Rate = R = k[A] x [B] where k is a rate constant and [Al and [B] are molar concentrations of the reactants A and B respectively.

Question 16.
Give examples of rate law with illustrations.
Answer:
Consider following examples :
(i) H_2(g) + I_2(g) → 2HI_(g)
R = k[H₂] [I₂]

(ii) 2H₂O_2(g) → 2H₂O_(I) + O_2(g)
Experimentally it is observed that the rate of the reaction is proportional to the concentration of H₂O₂.
∴ R = k [H₂O₂]

(iii) NO_2(g) + CO_(g) → NO_(g) + CO_2(g)
Experimentally it is observed that rate of the reaction does not depend on the concentration of CO but it is proportional to [NO₂]².
∴ R = k[NO₂]²

Question 17.
What are the applications of the rate law?
Answer:

The rate of any reaction at the given concentration can be measured by knowing the rate law and the rate constant.
The concentration of the reactants or the products at any instant during the progress of a reaction can be estimated with the help of rate law and the rate constant.
The mechanisms of simple or complex chemical reactions can be predicted and studied.

Question 18.
Define the rate constant. What are the factors which influence the rate constant of a chemical reaction?
Answer:
(A) Rate constant : The rate constant of a chemical reaction is defined as the rate of the chemical reaction when the concentration (or active masses) of each reactant has unit value, i.e., 1 mol dm^-3 in the case of solution and the pressure is 1 atm in case of gases, e.g., for a reaction, A → products, Rate R = k[A].

If [A] = 1 mol dm^-3, then k = R.

(B) The rate constant of a reaction depends on the following factors:

Nature of the reactants.
Temperature of the reaction. As the temperature increases, the velocity constant (rate constant) increases.
The conditions of the reactions like the presence of the catalyst, solvent, pH, etc.
It does not depend on the concentration of the reactants. But if one or more substances are in excess concentration, then the order of the reaction is independent of them.

Question 19.
What are the characteristics of rate constant?
Answer:
The characteristics of rate constant are as follows :

The rate constant depends upon the nature of the reaction.
Higher the value of the rate constant, faster is the reaction.
Lower the value of the rate constant, slower is the reaction.
By increasing the temperature, the magnitude of the rate constant increases.
For the given reaction, the rate constant has higher value in the presence of a catalyst than in the absence of the catalyst.
The reactions having lower activation energy have higher values for rate constants.

Solved Examples 6.2 – 6.3.2

Question 20.
Solve the following :

(1) Write the rate expressions for the following reactions in terms of rate of consumption of the reactants and the rate of formation of the products.
(i) 2NO_(g) + O_2(g) → 2NO_2(g)
(ii) H_2(g) + I_2(g) → 2HI_(g)
Solution :
(i) Given : 2NO_(g) + O_2(g) → 2NO_2(g)
Rate of consumption of NO at time \(t=\frac{-d[\mathrm{NO}]}{d t}\)
Rate of consumption of O₂ at time \(t=\frac{-d\left[\mathrm{O}_{2}\right]}{d t}\)
Rate of formation of NO₂ at time \(t=\frac{d\left[\mathrm{NO}_{2}\right]}{d t}\)
Rate of the reaction \(=-\frac{1}{2} \frac{d[\mathrm{NO}]}{d t}=\frac{-d\left[\mathrm{O}_{2}\right]}{d t}\)
\(=\frac{1}{2} \frac{d\left[\mathrm{NO}_{2}\right]}{d t}\)

(ii) Given : H_2(g) + I_2(g) → 2HI_(g)
Rate of consumption of H₂ at time \(t=\frac{-d\left[\mathrm{H}_{2}\right]}{d t}\)
Rate of consumption of I₂ at time \(t=\frac{-d\left[\mathrm{I}_{2}\right]}{d t}\)
Rate of formation of HI at time \(t=\frac{d[\mathrm{HI}]}{d t}\)
∴ Rate of reaction at any time t \(=-\frac{d\left[\mathrm{H}_{2}\right]}{d t}=-\frac{d\left[\mathrm{I}_{2}\right]}{d t}=\frac{1}{2} \frac{d[\mathrm{HI}]}{d t}\)

(2) The gas-phase reaction between NO and Br₂ is represented by the equation. 2NO_(g) + Br_2(g) → 2NOBr_(g)
(a) Write the expressions for the rate of consumption of reactants and formation of products.
(b) Write the expression for the rate of overall reaction in terms of rates of consumption of reactants and formation of products.
Solution :
Given : 2NO_(g) + Br_2(g) → 2NOBr_(g)
(a) Rate of consumption of NO at time t \(=-\frac{d[\mathrm{NO}]}{d t}\)
Rate of consumption of Br₂ at time t \(=\frac{-d\left[\mathrm{Br}_{2}\right]}{d t}\)
Rate of formation of NOBr at time \(t=\frac{d[\mathrm{NOBr}]}{d t}\)
(b) Rate of reaction \(=-\frac{1}{2} \frac{d[\mathrm{NO}]}{d t}=\frac{-d\left[\mathrm{Br}_{2}\right]}{d t}\)
\(=\frac{1}{2} \frac{d[\mathrm{NOBr}]}{d t}\)

(3) The decomposition of N2Os is represented by the equation
2N₂O_5(g) → 4NO_2(g) + O_2(g)
(a) How is the rate of formation of NO₂ related to the rate of formation of O₂?
(b) How is the rate of formation of O₂ related to the rate of consumption of N₂O₅?
Solution :
Given : 2N₂O_5(g) → 4NO_2(g) + O_2(g)
(a) Rate of formation of NO₂ at time \(t=\frac{d\left[\mathrm{NO}_{2}\right]}{d t}\)
Rate of formation of O₂ at time \(t=\frac{d\left[\mathrm{O}_{2}\right]}{d t}\)

They are related to each other through rate of reaction.
∴ Rate of reaction \(=\frac{1}{4} \frac{d\left[\mathrm{NO}_{2}\right]}{d t}=\frac{d\left[\mathrm{O}_{2}\right]}{d t}\)

(b) Rate of consumption of N₂O₅ at time t \(=-\frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}\)

Rate of reaction \(=-\frac{1}{2} \frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}=\frac{d\left[\mathrm{O}_{2}\right]}{d t}\)

In general,
Rate of reaction \(=-\frac{1}{2} \frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}=\frac{1}{4} \frac{d\left[\mathrm{NO}_{2}\right]}{d t}=\frac{d\left[\mathrm{O}_{2}\right]}{d t}\)

(4) Nitric oxide reacts with H₂ according to the reaction. 2NO_(g) + 2H_2(g) → N_2(g) + 2H₂O_(g)
What is the relationship among \(\frac{d[\mathrm{NO}]}{d t}=\frac{d\left[\mathrm{H}_{2}\right]}{d t}=\frac{d\left[\mathrm{~N}_{2}\right]}{d t} \text { and } \frac{d\left[\mathrm{H}_{2} \mathrm{O}\right]}{d t} ?\)
Solution :
Given : 2NO_(g) + 2H_2(g) → N_2(g) + 2H₂O_(g)The relationship among the rate of consumption of the reactants and the rate of formation of products is as follows :

Rate of reaction :
\(R=-\frac{1}{2} \frac{d[\mathrm{NO}]}{d t}=-\frac{1}{2} \frac{d\left[\mathrm{H}_{2}\right]}{d t}=\frac{d\left[\mathrm{~N}_{2}\right]}{d t}=\frac{1}{2} \frac{d\left[\mathrm{H}_{2} \mathrm{O}\right]}{d t}\)

(5) The rate of decomposition of N2Os was studied in liquid bromine,
2N₂O_5(g) → 4NO_2(g) + O_2(g)
If at a certain time, the rate of disappearance of N₂O₅ is 0.015 Ms^-1 find the rates of formation of NO₂ and O₂. What is the rate of the reaction at this instant?
Solution :
Given : 2N₂O_5(g) → 4NO_2(g) + O_2(g)
Rate of disappearance of N₂O₅ = 0.015 M s^-1
Rate of formation of NO₂ =?
Rate of formation of O₂ =?
Rate of reaction = ?
Rate of disappearance of \(\mathrm{N}_{2} \mathrm{O}_{5}=\frac{-d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}\)
= 0.015 M s^-1

Since 4 moles of NO₂ are formed from 2 moles of N₂O₅ Rate of formation of NO₂Answer:
Rate of formation of NO₂ = 0.03 Ms^-1
Rate of formation of O₂ = 0.0075 M s^-1
Rate of reaction = 0.0075 Ms^-1.

(6) In the reaction, PCl_5(g) → PCl_3(g) + CI_2(g), at a particular moment, the rate of disappearance of PCl₅ is 0.015 Ms^-1. What are the rates of formation of PCI₃ and Cl₂?
Solution :
Given : PCl_5(g) → PCl_3(g) + Cl_2(g)
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 22

Answer:
Rate of formation of PCl₃ = 0.015 Ms^-1
Rate of formation of Cl₂ = 0.015 Ms^-1

(7) In the reaction, 2N₃O_5(g) → 4NO_2(g) + O_2(g), at a certain time, the rate of formation of NO₂ is 0. 04 Ms^-1. Find the rate of consumption of N₂O₅, rate of formation of O₂ and the rate of the reaction.
Solution :
Given : 2N₂O_5(g) → 4NO_2(g) + O_2(g)
Rate of formation of NO₂ = \(\frac{d\left[\mathrm{NO}_{2}\right]}{d t}\) = 0.04 Ms^-1

From the reaction, rate of consumption of N₂O₅ is half the rate of formation of NO₂ since when 2 moles of N₂O₅ are consumed, 4 moles of NO₂ are formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 24
Rate of formation of O₂ is one-fourth rate of formation of NO₂.

Answer:
(i) Rate of consumption of N₂O₅
(ii) Rate of formation of O₂ = 0.01 Ms^-1
(iii) Rate of reaction = 0.01 Ms^-1

(8) Consider the reaction 2A + B → 2C. Suppose that at a particular moment during the reaction, rate of disappearance of A is 0.076 M/s,
(a) What is the rate of formation of C?
(b) What is the rate of consumption of B?
(c) What is the rate of the reaction?
Solution :
Given : 2A + B → 2C
Rate of disappearance of A = 0.076 Ms^-1
(a) Rate of formation of C =?
(b) Rate of consumption of B =?
(c) Rate of reaction = ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 26
Answer:
(a) Rate of formation of C = 0.076 Ms^-1
(b) Rate of consumption of B = 0.038 M s^-1
(c) Rate of reaction = 0.038 Ms^-1

(9) Consider the reation \(\mathbf{3 I}_{(\mathbf{a q})}^{-}+\mathbf{S}_{2} \mathbf{O}_{8(u q)}^{2-} \longrightarrow \mathbf{I}_{3(\mathrm{aq})}^{-}+2 \mathrm{SO}_{4}^{2-}\) At a particular time t, \(t, \frac{d\left[\mathrm{SO}_{4}^{2-}\right]}{d t}=2.2 \times 10^{-2} \mathrm{M} / \mathrm{s}\) What are the values of \(\text { (a) }-\frac{d\left[\mathrm{I}^{-}\right]}{d t}\) \(-\frac{d\left[\mathrm{~S}_{2} \mathrm{O}_{8}^{2-}\right]}{d t}\) \(\text { (c) } \frac{d\left[\mathbf{I}_{3}^{-}\right]}{d t}\) at the same time?
Solution :
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 27
(a) Rate of consumption of \(\mathrm{I}^{-}=-\frac{d\left[\mathrm{I}^{-}\right]}{d t}\)
When 2 moIes of \(\mathrm{SO}_{4}^{2-}\) are formed, 3 moves of I^– are consumed in the same time.

(b) In the formation of 2 moles of \(\mathrm{SO}_{4}^{2-}\), 1 mole of \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) is consumed in the same time.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 29
Answer:

(10) Ammonia and oxygen react at high temperature as :
4NH_3(g) + 5O_2(g) → 4NO_(g) + 6H₂O_(g)
In an experiment, rate of formation of NO_(g) is 3.6 x 10^-3 mol L^-1s^-1.
Calculate-
(a) Rate of disappearance of ammonia
(b) Rate of formation of water.
Solution :
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 31
Answer:
(a) Rate of disappearance of NH₃
= 3.6 x 10^-3 mol L^-1s^-1
(b) Rate of formation of H₂O
= 5.4 x 10^-3 mol L^-1s^-1

(11) The rate law for the reaction
C₂H₄Br₂ + 3I^– → C₂H₄ + 2Br^– +I^–₃is Rate = k [C₂H₄Br₂][I^–]. The rate of the reac-tion is found to be 1.1 x 10^-4 M/s when the concentrations of C₂H₄Br₂ and I^–– are 0.12M and 0.18 M respectively. Calculate the rate constant of the reaction.
Solution :
Given : C₂H₄Br₂ + 3I^– → C₂H₄ + 2Br^– +I^–₃
By rate law, Rate of reaction = R = k x [C₂H₄Br₂][I^–]
R = 1.1 x 10^-4 Ms^-1
[C₂H₄Br₂] = 0.12 M; [I^–] =0.18 M
Rate constant = k =?
R = k x [C₂H₄Br₂] x [I^–]
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 33
Answer:
Rate constant = k = 5.1 x 10^-3 M^-1s^-1

(12) For a reaction, 2A + B → C, the rate law is, rate =k x [A]² x [B]. If the rate constant of the reaction is 3.74 x 10^-2M^-2s^-1, calculate the rate of the reaction when the concentrations of A, B and C are 0.108 M, 0.132 M and 0.124 M respectively.
Solution :
Given : Rate constant of the reaction = k
= 3.74 x 10^-2M^-2s^-1
[A] =0.108 M, [B] = 0.132M, [C] = 0.124 M
Rate of the reaction = R = ?
By rate law,
R = k [A]² x [B] = (0.108)² x 0.132 = 1.54 x 10^-3 Ms^-1
(Concentration of C need not be considered since it is a product.)
Answer:
Rate of reaction = 1.54 x10^-3 Ms^-1

(13) For a reaction, A + B → C, if the concentration of A doubles, the rate of the reaction doubles. While if the concentration of B doubles the rate of the reaction increases by four fold. Write rate law. .
Solution :
Let x moles of A react with y moles of B. xA + yB → C
To write rate law, it is necessary to find x and y values.

(i) Initial rate \(=R_{1}=k[\mathrm{~A}]_{1}^{x}[\mathrm{~B}]_{1}^{y}\)
Final rate R₂ is doubled when the concentration of A is doubled, i.e., R₂ = 2R₁ when final concentration,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 174
(It is assumed that the concentration of B remains same.)

(ii) Initial rate \(=R_{1}=k[\mathrm{~A}]_{1}^{x}[\mathrm{~B}]^{y}\)
If the concentration of B is doubled keeping of A constant, rate becomes four times, i.e.,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 35
Hence the rate law is represented by an expression.
Rate = k[A] [B]²
Answer:
Rate law is. Rate = k [A] [B]²

(14) For the reaction, A2 + B + C → AC + AB, it is found that tripling the concentration of A₂ triples the rate, doubling the concentration of C doubles the rate and doubling the concentration of B has no effect,
(a) What is the rate law?
(b) Why the change in concentration of B has no effect?
Solution :
Given : A2 + B + C → AC + AB
(a) The rate law may be represented as,
Rate = k [A₂]^x [B]^y [C]^z
Let [A]₁, [B]₁ and [C]₁ represent initial concentration and [A]₂, [B]₂ and [C]₂ represent final concentrations, and let R₁ and R₂ be initial and final rates of the reaction when the concentrations are changed.

(i) If [A]₂ = 3[A]₁, R₂ = 3R₁
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 36
If the concentrations of B and C remain constant, then

(b) In the rate determining step, B may not be involved as the reactant, hence rate is independent of changes in concentration of B. (OR B may be in large excess as compared to the concentrations of A and C.)
Answer:
(a) Rate law : Rate = k [A] [C]

Question 21.
Define and explain the term order of a chemical reaction.
Answer:
Order of a chemical reaction : The order of a chemical reaction is defined as the number of molecules (or atoms) whose concentrations influence the rate of the chemical reaction.
OR
The order of a chemical reaction is defined as the sum of the powers (or exponents) to which the concentration terms of the reactants are raised in the rate law expression for the given reaction.

Explanation :
Consider a reaction,
n₁A + n₂B → Products
where n₁ moles of A react with n₂ moles of B.

The rate of this reaction can be expressed by the rate law equation as,
R = k [A]^n₁ [B]^n₂
where k is the rate constant of the reaction, hence, the order of the reaction is n – n₁ + n₂, (observed, experimentally).

If n = 1, the reaction is called the first order reaction, if n = 2, it is called the second order reaction, etc.

If n = 0, it is called the zero order reaction, e.g., photochemical reaction of H_2(g) and Cl_2(g).

Question 22.
What are the features (or key points) of order of a reaction?
Answer:
The features of order of reaction are as follows :

It represents the number of atoms, ions or molecules whose concentrations influence the rate of the reaction.
It is not related to the stoichiometric equation of the reaction, hence it cannot be predicted from stoichiometric balanced equation.
It is experimentally determined quantity.
It is defined only in terms of the concentrations of the reactants and not of products.
It may have values which are integers, fractional or zero.
Higher values are rare. Reactions of first and second order are in large number. Third order reactions are very few like,
2NO_(g) + Cl_2(g) → 2NOCl_(g).

Solved Examples 6.3.3

Question 23.
Solve the following :
(1) From the rate expressions for the following reactions, determine their order :
(a) 2N₂O_5(g) → 4NO_2(g) + O_2(g) : Rate = k [N₂O₅]
(b) CHCl_3(g) + Cl_2(g) → CCl_4(g) + HCl_(g) : Rate = k [CHL₃] [Cl₂]^1/2
(c) C₂H₅Cl_(g) → C₂H_4(g) + HCl_(g): Rate = k [C₂H₅Cl]
(d) 2NO_2(g) + F_2(g) → 2NO₂F_(g) → : Rate = k (NO₂] [F₂]
Solution :
(a) 2N₂O_5(g) → 4NO_2(g) + O_2(g)
The rate law expression given for the reaction is Rate = k x [N₂O₅]
Hence the reaction is of first order.

(b) CHCl_3(g) + Cl_2(g) → CCl_4(g) + HCl_(g)
The given rate law expression is, R = k [CHCl₃] x [Cl₂]^1/2 Here the order of a reaction is one with respect to CHCl_3(g) and half with respect to Cl_2(g). Therefore the overall order of the reaction is 1 + 1/2 = 1.5.

(c) C₂H₅Cl_(g) → C₂H_4(g) + HCl_(g)
The given rate law expression is, Rate = k [C₂H₅Cl]
Hence the reaction has order equal to one.

(d) 2NO_2(g) + F_2(g) → 2NO₂F_(g)
The given rate law expression for the reaction is Rate = k [NO₂] x [F₂]
Hence the reaction is first order with respect to NO2 and first order with respect to F₂. The overall order of the reaction is, n = n_NO2 + nF₁ = 1 + 1 = 2.

(2) Determine the order of following reactions from their rate expressions :
(a) 2H₂O₂ → 2H₂O + O₂ Rate = k [H₂O₂]
(b) NO₂ + CO → NO + CO₂ Rate = k [NO₂]²(c) 2NO + O₂ → 2NO₂ Rate = k [NO]² x [O₂]
(d) CHCl_3(g) + Cl_2(g) → CCl_4(g) + HCl_(g)Rate = k [CHCl₃] [Cl₂]
Solution :
(a) For the reaction,
2H₂O₂ → 2H₂O + O₂
Since the rate law expression given is,
Rate = k [H₂O₂]
Hence the reaction is of first order.

(b) For the reaction,
NO₂ + CO → NO + CO₂
Since the rate law given is Rate = k [NO₂]², the reaction is second order with respect to NO₂ and zero order with respect to CO. Hence the net order of the reaction is, n = nNO₂ + nco = 2 + 0 = 2

(c) For the reaction,
2NO + O₂ → 2NO₂
Since the rate law expression given is, Rate = k [NO]² x [O₂] the reaction is second order with respect to NO and first order with respect to O₂. Hence the overall order of reaction is n = nNO₂ + no₂ = 2 + 1 = 3.

(d) For the reaction, by rate law,
Rate = k [CHCl₃] x [Cl₂] reaction is first order with respect to CHCl₃ and first order with respect to Cl₂. Hence the overall order is, n = ncHcl₃ + ncl₂ = 1 + 1 = 2.

(3) Write the rate law expressions for the following reactions:
(1) 2N₂O_5(g) → 4NO₂ + O₂; order of the reaction is 1.
(2) CH₃CHO → CH₄ + CO; order of the reaction Is ³/₂.
Solution :
(1) For the given reaction, order is one hence the rate law expression is, Rate = k [N₂O₅].
(2) For the given reaction, order is ³/₂, hence the rate law expression is Rate = k x [CH₂CHO]³^/₂.

(4) The reaction \(\mathbf{H}_{2} \mathbf{O}_{2(\mathbf{a q})}+3 \mathbf{I}_{(\mathbf{a q})}^{-}+2 \mathbf{H}_{(\mathrm{aq})}^{+} \longrightarrow 2 \mathbf{H}_{2} \mathbf{O}_{(0)}+\mathbf{I}_{3(a q)}^{-}\) is first order in H₂O₂ and I^–, zero order in H⁺. Write the rate law.
Solution:
Given :
\(\mathrm{H}_{2} \mathrm{O}_{2(\mathrm{~g})}+3 \mathrm{I}_{(\mathrm{aq})}^{-}+2 \mathrm{H}^{+}{ }_{(\mathrm{aq})} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{i})}+\mathrm{I}_{3(\mathrm{aq})}^{-}\)
Since the reaction is first order in H₂O₂ and F and zero order in H⁺, the expression for rate law will be,
Rate =k [H₂O₂]¹ [I^–]¹ [H⁺]⁰
∴ Rate = k [H₂O₂] [I^–]
Answer:
Rate = k [H₂O₂] [I^–]

(5) The rate law for the gas-phase reaction
2NO_(g) + O_2(g) → 2NO_2(g)is rate = k [NO₂]² [O₂]. What is the order of the reaction with respect to each of the reactants and what is the overall order of the reaction?
Solution :
Given : 2NO_(g) + O_2(g) → 2NO_2(g)
Rate = k [NO]²[O₂]
Order of the reaction with respect to NO = n_No = 2
Order with respect to O₂ = n_O2 = 1
Overall order of the reaction = n = n_NO + n_O2
= 2 + 1
= 3
Answer:
Order with respect to NO = 2
Order with respect to O₂ = 1
Overall order = 3

(6) What is the order for the following reactions?
(a) 2NO_2(g) + F_2(g) → 2NO₂F_(g), rate = k [NO₂][F₂]
(b) CHCl_3(g) + Cl_2(g) → CCl_4(g) + HCl_(g), rate = k[CHCl₃][Cl₂]¹^/₂
Solution :
(a) Given : 2NO_2(g) + F_2(g) → 2NO₂F
Rate = k [NO₂][F₂]
Hence the reaction is first order with respect to NO₂ and first order with respect to F₂
∴ Order of reaction = n_NO2 + n_F2 = 1 + 1 = 2

(b) Given :
CHCl_3(g) + Cl_2(g) → CCl_4(g) + HCl_(g),
Rate = k [ CHCl₃] [Cl₂]¹^/₂
Hence the reaction is first order in CHCl₃ and half order in Cl₂.
∴ Order of reaction
= n_CHCl3 + n_Cl2 = 1 + \(\frac{1}{2}\) = \(\frac{3}{2}\)
Answer:
(a) Order of the reaction = 2
(b) The order of the reaction = \(\frac{3}{2}\)

(7) Write the rate law for the following reactions :
(a) A reaction that is zero order in A and second order in B.
(b) A reaction that is second order in NO and first order in Br₂.
Solution :
(a) Given : A + B → Products
The reaction is zero order in A and second order in B. Hence the rate law is represented as, Rate = k [A]^O[B]²
Rate = k[B]²

(b) Given : 2NO_(g) + Br_2(g) → 2NOBr_(g)
The reaction is second order in NO and first in Br₂. Hence the rate law is,
∴ Rate = k [NO]²[Br₂]
Answer: (a) Rate law : Rate = k[B]²
(b) Rate law : Rate = k [NO]²[Br₂]

(8) The reaction A + B → Products, is first order in each of the reactants, (a) Write the rate law.
(b) How does the reaction rate change if the concentration of B is decreased by a factor 3?
(c) What is the change in the rate if the concentration of each reactant is tripled? (d) What is the change in the rate, if the concentration of A is doubled and that of B is halved?
Solution :
(a) The reaction is first order in A and B. Hence the equation for rate law is,
Rate = k [A] [B]
(b) Before changing the concentration of B, Initial rate = R₁ – k [A]₁ [B]₁
After change in concentration of B,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 39
Hence the rate of the reaction will be decreased by a factor 3.

(c) When the concentration of each reactant is tripled, then the final concentrations will be, [A]₂ = 3[A]₁and [B]₂ = 3[B₁]
∴ R₂ = k x 3[A]₁ x 3 [B]₁
∴ R₂ = k x 3[A]₁ x 3 [B]₁
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 40
Hence the rate of the reaction will be increased by 9 times.

(d) When the concentration A is doubled and that of B is halved then the final concentrations will be,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 41
Rate of the reaction will remain unchanged.
Answer:
(a) Rate law is, Rate = k [A] [B],
(b) Rate is decreased by a factor 3,
(c) Rate is increased by 9 times,
(d) Rate remains unchanged.

(9) Consider the reaction A₂ + B → products. If the concentration of A₂ and B are halved, the rate of the reaction decreases by a factor of 8. If the concentration of A₂ is increased by a factor of 2.5, the rate increases by the factor of 2.5. What is the order of the reaction? Write the rate law.
Solution :
Given : A₂ + B → Products
(i) When concentration of A₂ and B are halved :
[A₂]_2(final) = 1/2 [A₂]_1(final) and [B]₂ = 1/2 [B]₁then, R_2(final) = 1/8R_1(intial).

(ii) When concentration of A₂ is increased by the factor 2.5,
[A₂]₂ = 2.5 [A₂]₁ (concentration of B is same) then, R₂ = 2.5 R₁
Now let the reaction be, XA₂ + yB → Products

From data in (ii),
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 43
Hence the reaction is of third order. The rate law can be represented as,
Rate = k [A₂] [B]²
Answer:
(i) Order of the reaction = 3
(ii) Rate law : Rate = k [A₂] [B]³

(10) Consider the reaction C + D → Products. The rate of the reaction increases by a factor of 4 when the concentration of C is doubled. The rate of the reaction is tripled when concentration of D is tripled. What is the order of the reaction? Write the rate law.
Solution :
Given : C + D → Products OR xC + yD → Products
(i) When the concentration of C is doubled, the rate of the reaction increases by 4.

[C]_2(final)= 2[C]_1(initial) then R_2(final) = 4R_1(initial)
(In this, the concentration of D is assumed to be constant.)
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 44
Hence, the reaction is second order in C.
∴ n_C = 2
(ii) When the concentration of D is tripled, rate is tripled. The concentration of C is assumed to be constant.

Rate law : Rate = A[C]²[D]
Answer:
(i) Order of the reaction = 3
(ii) Rate law : Rate = A[C]²[D]

(11) The reaction F_2(g) + 2ClO_2(g) → 2FClO_2(g)is first order in each of the reactants. The rate of the reaction is 4.88 x 10^-4 M/s when [F₂] = 0.015 M and [ClO₂]= 0.025 M. Calculate the rate constant of the reaction.
Solution :
Given :
F_2(g) + 2ClO_2(g) → 2FClO_2(g)
Order of reaction in F₂ = nF₂ = 1
Order of reaction in CIO₂ = n_ClO2 = 1
Rate = R = 4.88 x 10^-4 Ms^-1[F₂] = 0.015 M; [ClO₂] = 0.025 M
Rate = k = ?
By rate law,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 46
Answer:
Rate constant = 1 = 1.3 M^-2s^-1

(12) The reaction 2H_2(g) + 2NO_(g) → 2H₂O_(g) + N_2(g)is first order in H₂ and second order in NO. The rate constant of the reaction at a certain temperature is 0.42M^-2s^-1. Calculate the rate when [H₂] = 0.015 M and [NO] = 0.025 M.
Solution :
Given : 2H_2(g) + 2NO_(g) → 2H₂O_(g) + N_2(g)Order of reaction in H₂ = n_H1= 1
Order of reaction in NO = n_NO = 2
Rate constant = k = 0.42 M^-2s^-1[H₂] = 0.015 M; [NO] = 0.025 M
Rate of reaction = R = ?
By rate law,
Rate = R = k [H₂] [NO]²
= 0.42 x 0.015 x (0.025)²M^-2s^-1 M M
= 3.94 x 10^-6 Ms^-1Answer:
Rate of reaction = R = 3.94 x 10^-6 Ms^-1

(13) Find the order of following reactions whose rate laws are expressed as follows. C_A and C_B are the concentrations of reactants A and B respectively :
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 47
Solution :
Given :
(1) For, – \(\frac{d c}{d t}\) = k x \(\mathrm{C}_{A}^{0}\) the order of the reaction, n = 0. Hence it is a zero order reaction.

(2) For, – \(\frac{d c}{d t}\) = k x \(\mathrm{C}_{A}^{3 / 2}\), the overall order of the reaction is 3/2.

(3) For, –\(\frac{d c}{d t}\) = k x \(\mathrm{C}_{A}^{1 / 2} \mathrm{C}_{B}^{2}\), the reaction has order 1/2 with respect to A and 2 with respect to B.
∴ n = n_A+ n_B = \(\frac{1}{2}\) + 2 = \(\frac{5}{2}\).
Hence the (overall) order of the reaction is \(\frac{5}{2}\).

(4) For, \(-\frac{d c}{d t}=k \mathrm{C}_{A}^{5 / 2} \times \mathrm{C}_{B}^{0}\)
The reaction has order \(\frac{5}{2}\) with respect to A and zero with respect to B.
∴ n = n_A + n_B = \(\frac{5}{2}\) + 0 = \(\frac{5}{2}\)
Hence the order of the reaction is \(\frac{5}{2}\).

(5) For, \(-\frac{d c}{d t}=k \times \mathrm{C}_{A}^{1 / 3} \times \mathrm{C}_{B}^{2 / 3}\). The reaction has order \(\frac{1}{3}\) with respect to A and \(\frac{2}{3}\) with respect to B.
∴ n = n_A + n_B = \(\frac{1}{3}\) + \(\frac{2}{3}\) = 1
Hence the order of the reaction is 1.

(14) The rate of a reaction, 2A + B → Products is 3.78 x 10^-4M s^-1 when the concentrations of A and B are 0.3 M each. If the rate constant of the reaction is 4.2 x 10^-3s^-1 find the order of the reaction.
Solution :
Given : 2A + B → Products
Rate = R = 3.78 x 10^-4Ms^-1[A] = [B] = 0.3 M
Rate constant = 1 = 4.2 x 10^-3 s^-1Let the order of the reaction in A be x and in B be y.

Then, by rate law,
Rate = R = k [A]^x [B]^y3.78 x 10^-4
= 4.2 x 10^-3(0.3)^x(0.3)^y
= 4.2 x 10^-3 (0.3)^x+y∴ \(\frac{3.78 \times 10^{-4}}{4.2 \times 10^{-3}}\) = (0.3)^x+y
0.09 = (0.3)^x+y
(0.3)² = (0.3)^x+y .
∴ x + y = 2
Hence the order of overall reaction is 2.
Answer:
The order of the reaction is 2.

(15) The rate of the reaction, A → Products is 1.25 x 10^-2 M/s when concentration of A is 0. 45 M. Determine the rate constant if the reaction is
(a) first order in A
(b) second order in A.
Solution :
Given : A → Products
Rate = R = 1.25 x 10^-2 M/s
[A] = 0.45 M

(a) Rate constant, k = ? if order is one.
For first order, rate law is, R = k [A]
∴ \(k=\frac{R}{[\mathrm{~A}]}=\frac{1.25 \times 10^{-2}}{0.45}\)
= 2.78 x 10^-2s^-1

(b) Rate constant, k =? if order is two. For second order, rate law is, R = k [A]²
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 50
Answer:
(a) Rate constant, k = 2.78 x 10^-2
(b) Rate constant, k = 6.173 x 10^-2

Question 24.
Define and explain the term elementary reaction.
Answer:
Many reactions take place in a series of steps. Such reactions are called complex reactions. Each step taking place in a complex reaction is called an elementary reaction. This shows that a complex reaction is broken down in a series of elementary chemical reactions.

By adding all the elementary steps of a complex reaction we get the overall reaction.

The mechanism of a reaction is decided from the sequence of the elementary steps that are added to give overall reaction.

Elementary reaction : It is defined as the reaction which takes place in a single step and cannot be divided further into simpler chemical reactions.

The order and molecularity of the elementary reaction are same.

Some reactions take place in one step and cannot be broken down into simpler reactions. For example,

C₂H₅I_(g) → C₂H_4(g) + HI_(g)
O_3(g) → O_2(g) + O_(g)

Question 25.
Define and explain the term molecularity of a reaction. Give examples.
OR
Define the molecularity of a chemical reaction.
Answer:
Molecularity : The molecularity of an elementary reaction is defined as the number of molecules (or atoms or ions) which take part in a chemical reaction.

Explanation :

The molecularity of a reaction is always integral.
It cannot be determined experimentally.
The minimum value of the molecularity is one.
It cannot have fractional or zero values.
The reactions are classified according to the molecularity as follows :

(a) Unimolecular reaction (OR First order reaction) : In this only one molecule takes part in the reaction, e.g., N₂O₅₍g) → 2NO₂_(g) + \(\frac{1}{2}\)O₂_(g)

The rate law expression for this reaction is, Rate = k [N₂O₅]. Hence it is unimolecular and first order.

Other unimolecular reactions are,
O₃_(g) → O₂₍_g) + O_(g)
C₂H₅I_(g) → C₂H_2(g) + HI_(g)

(B) Bimolecular reaction In this two molecules take part in the reaction,
e.g., 2HI_(g) → H_2(g) + I_2(g)
O_3(g) + O_(g) → 2O_2(g)
2NO_2(g) → 2NO_(g) + O_2(g)

The higher molecularity is rare since the prob ability of simultaneous collisions between more molecules is very low.

Question 26.
Explain order and molecularity of elementary reactions.
Answer:
(1) The order and molecularity of elementary reaction are same.
(2) Consider second order bimolecular reaction,
2NO_2(g) → 2NO_(g) + O₂.
(3) The rate of the reaction is given by, Rate = k [NO₂]²(4) Similarly consider unimolecular first order reaction,
C₂H₅I_(g) → C₂H_4(g) + HI_(g)
Rate = k [C₂H₅I]

Question 27.
Define and explain the term rate-determining step.
Answer:
(1) Many chemical reactions take place in a series of elementary steps. Among many steps of the reaction, one of the steps is the slowest step compared to other steps.

Rate determining step : The slowest step in the reaction mechanism which involves many steps is called the rate-determining step.

(2) Example :
Consider decomposition of gaseous NO₂Cl.
2NO₂Cl_(g) → 2NO_2(g) + Cl_2(g)This reaction takes place in two steps :
Step I : \(\mathrm{NO}_{2} \mathrm{Cl}_{(g)} \stackrel{k_{1}}{\longrightarrow} \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{Cl}_{(\mathrm{g})}\) (slow, unimolecular)

Step II: \(\mathrm{NO}_{2} \mathrm{Cl}_{(g)} \stackrel{k_{2}}{\longrightarrow} \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{Cl}_{(\mathrm{g})}\) (fast, bimolecular)

2NO₂CI_(g) → 2NO_2(g) + CI_2(g)(overall reaction)

Among two steps, first step being slower represents rate-determining step. The rate law can be represented as, Rate = k₁ [NO₂CI]

Hence, the reaction is first order.

In this Cl_(g) is formed as a reaction intermediate.

Question 28.
What are the features of rate-determining step?
Answer:
Features of rate-determining step :

The overall reaction can never occur faster than its rate-determining step.
The rate-determining step can occur anywhere in the reaction mechanism and depends on nature of reactants, conditions of the reaction, etc.
The rate law of a rate-determining step can directly be obtained from its stoichiometric equation.
The rate law of a rate-determining step can directly be obtained from its stoichiometric equation.

Question 29.
What is reaction intermediate? Explain with an example.
Answer:
Reaction intermediate : The additional species other than the reactants or products formed in the mechanism during progress of the reaction is called reaction intermediate.

Features of reaction intermediate :

The reaction intermediate appears in the reaction mechanism but does not appear in the overall reaction or in the products.
It is always formed in one step and consumed in the subsequent step in the mechanism.
Its concentration is very small and cannot be determined easily.
Rate of the reaction is independent of concentration of this intermediate.
The life period of the reaction intermediate is extremely small, hence cannot be isolated.
The composition of the reaction intermediate, decides the mechanism of the reaction.
Consider decomposition of gaseous NO₂Cl. 2NO₂Cl_(g) → 2NO_2(g) + Cl_2(g)

This reaction takes place in two steps :
Step I : \(\mathrm{NO}_{2} \mathrm{Cl}_{(\mathrm{g})} \stackrel{k_{1}}{\longrightarrow} \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{Cl}_{(\mathrm{g})}\) (slow, unimolecular)

Step II : \(\mathrm{NO}_{2} \mathrm{Cl}_{(\mathrm{g})}+\mathrm{Cl}_{(\mathrm{g})} \stackrel{k_{2}}{\longrightarrow} \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}\) (fast, bimolecular)
2NO₂Cl_(g) → 2NO_2(g) + Cl_2(g)(overall reaction)
Cl formed in Step I is removed in Step II, Hence Cl is a reaction intermediate.

Question 30.
Identify the molecularity and write the rate law for each of the following elementary reactions :
(a) NO_(g) + O_3(g) → NO_3(g) + O_(g)
(b) H₂I_(g) + I_(g) → 2HI_(g)
(c) CI_(g) + Cl_(g)+ N_2(g) → N_2(g)
Answer:
NO_(g) + O_3(g) → NO_3(g) + O_(g)Molecularity is 2.
Rate law : Rate = k [NO] x [O₃]

(b) H₂I_(g) + I_(g) → 2HI_(g)Molecularity is 2.
Rate law : Rate = k [H₂I] x [I]

Question 31.
Write molecularity of the following reaction:
2NO_(g) + O_2(g) → 2NO_2(g).
Answer:
For the reaction, 2NO_(g) + O_2(g) → 2NO_2(g)Molecularity = 3.

Question 32.
How Is reaction intermediate predicted in the reaction?
Answer:
(1) When a reaction takes place in more than one steps, then a substance produced in one step is removed in the next step is called reaction intermediate.
(2) For example,
(I) NO_(g) + O_3(g) → NO_3(g) + O_(g)
(ii) NO_3(g) + O_(g) → NO_2(g) + O_(g)
In the reaction. NO₃ and O are reaction intermediates.

Question 33.
A certain reaction occurs in the following steps :
(i) Cl_(g) + O_3(g) → ClO_(g)+ O_2(g)
(ii) ClO_(g) + O_(g) → Cl_(g) + O_2(g)(a) Write the chemical equation for overall reaction.
(b) Identify the reaction intermediate.
(c) Identify the catalyst.
(d) What is the molecularity of each step?
Answer:
Step I : Cl_(g) + O_3(g) → ClO_(g)+ O_2(g)
Step II : ClO_(g) + O_(g) → Cl_(g) + O_2(g)
(a) Overall reaction is obtained by adding both the reactions.
O_3(g) + O_(g) → 2O_2(g)
(b) Reaction intermediate is ClO_(g) which is formed in the first step and removed in the second step.
(c) Cl_(g) acts as a catalyst. It is an example of homo-geneous catalysis in which catalyst Cl_(g) forms an intermediate ClO_(g) and again is released in the second step.
(d) Since both the steps involve two reactants each, both the steps are bimolecular.

Question 34.
The rate law for the reaction 2H_2(g) + 2NO_(g) → N_2(g) + 2H₂O_(g) is given by rate = k [H₂] [NO]².
The reaction occurs in the following two steps :
(i) H_2(g) + 2NO_(g) → N₂O_(g) + H₂O_(g)(ii) N₂O_(g) + H_2(g) → N_2(g) + H₂O_(g)What is the role of N₂O in the mechanism? What is the molecularity of each of the elementary steps?
Answer:
N₂O is a reaction intermediate which is formed in the first step and removed in the second step. Molecularity of the elementary steps :
(a) First step – Termolecular.
(b) Second step-Bimolecular.

Question 35.
What is the rate law for the reaction,
NO_2(g) + CO_(g) → NO_(g) + CO_2(g)The reaction occurs in the following steps :
NO₂ + NO₂ → NO₃ + NO (slow)
NO₃ + CO → NO₂ + CO₂ (fast)
What is the role of NO₃?
Answer:
Overall reaction :
NO_2(g) + CO_(g) → NO_(g) + CO_2(g)Step-I NO₂ + NO₂ → NO₃ + NO (slow) (slow)
Step-II NO₃ + CO → NO₂ + CO₂ (fast)

(A) From first rate determining slow step, rate law is, Rate = k[NO₂]²
(B) Role of NO₃ : In the reaction, NO₃ is the reaction intermediate which is formed in first step and removed in the second step.

Question 36.
The rate law for the reaction 2NO_(g) + Cl_2(g) → 2NOCl_(g) is given by rate = k[NO][Cl₂]. The reaction occurs in the following steps :
(i) NO_(g) + Cl_2(g) → NOCl_2(g)(ii) NOCl_2(g) + NO_(g)→ 2NOCl_(g)(a) Is NOCl₂ a catalyst or reaction intermedi-ate? Why?
(b) Identify the rate determining step.
Answer:
(a) NOCl₂ is a reaction intermediate since it is formed in the first step and removed in the second step. It is not a catalyst since it was not present in the first step or on reactant side nor in the second step on product side.
(b) Since rate law is, Rate = k[NO][Cl₂], and the sub-stances NO and Cl₂ are present in the first step as reactants, it is the slow and rate-determining step.

Question 37.
The rate law for the reaction 2H_2(g) + 2NO_(g) → N_2(g) + 2H₂O_(g) is given by rate = k[H₂][NO]². The reaction occurs in the following steps :
(i) H₂ + 2NO → N₂O + H₂O
(ii) N₂O + H₂→ N₂ + H₂O
What is the role of N₂O in the mechanism? Identify the slow step.
Answer:
(a) N₂O is the reaction intermediate since it is formed in the first step and removed in the second step.
(b) By rate law, Rate = k [H₂][NO]². Since the first step involves the substances H2 and NO, it is the slow and rate-determining step.

Question 38.
What are integrated rate laws?
Answer:
Integrated rate laws : The equations which are obtained by integrating the differential rate laws (expressions) and which provide direct relationship between the concentrations of the reactants and time are called integrated rate laws.

For example, integrated rate law for first order reaction is represented as,
\(k=\frac{2.303}{t} \log _{10} \frac{[\text { Reactant }]_{\text {final }}}{[\text { Reactant }]_{\text {initial }}}\)

Question 39.
Derive the expression for integrated rate law (equation) for the first-order reaction.
Answer:
Consider the following first-order reaction, A → B The rate of the chemical reaction is given by the rate law expression as, Rate, R = k [A] where [A] is the concentration of the reactant A and k is the velocity constant or specific rate of the reaction.
The instantaneous rate is given by,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 52

If [A₀] is the initial concentration of the reactant and [A]_t at time t, then by integrating the above equation,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 53

This is the integrated rate equation for the first order reaction. This is also called integrated rate law.

Question 40.
How is the integrated rate equation for the first order reaction represented by considering the concentration of the product?
Answer: The
integrated rate equation for the first order reaction can be represented as,
\(k=\frac{2.303}{t} \log _{10} \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{t}}\) where [A]₀ is the initial concentration of the reactant (at time, 1 = 0) and [A]_t is that at time t. Consider the reaction, A → B
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 54

If a is the initial concentration of the reactant A and x is the concentration of the product B after time t, then
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 55

Question 41.
Explain the exponential rate law expression for the first order reaction.
Answer:
The integrated rate equation for the first order reaction can be represented as,
\(k=\frac{1}{t} \log _{\mathrm{e}} \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{t}}\)
where k is a rate constant, [A]₀ and [A]_t are initial and final concentrations of the reactant after time t.
∴ k = \(-\frac{1}{t} \log _{\mathrm{e}} \frac{[\mathrm{A}]_{t}}{[\mathrm{~A}]_{0}}\)
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 56
where [A]₀ and [A]_t are the concentrations of the reactant when t = 0 and t = t respectively.

Thus, the concentration of the reactant decreases exponentially with time and the time required to complete the first order reaction will be infinity.

Another feature of the exponential behaviour is the time required to complete a definite fraction of the reaction is always constant. Therefore, the first order reactions are also described in terms of the half-life of the reaction ™.

Question 42.
What are the units of rate constant of first order reaction?
Answer:
The units of rate constant (k) for the first order reaction is per time (or s^-1).

Question 43.
Give three examples of first order reaction.
Answer:
The examples of first order reaction are :
(1) Decomposition of H₂O₂ :
2H₂O_2(I) → 2H₂O₍₁₎ + O_2(g) Rate = k[H₂O₂]
(2) Decomposition of N₂O_s :
2N₂O_5(g) → 4NO_2(g) + O_2(g) Rate = k[N₂O₅]
(3) Isomerisation of cyclopropane to propene :
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 67

Question 44.
Write a note on a zero order reaction.
OR
What is a zero order reaction? Explain.
Answer:
(1) Definition : Zero order reaction : A reaction in which the rate of the reaction does not depend on the concentration of any reactant taking part in the reaction is called zero order reaction.
(2) Explanation : For example, consider photochemical reaction between H₂ and Cl₂ gases.

In this the rate of the reaction remains constant throughout the progress of the reaction, even if the concentrations of the reactants decrease with time, until the reactant has reacted entirely.

Hence, by the rate law,
R = k [H2]° [Cl2]° = k (constant).

Question 45.
Derive the expression for integrated rate law for zero-order reaction A → Products.
Answer:
Consider a zero order reaction, A → Products
The rate of the reaction is, Rate \(=\frac{-d[\mathrm{~A}]}{d t}\)

By rate law,
Rate = k x [A]⁰ = k
∴ – d[A] = k x dt

If [A]₀ is the initial concentration of the reactant A at t = 0 and [A]_t is the concentration of A present after time t, then by integrating above equation,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 73
This is the integrated rate law expression for rate constant for zero order reaction.
∴ k x t = [A]₀ – [A]_t
∴ [A]_t = – kt + A₀

Question 46.
How would you obtain the unit of the velocity constant k for (i) the first order reaction (ii) the zero order reaction?
Answer:
(i) For a first order reaction :
Consider the reaction,
A → B
The rate (R) of the reaction will be, R = k [A] = kc, where [A] is concentration in mol dm^-3 Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 74
Hence, the SI unit of velocity constant for the first order reaction is second^-1.

(ii) For a zero order reaction :
The rate of reaction is R = k [A]⁰ = k
Hence, the velocity constant k has the unit of the rate of the reaction, i.e., mol dm^-3 s^-1.

Question 47.
Obtain an expression for half-life period of zero order reaction.
Answer:
The rate law expression for zero order reaction is, [A]_t = – kt + [A]₀
where [A]₀ and [A]_t are the concentrations of the reactant at time, t = 0 and after time t respectively, Half-life period, t_1/2 is the time when the concentration reduces from [A]₀ to [A]₀/2. i.e., at t = t_1/2, [A]_t = [A]₀/2.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 75
Hence for a zero-order reaction, the half-life period is directly proportional to the initial concentration of the reactant.

Question 48.
Give the examples of zero order reactions.
Answer:
Zero order reactions are not common. They take place under special conditions. They are hetero-geneous catalysed reactions generally involving metals as catalysts.

(1) Decomposition NH₃ on Pt surface :

(2) Decomposition of N₂O to N₂ and O₂ on Pt :
(3) Decomposition of PH₃ on hot tungsten catalyst at high pressure.

Question 49.
Decomposition of NH_3(g) on platinum surface at high temperature is a zero order reaction. Explain.
Answer:

The decomposition of NH_3(g) on platinum surface is represented as,
2NH_3(g) \(\frac{1130 \mathrm{~K}}{\mathrm{Pt}}\) N_2(g) + 3H_2(g)
Since it is a heterogeneous catalysed reaction, NH₃ gaseous molecules at high pressure are adsorbed on the metal surface covering the surface area.
The number of NH₃ molecules adsorbed is small compared to NH₃ molecules in the gaseous phase.
Only the molecules adsorbed on the surface get decomposed. Hence rate of the decomposition becomes independent of the concentration (pressure) of NH₃. Therefore the decomposition reaction is zero order.

Question 50.
The catalysed decomposition of nitrous oxide (N₂O) to nitrogen and oxygen is a zero order reaction. Explain.
Answer:

The decomposition of N₂O_(g) on platinum can be represented as, \(2 \mathrm{~N}_{2} \mathrm{O}_{(\mathrm{g})} \stackrel{\mathrm{Pt}}{\longrightarrow} 2 \mathrm{~N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}\)
Since it is heterogeneously catalysed reaction, N₂O gaseous molecules are adsorbed on the metal surface covering the surface area.
The number of N₂O molecules adsorbed is small compared to N₂O molecules in the gaseous phase.
Only the molecules adsorbed on the metal surface get decomposed. Hence rate of decomposition becomes independent of the concentration (pressure) of N₂O. Therefore the decomposition of N₂O is a zero order reaction.

Question 51.
Inversion of cane sugar (sucrose) is a pseudo-first-order reaction. Explain.
OR
The reaction,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 82
Can it be of pseudo-first-order type?
Answer:
The inversion of cane sugar (sucrose) is an acid catalysed hydrolysis reaction which can be represented as,

This is a bimolecular reaction. Hence, the true rate law for the reaction should be, Rate = k[C₁₂H₂₂O₁₁] [H₂O]. This shows that the reaction should be second order.

Since water (H₂O) is in large excess, its concentration remains constant and the rate depends only upon the concentration of cane sugar.

∴ Rate = k[C₁₂H₂₂O₁₁]

Therefore the second order true rate law becomes first order rate law. Hence the inversion of cane sugar is a pseudo first order reaction.

Solved Examples 6.4-6.5

Question 52.
Solve the following :

(1) For the reaction 2A + B → products, find the rate law from the following data :

[A]/M	[B]/M	rate/Ms^-1
0.3	0.05	0.15
0.6	0.05	0.30
0.6	0.2	1.20

Solution:
In steps (i) and (ii), the concentration of A is doubled but the concentration of B remains constant. Since the rate is doubled the rate is proportional to the concentration of A or R α [A] and hence with respect to A order of the reaction is 1 or n_A = 1.

In steps (ii) and (iii), the concentration of A is kept constant but the concentration of B is increased 4 times and rate of the reaction is increased 4 times. Hence the rate of reaction is proportional to concentration of B, R α [B] and hence with respect of B, order is 1 or n_B = 1. Hence rate law will be, Rate = k [A] x [B].

(2) In a first order reaction A → product, 80 % of the given sample of compound decomposes in 40 min. What is the half life period of the reaction ?
Solution :
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 91
Answer:
Half life period = 17.22 min

(3) The reaction A + B → products is first order in each of the reactants.
(a) How does the rate of reaction change if the concentration of A is increased by factor 3?
(b) What is the change in the rate of reaction if the concentration of A is halved and concentration of B is doubled?
Solution :
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 92
Hence the rate remains the same.
Answer:
(a) The rate increases by factor 3.
(b) The rate remains the same.

(4) Half-life period of a first order reaction is 41.09 min. Calculate rate constant in per second.
Solution :
Given : Half-life period = t_1/2
= 41.09 min = 41.09 x 60 s
= 2.465 x 10³s
Rate constant = k = ?
For a first order reaction,
\(\begin{aligned}
k &=\frac{0.693}{t_{1 / 2}} \\
&=\frac{0.693}{2.465 \times 10^{3}}
\end{aligned}\)
= 2.81 x 10^-4 s^-1
Answer:
Rate constant = k = 2.81 x 10^-4 s^-1

(5) A first order reaction takes 15 minutes to complete 25%. How much will it take to complete 65 %?
Solution:
(i) Given : For 25% completion, t₁= 15 min.
For 35 % completion, t₂ = ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 94
Answer:
Time required to complete 65 % reaction = 547 min

(6) Gaseous A₂ dissociates as, A_2(g) → 2A_(g). Initial pressure of A₂ is 0.8 atm. After 20 minutes the pressure is 1.1 atm. Calculate rate constant and half-life period for the reaction.
Solution :
Given : [A]₀ = Initial pressure = P₀ = 0.8 atm
Final pressure = Total pressure = P_T = 1.1 atm
Rate constant = k = ?
Half life period = t_1/2 = ?
A_2(g) → 2A_(g)
P₀– x 2x
Pressure of A₂ = P_t = P₀ – x
Total pressure of the mixture,
P_T = P₀ – x + 2x = P₀ + x
∴ x = P_T – P₀
∴ P_t = P₀ – X = P₀ – (P_T – P₀) – 2P₀ – P_T\(k=\frac{2.303}{t} \log _{10} \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{t}}\)
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 95
Answer:
Rate constant = k = 2.35 x 10^-2 min^-1Half-life period = t_1/2 = 29.5 min

(7) The decomposition of N₂O_5(g) at 320 K according to the following equation follows first order reaction :
N₂O_5(g) → 2NO_2(g)+ \(\frac{1}{2}\)O_2(g)The initial concentration of N₂O_5(g) is 1-24 x 10^-2 mol. L^-1 and after 60 minutes,
0.20 x 10^-2 mol. L^-1. Calculate the rate constant of the reaction at 320 K.
Solution :
Given :
Initial concentration
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 96

(8) From the following data for the liquid phase reaction A → B, determine the order of reaction and calculate its rate constant:

t/s	0	600	1200	1800
[A]/Mol L^-1	0.624	0.446	0.318	0.226

Solution:
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 97
Answer:
Rate constant = k = 5.618 x 10^-4 s^-1

(9) The concentration of a reactant in a first-order reaction A → products, varies with time as follows :

t/min	0	10	20	30	40
[AJ/M	0.0800	0.0536	0.0359	0.0241	0.0161

Show that the reaction is first order.
Solution :
Given : A → Products
[A]₀ = 0.08 M
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 98
Since all the values of rate constant using first order rate law equation come constant, the reaction is of first order.
Answer:
Order of the reaction is one.

(10) In a first order reaction x → y, 40% of the given sample of compound remains unreacted in 45 minutes. Calculate rate constant of the reaction.
Solution :
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 99
Answer:
k = 0.02036 min^-1

(11) If the half-life period of a zero order reaction with initial concentration 0.1 M is 21.3 min, what will be the half-life when the concentration is 0.3 M?
Solution :
Given : Reaction is zero order. t_1/2 = 21.3, when
initial concentration = [A]₁ x = 0.1 M t_1/2 = 2 when
initial concentration = [A]₂ = 0.3 M
For zero order reaction, t_1/2 = \(\frac{[\mathrm{A}]_{0}}{2 k}\)
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 100
Answer:
Half life period = 63.9 min

(12) Consider the reaction 2A + 2B → 2C + D.
From the following data, calculate the order and rate constant of the reaction.

[A]₀/M	[B]₀/M	r₀/Ms^_1
0.488	0.160	0.24
0.244	0.160	0.06
0.244	0.320	0.12

Write the rate law of the reaction.
Solution :
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 101

Hence the reaction is 2nd order in A.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 102
Hence the reaction is first order in B.
The order of overall reaction = n = n_A + n_B= 2 + 1 = 3
By rate law,
Rate = R = k[A]²[B]

Answer:
(i) Order of reaction = 3
(ii) Rate constant = k = 63M^-2s^-1(iii) Rate law : Rate = k [A]² [B]

(20) In acidic solution, sucrose is converted to a mixture of glucose and fructose in pseudo first order reaction. It has been found that the con-centration of sucrose decreased from 20 mmol L^-1 to 8 mmol L^-1 in 38 minutes. What is the half-life of the reaction?
Solution :
Given :
Initial concentration = [A]₀ = [sucrose]₀= 20 mmol L^-1= 20 x 10^-3 mol L^-1

Final concentration = [A]_t = [sucrose]_t= 8 mmol L^-1= 8 x 10^-3 mol L^-3time = t = 38 min
Half-life period = t_1/2 =?
For first order reaction,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 106
Answer:
Half-life period = t_1/2 = 28.74 min

(21) The half-life of a first order reaction is 1.7 hours. How long will it take for 20 % of the reactant to disappear?
Solution :
Given : Half-life period = t_1/2 = 1.7 hrs.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 107

Answer:
Time required for 20% reaction = 32.86 min

(22) The gaseous reaction A₂→ 2A is first order in A₂. After 12.3 minutes, 65% of A₂ remains undecomposed. How long will it take to decompose 90% of A₂? What is the half-life of the reaction?
Solution :
Given : A₂ → 2A
t₁ = 12.3 min
[A]₀ = 100, [A], = 65
t₂ = ? for 90 % decomposition Half-life period = t_1/2 = ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 109
Answer:
(i) Time required for 90% reaction = 65.8 min
(ii) Half-life periods = t_1/2= 19.8 min

(23) Sucrose decomposes in acid solution to give glucose and fructose according to the first-order rate law. The half-life of the rection is 3 hours. Calculate the fraction of sucrose which will remain after 8 hours.
Solution :
Given : Half-life period = t_1/2 = 3 hrs
Time = t = 8 hrs
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 110
Answer:
Fraction of sucrose left = 0.1576

(24) The rate constant of a first order reaction is 6.8 x 10^-4s^-1. If the initial concentration of the reactant is 0.04 M, what is its molarity after 20 minutes? How long will it take for 25% of the reactant to react?
Solution :
Given : Rate constant = k = 6.8 x 10^-4s^-1 Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 111

Answer:
(i) Molarity of reactant after 20 min = 0.0177 M
(ii) Time for 25 % of the reaction = 7.05 min

(25) The rate constant of a certain first-order reaction is 3.12 x 10^-3 min^-1,
(a) How many minutes does it take for the reactant concentration to drop to 0.02 M if the initial concentration of the reactant is 0.045 M?
(b) What is the molarity of the reactant after 1.5 hr?
Solution :
Given : Rate constant = k = 3.12 x 10^-3 min^-1
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 113

Answer:
(i) Time required to drop the concentration to 0.02 M = 260 min
(ii) Molarity after 1.5 hr = 0.034 M

(26) From the following data for the decomposition of azoisopropane,
(CH3₂)₂CHN = NCH(CH₃)₂ → N₂ + C₆H₁₄estimate the rate of the reaction when total pressure is 0.75 stm.

Time/s	Total pressure/atm
0	0.65
200	1.0

Solution :
Given :
(CH₃)₂CHN = NCH(CH₃)_2(g)→ N_2(g) + C₆H_14(g)
At time t P₀ – x x x
At t = 0, [A]₀ = P₀ = 0.65 atm
At t = 200 s,
Total pressure = P_T = 0.75 atm, Rate =?
From the reaction,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 115

Answer:
Rate of the reaction = 2.13 x 10^-3 atm s^-1

(27) The rate constant for a zero order reaction is 0.04 Ms^-1. Calculate the half-life period of the reaction, when the initial concentration of the reactant is 0.01 M.
Solution :
Given : Order of the reaction = 0
Rate constant = k = 0.04 Ms^-1
Concentration = [A]₀ = 0.01 M
Half-life period = t_1/2 =?
For zero order reaction,
\(t_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 k}=\frac{0.01}{0.04}=0.25 \mathrm{~s}\)
Answer:
Half-life period = t_1/2 = 0.25 s

(28) A flask contains a mixture of A and B. Both the compounds decompose by first order kinetics. The half-lives are 60 min for A and 15 min for B. If the initial concentrations of A and B are equal, how long will it take for the concentration of A to be three times that of B?
Solution :
Given :
For A : tm = 60 min For B : t_1/2 = 15 min
Let initial concentrations of
[A]₀ = [B]₀ = M mol dm^-3
After time t, let the concentrations be, [B]_t = x, then [A]_t = 3x
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 117
Answer:
After 31.8 min, concentration of A will be three time that of B. ‘

Question 53.
Obtain Arrhenius equation from collision theory of bimolecular reactions.
Answer:
Consider a bimolecular reaction,
A – B + C → A + B – C
(i) Collisions of reactant molecules : The basic
requirement for a reaction to occur is reacting species A – B and C must come together and collide. The rate of reaction will depend on the rate and frequency of collisions between them. As the i concentration and temperature increase, rate of collisions increases, hence the rate of reaction increases. But the rate of reaction is low as com-pared to the rate of collisions.

(ii) Energy of activation : For fruitful collisions, the colliding molecules must possess a certain amount of energy called activation energy Ea. Due to collisions between A – B and C, there is a change in electron distribution about three nuclei namely A, B and C so that old A – B bond is weakened while new bond is partially formed between B and C, and results in the formation of an activated complex or a transition state.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 119

Therefore transition state always has higher energy than reactants or products. Due to high energy, activated complex is unstable, short lived and decomposes into the products.

To form activated complex, the reactant mol-ecules have to climb the potential energy barrier i. e., activation energy level, hence molecular collision energy of colliding molecules must be high so that reactant molecules form activated complex and further decompose into products.

The fraction (f) of molecules at temperature T having activation energy E_a is given by f = e^-E_a/RT.

If P represents the probability of Z collisions with proper orientation then,
Reaction rate = P x Z x e^-E_a/RT,

Hence the rate constant k of the reaction may be represented as, k = A x e^-E_a/RTwhere A is called frequency factor or pre-exponential factor and ΔH is the enthalpy change of the reaction. This equation is called Arrhenius equation.

Question 54.
Define :
(i) Transition state or activated complex.
Answer:
Transition state or activated complex : The configuration of atoms formed from reactant molecules and which is at the peak of barrier in energy profile diagram having maximum potential energy compared to reactants and products is called transition state or activated complex.

Question 55.
If a gaseous reaction has activation energy 75k J mol^-1 at 298 K, find the fraction of successful collisions.
Answer:
Activation energy = E_a = 75 kJ mol^-1 = 75000 mol^-1; Temperature = T = 298 K The fraction (f) of successful collisions between the molecules with an energy equal to E_a is given by,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 120

This shows that only 7 collisions out of 1014 collisions are sufficiently energetic to convert reactants into products.

Question 56.
Draw energy profile diagram and show
(i) Activated complex
(ii) Energy of activation for forward reaction
(iii) Energy of activation for backward reaction
(iv) Heat of reaction.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 122
(i) B → Activated complex
(ii) E_f → Energy of activation for forward reaction
(iii) E_b → Energy of activation for backward reaction
(iv) ΔH → Heat of reaction.

Question 57.
Obtain Arrhenius equation, k = A x e^-E_a/RTAnswer:
(i) From experimental observations of variation in rate constants with temperature, Arrhenius developed a mathematical equation between reaction rate constant (k), activation energy (E_a) and temperature T.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 123

When a graph of Ink is plotted against reciprocal of temperature (1/T) a straight line with a negative slope is obtained. This is described by a mathematical equation as,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 124

where k is a rate constant, R is the gas constant, E.a is activation energy, T is absolute temperature and the parameter A is called frequency factor or preexponential factor. This is Arrhenius equation.

Question 58.
What is a frequency factor or pre-exponential factor?
Answer:
In Arrhenius equation, k=A x e^-E_a/RT the factor A is called frequency factor and since it is a coefficient of exponential expression, e~^Ea/RT it is also called a pre-exponential factor.

In the above equation k is a rate constant at temperature T, E_a is the energy of activation and R is a gas constant.

A is related to frequency of collisions (Z) or rate of collisions. It is represented as, A = P x Z where P is the probability of collisions with proper orientations and Z is the frequency of collisions of reacting molecules.

The units of A are same as that of k.

Question 59.
Obtain a relation, \(\log _{10} \frac{k_{2}}{k_{1}}=\frac{E_{\mathrm{a}}\left(T_{2}-T_{1}\right)}{2.303 R \times T_{1} \times T_{2}}\),
OR
Obtain a relation showing variation in rate constant with temperature.
Answer:
By arrhenius equation, the rate constant k of the reaction at a temperature T is represented as, k = A x e^-E_a/RT where A is a frequency factor, R is a gas constant and Ed is the energy of activation.

By taking logarithm to the base e, we get,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 126

If k_t and k₂ are the rate constants at temperatures T₁ and T₂ respectively, then
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 127
By measuring the rate constants k₁ and k₂ at two different temperatures T₁ and T₂, the energy of activation Ea of the reaction can be obtained.

Question 60.
How is the energy of activation determined from rate constants at two different temperatures?
Answer:
For the given reaction, rate constants k₁ and k₂ are measured at two different temperatures T₁ and T₂ respectively. Then \(\log _{10} \frac{k_{2}}{k_{1}}=\frac{E_{\mathrm{a}}\left(T_{2}-T_{1}\right)}{2.303 R \times T_{1} \times T_{2}}\) where E_a is the energy of activation.

Hence by substituting appropriate values, energy of activation E_a for the reaction is determined.

Question 61.
Obtain a relation, \(\frac{k_{2}}{k_{1}}=\frac{\left(t_{1 / 2}\right)_{2}}{\left(t_{1 / 2}\right)_{1}}\), where k₁ and k₂are rate constants while (t_1/2)₁ and (t_1/2)₂ are halflife periods of the first order reaction at temperatures T₁ and T₂ respectively. Write the relation for activation energy.
Answer:
The rate constant k and half-life period t_1/2 are related as
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 129

Question 62.
How does a catalyst differ from reaction intermediate?
Answer:

A catalyst accelerates the rate of reaction, while reaction intermediate has no effect on the rate of the reaction.
The catalyst is always present at the start of the reaction whereas reaction intermediate is produced during the mechanism of the reaction.
A catalyst is consumed in one of the steps of mechanism and regenerated in a subsequent step while the reaction intermediate is formed in one step and consumed in subsequent step.
The catalyst is stable but the reaction intermediate is unstable and short lived.

Question 63.
How is lowering of activation energy in the presence of a catalyst obtained?
Answer:

In the presence of a catalyst, activation energy of a reaction is lowered, hence rate and rate constant increase.
If ΔE_a is lowering of activation energy, while k₁ and k₂ are the rate constants of the reaction in the absence and presence of the catalyst respectively then,

Question 64.
The rate constant of a reaction of 400 K is 1.35 x 10²s^-1. When a nickel catalyst is used, the rate constant of the reaction becomes 3.8 x 10²s^-1. Find activation energy. If the initial activation energy is 20 KJ, what will be activation energy in the presence of the catalyst?
Answer:
In the presence of a catalyst, the activation energy is lowered and rate constant is increased.
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 136
The decrease activation energy of the reaction in the presence of a catalyst will be E_a = 20 – 3.446 = 16.554 kJ.

Solved Examples 6.6-6.7

Question 65.
Solve the following :

(1) Calculate activation energy for a reaction of which rate constant becomes four times when temperature changes from 30 °C to 50 °C. (Given : R = 8.314 K^-1mol^-1)
Solution :
Given : k₂= 4k₁
T₁ = 273 + 30 = 303 K
T₂ = 273 + 50 = 323 K
Activation energy = Ea =?
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 141
Answer:
Activation energy = E_a = 56.41 kJ

(2) The rate constant of a first order reaction are 0.58 s^-1 at 313 K and 0.045 s^-1 at 293 K. What is the energy of activation for the reaction?
Solution :
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 142
Answer:
Energy of activation = Ea = 97.46 kJ mol^-1

(3) The energy of activation for a first order reaction is 104 kJ mol^-1. The rate constant at 25°C is 3.7 x 10^-5s^-1. What is the rate constant at 30 °C?
Solution :
Given : Energy of activation = E_a = 104 kJ mol^-1= 104 x 10³ mol^-1
Initial rate constant – k₁= 3.7 x 10^-5 s^-1
Initial temperature = T₁ = 273 + 25 = 298 K
Final temperature = T₂ = 273 + 30 = 303 K
Final rate constant = k₂ =?

Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 144
Answer:
Rate constant at 30 ⁰C = 7.4 x 10^-4 s^-1

(4) What is the activation energy for a reaction whose rate constant doubles when temperature changes from 30 °C to 40 °C?
Solution :
Given :
Initial rate constant = k₁and final rate constant = k₂; \(\frac{k_{2}}{k_{1}}\) = 2
Initial temperature = T₁ = 273 + 30 = 303 K
Final temperature = T₂ = 273 + 40 = 313 K
Energy of activation = E_a = ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 145
Answer:
Activation energy = E_a = 54.66 kj mol^-1

(5) The activation energy for a certain reaction is 334.4 kj mol^-1. How many times larger is the rate constant at 610 K than the rate constant at 600 K?
Solution :
Given :
Activating energy = E_a = 334.4 kJ mol^-1= 334.4 x 10³ J mol^-1Initial temperature = T₁ = 600 K
Final temperature = T₂ = 610 K
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 146

Answer:
Rate constant increase three time.

(6) The rate of a reaction at 600 K is 7.5 x 10⁵ times the rate of the same reaction at 400 K. Calculate the energy of activation for the reaction. (Hint: The ratio of rates is equal to the ratio of rate constants.)
Solution :
Given : \(\frac{R_{2}}{R_{1}}\) = 7.5 x 10⁵.
From the hint, \(\frac{R_{2}}{R_{1}}=\frac{k_{2}}{k_{1}}\) = 7.5 x 10^sInitial temperature = T₁ = 400 K
Final temperature = T₂ = 600 K
Energy of activation = E_a = ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 148
Answer:
Activation energy = E_a = 135 kj mol^-1

(7) The rate constant of a first order reaction at 25 °C is 0.24 s’. If the energy of activation of the reaction is 88 kJmol^-1, at what temperature would this reaction have rate constant of 4 x 10^-2s^-1?
Solution :
Given : k₂ =0.24s^-1; k₂ =4 x 10^-2s^-1T₁ = 273 + 25 = 298 K
Energy of activation = E_a= 88 kJ mol^-1 = 88000 J mol^-1T₂ = ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 149
Answer:
Temperature = 283.6 K

(8) The half-life of a first order reaction is 900 min at 820 K. Estimate its half-life at 720 K if the energy of activation ot the reaction is 250 kJ mol^-1 (1.464 x 10⁵ mm).
Solution:
Given: Initial half-life period = (t_1/2)₁ = 900 min
Energy of activation = 250 kJ mol^-1
= 250 x 10³ kJ mol^-1
Initial temperature = T₁ = 820 K
Final temperature = T₂ = 720 K
Final half-life period = (t_1/2)₂ = ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 150

Answer:
Half-life period = 1.46 x 10⁵ min

(9) The rate of a gaseous reaction is 6.08 x 10^-2Ms^-1 at 50°C. What will be its rate at 60°C? Energy of activation of the reaction is 18.26 kj mol^-1. (R = 8.314k^-1 mol^-1)
Solution :
Given : Initial rate = R₁ = 6.08 x 10″2Ms^-1
Energy of activation = E_a = 18.26 kJmol^-1 = 18260 mol^-1
Initial temperature = T₁ = 273 + 50 = 323 K
Final temperature = T₂ = 273 + 60 = 333 K
Final rate of the reaction = R₂ = ?
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 152
Answer:
Rate of reaction at 37°C = 7.46 x 10^-2 Ms^-1

(10) A first order gas-phase reaction has an energy of activation of 240 kj mol^-1. If the frequency factor of the reaction is 1.6 x 10¹³ s^-1, calculate its rate constant at 600 K.
Solution :
Given : Energy of activation = E_a = 240 kJ mol^-1 = 240 x 10³ mol^-1
Frequency factor = A = 1.6x 10¹³ s^-1
Temperature = T= 600 K
Rate constant = k = ?
By Arrhenius equation,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 153
Answeer:
Rate constant = k = 2.01 x 10^-8 s^-1

(11) In the Arrhenius equation for a first order reaction, the values of ‘A’ and ‘E_a’ are 4 x 10¹³sec^-1 and 98.6 kJ mol^-1 respectively. At what temperature will its half-life period be 10 minutes? [R = 8.314 JK^-1 mol^-2]
Solution :
Given
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 154
= 311.3 K
Answer:
Temperature = T = 311.3 K

(12) The frequency factor for a second-order reaction is 4.83 x 10¹²M^-1s^-1 at 27°C. If the rate constant of the reaction is 1.37 x 10^-3M^-1s^-1, find the energy of activation.
Solution :
Given : Frequency factor = A
= 4.83 x 10¹² M^-1s^-1Rate constant = k= 1.37 x 10^-3 M^-1s^-1Temperature = T = 273 + 27 = 300 K
Energy of activation = E_a = ?
By Arrhenius equation,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 155
Answer:
Energy of activation = E_a = 89.305 kJ mol^–¹

(13) Rate constants (k) for a reaction were measured at different temperatures. When log10ft was plotted against 1/T, the slope of the graph was 3.28 x 10³. Calculate the energy of activation.
Solution :
Given : Slope of a graph = 3.28 x 10³
Activation energy = Ea = ?
From Arrhenius equation, k = A x e^-E_a/RT\(\log _{10} k=\frac{-E_{\mathrm{a}}}{2.303 R} \times \frac{1}{T}+\log _{10} A\)

The graph is a straight line with slope equal to E_a/2.303R
∴ \(\frac{E_{\mathrm{a}}}{2.303 R}\) = 3.28 x 10³
∴ E_a = 2.303/? x 3.28 x 10³
= 2.303 x 8.314 x 3.28 x 10³
= 62.8 x 10³ mol^-1
= 62.8 kJ mol^-1
Answer:
Activation energy = E_a = 62.8 kj mol^-1

Multiple Choice Questions

Question 66.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. The rate of a reaction is expressed in the units
(a) L mol^-1t^-1
(b) mol dm^-3t^-1
(c) Ms
(d) M^-1s^-1
Answer:
(b) mol dm^-3t^-1

2. For a gaseous reaction the unit of rate of reaction is
(a) L atm s^-1
(b) atm mol^-1s^-1
(c) atm s^-1(d) mol s
Answer:
(c) atm s^-1

3. In the reaction A 4- 3B → 2C, the rate of formation of C is
(a) the same as rate of consumption of A
(b) the same as the rate of consumption of B
(c) twice the rate of consumption of A
(d) 3/2 times the rate of consumption of B
Answer:
(c) twice the rate of consumption of A

4. The units of rate of a reaction and rate constant are same for a reaction of order.
(a) zero
(b) one
(c) two
(d) fractional
Answer:
(a) zero

5. During the progress of a reaction, the rate constant of a reaction
(a) increases
(b) decreases
(c) remains unchanged
(d) first increases and then decreases
Answer:
(a) increases

6. For the reaction, 2A → 3C, the reaction rate is equal to
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 158
Answer:
(c)

7. For the reaction, 2X + 3Y → 4Z, reaction may be represented as
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 159
Answer:
(b)

8. For the reaction 2N₂O_5(g)→ 4NO_2(g) + O_2(g)liquid bromine, which of the following rate equation is ‘incorrect’?
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 160
Answer:
(b)

9. The rate of reaction for certain reaction is expressed as :

The reaction is
(a) 3A → 2B + C
(b) 2B → 3A + C
(c) 2B+C → 3A
(d) 3A + 2B → C
Answer:
(c) 2B+C → 3A

10. Order of a reaction is
(a) number of molecules reacting in a reaction
(b) the number of molecules whose concentration changes during a reaction
(c) the number of molecules of reactants whose concentration determine the rate
(d) increase in number of molecules of products
Answer:
(c) the number of molecules of reactants whose concentration determine the rate

11. The unit of rate constant for zero order reaction is
(a) t^-1
(b) mol dm^-3 t^-1
(c) mol^-1 dm³ t^-1
(d) mol^-2 dm⁶ t^-1
Answer:
(b) mol dm^-3 t^-1

12. A → B is a first order reaction with rate 6.6 x 10^-5 ms^-1. When [A] is 0.6 m, rate constant of the reaction is-
(a) 1.1 x 10^-5 s^-1
(b) 1.1 x 10^-4 s^-1
(c) 9 x 10^-5 s^-1
(d) 9 x 10^-4 s^-1
Answer:
(b) 1.1 x 10^-4 s^-1

13. For a first order reaction, when the rate of a reaction is plotted against concentration of the reactant, then the graph obtained is
(a) a curve
(b) a straight line with negative slope
(c) a straight line with a positive slope
(d) a straight line with positive intercept
Answer:
(c) a straight line with a positive slope

14. For a chemical reaction, A → products, the rate of reaction doubles when the concentration of ‘A’ is increased by a factor of 4, the order of reaction is
(a) 2
(b) 0.5
(c) 4
(d) 1
Answer:
(b) 0.5

15. The order of reaction between equimolar mixture of H₂ and Cl₂ in the presence of sunlight is
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(a) 0

16. Molecularity of reaction can be
(a) zero
(b) integral
(c) fractional
(d) negative
Answer:
(b) integral

17. The reaction,
CH₃COOC₂H₅ + H₂O \(\stackrel{\mathrm{H}^{+}}{\longrightarrow}\) CH₃COOH + C₂H₅OH is of
(a) zero order
(b) first order
(c) second order
(d) pseudo first order reaction
Answer:
(d) pseudo first order reaction

18. A reaction is first order with respect to reactant A and second order with respect to reactant B. The rate law for the reaction is given by
(a) rate = k[A][B]²(b) rate = [A][B]²(c) rate = k [A]²[B]
(d) rate = k[A]⁰[B]²Answer:
(a) rate = k[A][B]²

19. Molecularity of an elementary reaction
(a) may be zero
(b) is always integral
(c) may be semi-integral
(d) may be integral, fractional or zero.
Answer:
(b) is always integral

20. The unit of rate constant for first order reaction is
(a) min^-2
(b) s
(c) s^-1
(d) min
Answer:
(c) s^-1

21. The integrated rate equation for first order reaction A → products is given by
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 162
Answer:
(b)

22. Time required to complete 90% of the first order reaction is
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 163
Answer:
(a)

23. The rate constant of a first order reaction is given by
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 164
Answer:
(d)

24. The half-life of a first order reaction is 30 min and the initial concentration of the reactant is 0.1M. If the initial concentration of reactant is doubled, then the half-life of the reaction will be
(a) 1800s
(c) 15 min
(b) 60 min
(d) 900s
Answer:
(a) 1800s

25. The rate constant for a first order reaction is loos the time required for completion of 50% of reaction is-
(a) 0.0693 milliseconds
(b) 0.693 milliseconds
(c) 6.93 milliseconds
(d) 69.3 milliseconds
Answer:
(c) 6.93 milliseconds

26. The slope of the straight line obtained by plotting rate versus concentration of reactant for a first order reaction is
(a) – k
(b) – k/2.303
(c) k/2.303
(d) k
Answer:
(d) k

27. If C₀ and C are the concentrations of a reactant initially and after time t then, for a first order reaction
(a) C = C₀e^kr
(b) C₀ = 1/C e^-kr
(c) C = C₀e^-kr
(d) CO = C e^kr
Answer:
(b) C₀ = 1/C e^-kr

28. A graph corresponding to a first order reaction is
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 165
Answer:
(b)

29. For two first order reactions, A → products and B → products, k₁ and k₂ are the rate constants. The fIrst reaction (A) is slower than the second reaction (B). The graphical observation corresponding to this observation will be
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 166

Answer:
(b)

30. Half-life (t_1/2) of first order reaction is
(a) dependent of concentration
(b) independent of concentration
(c) dependent of time
(d) dependent of molecularity
Answer:
(b) independent of concentration

31. For a first order reaction, the half-life period is
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 168
Answer:
(c)

32. When half-life period of a zero order reaction is plotted against concentration of the reactant at constant temperature, the graph obtained is
(a) a curve
(b) a straight line with a positive slope
(c) a straight line with a negative slope
(d) an exponential graph
Answer:
(b) a straight line with a positive slope

33. The rate of a reaction between A and B is R = k [A]ⁿ x [B]^mOn doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be
(a) m + n
(b) n – m
(c) 2^(n-m)
(d) \(\frac{1}{{ }_{2} n+m}\)
Answer:
(c) 2^(n-m)

34. Consider the reaction
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 169
(a) 0,052 M/s
(b) 0.114 M/s
(c) 0.026 M/s
(d) -0.026 M/s
Answer:
(c)

35. The rate of the first order reaction A → products is 0.01 M/s, when reactant concentration is 0.2 M. The rate constant for the reaction will be
(a) 0.05 s^-1
(b) 0.05 min^-1
(c) 0.1 s^-1
(d) 0.01 s^-1Answer:
(a) 0.05 s^-1

36. The rate constant of a reaction
(a) decreases with increasing E_a
(b) decreases with decreasing E_a
(c) is independent of E_a
(d) decreases with increasing temperature
Answer:
(a) decreases with increasing E_a

37. The slope of a graph In [A]_t versus t for a first order reaction is -2.5 x 10^-3s^-1. The rate constant for the reaction will be
(a) 5.76 x 10^-3s^-1
(b) 1.086 x 10^-3s^-1
(c) -2.5 x 10^-3s^-1
(d) 2.5 x 10^-3s^-1
Answer:
(d) 2.5 x 10^-3s^-1

38. For the reaction, Cl₂ + 2I^– → 2CI^– + I₂, the initial concentration of I^– was 0.2 mol L^– and the concentration after 20 minutes was 0.18 mol L^-1. Then the rate of formation of I₂ in mol L^– min^-1 will be
(a) 1 x 10^-3
(b) 5 x 10^-4
(c) 1 x 10^-4
(d) 2 x 10^-3
Answer:
(b) 5 x 10^-4

39. A catalyst increases the rate of the reaction by
(a) increasing E_a
(b) increasing T
(c) decreasing E_a
(d) decreasing T
Answer:
(c) decreasing E_a

40. The Arrhenius equation is
(a) A = ke^-E_a/RT
(b) A/k = e^-E_a/RT
(c) k = Ae^E_a/RT
(d) k = Aee^-RT/E_a
Answer:
(b) A/k = e^-E_a/RT

41. The Arrhenius equation is
(a) k = Ae^-RT/E_a
(b) A = ke^–^E_a/RT
(c) k = Ae^-RT/E_a
(d) A = ke^E_a/RT
Answer:
(d) A = ke^E_a/RT

42. When the initial concentration of the reactant is doubled, the half-life period of the reaction is also doubled. Hence the order of the reaction is
(a) one
(b) two
(c) fraction
(d) zero
Answer:
(d) zero

43. If k₁ and k₂ are the rate constants of the given reaction in the presence and absence of the catalyst, then
(a) k₁ = k₂
(b) k₁ > k₂
(c) k₁ < k₂
(d) k₁> k₂
Answer:
(b) k₁ > k₂

44. If the ratio of rate constants at two temperatures for the given reaction is 2.5, the ratio of corresponding half-life periods is
(a) 2.5
(b) 4
(c) 5
(d) 0.4
Answer:
(d) 0.4

45. For a zero order reaction, if Co is the initial concentration, then the half life period will be
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 170
Answer:
(c)

46. The order of nuclear disintegration reaction is
(a) zero
(b) one
(c) two
(d) fraction
Answer:
(b) one

47. The unit of rate constant for zero order reaction is
(a) mol L^-2 s^-1
(b) mol^-1Ls^-1
(c) mol²L^-2s^-1
(d) mol L^-1 s^-1
Answer:
(d) mol L^-1 s^-1

48. When a graph of log₁₀k is plotted against 1 /T, the slope of the line is,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 171
Answer:
(d)

49. The slope of a graph obtained by plotting half-life period and initial concentration of the reactant in zero order reaction is
\((a) \frac{2.303}{k}
(b) \frac{1}{k}
(c) \frac{1}{2 k}
(d) \frac{k}{2.303}\)
Answer:
(c)

50. When a graph of log, 0k against 1/T is plotted, for reaction, a graph with slope equal to 1 x 10³ is obtained. Hence the activation energy is
(a) 8.314 x 10³ Jmor^-1
(b) 3.61 kJ mol^-1
(c) 4.85 x 10³ Jmol^-1
(d) 19.1 kJ mol^-1
Answer:
(d) 19.1 kJ mol^-1

51. The correct expression for activation energy is,
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 172
Answer:
(c)

52. In the reaction, 2A_(g) → B_(g), the initial pressure of A is 2.5 atm. After 10 minutes the pressure becomes 2.2 atm. Hence the pressure of A is
(a) 1.2 atm
(b) 1.9 atm
(c) 2.3 atm
(d) 0.3 atm
Answer:
(b) 1.9 atm

53. The half-life period of zero order reaction A → product is given by –
Maharashtra Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics 173
Answer:
(c)

Maharashtra Board Class 11 Biology Important Questions Chapter 7 Cell Division

July 22, 2024July 21, 2024 by Bhagya

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 7 Cell Division Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 7 Cell Division

Question 1.
Why interphase is known as preparatory phase.
Answer:
1. During interphase, the cell is metabolically very active.
2. In this phase, a cell grows to its maximum size, chromosomal material (DNA and histone proteins) duplicates and the cell prepares itself for next mitotic division. Hence, inteiphase is known as preparatory phase.

Question 2.
Name the following.
1. In which phase the amount of DNA per cell doubles?
2. Which types of RNA are synthesized during first growth phase?
Answer:
1. S-phase.
2. m-RNA, t-RNA and r-RNA

Question 3.
Match the Column I (Phases of Cell cycle) with Column II (Approximate time for completion).

Column I	Column II
1. G: Phase	(a) 1-3 Hours
2. G_i Phase	(b) 2-5 Hours
3. M Phase	(c) 6-8 Hours
4. S Phase	(d) 8 Hours

Answer:

Column I	Column II
1. G: Phase	(b) 2-5 Hours
2. G_i Phase	(d) 8 Hours
3. M Phase	(a) 1-3 Hours
4. S Phase	(c) 6-8 Hours

Question 4.
What is cell division? Mention the types of cell division.
Answer:
The division of cells into two (or more) daughter cells with same (or different) genetic material is called cell division. There are three types of cell divisions:
1. Amitosis:
a. It is the simplest form of cell division. The nucleus elongates and a constriction appears along its length.
b. This constriction deepens and divides nucleus into two daughter nuclei followed by division of cytoplasm resulting in formation of two daughter cells.
c. This type of division is observed in unicellular organisms, abnormal cells, old cells and in foetal membrane cells.

2. Mitosis:
a. In this type of cell division, the cell divides and forms two similar daughter cells which are identical to the parent cell.
b. It is completed in two steps as karyokinesis and cytokinesis.

3. Meiosis:
a. In this type of cell division, the number of chromosomes is reduced to half. Hence, this type of cell division is also called reductional division.
b. Meiosis produces four haploid daughter cells from a diploid parent cell.

Question 5.
With the help of suitable diagrams, explain karyokinesis in brief.
Answer:
Karyokinesis is the nuclear division which is divided into prophase, metaphase, anaphase and telophase.
1. Prophase:
a. In this phase, condensation of chromatin material, migration of centrioles, appearance of mitotic apparatus and disappearance of nuclear membrane takes place.
b. Due to condensation, each chromosome with its sister chromatids connected by centromere is clearly visible under light microscope.
c. The nucleolus starts to disappear.
d. Centrosome start moving towards the opposite poles of the cell.
e. Mitotic apparatus is almost completely formed.

2. Metaphase:
a. Chromosomes are completely condensed and appear short.
b. Centromere and sister chromatids become very prominent.
c. All the chromosomes are arranged at equatorial plane of cell. This is called metaphase plate.
d. Mitotic spindle is fully formed in this phase.
e. Centromere of each chromosome divides horizontally into two, each being associated with a chromatid. [Note: The centromeres divide at the beginning of anaphase so that the two chromatids of each chromosome become separated from each other.
Source: Cell Division, Donald B. McMillan, Richard J. Harris, in An Atlas of Comparative Vertebrate Histology, 2018.]

3. Anaphase:
a. In this phase, chromatids of each chromosome separate and form two chromosomes called daughter chromosomes.
b. The chromosomes which are formed are pulled away in opposite direction by spindle apparatus.
c. Anaphase ends when each set of chromosomes reach at opposite poles of the cell.

4. Telophase:
a. This is the final stage of karyokinesis.
b. The chromosomes with their centromeres begin to uncoil at the poles.
c. The chromosomes lengthen and lose their individuality.
d. The nucleolus reappears and the nuclear membrane appear around the chromosomes.
e. Spindle fibres breakdown and get absorbed in the cytoplasm. Thus, two daughter nuclei are formed.

These are small disc-shaped structures at the surface of the centromeres which serve as the sites of attachment of spindle fibres to the chromosomes.

Question 6.
Draw neat and labelled diagram of Anaphase.
Answer:
Anaphase:
a. In this phase, chromatids of each chromosome separate and form two chromosomes called daughter chromosomes.
b. The chromosomes which are formed are pulled away in opposite direction by spindle apparatus.
c. Anaphase ends when each set of chromosomes reach at opposite poles of the cell.

Question 7.
Match the following.

Column I	Column II
1. Prophase	(a) Chromatids moving to opposite poles.
2. Metaphase	(b) Nuclear membrane starts disappearing.
3. Anaphase	(c) Chromosomes at equatorial plane of the cell.
4. Telophase	(d) Nuclear membrane reappears

Answer:

Column I	Column II
1. Prophase	(b) Nuclear membrane starts disappearing.
2. Metaphase	(c) Chromosomes at equatorial plane of the cell.
3. Anaphase	(a) Chromatids moving to opposite poles.
4. Telophase	(d) Nuclear membrane reappears

Question 8.
Observe the given diagram and explain the depicted process in your own words.
Maharashtra Board Class 11 Biology Important Questions Chapter 7 Cell Division 1
Answer:

The process depicted in the given diagram is cytokinesis in animal cell.
This step takes place at the end of karyokinesis (nuclear division) of mitosis.
It depicts the division of the cytoplasmic material in order to form two daughter cells that resemble each other.
The division starts with a constriction generally at the equator. This constriction gradually deepens and ultimately joins in the centre dividing into two cells.
At the time of cytoplasmic division, organelles like mitochondria and plastids get distributed between the two daughter cells.

Question 9.
Diagrammatically differentiate between cytokinesis in animal cell and plant cell.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 7 Cell Division 2

Question 10.
How cell wall is formed in plant cell?
Answer:
The formation of the new cell wall begins with the formation of a simple precursor, called the ‘cell-plate’ that represents the middle lamella between the walls of two adjacent cells.

Question 11.
What is necrosis?
Answer:
Necrosis is a form of cell injury which leads to the premature death of cells. For example: due to scrape or a harmful chemical.

Question 12.
What is apoptosis? Write its significance.
Answer:

Apoptosis also known as programmed cell death or cellular suicide. In apoptosis cells die in controlled way.
For example: during embryonic development the cells between the embryonic fingers die in a normal process called apoptosis to give a definite shape to the fingers.
Apoptosis also helps in eliminating potential cancer cells.

Question 13.
Which type of cell division is known as reductional division? Why?
Answer:
1. Meiosis is known as reductional division.
2. The number of chromosome is reduced to half, hence, meiosis is known as reductional division.

Question 14.
Describe the various phases of heterotypic division.
Answer:
Heterotypic division is first meiotic division, during which a diploid cell is divided into two haploid cells. The daughter cells resulting from this division are different from the parent cell in chromosome number. Hence the division is called heterotypic division.
It consists of following phases:
1. Prophase -I:
It is the most complicated and longest phase of meiotic division.
It is further divided into five sub-phases viz. leptotene, zygotene, pachytene, diplotene and diakinesis.

a. Leptotene:

The volume of the nucleus increases.
The chromosomes become long distinct and coiled.
They orient themselves in a specific fonn known as bouquet stage. This is characterized with the ends of chromosomes converged towards the side of nucleus where the centrosome lies.
The centriole duplicates into two and migrates to opposite poles. [Note: Centrioles divide during Gj phase of interphase.]

b. Zygotene:

Pairing of non-sister chromatids of homologous chromosomes takes place by formation of synaptonemal complex. This pairing is called synapsis.
Each pair consists of a maternal chromosome and a paternal chromosome. Chromosomal pairs are called bivalents or tetrads.

c. Pachytene:

Each individual chromosome begins to split longitudinally into two similar chromatids. Therefore, each bivalent now appears as a tetrad consisting of four chromatids.
The homologous chromosomes begin to separate but they do not separate completely and remain attached to one or more points. These points are called chiasmata (Appear like a cross-X).
Chromatids break at these points and broken segments are exchanged between non-sister chromatids of homologous chromosomes resulting in recombination.

d. Diplotene:
The chiasma becomes clearly visible in diplotene due to beginning of repulsion between synapsed homologous chromosomes. This is known as desynapsis. Synaptonemal complex also starts to disappear.

e. Diakinesis:

The chiasmata begin to move along the length of chromosomes from the centromere towards the ends of chromosomes. The displacement of chiasmata is termed as terminalization.
The terminal chiasmata exist till the metaphase.
The nucleolus and nuclear membrane completely disappear and spindle fibres begin to appear.

2. Metaphase -1:
a. The spindle fibres are well developed.
b. The tetrads orient themselves on equator in such a way that centromeres of homologous tetrads lie towards the poles and arms towards the equator.
c. They are ready to separate as repulsive force increases.
a. Homologous chromosomes are carried towards the opposite poles by spindle apparatus. This is known as disjunction.
b. The two sister chromatids of each chromosome do not separate in meiosis -I. This is reductional division.
c. The sister chromatids of each chromosome are connected by a common centromere.
d. Both sister chromatids of each chromosome are now different in genetic content as one of them has undergone recombination.
Maharashtra Board Class 11 Biology Important Questions Chapter 7 Cell Division 3

3. Anaphase – I:
1. Homologous chromosomes are carried towards the opposite poles by spindle apparatus. This is known as disjunction.
2. The two sister chromatids of each chromosome do not separate in meiosis -I. This is reductional division.
3. The sister chromatids of each chromosome are connected by a common centromere.
4. Both sister chromatids of each chromosome are now different in genetic content as one of them has undergone recombination.
Maharashtra Board Class 11 Biology Important Questions Chapter 7 Cell Division 4

4. Telophase-I:
a. The haploid number of chromosomes becomes uncoiled and elongated after reaching their respective poles.
b. The nuclear membrane and nucleolus reappear and thus two daughter nuclei are formed.
Maharashtra Board Class 11 Biology Important Questions Chapter 7 Cell Division 5

Cytokinesis -1:
Cytokinesis occurs after karyokinesis and two haploid cells are formed. In many cases, these daughter cells pass through interkinesis.
Maharashtra Board Class 11 Biology Important Questions Chapter 7 Cell Division 6
[Note: The association between the homologous chromosomes i.e. chiasmata remain till metaphase I. During metaphase /, the paired homologous chromosomes move to the metaphase plate. In anaphase [ the spindle fibers begin to shorten. As these spindle fibres shorten, the association between homologous chromosomes (chiasmata) are broken, allowing homologous chromosomes to be pulled to opposite poles.]

Question 15.
What is Homotypic Division? Explain its phases.
Answer:
Two haploid cells formed during first meiotic division divide further into four haploid cells this division is called homotypic division. It consists of five phases: prophase – II, metaphase – II, anaphase – II, telophase – II, and Cytokinesis – II.
Maharashtra Board Class 11 Biology Important Questions Chapter 7 Cell Division 7

1. Prophase-II:
a. The chromosomes are distinct with two chromatids.
b. Each centriole divides into two resulting in formation of two centrioles which migrate to opposite poles and form asters.
c. Spindle fibres are formed between the centrioles.
d. The nuclear membrane and nucleolus disappears in this phase.

2. Metaphase -II:
a. Chromosomes are arranged at the equator.
b. The two chromatids of each chromosome are separated by division of the centromere.
c. Some of the spindle fibres are attached to the centromeres and some are arranged end to end between two opposite centrioles.

3. Anaphase – II:
In this phase, the separated chromatids become daughter chromosomes and move to opposite poles due to the contraction of the spindle fibres attached to centromeres.

4. Telophase – II:
a. In this stage daughter chromosomes starts to uncoil.
b. The nuclear membrane surrounds each group of chromosomes.
c. Nucleolus reappears in this phase.

5. Cytokinesis – II
a. Cytokinesis takes place after the nuclear division.
b. Two haploid cells are formed from each haploid cell.
c. Thus, four haploid daughter cells are formed.
d. These cells then undergo changes to form gametes.

Question 16.
Why meiosis is important?
Answer:

Meiotic division produces gametes or spores.
If it is absent, the number of chromosomes would double or quadruple resulting in the formation of monstrosities (abnormal gametes).
The constant number of chromosomes in a given species across generations is maintained by meiosis and fertilization.
Because of crossing over, exchange of genetic material takes place leading to genetic variations, which are the raw materials for evolution.

Question 17.
Observe the diagram and answer the questions given below it.
Maharashtra Board Class 11 Biology Important Questions Chapter 7 Cell Division 8
1. Identify the type of cell division shown in the diagram.
2. Write its significance of meiosis.
Answer:
1. The type of cell division shown in diagram is meiosis II.
2. Meiotic division produces gametes or spores.

If it is absent, the number of chromosomes would double or quadruple resulting in the formation of monstrosities (abnormal gametes).
The constant number of chromosomes in a given species across generations is maintained by meiosis and fertilization.
Because of crossing over, exchange of genetic material takes place leading to genetic variations, which are the raw materials for evolution.

Question 18.
Explain Anaphase-I with a neat labelled diagram.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 7 Cell Division 9
Anaphase:
a. In this phase, chromatids of each chromosome separate and form two chromosomes called daughter chromosomes.
b. The chromosomes which are formed are pulled away in opposite direction by spindle apparatus.
c. Anaphase ends when each set of chromosomes reach at opposite poles of the cell.

Question 19.
What is crossing over? Give its significance.
Answer:
Crossing over:
The process of exchange of genetic material between non-sister chromatids of homologous chromosomes is known as crossing over.
Significance of crossing over:
Crossing over results in genetic recombination of parental characters that leads to variations.

Question 20.
What happens during diakinesis?
Answer:

In diakinesis, the chiasmata begin to move along the length of chromosomes from the centromere towards the ends of chromosomes.
The displacement of chiasmata is termed as terminalization. The terminal chiasmata exist till the metaphase.
The nucleolus disappears and the nuclear membrane also begins to disappear.
Spindle fibres starts to appear in the cytoplasm.

Question 21.
Differentiate between anaphase of mitosis and anaphase – I of meiosis.
Answer:

Anaphase of mitosis	Anaphase – I of meiosis
1. Centromere divides into two, resulting in the separation of chromatids.	Centromere does not divide.
2. Homologous chromosomes are not involved.	Homologous chromosomes are involved.
3. Disjunction does not occur.	Disjunction occurs.
4. Same number of chromosomes gather at each pole.	Half the chromosome number gather at respective pole.

Question 22.
Give reasons: Meiosis is known as reductional division.
Answer:
Meiosis is known as reductional division because the parent cell produces four daughter cells each having half the number of chromosomes present in the parent cell.

Question 23.
Fill in the blanks:

The process of mitosis maintains the _______.
________ involves the cell death, but it benefits the organism as a whole.
Crossing over takes place in _______ phase of Prophase-I.

Answer:

The process of mitosis maintains the nucleo-cytoplasmic ratio.
Apoptosis involves the cell death, but it benefits the organism as a whole.
Crossing over takes place in pachytene phase of Prophase-I.

Question 24.
1. Complete the following flowchart.
2. Explain the type of cell division in which chromosome number remain the same as that of the parent cell.
Maharashtra Board Class 11 Biology Important Questions Chapter 7 Cell Division 10
Answer:
Karyokinesis is the nuclear division which is divided into prophase, metaphase, anaphase and telophase.
1. Prophase:
a. In this phase, condensation of chromatin material, migration of centrioles, appearance of mitotic apparatus and disappearance of nuclear membrane takes place.
b. Due to condensation, each chromosome with its sister chromatids connected by centromere is clearly visible under light microscope.
c. The nucleolus starts to disappear.
d. Centrosome start moving towards the opposite poles of the cell.
e. Mitotic apparatus is almost completely formed.

These are small disc-shaped structures at the surface of the centromeres which serve as the sites of attachment of spindle fibres to the chromosomes.

Question 25.
While studying mitosis, different teams of students made following observations in the cells focused under microscope.
1. In certain cells chromosomes were arranged at equatorial plane with fibres originating from cylindrical structures at both the poles.
2. Few cells showed chromatids moving towards opposite poles.
a. In first observation which stage of mitosis is seen by students and what is the scientific term used to represent cylindrical structures?
b. Which stage is seen in the second observation?
Answer:
a. The stage observed in the first case is metaphase. The scientific term used to represent the cylindrical structures are centrioles.
b. The other stage seen in second observation is anaphase.

Question 26.
During biology practical students were asked to see the slide mounted under microscope and note down their observations. Few students noted that the stage observed is anaphase of mitosis and others said that it is anaphase I of meiosis. Later while explaining about experiments teacher said that it is anaphase I of meiosis. On what basis teacher confirmed that it is anaphase I of meiosis?
Answer:
Chromosomes moving towards opposite poles during anaphase I do not separate at the centromeres.

Question 27.
Colchicine is an alkaloid extracted from plants. It prevents the formation of spindle fibres. In the presence colchicine, if a cell enters mitosis what would be the outcome?
Answer:
The spindle fibres are necessary for segregating the sister chromatids to opposite poles of the cell during anaphase. In the presence of colchicine, no spindle fibres will form to attach to the kinetochores (small disc¬like structures present on chromosomes). Therefore, the cell will be stuck in mitosis with the condensed pairs of sister chromatids in an unorganized array.

Question 28.
Read the following statements and mention whether they are TRUE or FALSE in respective boxes.
1. Life of all multicellular organisms starts from single cell which is known as zygote.
2. Spindle fibres present between centriole and centromere are known as polar fibres which can contract.
3. Growth of every living organism depends on cell division.
4. Spindle fibres present between opposite centrioles are called as kinetochore fibres which can elongate.

	(i)	(ii)	(iii)	(iv)
(A)	T	T	F	T
(B)	F	F	T	F
(C)	T	F	T	F
(D)	1	T	F	F ‘

Answer:
(C)

Question 29.
Quick Review

Maharashtra Board Class 11 Biology Important Questions Chapter 7 Cell Division 12

Question 30.
Exercise:

Question 1.
Define cell cycle.
Answer:

Sequential events occurring in the life of a cell is called cell cycle.
Interphase and M – phase are the two phases of cell cycle.
Cell undergoes growth or rest during interphase and divides during M – phase.

Question 2.
Observe the following diagram and the questions based on it.
Maharashtra Board Class 11 Biology Important Questions Chapter 7 Cell Division 13
1. If the initial amount of DNA in a cell is 2C then in which phase of cell cycle the amount of this DNA would become 4C? Also name the process.
2. Which sub-phase of the interphase is of short duration?
3. Enlist the phases of karyokinesis in proper order.
Answer:
S – phase (Synthesis phase):
In this phase DNA is synthesized (replicated), so that amount of DNA per cell doubles.
Synthesis of histone proteins takes place in this phase.
Karyokinesis is the nuclear division which is divided into prophase, metaphase, anaphase and telophase.
1. Prophase:
a. In this phase, condensation of chromatin material, migration of centrioles, appearance of mitotic apparatus and disappearance of nuclear membrane takes place.
b. Due to condensation, each chromosome with its sister chromatids connected by centromere is clearly visible under light microscope.
c. The nucleolus starts to disappear.
d. Centrosome start moving towards the opposite poles of the cell.
e. Mitotic apparatus is almost completely formed.

These are small disc-shaped structures at the surface of the centromeres which serve as the sites of attachment of spindle fibres to the chromosomes.

Question 3.
During which stage of Prophase-I synapsis occurs?
Answer:
b. Zygotene:
Pairing of non-sister chromatids of homologous chromosomes takes place by formation of synaptonemal complex. This pairing is called synapsis.
Each pair consists of a maternal chromosome and a paternal chromosome. Chromosomal pairs are called bivalents or tetrads.

Question 4.
During which stage disjunction takes place?
Answer:
Telophase-I:
a. The haploid number of chromosomes becomes uncoiled and elongated after reaching their respective poles.
b. The nuclear membrane and nucleolus reappear and thus two daughter nuclei are formed.

Question 5.
What is disjunction?
Answer:
Telophase-I:
a. The haploid number of chromosomes becomes uncoiled and elongated after reaching their respective poles.
b. The nuclear membrane and nucleolus reappear and thus two daughter nuclei are formed.

Question 6.
Why meiosis is known as reductional division?
Answer:
In this type of cell division, the number of chromosomes is reduced to half. Hence, this type of cell division is also called reductional division.

Question 7.
Sketch and label the phase of cell division in which synaptonemal complex is formed?
Answer:
Zygotene:
Pairing of non-sister chromatids of homologous chromosomes takes place by formation of synaptonemal complex. This pairing is called synapsis.
Each pair consists of a maternal chromosome and a paternal chromosome. Chromosomal pairs are called bivalents or tetrads.

Question 8.
Make a schematic representation of a type of cell division in which chromosome number is reduced to half.
Answer:
Karyokinesis is the nuclear division which is divided into prophase, metaphase, anaphase and telophase.
1. Prophase:
a. In this phase, condensation of chromatin material, migration of centrioles, appearance of mitotic apparatus and disappearance of nuclear membrane takes place.
b. Due to condensation, each chromosome with its sister chromatids connected by centromere is clearly visible under light microscope.
c. The nucleolus starts to disappear.
d. Centrosome start moving towards the opposite poles of the cell.
e. Mitotic apparatus is almost completely formed.

These are small disc-shaped structures at the surface of the centromeres which serve as the sites of attachment of spindle fibres to the chromosomes.

Question 9.
Describe mitosis and its stages in brief.
Answer:
Karyokinesis is the nuclear division which is divided into prophase, metaphase, anaphase and telophase.
1. Prophase:
a. In this phase, condensation of chromatin material, migration of centrioles, appearance of mitotic apparatus and disappearance of nuclear membrane takes place.
b. Due to condensation, each chromosome with its sister chromatids connected by centromere is clearly visible under light microscope.
c. The nucleolus starts to disappear.
d. Centrosome start moving towards the opposite poles of the cell.
e. Mitotic apparatus is almost completely formed.

These are small disc-shaped structures at the surface of the centromeres which serve as the sites of attachment of spindle fibres to the chromosomes.

Question 10.
Describe chiasmata. Draw diagram to illustrate your answer.
Answer:
Pachytene:
Each individual chromosome begins to split longitudinally into two similar chromatids. Therefore, each bivalent now appears as a tetrad consisting of four chromatids.
The homologous chromosomes begin to separate but they do not separate completely and remain attached to one or more points. These points are called chiasmata (Appear like a cross-X).
Chromatids break at these points and broken segments are exchanged between non-sister chromatids of homologous chromosomes resulting in recombination.

Diplotene:
The chiasma becomes clearly visible in diplotene due to beginning of repulsion between synapsed homologous chromosomes. This is known as desynapsis. Synaptonemal complex also starts to disappear.

Question 11.
Correct the following diagram and write a short note on it.
Maharashtra Board Class 11 Biology Important Questions Chapter 7 Cell Division 14
Answer:
b. Zygotene:
Pairing of non-sister chromatids of homologous chromosomes takes place by formation of synaptonemal complex. This pairing is called synapsis.
Each pair consists of a maternal chromosome and a paternal chromosome. Chromosomal pairs are called bivalents or tetrads.

Question 12.
Explain prophase I in your own words.
Answer:
Prophase -I:
It is the most complicated and longest phase of meiotic division.
It is further divided into five sub-phases viz. leptotene, zygotene, pachytene, diplotene and diakinesis.

Question 13.
Explain homotypic division.
Answer:
Two haploid cells formed during first meiotic division divide further into four haploid cells this division is called homotypic division. It consists of five phases: prophase – II, metaphase – II, anaphase – II, telophase – II, and Cytokinesis – II.

3. Anaphase – II:
In this phase, the separated chromatids become daughter chromosomes and move to opposite poles due to the contraction of the spindle fibres attached to centromeres.

4. Telophase – II:
a. In this stage daughter chromosomes starts to uncoil.
b. The nuclear membrane surrounds each group of chromosomes.
c. Nucleolus reappears in this phase.

Question 14.
How does cytokinesis in plant cells differ from animal cells?
Answer:
The formation of the new cell wall begins with the formation of a simple precursor, called the ‘cell-plate’ that represents the middle lamella between the walls of two adjacent cells.

Question 15.
What is the significance of meiosis in sexually reproducing animals?
Answer:

Meiotic division produces gametes or spores.
If it is absent, the number of chromosomes would double or quadruple resulting in the formation of monstrosities (abnormal gametes).
The constant number of chromosomes in a given species across generations is maintained by meiosis and fertilization.
Because of crossing over, exchange of genetic material takes place leading to genetic variations, which are the raw materials for evolution.
Gametes are produced by the process of meiosis which are essential for sexual reproduction.
Diploid organisms have two set of chromosomes (one paternal and one maternal).
For a diploid organism to undergo sexual reproduction it needs to create gametes that contain only one set of chromosomes so the number of chromosomes remains same in the next generation.
In absence of meiosis, the chromosome number of parents and their offsprings will differ in every generation; hence no species will hold its characters.
Also, there will be no crossing over of homologous chromosomes. Thus, there will be no variations with respect to the changing environment in progeny to maintain their existence, which may lead to extinction of species.

Question 16.
Explain the first three stages of Meiosis II.
Answer:
Two haploid cells formed during first meiotic division divide further into four haploid cells this division is called homotypic division. It consists of five phases: prophase – II, metaphase – II, anaphase – II, telophase – II, and Cytokinesis – II.

3. Anaphase – II:
In this phase, the separated chromatids become daughter chromosomes and move to opposite poles due to the contraction of the spindle fibres attached to centromeres.

4. Telophase – II:
a. In this stage daughter chromosomes starts to uncoil.
b. The nuclear membrane surrounds each group of chromosomes.
c. Nucleolus reappears in this phase.

Question 17.
Sketch, label and describe telophase in mitosis.
Answer:
Telophase:
a. This is the final stage of karyokinesis.
b. The chromosomes with their centromeres begin to uncoil at the poles.
c. The chromosomes lengthen and lose their individuality.
d. The nucleolus reappears and the nuclear membrane appear around the chromosomes.
e. Spindle fibres breakdown and get absorbed in the cytoplasm. Thus, two daughter nuclei are formed.

These are small disc-shaped structures at the surface of the centromeres which serve as the sites of attachment of spindle fibres to the chromosomes.

Question 18.
Explain the process recombination.
Answer:
a. Recombination is exchange of genetic material between paternal and maternal chromosomes during gamete formation.
b. The points where crossing over takes place is known as chiasmata.
c. Chromatids acquire new combinations of alleles by physically exchanging segments in crossing-over.
d. The exchange of genetic material between homologous chromosomes involves accurate breakage and joining of DNA molecules through a complex mechanism.
e. It is catalyzed by enzymes.

Question 19.
1. What is necrosis?
2. What is apoptosis? Mention its significance.
Answer:
1. Necrosis is a form of cell injury which leads to the premature death of cells. For example: due to scrape or a harmful chemical.
(2) 1. Apoptosis also known as programmed cell death or cellular suicide. In apoptosis cells die in controlled way.
2. For example: during embryonic development the cells between the embryonic fingers die in a normal process called apoptosis to give a definite shape to the fingers.
3. Apoptosis also helps in eliminating potential cancer cells.

Question 20.
Multiple Choice Questions:

Question 1.
Replication of DNA takes place during
(A) prophase
(B) S-phase
(C) G₂ phase
(D) Interkinesis
Answer:
(B) S-phase

Question 2.
During cell division, spindle fibers are attached to
(A) telomere
(B) centromere
(C) chromomeres
(D) chromosome
Answer:
(B) centromere

Question 3.
Which of the following is the shortest phase?
(A) metaphase
(B) anaphase
(C) interphase
(D) S-phase
Answer:
(B) anaphase

Question 4.
Reappearance of nucleolus is during
(A) telophase
(B) prophase
(C) cytokinesis
(D) inter-kinesis
Answer:
(A) telophase

Question 5.
During telophase,
(A) nuclear membrane is formed.
(B) nucleolus appears.
(C) astral rays disappear.
(D) all the above
Answer:
(D) all the above

Question 6.
Cytokinesis in plant cell takes place by
(A) furrowing
(B) cell plate formation
(C) any one of (A) or (B)
(D) none of these
Answer:
(B) cell plate formation

Question 7.
Meiosis is a
(A) homotypic division
(B) equatorial division
(C) reductional division
(D) none of the above
Answer:
(C) reductional division

Question 8.
Formation of Synaptonemal complex during meiosis occurs at
(A) Leptotene
(B) Zygotene
(C) Diplotene
(D) Pachytene
Answer:
(B) Zygotene

Question 9.
Crossing over takes place in the ________ stage.
(A) leptotene
(B) zygotene
(C) pachytene
(D) diplotene
Answer:
(C) pachytene

Question 10.
Crossing over takes place between
(A) sister chromatids
(B) non-homologous chromosomes
(C) non-sister chromatids of homologues
(D) any two chromatids
Answer:
(C) non-sister chromatids of homologues

Question 11.
Crossing over of chromosomes during meiosis leads to
(A) mutation
(B) sex determination
(C) new gene combination
(D) loss of chromosomes
Answer:
(C) new gene combination

Question 12.
Points at which crossing over has taken place between homologous chromosomes are called
(A) chiasmata
(B) synaptonemal complexes
(C) centromeres
(D) telomere
Answer:
(A) chiasmata

Question 13.
Which of the following events take place during diplotene stage of prophase I of meiosis?
(A) Compaction of chromosomes
(B) Formation of synapsis
(C) Process of crossing over
(D) Repulsion of homologues
Answer:
(D) Repulsion of homologues

Question 14.
The correct sequence of stages in prophase I of meiosis is
(A) Leptotene, Pachytene, Zygotene, Diakinesis, Diplotene
(B) Zygotene, Leptotene, Pachytene, Diakinesis, Diplotene
(C) Leptotene, Zygotene, Pachytene, Diplotene, Diakinesis
(D) Diplotene, Diakinesis, Pachytene, Zygotene, Leptotene
Answer:
(C) Leptotene, Zygotene, Pachytene, Diplotene, Diakinesis

Question 15.
In which phase of meiosis are homologous chromosomes separated?
(A) Anaphase I
(B) Prophase II
(C) Anaphase II
(D) Prophase I
Answer:
(A) Anaphase I

Question 16.
Mitosis differs from meiosis in not having
(A) Long prophase
(B) duplication of DNA
(C) Synapsis and crossing over
(D) interphase
Answer:
(C) Synapsis and crossing over

Question 17.
How many divisions are required to produce 128 gametes?
(A) 64
(B) 16
(C) 32
(D) 12
Answer:
(C) 32

Question 18.
Number of cells undergoing meiotic divisions to produce 124 microspores in angiosperm is
(A) 62
(B) 31
(C) 124
(D) 8
Answer:
(B) 31

Question 19.
How many haploid daughter cells are produced at the end of meiosis-II?
(A) 2
(B) 4
(C) 6
(D) 8
Answer:
(B) 4

Question 21.
Competitive Corner:

Question 1.
Crossing over takes place between which chromatids and in which stage of the cell cycle?
(A) Non-sister chromatids of nonhomologous chromosomes at Pachytene stage of prophase I
(B) Non-sister chromatids of nonhomologous chromosomes at Zygotene stage of prophase I
(C) Non-sister chromatids of homologous chromosomes at Pachytene stage of prophase I
(D) Non-sister chromatids of homologous chromosomes at Zygotene stage of prophase I
Answer:
(C) Non-sister chromatids of homologous chromosomes at Pachytene stage of prophase I

Question 2.
After meiosis I, the resultant daughter cells have
(A) four times the amount of DNA in comparison to haploid gamete.
(B) same amount of DNA as in the parent cell in S phase.
(C) twice the amount of DNA in comparison to haploid gamete.
(D) same amount of DNA in comparison to haploid gamete.
Answer:
(C) twice the amount of DNA in comparison to haploid gamete.

Question 3.
Cells in G₀ phase
(A) suspend the cell cycle
(B) terminate the cell cycle
(C) exit the cell cycle
(D) enter the cell cycle
Answer:
(C) exit the cell cycle

Question 4.
The CORRECT sequence of phases of cell cycle is: [NEET (UG) 2019]
(A) S → G₁ → G₂ → M
(B) G₁ → S → G₂ → M
(C) M → G₁ → G₂ → S
(D) G₁ → G₂ → S → M
Answer:
(B) G₁ → S → G₂ → M

Question 5.
The stage during which separation of the paired homologous chromosomes begins is
(A) Diakinesis
(B) Diplotene
(C) Pachytene
(D) Zygotene
Answer:
(B) Diplotene

Question 6.
Which of the following options gives the correct sequence of events during mitosis?
(A) Condensation → nuclear membrane disassembly → crossing over – segregation → telophase
(B) Condensation → nuclear membrane disassembly → arrangement at equator → centromere division → segregation → telophase
(C) Condensation → crossing over → nuclear membrane disassembly → segregation → telophase
(D) Condensation → arrangement at equator → centromere division → segregation → telophase
Answer:
(B) Condensation → nuclear membrane disassembly → arrangement at equator → centromere division → segregation → telophase

Question 7.
Which of the following is not a characteristic feature during mitosis in somatic cells?
(A) Chromosome movement
(B) Synapsis
(C) Spindle fibres
(D) Disappearance of nucleolus
Answer:
(B) Synapsis

Question 8.
Arrange the following events of meiosis in correct sequence. [AIPMT Retest 2015]
(a) Crossing over
(b) Synapsis
(c) Terminalisation of chiasmata
(d) Disappearance of nucleolus
(A) (b), (c), (d), (a)
(B) (b), (a), (d), (c)
(C) (b),(a), (c), (d)
(D) (a), (b), (c), (d)
Answer:
(C) (b),(a), (c), (d)

Maharashtra Board 11th BK Important Questions Chapter 4 Ledger

July 22, 2024July 21, 2024 by Bhagya

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 4 Ledger Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 4 Ledger

1. Answer in one sentence only.

Question 1.
State the meaning of ‘Debit Balance’.
Answer:
Excess of the debit total of an account over its credit total is called debit balance.

Question 2.
What do you mean by Credit Balance?
Answer:
Excess of credit total of an account over its debit total is called credit balance.

Question 3.
What is the balance of Outstanding Salary A/c?
Answer:
The balance of Outstanding Salary A/c is a credit balance.

Question 4.
What is the balance of Cash A/c?
Answer:
The balance of Cash A/c is always a debit balance.

Question 5.
Where is the balance of Salaries A/c transferred at the end of the year?
Answer:
At the end of the year, the balance of Salaries A/c is transferred to Profit and Loss A/c.

Question 6.
From which books of accounts posting is made in the Ledger?
Answer:
From Journal and Subsidiary books posting is made in the ledger.

Question 7.
What do you mean by the head of an Account?
Answer:
The name of the account which is written on the top of the ledger account in the middle is called “Head of an account”.

2. Write the word, term, phrase, which can substitute each of the statements.

Question 1.
A bound book of Account.
Answer:
Ledger

Question 2.
The credit balance of the Bank Account.
Answer:
Bank Overdraft

Question 3.
Recording of transactions from Journal to Ledger.
Answer:
Ledger Posting

Question 4.
Page number of Ledger.
Answer:
L.F. No.

Question 5.
The right-hand side of an account.
Answer:
Credit Side

Question 6.
The left-hand side of an account.
Answer:
Debit Side

Question 7.
Excess of a debit total of an account over its credit total.
Answer:
Debit Balance

Question 8.
Excess of credit total of an account over its debit total.
Answer:
Credit Balance

Question 9.
A secondary book of account in which entries are recorded from Journal or Subsidiary books.
Answer:
Ledger

Question 10.
Page of the account book.
Answer:
Folio

Question 11.
A book of accounts in which all ledger accounts are maintained.
Answer:
Ledger

Question 12.
Type of accounts that are not balanced but transferred to Trading Account or P/L A/c.
Answer:
Nominal Accounts

Question 13.
An account where the total purchase book is posted.
Answer:
Purchases Account

Question 14.
A statement of accounts’s prepared from the balances of the ledger account.
Answer:
Trial Balance

3. Select appropriate alternatives from those given below and rewrite the sentences.

Question 1.
_______________ is the act of transferring an entry from journal to ledger.
(a) Journalising
(b) Casting
(c) Balancing
(d) Posting
Answer:
(d) Posting

Question 2.
Cash account always shows _______________ balance.
(a) Credit
(b) Negative
(c) Minus
(d) Debit
Answer:
(d) Debit

Question 3.
Bank overdraft means _______________ balance of Bank A/c.
(a) Debit
(b) Credit
(c) Positive
(d) Nil
Answer:
(b) Credit

Question 4.
Real accounts generally shows _______________ balance.
(a) Debit
(b) Credit
(c) Negative
(d) Nil
Answer:
(a) Debit

Question 5.
Excess of debit total of an account over it’s credit total indicates _______________ balance.
(a) Negative
(b) Positive
(c) Debit
(d) Credit
Answer:
(c) Debit

Question 6.
When goods are sold on credit _______________ account is credited.
(a) Buyer’s A/c
(b) Cash A/c
(c) Sales A/c
(d) Seller’s A/c
Answer:
(c) Sales A/c

Question 7.
When the total of debit side of an account exceeds credit side, it is called _______________ balance.
(a) Debit
(b) Credit
(c) Nil
(d) Real
Answer:
(a) Debit

Question 8.
Machinery A/c shows _______________ balance.
(a) Credit
(b) Debit
(c) Positive
(d) Negative
Answer:
(b) Debit

Question 9.
Left hand side of an account is called _______________
(a) Credit
(b) Debit
(c) Middle
(d) Centre
Answer:
(b) Debit

Question 10.
_______________ is prepared from the balances in the ledger accounts.
(a) List
(b) Journal
(c) Book
(d) Trial Balance
Answer:
(d) Trial Balance

Question 11.
Total of Sales book is _______________ to sales account.
(a) entered
(b) moved
(c) posted
(d) given
Answer:
(c) posted

Question 12.
Wages A/c is transferred to _______________ A/c.
(a) Trading A/c
(b) Profit & Loss A/c
(c) Trial Balance
(d) Any
Answer:
(a) Trading A/c

Question 13.
All entries are posted from Journal to _______________
(a) Ledger
(b) Balance Sheet
(c) Trial Balance
(d) Cash A/c
Answer:
(a) Ledger

4. State whether the following statements are ‘True or False’ with reasons.

Question 1.
Ledger accounts are balanced every day.
Answer:
This statement is False.
Ledger accounts are balanced on a particular period of time or on the accounting year of the firm. They are not balanced every day.

Question 2.
Every account is divided into two sides.
Answer:
This statement is True.
Every account is divided into two sides i.e. Debit and Credit to do ledger pasting from a journal which follows the rules of double-entry book-keeping.

Question 3.
The discount allowed is a real account.
Answer:
This statement is False.
Discount allowed is an expense for the business. All expenses come under the Nominal account. So the discount allowed is a Nominal account.

Question 4.
The bank is a Nominal A/c.
Answer:
This statement is False.
A bank is an Artificial person created by law. The bank is personal accounts.

Question 5.
Accounts of expenses may show a credit balance.
Answer:
This statement is False.
Accounts of expenses always show debit balance. They fall under the nominal account and its Golden rule is to Debit all Expenses.

Question 6.
A Debtors A/c always show a credit balance.
Answer:
This statement is False.
A Debtor A/c always shows a Debit balance. The debtor is a personal account and as per the rule of personal account Debit the receiver.

Question 7.
When goods are sold for cash, the personal a/c of the buyer is debited.
Answer:
This statement is False.
In a cash transaction, there is no Personal A/c. The personal account comes in a credit transaction. So when goods are sold for cash, the cash account is debited. A cash account is a real account and as per rule debit what comes in.

Question 8.
Capital a/c is a nominal a/c.
Answer:
This statement is False.
Capital is a Personal account. Capital is the amount invested by the proprietor in his business. A proprietor is a person. Capital is also personal A/c.

Question 9.
Drawing a/c is a personal a/c.
Answer:
This statement is True.
Drawings mean cash or goods withdrawn by the proprietor for his personal or family use he is a person. So drawing is personal A/c.

Question 10.
Narration is not necessary for a ledger.
Answer:
This statement is False.
Narration is necessary for a Journal book and not in Ledger. Ledgers are self-narrative as every ledger has headed.

Question 11.
All a/c’s are closed down at the end of the accounting year.
Answer:
This statement is True.
All ledger accounts are closed an accounting year or Financial year to prepare the Trial balance and Financial statements of the year.

Question 12.
Recording of a transaction in the journal is called posting.
Answer:
This statement is False.
The process of recording a transaction in the Journal is called Journalising. Posting means transferring journal entries to respective ledger accounts.

Question 13.
Balance of personal a/c is brought down for the next year.
Answer:
This statement is True.
Personal accounts are the account of Debtors or Creditors. The business has to collect or pay amounts on personal accounts. Current year collections or payments are done in the following year. So balances of personal accounts are brought for the next year.

Question 14.
All transactions are recorded directly in the ledger.
Answer:
This statement is False.
First, every transaction is recorded in a Journal book or Subsidiary book and from there they are posted to respective ledger accounts. No transaction can be recorded directly in the ledger.

Question 15.
Ledger Folio and index are necessary for the ledger.
Answer:
This statement is True.
The index is necessary to final particular account on its page number and ledger folio number is also necessary for cross-checking with the journal book. This makes handling convenient and easier.

Question 16.
Ledger posting is not necessary for journal proper.
Answer:
This statement is False.
Ledger posting is necessary for journal proper. From Ledger trial balance and financial statements are prepared for the accounting year. Without posting of journal proper Trial balance and financial statements will not be tally and can’t give true and fair accounting.

Question 17.
Ledger Folio is recorded in the journal.
Answer:
This statement is True.
Ledger folio is a page of the ledger, where posting is made from a journal book. For cross-reference, the ledger folio is recorded in the journal.

Question 18.
Ledger posting is made before journal entry.
Answer:
This statement is False.
Ledger posting is made after journal entry. First, all transactions are recorded to Journal and from the journal, posting is made to Ledger.

5. Fill in the blanks.

Question 1.
Debtors A/c shows _______________ balance.
Answer:
Debit

Question 2.
Left hand side of an account is called _______________ side.
Answer:
Debit

Question 3.
_______________ is prepared from the balances in the Ledger A/c.
Answer:
Trial Balance

Question 4.
Total of salary A/c is transfered to _______________ A/c.
Answer:
Profit and loss A/c

Question 5.
b/d (brought down) balance indicates _______________ balance.
Answer:
Opening

Question 6.
_______________ Balance on Nominal Account shows Inconies or gains.
Answer:
Credit

Question 7.
Debtors A/c shows _______________ balance.
Answer:
Debit

Question 8.
Drawing A/c shows _______________ balance.
Answer:
Debit

Question 9.
Wages A/c balance transferred to _______________ A/c.
Answer:
Profit and Loss

Question 10.
Discount allowed shows _______________ Balance.
Answer:
Debit

Question 11.
Dividend received shows _______________ Balance.
Answer:
Credit

Question 12.
Ledger is the _______________ book of Accounts.
Answer:
Principal

6. Complete the following table.

Question 1.
Maharashtra Board 11th BK Important Questions Chapter 4 Ledger 6 Q1
Answer:
Credit balance

Question 2.
Maharashtra Board 11th BK Important Questions Chapter 4 Ledger 6 Q2
Answer:
Ledger

Question 3.
Maharashtra Board 11th BK Important Questions Chapter 4 Ledger 6 Q3
Answer:
Credit Balance

Question 4.
Maharashtra Board 11th BK Important Questions Chapter 4 Ledger 6 Q4
Answer:
Livestock A/c

Question 5.
Maharashtra Board 11th BK Important Questions Chapter 4 Ledger 6 Q5
Answer:
Output GST

7. Put ‘✓’ mark for the nature of balance for the following.

Question 1.

Account	Dr. Balance	Cr. Balance
1. Salary A/c
2. Purchase A/c
3. Sales Return A/c
4. Machinery A/c
5. Bank Balance A/c
6. Cash Balance A/c
7. Sale A/c
8. Purchase Return A/c
9. Bills Payable A/c
10. Debtors A/c
11. Creditors A/c
12. Stationery A/c
13. Furniture A/c
14. Amit’s Loan A/c

Answer:

Account	Dr. Balance	Cr. Balance
1. Salary A/c	✓
2. Purchase A/c	✓
3. Sales Return A/c	✓
4. Machinery A/c	✓
5. Bank Balance A/c	✓
6. Cash Balance A/c	✓
7. Sale A/c		✓
8. Purchase Return A/c		✓
9. Bills Payable A/c		✓
10. Debtors A/c	✓
11. Creditors A/c		✓
12. Stationery A/c	✓
13. Furniture A/c	✓
14. Amit’s Loan A/c		✓

Maharashtra Board Class 11 Biology Important Questions Chapter 6 Biomolecules

July 20, 2024July 19, 2024 by Bhagya

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 6 Biomolecules Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 6 Biomolecules

Question 1.
How are living organisms classified? Give examples of each.
Answer:
1. Living organisms are classified as unicellular (consisting of single cell) and multicellular (having many cells).
2. Example of unicellular organisms: bacteria, yeast.
Example of multicellular organisms: plants, animals.

Question 2.
What is biochemistry?
Answer:
1. Biochemistry is biological chemistry that provides us the idea of the chemistry of living organisms and molecular basis for changes taking place in plants, animals and microbial cells.
2. It develops the foundation for understanding all biological processes and communication within and between cells as well as chemical basis of inheritance and diseases in animals and plants.

Question 3.
What does chemical analysis of living organisms indicate?
Answer:
Chemical analysis of all living organisms indicates the presence of the most common elements like carbon, hydrogen, nitrogen, oxygen, sulphur, calcium, phosphorus, magnesium and others with their respective content per unit mass of a living tissue.

Question 4.
Name the basic macromolecules present in the living organisms.
Answer:
Polysaccharides (carbohydrate) polymer of monosaccharide, polypeptides (proteins) polymer of amino acids and polynucleotides (nucleic acids) polymer of nucleotides are the three basic macromolecule present in the living organisms.

Question 5.
Draw a flowchart showing classification of carbohydrates.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 6 Biomolecules 1

Question 6.
Write a short note on
1. Glucose
2. Galactose and
3. Fructose.
Answer:
1. Glucose:
a. It is the most important fuel in living cells.
b. Its concentration in the human blood is about 90mg per 100ml of blood.
c. The small size and solubility in water of glucose molecules allows them to pass through the cell membrane into the cell.
d. Energy is released when the molecules are metabolized by cellular respiration.

2. Galactose:
a. It looks very similar to glucose molecules.
b. They can also exist in a and p forms.
c. Galactose react with glucose to form the disaccharide lactose.
d. However, glucose and galactose cannot be easily converted into one another.
e. Galactose cannot play the same role in respiration as glucose.

3. Fructose:
a. It is the fruit sugar and chemically it is ketohexose but it has a five-atom ring rather than a six-atom ring.
b. Fructose reacts with glucose to form the sucrose, a disaccharide.

Question 7.
How are disaccharides absorbed through the cell membrane?
Answer:
1. Disaccharides are soluble in water but they are too big to pass through the cell membrane by diffusion.
2. They are broken down in the small intestine during digestion.
3. Thus, formed monosaccharides then pas into the blood and through cell membranes into the cells.

Question 8.
Identify the X and Y in the following structure of a disaccharide.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 6 Biomolecules 3
X: Glycosidic bond Y: Glucose

Question 9.
Distinguish between monosaccharides and disaccharides.
Answer:

Monosaccharides	Disaccharides
1. They are composed of 3-6 carbon atoms.	They are composed of two monosaccharide units covalently linked to each other.
2. They cannot be hydrolyzed into smaller units.	They can be hydrolysed into monosaccharides.
3. Glucose, Fructose	Sucrose and Lactose

Question 10.
Which macromolecules are too big to escape from the cell?
Answer:
Polysaccharides are too big to escape from the cell.

Question 11.
Write a short note on
1. Starch
2. Glycogen
3. Cellulose.
Answer:
1. Starch:
a. Starch is a stored food in the plants.
b. Starch contains two types of glucose polymer: amylose and amylopectin.
c. Both are made from a-glucose.
d. Amylose is an unbranched polymer of a-glucose.
e. The molecules coil into a helical structure.
f. It foims a colloidal suspension in hot water.
g. Amylopectin is a branched polymer of a-glucose.
h. It is completely insoluble in water.
Maharashtra Board Class 11 Biology Important Questions Chapter 6 Biomolecules 4

2. Glycogen:
a. It is amylopectin with very short distances between the branching side-chains.
b. Glycogen is stored in animal body particularly in liver and muscles from where it is hydrolyzed as per need to produce glucose.

3. Cellulose:
a. It is a polymer made from P-glucosc molecules and the polymer molecules are ‘straight’.
b. Cellulose serves to form the cell walls in plant cells.
c. These are much tougher than cell membranes.
d. This toughness is due to the arrangement of glucose units in the polymer chain and the hydrogen-bonding between neighbouring chains.

Question 12.
Why plant fats are liquid at room temperature while animal fats are solid?
Answer:

Plant fats are unsaturated fatty acids, whereas animal fats are saturated fatty acids.
Fats having unsaturated fatty acids are liquid at room temperature.
Saturated fatty acids are solid at room temperature. Hence, plant fats are liquid at room temperature, while animal fats are solid.

Question 13.
Draw the structure of triglyceride.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 6 Biomolecules 5

Question 14.
Observe the following diagram and answer the questions based on it.
1. Identify the part ‘X’ in the given diagram.
2. What is the chemical property of the part ‘X’.
Answer:
1. The part labelled as ‘X’ is non-polar tail.
2. Non-polar tails are hydrophobic in nature.

Question 15.
Give two examples of unsaturated fatty acids.
Answer:
Oleic acid and linoleic acid are the examples of unsaturated fatty acids.

Question 16.
Explain primary structure of protein.
Answer:
The linear sequence of amino acids in polypeptide chain of a protein forms the primary structure of a protein.

Question 17.
Explain the secondary structure of protein with examples.
Answer:

There are two types of secondary structure of protein: a-helix and P-pleated sheets.
The polypeptide chain is arranged in a spiral helix. These spiral helices are of two types: a-helix (right handed) and P-helix (left handed).
This spiral configuration is held together by hydrogen bonds.
The sequence of amino acids in the polypeptide chain determines the location of its bend or fold and the position of formation of hydrogen bonds between different portions of the chain or between different chains. Thus, peptide chains form an a-helix structure.
Example of a-helix structure is keratin.
In some proteins two or more peptide chains are linked together by intermolecular hydrogen bonds. Such structures are called P-pleated sheets.
Example of P-pleated sheet is silk fibres.
Due to formation of hydrogen bonds peptide chains assume a secondary structure.

Question 18.
Explain the tertiary and quaternary structure of protein with example.
Answer:
Tertiary structure:

In tertiary structure the peptide chains are much looped, twisted and folded back on themselves due to formation of disulphide bonds.
Such loops and bends give the protein a tertiary structure.
E.g. Myoglobin, enzymes.

Quaternary structure:
1. When a protein has more than two polypeptide subunits their arrangement in space is called quaternary structure.
2. E.g. Haemoglobin.

Question 19.
Write a note on properties of protein.
Answer:
Properties of proteins are as follows:

Proteins are extremely reactive and highly specific in behaviour.
Proteins are amphoteric in nature i.e. they act as both acids and bases.
The behaviour of proteins is strongly influenced by pH.
Like amino acids, proteins are dipolar ions at the isoelectric point i.e. the sum of the positive charges is equal to the sum of the negative charges and the net charge is zero.
The ionic groups of a protein are contributed by the side chains of the polyfunctional amino acids.
A protein consists of more basic amino acids such as lysine and arginine exist as a cation at the physiological pH of 7.4. Such proteins are called basic proteins.
Histones of nucleoproteins are basic proteins.
A protein rich in acidic amino acids exists as an anion at the physiological pH. Such proteins are called acidic proteins.
Most of the blood proteins are acidic proteins.

Question 20.
Mention the findings of Feulgen.
Answer:
1. In 1924, Feulgen showed that chromosomes contain DNA.
2. He found that nucleic acids contain two pyrimidine (cytosine and thymine) and two purine (adenine and guanine) bases.

Question 21.
What were the findings of Wilkins and coworkers?
Answer:
The findings of Wilkins and coworkers were as follows:

Purine and pyrimidine bases are placed regularly along the DNA molecules at a distance of 3.4 A.
DNA (Deoxyribonucleotide) is composed of sugar molecule (a pentose sugar of deoxyribose type), phosphoric acid (phosphates when in chemical combination), nitrogen containing bases (nitrogen containing organic ring compounds).
Bases are of two types: Pyrimidine bases and purine bases.
Pyrimidine bases are single ring (monocyclic) nitrogenous bases. Cytosine, Thymine and uracil are pyrimidines.
Purine are double ring (dicyclic) nitrogenous bases. Adenine and guanine are purines.

Question 22.
Chargaff analyzed the composition of DNA from various sources. Mention what were his implications from all his experiments.
Answer:
Implications proposed by Erwin Chargaff:
1. Purine and pyrimidine always occur in equal amount in DNA.
2. The base ratio i.e. A+T/G+C may vary in the DNA of different groups of animals and plants but the ratio remains constant for particular species.

Question 23.
Describe the structure of DNA.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 6 Biomolecules 6

DNA is a long chain made up of alternate sugar and phosphate groups. The sugar present in DNA is always a deoxyribose attached to a phosphate group. So, it forms a regular, repeating phosphate sugar sequence.
A base is attached to sugar -phosphate chain. Together this unit which consist of sugar, phosphate and a base is called nucleotide.
The nitrogenous base and a sugar of a nucleotide form a molecule called nucleoside. It lacks phosphate group. Four types of nucleoside are found in DNA molecule.
In a nucleoside, nitrogenous base is attached to the first carbon atom (C-1) of the sugar and when a phosphate group gets attached with that of the carbon (C-5) atom of the sugar molecule a nucleotide molecule is formed.
A single strand of DNA consists of several thousands of nucleotides one above the other.
The phosphate group of the lower nucleotide attached with the 5th carbon atom of the deoxyribose sugar forms phospho-di-ester bond with that of the 3rd carbon atom of the deoxyribose sugar of the nucleotide placed just above it.
Single long chain of polynucleotides of DNA consists of one end with sugar molecules not connected with another nucleotide having C-3 carbon which is not connected with phosphate group, similarly the other end having C-5 of the sugar is not connected with any phosphate group. These two ends of the polynucleotide chain are called as 3′ and 5′ ends respectively.
The single polynucleotide strand of DNA is not straight but helical in shape.
The DNA molecule consists of such two helical polynucleotide chains which are complementary to each other.
The two complementary polynucleotide chains of DNA are held together by the weak hydrogen bonds.
Adenine always pairs with thymine, and guanine with cytosine (a pyrimidine with a purine).
Adenine-thymine pair consists of two hydrogen bonds and guanine-cytosine pair consists of three hydrogen bonds (Thus, if the sequence of bases of a polynucleotide chain is known, that of the other can be determined).

Question 24.
Draw the structures of nitrogen bases in nucleic acid.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 6 Biomolecules 7

Question 25.
Describe the structure of RNA.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 6 Biomolecules 8

The other nucleic acid found in living organisms is Ribose nucleic acid.
In most of the organisms it is not found to be hereditary material but in certain organisms like tobacco mosaic virus, it is the hereditary material.
Like DNA, ribose nucleic acid also consists of polynucleotide chain with the difference that it consists of single strand. Exceptions are Reovirus and wound tumor virus where RNA is double stranded.
The nucleotides of RNA have ribose sugar instead of the deoxyribose sugar as in the case of DNA.
In case of RNA, Uracil substitutes thymine of DNA.
Purine, pyrimidine equality is not found in RNA molecule because of its single stranded structure.
RNA strand is usually found folded upon itself in certain regions or entirely. These folding helps in stability of the RNA molecule.
Most of the RNA polynucleotide chains start either with adenine or guanine.
Three types of cellular RNAs have been distinguished:
1. messenger RNA (mRNA) or template RNA,
2. ribosomal RNA (rRNA),
3. transfer RNA (tRNA) or soluble RNA.

Question 26.
Observe the following figure and name the type of bond shown by arrow in the structure.
Maharashtra Board Class 11 Biology Important Questions Chapter 6 Biomolecules 9
Answer:
The type of bond shown in the diagram is hydrogen bond.

Question 27.
What would have happened if there were no enzymes in the body?
Answer:
If enzymes were absent in the body, either the reactions would not occur or if they occur they would occur at a very slow rate.

Question 28.
How many reactions are catalyzed by an enzyme?
Answer:
Each enzyme catalyzes only one reaction.

Question 29.
What is a substrate?
Answer:
The substance upon which an enzyme acts is termed as the substrate.

Question 30.
What is endo-enzyines? Give examples.
Answer:
The enzymes which act within the cell in which they are synthesized are known as endo-enzymes E.g., enzymes produced in the chloroplast and mitochondria.

Question 31.
What are exo-enzymes?
Answer:
1. The enzymes which act outside the cell of which they are synthesized are known as exo-enzymes. E.g. enzymes released by many fungi.
2. These enzymes, synthesized by living cell, retain their catalytic property even when extracted from cells.

Question 32.
How are enzymes categorised?
Answer:
On basis of chemical composition enzymes are categorised:
1. Purely proteinaceous enzymes: e.g. Proteases that spilt protein
2. Conjugated enzymes: enzymes are made up of a protein to which a non-protein prosthetic group is attached.

Question 33.
What is a prosthetic group ? What w ill happen if it is removed?
Answer:
1. Prosthetic group is non-protein in nature and is attached to the protein component of enzyme by chemical bonds.
2. It is not removed by hydrolysis.
3. If the prosthetic group is removed the protein part of the enzyme becomes inactive.

Question 34.
What are coenzymes?
Answer:
1. Enzymes require certain organic compounds for their activity.
2. The organic compounds that are tightly attached to the protein part are called coenzymes.
3. E.g. Nicotinamaide adenine dinucleotide (NAD), Flavin mononucleotide (FMN).

Question 35.
What are co-factors?
Answer:
1. Enzymes require certain inorganic ions for their activity.
2. The inorganic ions which are loosely attached to the protein part are called co-factors.
E.g. Magnesium, copper, zinc, iron, manganese etc.
[Note: Apoenzyme (protein part) and co-factor together form a complete catalytically active enzyme which is known as holoenzyme. The co-factors are of two types; metal ions and coenzymes. A coenzyme or metal ion that is very tightly or even covalently bound to the enzyme protein is known as a prosthetic group.]

Question 36.
Complete the analogy.
Iron (Fe): Catalase: Manganese (Mn):
Answer:
Peptidase

Question 37.
Give examples of coenzymes and cofactors.
Answer:
1. Nicotinamaide adenine dinucleotide (NAD), Flavin mononucleotide (FMN).
2. Magnesium, copper, zinc, iron, manganese etc.

Question 38.
How are enzymes named?
Answer:

Enzymes are named by adding the suffix- ‘ase’ to the name of the substrate on which they act e.g. protease, sucrase, nuclease etc. which break up proteins, sucrose and nucleic acids respectively.
The enzymes can be named according to the type of function they perform.
For e.g., dehydrogenase remove hydrogen, carboxylase add CO; decarboxylases remove C0₂, oxidases helping in oxidation.
Some enzymes are named according to the source from which they are obtained.
For e.g., papain from papaya, bromelain from the member of Bromeliaceae family, pineapple.
According to international code of enzyme nomenclature, the name of each enzyme ends with an -ase and consists of double name!
The first name indicates the nature of substrate upon which the enzyme acts and the second name indicates the reaction catalyzed.

For e.g., pyruvic decarboxylase catalyses the removal of C0₂ from the substrate pyruvic acid.
Similarly, the enzyme glutamate pyruvate transaminase catalyses the transfer of an amino group from the substrate glutamate to another substrate pyruvate.

Question 39.
Explain in detail the mechanism of enzyme action. Write a note on model proposed by Emil Fischer for mechanism of enzyme action.
Answer:
1. The basic mechanism by which enzymes catalyze chemical reactions begins with the binding of the substrate (or substrates) to the active site on the enzyme.
2. The active site is the specific region of the enzyme which combines with the substrate.
3. The binding of the substrate to the enzyme causes changes in the distribution of electrons in the chemical bonds of the substrate and ultimately causes the reactions that lead to the formation of products.
4. The products are released from the enzyme surface to regenerate the enzyme for another reaction cycle.
Maharashtra Board Class 11 Biology Important Questions Chapter 6 Biomolecules 10
5. Lock and Key model proposed by Emil Fischer: i. Proteinaceous Nature:
All enzymes are basically made up of protein.

Question 40.
Describe the concept of metabolism.
Answer:

Metabolism is the sum of the chemical reactions that take place within each cell of a living organism and provide energy for vital processes and for synthesizing new organic material.
It involves continuous process of breakdown and synthesis of biomolecules through chemical reactions.
Each of the metabolic reaction results in a transformation of biomolecules.
Most of these metabolic reactions do not occur in isolation but are always linked with some other reactions.
In cells, metabolism involves two following types of pathways:

a. Catabolic pathways
This involves formation of simpler structure from a complex biomolecule.
For e.g. when we eat wheat, bread or chapati, our gastrointestinal tract digests (hydrolyses) the starch to glucose units with help of enzymes and releases energy in form of ATP (Adenosine triphosphate).

b. Anabolic pathway
It is also called as biosynthetic pathway that involves formation of a more complex biomolecules from a simpler structure
Fone.g., synthesis of glycogen from glucose and protein from amino acids. These pathways consume energy.

Question 41.
Draw a flowchart showing catabolic and anabolic reactions.
Answer:
Maharashtra Board Class 11 Biology Important Questions Chapter 6 Biomolecules 11

Question 42.
Write a short note on secondary metabolites.
Answer:

Secondary metabolites are small organic molecules produced by organisms that are not essential for their growth, development and reproduction.
Several types of bacteria, fungi and plants produce secondary metabolites.
Secondary metabolites can be classified on the basis of chemical structure (e.g. SMs containing rings, sugar), composition (with or without nitrogen), their solubility in various solvents, or the pathway by which they are synthesized (e.g. phenylpropanoid produces tannins).
A simple way of classifying secondary metabolites includes three main groups such as:
- Terpenes: Made from mevalonic acid that is composed mainly of carbon and hydrogen
- Phenolics: Made from simple sugars containing benzene rings, hydrogen and oxygen.
- Nitrogen-containing compounds: Extremely diverse class may also contain sulphur.

Question 43.
Fill in the blanks.

Living organism have _________ as the basic structural and functional unit.
The cells have _______ containing numerous chemical molecules, the biomolecules.
________ are used very quickly by cells but if a cell is not in need of all the energy released immediately then it may get stored.
By ________ reaction monosaccharide is converted to disaccharide.
The balance between catabolism and anabolism maintain _______ in the cell as well as in the whole body.

Answer:

Cell
Protoplasm
monosaccharides
Condensation reaction
Homeostasis

Question 44.
Apply Your Knowledge:

Question 1.
While performing an experiment, to understand effect of pH on enzyme activity, a student prepared
solution of varied pH. He observed that enzyme activity is maximum at a particular range of pH.
What is the reason for its maximum activity at a particular range of pH? What would be the effect on enzyme activity if strong acid or strong base is added?
Answer:
The enzymes are highly specific to pH and remain active within particular range of pH only. Hence, exhibit maximum activity only at particular range of pH. When strong acid or strong base is added in the reaction the enzyme activity is inhibited as most of the enzymes are denatured.

Question 2.
When a compound ‘x’ is added to a chemical solution containing enzyme and substrate, the enzymatic activity stops. What could be the nature of compound ‘x’?
Answer:
Compound ‘x’ could be either competitive or non-competitive inhibitor.

Question 45.
Quick Review

Maharashtra Board Class 11 Biology Important Questions Chapter 6 Biomolecules 12

Question 46.
Exercise:

Question 1.
Draw a flow chart of biomolecules in living system.
Answer:
Refer Quick Review

Question 2.
Explain what is biochemistry?
Answer:
1. Biochemistry is biological chemistry that provides us the idea of the chemistry of living organisms and molecular basis for changes taking place in plants, animals and microbial cells.
2. It develops the foundation for understanding all biological processes and communication within and between cells as well as chemical basis of inheritance and diseases in animals and plants.

Question 3.
Mention the basic macromolecules present in the living organism.
Answer:
Polysaccharides (carbohydrate) polymer of monosaccharide, polypeptides (proteins) polymer of amino acids and polynucleotides (nucleic acids) polymer of nucleotides are the three basic macromolecule present in the living organisms.

Question 4.
Write a note on monosaccharides.
Answer:
Monosaccharides:
a. Monosaccharides are the simplest sugars having crystalline structure, sweet taste and soluble in water.
b. They cannot be further hydrolyzed into smaller molecules.
c. They are the building blocks or monomers of complex carbohydrates.
d. They have the general molecular formula (CH20)n, where n can be 3, 4, 5, 6 and 7.
e. They can be classified as triose, tetrose, pentose, etc.
f. Monosaccharides containing the aldehyde (-CHO) group are classified as aldoses e.g. glucose, xylose, and those with a ketone(-C=0) group are classified as ketoses. E.g. ribulose, fructose.

Question 5.
Explain the absorption of disaccharides through the cell membrane.
Answer:
1. Disaccharides are soluble in water but they are too big to pass through the cell membrane by diffusion.
2. They are broken down in the small intestine during digestion.
3. Thus, formed monosaccharides then pass into the blood and through cell membranes into the cells.

Question 6.
Draw the structure of amylose.
Answer:
Starch:
a. Starch is a stored food in the plants.
b. Starch contains two types of glucose polymer: amylose and amylopectin.
c. Both are made from a-glucose.
d. Amylose is an unbranched polymer of a-glucose.
e. The molecules coil into a helical structure.
f. It foims a colloidal suspension in hot water.
g. Amylopectin is a branched polymer of a-glucose.
h. It is completely insoluble in water.

Question 7.
Write the significance of carbohydrates.
Answer:
Significances of carbohydrates are as follows:

Carbohydrates provide energy for metabolism.
Glucose is the main substrate for ATP synthesis.
Lactose, a disaccharide present in the milk provides energy to babies.
Polysaccharide serves as a structural component of cell membrane, cell wall and reserved food as starch and glycogen.

Question 8.
What is glycosidic bond?
Answer:
Oligosaccharides:
a. A carbohydrate polymer comprising of two to six monosaccharide molecules is called oligosaccharide.
b. They are linked together by glycosidic bond.
c. They are classified on the basis of monosaccharide units:
Disaccharides: These are the sugars containing two monosaccharide units and can be further hydrolysed into smaller components. E.g.: Sucrose, maltose, lactose, etc.
Trisaccharides: These contain three monomers. E.g. Raffmose.
Tetrasaccharides: These contain four monomers. E.g.: Stachyose.

Glycosidic bond:
a. Glycosidic bond is a covalent bond that forms a linkage between two monosaccharides by a dehydration reaction.
b. It is formed when a hydroxyl group of one sugar reacts with the anomeric carbon of the other.
c. Glycosidic bonds are readily hydrolyzed by acid but resist cleavage by base.
d. There are two types of glycosidic bonds: a-glycosidic bond and P-glycosidic bond.

Question 9.
What are saturated fatty acids?
Answer:
1. Saturated fatty acids: They contain single chain of carbon atoms with single bonds.
E.g. Palmitic acid, stearic acid
2. Unsaturated fatty acids: They contain one or more double bonds between the carbon atoms of the hydrocarbon chain.
a. Simple lipids: These are esters of fatty acids with various alcohols.
E.g. Fats, wax.
b. Compound lipids: These are ester of fatty acids containing other groups like phosphate (Phospholipids), sugar (glycolipids), etc.
E.g. Lecithin
c. Sterols: They are derived lipids. They are composed of fused hydrocarbon rings (steroid nucleus) and a long hydrocarbon side chain.
E.g. Cholesterol, phytosterols.

Question 10.
Write a note on simple lipids.
Answer:
Lipids are classified into three main types:
Simple lipids:
a. These are esters of fatty acids with various alcohols. Fats and waxes are simple lipids.
b. Fats are esters of fatty acids with glycerol (CH₂OH-CHOH-CH₂OH).
c. Triglycerides are three molecules of fatty acids and one molecule of glycerol.
d. Unsaturated fats are liquid at room temperature and are called oils. Unsaturated fatty acids are hydrogenated to produce fats e.g. Vanaspati ghee.

Question 11.
Write a note on derived lipids.
Answer:
Derived Lipids:
a. They are composed of fused hydrocarbon rings (steroid nucleus) and a long hydrocarbon side chain.
b. One of the most common sterols is cholesterol.
Biological significance:
a. It is widely distributed in all cells of the animal body, but particularly in nervous tissue.
b. Cholesterol exists either free or as cholesterol ester.
c. Adrenocorticoids, sex hormones (progesterone, testosterone) and vitamin D are synthesized from cholesterol.
d. Cholesterol is not found in plants.
e. Sterols exist as phytosterols in plants.
f. Yam Plant (Dioscorea) produces a steroid compound called diosgenin. It is used in the manufacture of antifertility pills, i.e. birth control pills.

Question 12.
What are compound lipids? Mention their biological significance.
Answer:
Compound lipids:
a. These are ester of fatty acids containing other groups like phosphate (Phospholipids), sugar (glycolipids), etc.
b. They contain a molecule of glycerol, two molecules of fatty acids and a phosphate group or simple sugar.
c. Some phospholipids such as lecithin also have a nitrogenous compound attached to the phosphate group.
d. Phospholipids have both hydrophilic polar groups (phosphate and nitrogenous group) and hydrophobic non-polar groups (hydrocarbon chains of fatty acids).
e. Glycolipids contain glycerol, fatty acids, simple sugars such as galactose. They are also called cerebrosides.
Biological significance:
a. Phospholipids contribute in the formation of cell membrane.
b. Large amounts of glycolipids are found in the brain white matter and myelin sheath.

Question 13.
Explain the classification of proteins based on their chemical composition.
Answer:
On the basis of structure, proteins are classified into three categories:
1. Simple proteins:
a. Simple proteins on hydrolysis yield only amino acids.
b. These are soluble in one or more solvents.
c. Simple proteins may be soluble in water.
d. Histones of nucleoproteins are soluble in water.
e. Globular molecules of histones are not coagulated by heat.
f. Albumins are also soluble in water but they get coagulated on heating.
g. Albumins are widely distributed e.g. egg albumin, serum albumin and legumelin of pulses are albumins.
Importance: They are involved in structural components; they also act as a storage kind of protein.
Some are associated with nucleic acids in nucleoproteins of cell.

2. Conjugated proteins:
a. Conjugated proteins consist of a simple protein united with some non-protein substance.
b. The non-protein group is called prosthetic group e.g. haemoglobin.
c. Globin is the protein and the iron containing pigment haem is the prosthetic group.
d. Similarly, nucleoproteins have nucleic acids.
e. Proteins are classified as glycoproteins and mucoproteins.
f. Mucoproteins are carbohydrate-protein complexes e.g. mucin of saliva and heparin of blood.
g. Lipoproteins are lipid-protein complexes e.g. conjugate protein found in brain, plasma membrane, milk etc. Importance: They are involved in structural components of cell membranes and organelles.
They also act as a transporter.
Some conjugated proteins are important in electron transport chain in respiration.

3. Derived proteins:
a. These proteins are not found in nature as such.
b. These proteins are derived from native protein molecules on hydrolysis.
c. Metaproteins, peptones are derived proteins.
Importance: They act as a precursor for many molecules which are essential for life.

Question 14.
What is peptide bond? Explain its formation.
Answer:
1. The covalent bond that links the two amino acids is called a peptide bond.
2. Peptide bond is formed by condensation reaction.

Question 15.
Mention the examples of simple proteins and write their significance.
Answer:
Examples of simple proteins are: E.g.: Albumins and histones.
Significance:
1. Albumin:
a. % It is the main protein in the blood.
b. It maintains the pressure in the blood vessels.
c. It helps in transportation of substances like hormone and drugs in the body.
2. Histones:
a. It is the chief protein of chromatin.
b. They are involved in packaging of DNA into structural units called nucleosomes.

Question 16.
What is nucleotide?
Answer:
Nucleotide is a unit which consists of a sugar, phosphate and a base. Nucleotides are basic units of nucleic acids.

Question 17.
Write a note on structure of DNA molecule proposed by Watson and Crick.
Answer:
1. DNA is a long chain made up of alternate sugar and phosphate groups. The sugar present in DNA is always a deoxyribose attached to a phosphate group. So, it forms a regular, repeating phosphate sugar sequence.
2. A base is attached to sugar -phosphate chain. Together this unit which consist of sugar, phosphate and a base is called nucleotide.
3. The nitrogenous base and a sugar of a nucleotide form a molecule called nucleoside. It lacks phosphate group. Four types of nucleoside are found in DNA molecule.
4. In a nucleoside, nitrogenous base is attached to the first carbon atom (C-1) of the sugar and when a phosphate group gets attached with that of the carbon (C-5) atom of the sugar molecule a nucleotide molecule is formed.
5. A single strand of DNA consists of several thousands of nucleotides one above the other.
6. The phosphate group of the lower nucleotide attached with the 5th carbon atom of the deoxyribose sugar forms phospho-di-ester bond with that of the 3rd carbon atom of the deoxyribose sugar of the nucleotide placed just above it.
7. Single long chain of polynucleotides of DNA consists of one end with sugar molecules not connected with another nucleotide having C-3 carbon which is not connected with phosphate group, similarly the other end having C-5 of the sugar is not connected with any phosphate group. These two ends of the polynucleotide chain are called as 3′ and 5′ ends respectively.
8. The single polynucleotide strand of DNA is not straight but helical in shape.
9. The DNA molecule consists of such two helical polynucleotide chains which are complementary to each other.
10. The two complementary polynucleotide chains of DNA are held together by the weak hydrogen bonds.
11. Adenine always pairs with thymine, and guanine with cytosine (a pyrimidine with a purine).
12. Adenine-thymine pair consists of two hydrogen bonds and guanine-cytosine pair consists of three hydrogen bonds (Thus, if the sequence of bases of a polynucleotide chain is known, that of the other can be determined).

According to Watson and Crick, DNA molecule consists of two strands twisted around each other in the form of a double helix.
The two strands i.e. polynucleotide chains are supposed to be in opposite direction so end of one chain having 3′ lies beside the 5′ end of the other.
One turn of the double helix of the DNA measures about 34A.
It consists paired nucleotides and the distance between two neighboring pair nucleotides is 3.4A.
The diameter of the DNA molecule has been found be 20A.

Question 18.
What is the function of ribosomal RNA?
Answer:
Ribosomal RNA (rRNA):
a. rRNA was discovered by Kurland in 1960.
b. It forms 50-60% part of ribosomes.
c. It accounts 80-90% of the cellular RNA.
d. It is synthesized in nucleus.
e. It gets coiled at various places due to intrachain complementary base pairing.
Role of ribosomal RNA: It provides proper binding site for m-RNA during protein synthesis.

Question 19.
Write a short note on m-RNA.
Answer:
Messenger RNA (mRNA):
a. It is a linear polynucleotide.
b. It accounts 3% of cellular RNA.
c. Its molecular weight is several million. , d. mRNA molecule carrying information to form a complete polypeptide chain is called cistron.
e. Size of mRNA is related to the size of message it contains.
f. Synthesis of mRNA begins at 5’ end of DNA strand and terminates at 3’ end.

Role of messenger RNA:
It carries genetic information from DNA to ribosomes, which are the sites of protein synthesis.
Maharashtra Board Class 11 Biology Important Questions Chapter 6 Biomolecules 14

Question 20.
Write a note on types of non-genetic RNA.
Answer:
There are three types of cellular RNAs:
1. messenger RNA (mRNA),
2. ribosomal RNA (rRNA),
3. transfer RNA (tRNA). ‘

1. Messenger RNA (mRNA):
a. It is a linear polynucleotide.
b. It accounts 3% of cellular RNA.
c. Its molecular weight is several million. , d. mRNA molecule carrying information to form a complete polypeptide chain is called cistron.
e. Size of mRNA is related to the size of message it contains.
f. Synthesis of mRNA begins at 5’ end of DNA strand and terminates at 3’ end.

Role of messenger RNA:
It carries genetic information from DNA to ribosomes, which are the sites of protein synthesis.

2. Ribosomal RNA (rRNA):
a. rRNA was discovered by Kurland in 1960.
b. It forms 50-60% part of ribosomes.
c. It accounts 80-90% of the cellular RNA.
d. It is synthesized in nucleus.
e. It gets coiled at various places due to intrachain complementary base pairing.
Role of ribosomal RNA: It provides proper binding site for m-RNA during protein synthesis.

3. Transfer RNA (tRNA):
a. These molecules are much smaller consisting of 70-80 nucleotides.
b. Due to presence of complementary base pairing at various places, it is shaped like clover-leaf.
c. Each tRNA can pick up particular amino acid.
d. Following four parts can be recognized on tRNA
1. DHU arm (Dihydroxyuracil loop/ amino acid recognition site
2. Amino acid binding site
3. Anticodon loop / codon recognition site
4. Ribosome recognition site.
e. In the anticodon loop of tRNA, three unpaired nucleotides are present called as anticodon which pair with codon present on mRNA.
f. The specific amino acids are attached at the 3’ end in acceptor stem of clover leaf of tRNA.
Role of transfer RNA: It helps in elongation of polypeptide chain during the process called translation.

Question 21.
What are co-factors? Give examples.
Answer:

Enzymes require certain inorganic ions for their activity.
The inorganic ions which are loosely attached to the protein part are called co-factors.
E.g. Magnesium, copper, zinc, iron, manganese etc.

[Note: Apoenzyme (protein part) and co-factor together form a complete catalytically active enzyme which is known as holoenzyme. The co-factors are of two types; metal ions and coenzymes. A coenzyme or metal ion that is very tightly or even covalently bound to the enzyme protein is known as a prosthetic group.]

Question 22.
Describe the important properties of enzymes.
Answer:
1. Proteinaceous Nature:
All enzymes are basically made up of protein.

2. Three-Dimensional conformation:
a. All enzymes have specific 3-dimensional conformation.
b. They have one or more active sites to which substrate (reactant) combines.
c. The points of active site where the substrate joins with the enzyme is called substrate binding site.

3. Catalytic property:
a. Enzymes are like inorganic catalysts and influence the speed of biochemical reactions but themselves remain unchanged.
b. After completion of the reaction and release of the product they remain active to catalyze again.
c. A small quantity of enzymes can catalyze the transformation of a very large quantity of the substrate
into an end product.
d. For example, sucrase can hydrolyze 100000 times of sucrose as compared with its own weight.

4. Specificity of action:
a. The ability of an enzyme to catalyze one specific reaction and essentially no other is perhaps its most significant property. Each enzyme acts upon a specific substrate or a specific group of substrates.
b. Enzymes are very sensitive to temperature and pH.
c. Each enzyme exhibits its highest activity at a specific pH i.e. optimum pH.
d. Any increase or decrease in pH causes decline in enzyme activity e.g. enzyme pepsin (secreted in stomach)shows highest activity at an optimum pH of 2 (acidic)

5. Temperature:
a. Enzymes are destroyed at higher temperature of 60-70°C or below, they are not destroyed but become inactive.
b. This inactive state is temporary and the enzyme can become active at suitable temperature.
c. Most of the enzymes work at an optimum temperature between 20°C and 35°C.

There are two types of models:
1. Lock and Key model:
a. Lock and Key model was first postulated in 1894 by Emil Fischer.
b. This model explains the specific action of an enzyme with a single substrate.
c. In this model, lock is the enzyme and key is the substrate.
d. The correctly sized key (substrate) fits into the key hole (active site) of the lock (enzyme).

2. Induced Fit model (Flexible Model):
a. Induced Fit model was first proposed in 1959 by Koshland.
b. This model states that approach of a substrate induces a conformational change in the enzyme.
c. It is the more accepted model to understand mode of action of enzyme.
d. The induced fit model shows that enzymes are rather flexible structures in which the active site continually reshapes by its interactions with the substrate until the time the substrate is completely bound to it.
e. It is also the point at which the final form and shape of the enzyme is determined.
[Note: Temperature is a factor affecting enzyme activity and not a property of enzyme.]

Question 23.
Explain the classification enzymes and mention the example of each class.
Answer:
1. Enzymes are biological macromolecules which act as a catalyst and accelerates the reaction in the body.
2. Enzymes are classified into six classes:
a. Oxidoreductases: These enzymes catalyze oxidation and reduction reactions by the transfer of hydrogen and/or oxygen, e.g. alcohol dehydrogenase
b. Transferases: These enzymes catalyse the transfer of certain groups between two molecules, e.g. glucokinase
c. Hydrolases: These enzymes catalyse hydrolytic reactions. This class includes amylases, proteases, lipases etc. e.g. Sucrase
d. Lyases: These enzymes are involved in elimination reactions resulting in the removal of a group of atoms from substrate molecule to leave a double bond. It includes aldolases, decarboxylases, and dehydratases, e.g. fumarate hydratase.
e. Isomerases: These enzymes catalyze structural rearrangements within a molecule. Their nomenclature is based on the type of isomerism. Thus, these enzymes are identified as racemases, epimerases, isomerases, mutases, e.g. xylose isomerase.
f. Ligases or Synthetases: These are the enzymes which catalyze the covalent linkage of the molecules utilizing the energy obtained from hydrolysis of an energy-rich compound like ATP, GTP e.g. glutathione synthetase, Pyruvate carboxylase.

Question 24.
Enlist the factors affecting the activity of enzymes.
Answer:
The factors affecting the enzyme activity are as follows:
1. Concentration of substrate:
a. Increase in the substrate concentration gradually increases the velocity of enzyme activity within the limited range of substrate levels.
b. A rectangular hyperbola is obtained when velocity is plotted against the substrate concentration.
c. Three distinct phases (A, B and C) of the reaction are observed in the graph.
Where V = Measured velocity, V_max = Maximum velocity, S = Substrate concentration,
K_m = Michaelis-Menten constant.
d. K_m or the Michaelis-Menten constant is defined as the substrate concentration (expressed in moles/lit) to produce half of maximum velocity in an enzyme catalyzed reaction.
e. It indicates that half of the enzyme molecules (i.e. 50%) are bound with the substrate molecules when the substrate concentration equals the Km value.
f. K_m value is a constant and a characteristic feature of a given enzyme.
g. It is a representative for measuring the strength of ES complex.
h. A low K_m value indicates a strong affinity between enzyme and substrate, whereas a high Km value reflects a weak affinity between them.
i. For majority of enzymes, the K_m values are in the range of 10_-5 to 10_-2 moles.

2. Enzyme Concentration:
a. The rate of an enzymatic reaction is directly proportional to the concentration of the substrate.
b. The rate of reaction is also directly proportional to the square root of the concentration of enzymes.
c. It means that the rate of reaction also increases with the increasing concentration of enzyme and the rate of reaction can also decrease by decreasing the concentration of enzyme.

3. Temperature:
a. The temperature at which the enzymes show maximum activity is called Optimum temperature.
b. The rate of chemical reaction is increased by a rise in temperature but this is true only over a limited range of temperature.
c. Enzymes rapidly denature at temperature above 40°C.
d. The activity of enzymes is reduced at low temperature.
e. The enzymatic reaction occurs best at or around 37°C which is the average normal body temperature in homeotherms.

4. Effect of pH:
a. The pH at which an enzyme catalyzes the reaction at the maximum rate is known as optimum pH.
b. The enzyme cannot perform its function beyond the range of its pH value.

5. Other substances:
a. The enzyme action is also increased or decreased in the presence of some other substances such as co-enzymes, activators and inhibitors.
b. Most of the enzymes are combination of a co-enzyme and an apo-enzyme.
c. Activators are the inorganic substances which increase the enzyme activity.
d. Inhibitor is the substance which reduces the enzyme activity.

Question 25.
With the help of lock and key theory explain the mechanism of enzyme action.
Answer:
1. Proteinaceous Nature:
All enzymes are basically made up of protein.

Question 26.
With the help of suitable examples give any three classes of enzymes.
Answer:
a. Oxidoreductases: These enzymes catalyze oxidation and reduction reactions by the transfer of hydrogen and/or oxygen, e.g. alcohol dehydrogenase
b. Transferases: These enzymes catalyse the transfer of certain groups between two molecules, e.g. glucokinase
c. Hydrolases: These enzymes catalyse hydrolytic reactions. This class includes amylases, proteases, lipases etc. e.g. Sucrase

Question 27.
Following graph represents the effect of substrate concentration on enzyme activity. Identify ‘X’ and ‘Y’ Write proper explanation of the process.
Maharashtra Board Class 11 Biology Important Questions Chapter 6 Biomolecules 17
Answer:
The factors affecting the enzyme activity are as follows:
1. Concentration of substrate:
a. Increase in the substrate concentration gradually increases the velocity of enzyme activity within the limited range of substrate levels.
b. A rectangular hyperbola is obtained when velocity is plotted against the substrate concentration.
c. Three distinct phases (A, B and C) of the reaction are observed in the graph.
Where V = Measured velocity, V_max = Maximum velocity, S = Substrate concentration,
K_m = Michaelis-Menten constant.
d. K_m or the Michaelis-Menten constant is defined as the substrate concentration (expressed in moles/lit) to produce half of maximum velocity in an enzyme catalyzed reaction.
e. It indicates that half of the enzyme molecules (i.e. 50%) are bound with the substrate molecules when the substrate concentration equals the Km value.
f. K_m value is a constant and a characteristic feature of a given enzyme.
g. It is a representative for measuring the strength of ES complex.
h. A low K_m value indicates a strong affinity between enzyme and substrate, whereas a high Km value reflects a weak affinity between them.
i. For majority of enzymes, the K_m values are in the range of 10_-5 to 10_-2 moles.

Question 28.
Explain the concept of metabolism.
Answer:

Metabolism is the sum of the chemical reactions that take place within each cell of a living organism and provide energy for vital processes and for synthesizing new organic material.
It involves continuous process of breakdown and synthesis of biomolecules through chemical reactions.
Each of the metabolic reaction results in a transformation of biomolecules.
Most of these metabolic reactions do not occur in isolation but are always linked with some other reactions.
In cells, metabolism involves two following types of pathways:

Question 29.
Distinguish between Catabolic pathways and anabolic pathways.
Answer:
In cells, metabolism involves two following types of pathways:

Question 30.
Write the application of secondary metabolites.
Answer:

Drugs developed from secondary metabolites have been used to treat infectious diseases, cancer, hypertension and inflammation.
Morphine, the first alkaloid isolated from Papaver somniferum is used as pain reliver and cough suppressant.
Secondary metabolites like alkaloids, nicotine, cocaine and the terpenes, cannabinol are widely used for recreation and stimulation.
Flavours of secondary metabolites improve our food preferences.
Tannins are added to wines and chocolate for improving astringency.
Since most secondary metabolites have antibiotic property, they are also used as food preservatives.
Glucosinolates is a secondary metabolite which is naturally present in cabbage imparts a characteristic flavour and aroma because of nitrogen and sulphur-containing chemicals. It also offers protection to these plants from many pests.

Question 31.
Explain the formation of metabolic pool.
Answer:

Metabolism is the sum of the chemical reactions that take place within each cell of a living organism and provide energy for vital processes and for synthesizing new’ organic material.
Metabolic pool in the cell is formed due to glycolysis and Krebs cycle.
The catabolic chemical reaction of glycolysis and Krebs cycle provides ATP and biomolecules. These biomolecules form the metabolic pool of the cell.
These biomolecules can be utilized for synthesis of many important cellular components.
The metabolites can be added or withdrawn from the pool according to the need of the cell.

Question 32.
Explain the concept of metabolic pool.
Answer:
1. Metabolic pool is the reservoir of biomolecules in the cell on which enzymes can act to produce useful products as per the need of the cell.
2. The concept of metabolic pool is significant in cell biology because it allows one type of molecule to change into another type E.g. Carbohydrates can be converted to fats and vice-versa.

Question 33.
Multiple Choice Questions:

Question 1.
Most common constituents of organic compounds found in organims are
(A) C, H, O, P
(B) C, H, O
(C) C, H, N, P
(D) C, H, O, N, P
Answer:
(B) C, H, O

Question 2.
Carbohydrates are composed of
(A) carbon
(B) hydrogen
(C) oxygen
(D) all of these
Answer:
(D) all of these

Question 3.
In which of the following, the ratio of hydrogen and oxygen atoms is 2:1?
(A) proteins
(B) fats
(C) oil
(D) carbohydrates
Answer:
(D) carbohydrates

Question 4.
Which of the following do not give smaller sugar units on hydrolysis?
(A) Monosaccharides
(B) Disaccharides
(C) Polysaccharides
(D) Glycogen
Answer:
(A) Monosaccharides

Question 5.
The simplest monosaccharide made up of three carbons amongst the following is
(A) erythrose
(B) glucose
(C) glyceraldehyde
(D) ribose
Answer:
(C) glyceraldehyde

Question 6.
Deoxyribose sugar is an example of
(A) monosaccharide
(B) disaccharide
(C) polysaccharide
(D) simple protein
Answer:
(A) monosaccharide

Question 7.
Common examples of hexose sugar is/are
(A) glucose
(B) fructose
(C) erythrose
(D) both (A) and (B)
Answer:
(D) both (A) and (B)

Question 8.
If a compound contains 2 monosaccharides, then it is described as
(A) derived monosaccharide
(B) disaccharide
(C) polysaccharide
(D) pentose sugar
Answer:
(B) disaccharide

Question 9.
In a disaccharide, monomers are linked with each other through ________ bonds.
(A) peptide
(B) hydrogen
(C) glycosidic
(D) ester
Answer:
(C) glycosidic

Question 10.
A disaccharide that gives two molecules of glucose on hydrolysis is
(A) sucrose
(B) maltose
(C) lactose
(D) none of these
Answer:
(B) maltose

Question 11.
Sugar present in milk is
(A) fructose
(B) lactose
(C) galactose
(D) sucrose
Answer:
(B) lactose

Question 12.
Polysaccharides consist of
(A) two monosaccharide units
(B) eight monosaccharide units
(C) many monosaccharide units
(D) amino acids
Answer:
(C) many monosaccharide units

Question 13.
________ are water insoluble and small molecular weight compounds as compared to macromolecules.
(A) Lipids
(B) proteins
(C) carbohydrates
(D) nucleic acids.
Answer:
(A) Lipids

Question 14.
Simple lipids are esters of
(A) amino acids
(B) proteins
(C) phosphorus
(D) fatty acids with glycerol
Answer:
(D) fatty acids with glycerol

Question 15.
Fatty acids which do not contain double bond between carbon atoms are
(A) saturated fatty acids
(B) unsaturated fatty acids
(C) oleic and linoleic acids
(D) linoleic and linolenic acids
Answer:
(A) saturated fatty acids

Question 16.
Proteins are linear polymers of
(A) amino acids
(B) fatty acids
(C) monosaccharides
(D) nucleic acids
Answer:
(A) amino acids

Question 17.
Proteins are formed by the condensation of
(A) nucleic acids
(B) amino acids
(C) fatty acids
(D) carbohydrates
Answer:
(B) amino acids

Question 18.
Protein is
(A) micromolecule
(B) macromolecule
(C) soluble
(D) specific
Answer:
(B) macromolecule

Question 19.
Keratin is a ________ protein.
(A) transport
(B) protective
(C) structural
(D) storage
Answer:
(C) structural

Question 20.
A nucleotide contains
(A) sugar + phosphate
(B) N-base + phosphate
(C) sugar + nitrogenous base
(D) sugar + N-base + phosphate
Answer:
(D) sugar + N-base + phosphate

Question 21.
Nucleotides, the polymers of nucleic acid are joined together by __________ bond.
(A) Peptide
(B) Ester
(C) Phosphodiester
(D) Glycosidic
Answer:
(C) Phosphodiester

Question 22.
Find the odd one.
(A) Adenine
(B) Cytosine
(C) Thymine
(D) Uracil
Answer:
(D) Uracil

Question 23.
The two strands of DNA are
(A) similar in nature and complementary
(B) anti-parallel and complementary
(C) parallel and complementary
(D) basically, different in nature
Answer:
(B) anti-parallel and complementary

Question 24.
RNA is genetic material in
(A) bacteria
(B) cyanobacteria
(C) bacteriophages
(D) plant viruses
Answer:
(D) plant viruses

Question 25.
Which RNA is present in more amount in the cell?
(A) m-RNA
(B) t-RNA
(C) r-RNA
(D) not certain
Answer:
(C) r-RNA

Question 26.
Smallest RNA is
(A) t-RNA
(B) m-RNA
(C) r-RNA
(D) not specific
Answer:
(A) t-RNA

Question 27.
________ catalyze hydrolysis of ester, ether etc.
(A) Lyases
(B) Ligases
(C) Hydrolases
(D) Transferases
Answer:
(C) Hydrolases

Question 28.
_______ catalyze interconversions of geometric, optical and positional isomers.
(A) Transferases
(B) Ligases
(C) Oxidoreductase
(D) Isomerases
Answer:
(D) Isomerases

Question 29.
Metal cofactors are also known as?
(A) prosthetic group
(B) coenzyme
(C) activators
(D) inhibitors
Answer:
(C) activators

Question 30.
________ are also known as dehydrogenases.
(A) Oxidoreductases
(B) Ligases
(C) Lyases
(D) Transferases
Answer:
(A) Oxidoreductases

Question 31.
The enzyme functions best at temperature
(A) 30°C to 50°C
(B) 15°C to 25°C
(C) 20°C to 35°C
(D) 40°C to 50°C
Answer:
(C) 20°C to 35°C

Question 32.
As temperature changes from 30° to 45° C, the rate of enzyme activity will
(A) decrease
(B) increase
(C) first increase and then decrease
(D) first decrease and then increase
Answer:
(C) first increase and then decrease

Question 33.
Out of the following, which is not a property of enzymes?
(A) Specific in nature
(B) Proteinaceous
(C) Used up in reaction
(D) Increased rate of biochemical reaction
Answer:
(C) Used up in reaction

Question 34.
Majority of cellular enzymes function best at _______ PH.
(A) acidic
(B) basic
(C) neutral
(D) strong base
Answer:
(B) basic

Question 35.
The _______ action of enzyme with a substrate is explained by lock and key theory.
(A) relative
(B) specific
(C) random
(D) abstract
Answer:
(B) specific

Question 36.
Morphine, the first alkaloid isolated from ________
(A) Pisum sativum
(B) Hibiscus rosa sinensis
(C) Papaver somniferum
(D) Azadirachta indica
Answer:
(C) Papaver somniferum

Question 34.
Competitive Corner:

Question 1.
Prosthetic groups differ from co-enzymes in that –
(A) They can serve as co-factors in a number of enzyme – catalyzed reactions
(B) They require metal ions for their activity
(C) They (prosthetic groups) are tightly bound to apoenzymes
(D) Their association with apoenzymes is transient
Hint: Apoenzyme (protein part) and co-factor together form a complete catalytically active enzyme which is known as holoenzyme. The co-factors are of two types; metal ions and coenzymes. A coenzyme or metal ion that is very tightly or even covalently bound to the enzyme protein is known as a prosthetic group.
Answer:
(C) They (prosthetic groups) are tightly bound to apoenzymes

Question 2.
Consider the following statements:
1. Coenzyme or metal ion that is tightly bound to enzyme protein is called prosthetic group.
2. A complete catalytic active enzyme with its bound prosthetic group is called apoenzyme. Select the correct option.
(A) Both (i) and (ii) are false.
(B) (i) is false but (ii) is true.
(C) Both (i) and (ii) are true.
(D) (i) is true but (ii) is false.
Answer:
(D) (i) is true but (ii) is false.

Question 3.
Concanavalin A is:
(A) a lectin
(B) a pigment
(C) an alkaloid
(D) an essential oil
Answer:
(A) a lectin

Question 4.
Which one of the following carbohydrates is a heteropolysaccharide?
(A) Cellulose
(B) Starch
(C) Glycogen
(D) Hyaluronic acid
Answer:
(D) Hyaluronic acid

Question 5.
The two functional groups characteristic of sugars are
(A) Carbonyl and phosphate
(B) Carbonyl and methyl
(C) Hydroxyl and methyl
(D) Carbonyl and hydroxyl
Answer:
(D) Carbonyl and hydroxyl

Question 6.
Which one of the following statements is correct with reference to enzymes?
(A) Apoenzyme = Holoenzyme + coenzyme
(B) Holoenzyme = Apoenzyme + Coenzyme
(C) Coenzyme = Apoenzyme + Holoenzyme
(D) Holoenzyme = Coenzyme + Co-factor
Answer:
(B) Holoenzyme = Apoenzyme + Coenzyme

Question 7.
Which of the following are NOT polymeric?
(A) Nucleic acids
(B) Proteins
(C) Polysaccharides
(D) Lipids
Answer:
(D) Lipids

Maharashtra Board 11th BK Important Questions Chapter 3 Journal

July 20, 2024July 19, 2024 by Bhagya

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 3 Journal Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 3 Journal

1A. Answer in One Sentence:

Question 1.
What do you mean by Journalising?
Answer:
Journalisation means a process of recording two-fold effects of business transactions in the summarized form of debit and credit in the journal.

Question 2.
Which column in a journal is not filled in at the time of journalising?
Answer:
Ledger Folio column in a journal is not filled in at the time of journalising.

Question 3.
Which account is credited when Salary is paid by cheque?
Answer:
Bank Account is credited when salary is paid by cheque.

Question 4.
Why is a journal called the book of prime entry?
Answer:
Journal is called a book of prime entry because all the business transactions are recorded first in the journal in a chronological order i.e. in the order of their occurrence.

Question 5.
In which order the transactions are recorded in a journal.
Answer:
The transactions are recorded in a journal in chronological order i.e. in the order of their occurrence or taking place.

Question 6.
Which account is debited, when goods are destroyed by fire?
Answer:
When goods are destroyed by fire, the Loss by Fire Account is debited.

2. Correct the following statements and rewrite the statements.

Question 1.
Transactions can be recorded in any order in the Journal.
Answer:
Transactions must be recorded in chronological order in the Journal.

Question 2.
Trade discount is recorded in the books of accounts.
Answer:
A cash discount is recorded in the books of accounts.

Question 3.
Trade discount is calculated after cash discount while calculating discount on purchase or sales.
Answer:
Trade discount is calculated before cash discount while calculating discount on purchase or sales.

Question 4.
Trade discount is allowed for prompt payments.
Answer:
A cash discount is allowed for prompt payments.

Question 5.
The process of entering or recording the transactions in the Journal is called posting.
Answer:
The process of entering or recording the transaction in a Journal is called Journalising.

3. Do you agree or disagree with the following statements

Question 1.
Purchase of Assets should be debited to purchase A/c.
Answer:
Disagree

Question 2.
A cash discount is recorded in the books of Accounts.
Answer:
Agree

Question 3.
GST is imposed by the local body.
Answer:
Disagree

Question 4.
Capital is an asset for business organisations.
Answer:
Disagree

Question 5.
With NEFT, RTGS Transaction Cash A/c is affected.
Answer:
Disagree

Question 6.
On purchase of goods or assets output, GST A/c is credited.
Answer:
Disagree

Question 7.
5% GST Charge on luxury cars.
Answer:
Disagree

Question 8.
GST Generates income to Central Government only.
Answer:
Disagree

Question 9.
Ledger Folio column in Journal filled while passing Journal entry only.
Answer:
Disagree

Question 10.
Purchase of shares of TATA Ltd should be debited to TATA Ltd’s A/c.
Answer:
Disagree

Maharashtra Board 11th BK Important Questions Chapter 2 Meaning and Fundamentals of Double Entry Book-Keeping

July 20, 2024July 19, 2024 by Bhagya

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 2 Meaning and Fundamentals of Double Entry Book-Keeping Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 2 Meaning and Fundamentals of Double Entry Book-Keeping

1. Answer in one sentence only.

Question 1.
State the meaning of the accounting equation.
Answer:
An equation that indicates or shows that the total assets of a business are always equal to the total liabilities of a business plus capital is called the accounting equation.

Question 2.
What do you mean by debt?
Answer:
To debit an account means to enter the entry or write on the left-hand side of an account.

Question 3.
Give two examples of personal accounts.
Answer:
Two examples of personal accounts are stated below:

Mr. Raghuvir Sharma’s Account
The Bank of India’s Account.

Question 4.
What is a Nominal Account?
Answer:
The account relating to business expenses, incomes, and gains is called a nominal account.
e.g. Rent A/c.

Question 5.
Give two examples of real accounts.
Answer:
Two examples of real accounts are:

Cash A/c
Goodwill A/c

Question 6.
State whether drawings increase or decrease owner’s equity.
Answer:
Drawings made by the owner of the business decrease its equity.

Question 7.
What is a conventional cash book?
Answer:
A cash book that is prepared to record not only cash transactions but all types of transactions such as credit purchase or sale, banking transactions, opening, and closing entries, adjustments entries is called a conventional cash book.

Question 8.
Explain the term dual aspect.
Answer:
Every transaction has two aspects i.e. for every debit there is corresponding and equal credit.

Question 9.
State the meaning of impersonal account.
Answer:
The account which is not of a person is called an impersonal account.

Question 10.
Write the rule of Real account.
Answer:
The rule of real account states that Debit what comes in and Credit what goes out.

Question 11.
Which account will be debited when goods are sold to Ram on credit?
Answer:
When goods are sold to Ram, Ram’s A/c will be debited.

Question 12.
Which account will be debited when Mr. Shyam has paid cash to you?
Answer:
Cash A/c will be debited when Mr. Shyam has paid cash to us.

2. Write one word/term or phrase which can substitute each of the following statements.

Question 1.
Expenses are paid before it is due.
Answer:
Prepaid Expenses

Question 2.
Income due but not yet received.
Answer:
Accrued Income

Question 3.
Carriage paid on the sale of goods.
Answer:
Carriage Outward

Question 4.
Statement of Assets & Liabilities.
Answer:
Balance Sheet

Question 5.
Account prepared to know Net Profit or Net Loss.
Answer:
Profit & Loss Account

Question 6.
Value of goods remaining unsold at the end of the year.
Answer:
Closing Stock

Question 7.
The provision was made to compensate the loss on account of likely debts.
Answer:
Provision for Bad & Doubtful Debts

Question 8.
The accounts are prepared at the end of the accounting year to know the profit or loss and financial position of the business.
Answer:
Final Accounts

Question 9.
An amount spent on promoting the sale of goods.
Answer:
Selling Expenses

Question 10.
Additional information is provided below the Trial Balance.
Answer:
Adjustments

3. Select the most appropriate alternatives from those given below and rewrite the statements.

Question 1.
___________ is excess of assets over liabilities.
(a) Goodwill
(b) Capital
(c) Investments
(d) Drawings
Answer:
(b) Capital

Question 2.
Discount earned is transferred to credit side of ___________ account.
(a) Current A/c
(b) Profit & Loss Account
(c) Trading
(d) Capital
Answer:
(b) Profit & Loss Account

Question 3.
___________ is a statement that shows the financial position of a business on a specific date.
(a) Trading account
(b) Trial Balance
(c) Profit & Loss A/c
(d) Balance Sheet
Answer:
(d) Balance Sheet

Question 4.
Outstanding expenses are shown on the ___________ side of Balance Sheet.
(a) Assets
(b) Liability
(c) Both
(d) None of these
Answer:
(b) Liability

Question 5.
Interest on Drawing is credited to ___________ Account.
(a) Trading
(b) Profit & Loss Account
(c) Capital
(d) All
Answer:
(b) Profit & Loss Account

Question 6.
Debit balance of Trading Account means ___________
(a) Gross Loss
(b) Net Loss
(c) Net Profit
(d) Gross Profit
Answer:
(a) Gross Loss

Question 7.
Carriage Inward is debited to ___________ Account.
(a) Trading A/c
(b) Profit & Loss
(c) Capital
(d) Bank
Answer:
(a) Trading A/c

Question 8.
Excess of credit over to debit in Profit & Loss Account indicates ___________
(a) Net Profit
(b) Gross Profit
(c) Gross Loss
(d) Net Loss
Answer:
(a) Net Profit

Question 9.
Closing stock is always valued at cost or market price which is ___________
(a) more
(b) less
(c) zero
(d) equal
Answer:
(b) less

Question 10.
When a specific date is not given, in that case, interest on the drawing is charged for ___________ month.
(a) Four
(b) Six
(c) Eight
(d) Nine
Answer:
(b) Six

4. Fill in the blanks.

Question 1.
Gross Profit is transferred to ___________ account.
Answer:
Profit & Loss Account

Question 2.
Debit Balance of Trading Account indicates ___________
Answer:
Gross Loss

Question 3.
Income Receivable appears on ___________ side of Balance Sheet.
Answer:
Asset

Question 4.
Interest on Bank Loan is debited to ___________ A/c.
Answer:
Profit & Loss Account

Question 5.
Profit and Loss Account is prepared to find out ___________ results of the business.
Answer:
Networking

Question 6.
All indirect/operating expenses are transferred to ___________ account.
Answer:
Profit and Loss Account

Question 7.
Interest of proprietor’s drawing is credited to ___________ account.
Answer:
Profit & Loss Account

Question 8.
An excess of debit over credit in the Profit & Loss A/c represents the ___________
Answer:
Net Loss

Question 9.
All direct expenses are transferred to ___________ account.
Answer:
Trading A/c

Question 10.
Balance Sheet is ___________ of assets & liabilities.
Answer:
Statement

5. Classify the following accounts under the types of Personal, Real, and Nominal accounts.

Question 1.
Investments A/c
Answer:
Real Account

Question 2.
Creditors A/c
Answer:
Personal Account

Question 3.
Land A/c
Answer:
Real Account

Question 4.
Purchase Returns A/c
Answer:
Personal Account

Question 5.
Cash A/c
Answer:
Real Account

Question 6.
Building A/c
Answer:
Real Account

Question 7.
Capital A/c
Answer:
Personal Account

Question 8.
Goodwill A/c
Answer:
Real Account

Question 9.
Interest received A/c
Answer:
Nominal Account

Question 10.
Depreciation A/c
Answer:
Nominal Account

Question 11.
Stationery A/c
Answer:
Nominal Account

Question 12.
Salary A/c
Answer:
Nominal Account

Question 13.
Excise duty A/c
Answer:
Nominal Account

Question 14.
Bank Loan A/c
Answer:
Personal Account

Question 15.
Bank Overdraft A/c
Answer:
Personal Account

Question 16.
Sales A/c
Answer:
Nominal Account

Question 17.
Return Inwards A/c
Answer:
Personal Account

Question 18.
Rent received A/c
Answer:
Nominal Account

Question 19.
Wages A/c
Answer:
Nominal Account

Question 20.
Discount received A/c
Answer:
Nominal Account

Question 21.
Debtors A/c
Answer:
Personal Account

Question 22.
Furniture & Fixtures A/c
Answer:
Nominal Account

Question 23.
Purchases A/c
Answer:
Nominal Account

Question 24.
Bad Debts A/c
Answer:
Nominal Account

Question 25.
Dadar Library A/c
Answer:
Personal Account

Question 26.
Rent paid A/c
Answer:
Nominal Account

Question 27.
Prepaid Insurance A/c
Answer:
Personal Account

Question 28.
Carriage Outwards A/c
Answer:
Nominal Account

Question 29.
Rent Receivable A/c
Answer:
Personal Account

Question 30.
Profit on sale of machinery A/c
Answer:
Nominal Account

Question 31.
Bills Payable A/c
Answer:
Personal Account

Question 32.
Bank of India A/c
Answer:
Personal Account

Question 33.
Carriage Inwards A/c
Answer:
Nominal Account

Question 34.
Stock A/c
Answer:
Real Account

Question 35.
Accrued Interest A/c
Answer:
Personal Account

Question 36.
Bank A/c
Answer:
Personal Account

Question 37.
12% Government Bonds A/c
Answer:
Real Account

Question 38.
Carriage A/c
Answer:
Nominal Account

Question 39.
Advertisement A/c
Answer:
Nominal Account

Question 40.
Conveyance A/c
Answer:
Nominal Account

Question 41.
Premises A/c
Answer:
Real Account

Question 42.
Octroi A/c
Answer:
Nominal Account

Question 43.
Postage A/c
Answer:
Nominal Account

Question 44.
Electricity Charges A/c
Answer:
Nominal Account

Maharashtra Board 11th BK Important Questions Chapter 1 Introduction to Book Keeping and Accountancy

July 19, 2024July 18, 2024 by Bhagya

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 1 Introduction to Book Keeping and Accountancy Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 1 Introduction to Book Keeping and Accountancy

1. Answer in One Sentence:

Question 1.
What is Transaction?
Answer:
The exchange of goods or services for money or money’s worth is called a transaction.

Question 2.
What is Cash Discount?
Answer:
The amount which is deducted by the seller from the amount due at the time of the receipt is called cash discount.

Question 3.
What is the entity concept?
Answer:
The business entity concept states that a business has a separate entity and has an independent legal existence distinct from the person who owns it.

Question 4.
What is the money measurement concept?
Answer:
The money measurement concept implies that every business transaction must be recorded in a common unit of measurement i.e. in terms of money only.

Question 5.
What is the consistency concept?
Answer:
The consistency concept implies that any policy adopted for accounting should be continuous or consistent throughout the business and it need not be changed generally unless and until circumstances demand.

Question 6.
What is conservatism?
Answer:
The concept of conservatism states that while deciding the policy of the enterprise the businessman has to anticipate no profit but provide for all possible losses.

Question 7.
What is accounting?
Answer:
Accounting is a process of recording, classifying, summarising, analyzing, and interpreting financial transactions and communicating the result thereof to the users of such information.

Question 8.
Name the financial statements prepared by a business.
Answer:
A businessman prepares the financial statements such as income statements i.e. Trading Account and Profit and Loss Account and Position Statement i.e. Balance Sheet.

Question 9.
Explain the term ‘Profit’.
Answer:
Excess of revenue income over revenue expenditure is termed as profit.

Question 10.
Explain the term ‘Cost Concept’.
Answer:
The concept according to which assets are recorded in the books of accounts at the price at which they are acquired or purchased is called the cost concept.

2. Give the word term or phrase which can substitute each of the following statements:

Question 1.
Dealings between two persons.
Answer:
Transaction

Question 2.
Business transaction in which cash is not paid or received immediately.
Answer:
Credit Transaction

Question 3.
An allowance is given by the receiver of the cash to the giver of cash at the time of payment.
Answer:
Cash Discount

Question 4.
Liability depends on the happening or not happening a certain event.
Answer:
Contingent Liability

Question 5.
Expenditure on fixed assets increases the earning capacity of the business.
Answer:
Capital Expenditure

Question 6.
System in which entry is recorded for cash as well as credit transactions.
Answer:
Accrual System

Question 7.
The concept under which comparison of one accounting period with the other period is possible.
Answer:
Consistency Concept

Question 8.
A science and art of correctly recording in the books of accounts, all those business transactions that result in the transfer of money or money’s worth.
Answer:
Bookkeeping

Question 9.
Exchange of goods and services either for cash or any other goods or services.
Answer:
Transaction

Question 10.
Sale or purchase of goods or services for immediate cash payment.
Answer:
Cash Transaction

Question 11.
Sale or purchase of goods or services for a certain value to be receivable or payable in the future.
Answer:
Credit Transaction

Question 12.
Commodity purchased or produced for sale.
Answer:
Goods

Question 13.
Excess of assets over liabilities.
Answer:
Capital

Question 14.
Excess of profit over normal profit.
Answer:
Super Profit

Question 15.
The total amount of goods and services are withdrawn by the proprietor for self-use.
Answer:
Drawings

Question 16.
Amount of discount deducted from the invoice price.
Answer:
Trade discount

Question 17.
The reputation of a business is expressed in terms of money.
Answer:
Goodwill

Question 18.
The concept states that assets when purchased should be recorded at cost price.
Answer:
Cost concept

Question 19.
The concept states that business operations will continue forever.
Answer:
Concept of Going Concern

Question 20.
A concept on which a double-entry bookkeeping system is based.
Answer:
Dual Aspect Concept

Question 21.
Rules of conduct are accepted universally to record business transactions.
Answer:
Accounting Principles

Question 22.
General ideas which convey certain meanings.
Answer:
Concepts

Question 23.
The general notion or abstract ideas on which accounting is based.
Answer:
Accounting Concept

Question 24.
The accounting concept suggests that a business has a separate identity from its owner.
Answer:
Business Entity Concept

Question 25.
The accounting concept states that monetary transactions are only recorded in the books of accounts.
Answer:
Money Measurement Concept

Question 26.
The money value for which assets are acquired or manufactured.
Answer:
Cost

Question 27.
Proprietor’s contribution to the business.
Answer:
Capital

Question 28.
Valuable things are owned by the business.
Answer:
Assets

Question 29.
Period of time for which accounts of the business are prepared.
Answer:
Financial year/Accounting year

Question 30.
Amount received after selling of goods or services.
Answer:
Revenue

Question 31.
Excess of revenue over its cost.
Answer:
Profit

Question 32.
Name the accounting concept on the basis of which the income statement is prepared.
Answer:
Accrual Accounting Concept

3. Select the most appropriate alternatives from those given below and rewrite the statements.

Question 1.
Money value of the reputation of business is known as ____________
(a) Copyright
(b) Goodwill
(c) Patents
(d) Trademark
Answer:
(b) Goodwill

Question 2.
Heavy advertising expenditure for launching a new product is called as ____________
(a) Capital expenditure
(b) Revenue Expenditure
(c) Deferred Revenue Expenditure
(d) None of these
Answer:
(c) Deferred Revenue Expenditure

Question 3.
Expenditure incurred on purchase of fixed asset is ____________
(a) Revenue Expenditure
(b) Capital Expenditure
(c) Deferred Revenue Expenditure
(d) None of these
Answer:
(b) Capital Expenditure

Question 4.
Concept which provides a line between present and future is known as ____________
(a) Going Concern Concept
(b) Cost Concept
(c) Accrual Concept
(d) Entity Concept
Answer:
(a) Going Concern Concept

Question 5.
Totalling of Journal or Ledger is called as ____________
(a) Posting
(b) Folio
(c) Casting
(d) Journalising
Answer:
(c) Casting

Question 6.
In cash transaction, goods or services are exchanged for ____________
(a) other goods
(b) other services
(c) immediate cash
(d) grains
Answer:
(c) immediate cash

Question 7.
The work of book-keeping is of ____________ nature.
(a) competitive
(b) primary/basic
(c) secondary
(d) none of these
Answer:
(b) primary/basic

Question 8.
Capital is ____________ of the business.
(a) asset
(b) liability
(c) property
(d) goodwill
Answer:
(b) liability

Question 9.
The work of accounting depends upon ____________
(a) book-keeping
(b) cash book
(c) subsidiary books
(d) ledger
Answer:
(a) book-keeping

Question 10.
____________ is the amount invested by the owner of a business.
(a) Cash
(b) Money
(c) Asset
(d) Capital
Answer:
(d) Capital

Question 11.
Revenue expenditure is ____________ in nature.
(a) capital
(b) outstanding
(c) recurring
(d) contingent
Answer:
(c) recurring

Question 12.
Amount withdrawn by the owner for his personal expenses is called ____________
(a) Drawings
(b) Personal expenses
(c) Cash
(d) Assets
Answer:
(a) Drawings

Question 13.
Financial statements are a part of ____________
(a) Book keeping
(b) Planning
(c) Accounting
(d) None of these
Answer:
(c) Accounting

Question 14.
A ____________ liability is an uncertain liability.
(a) Long term
(b) Contingent
(c) Current
(d) Fixed
Answer:
(b) Contingent

Question 15.
A Derson who is unable to nay his debts, is called ____________
(a) insolvent
(b) solvent
(c) well to do
(d) poor
Answer:
(a) insolvent

Question 16.
According to ____________ concept business shall go on for a long time.
(a) going concerned
(b) consistency
(c) materiality
(d) dual aspects
Answer:
(a) going concerned

Question 17.
According to ____________ concept, every business transaction has two aspects.
(a) going concerned
(b) materiality
(c) business entity
(d) dual aspects
Answer:
(d) dual aspects

Question 18.
According to ____________ concept revenue is recognised when it is earned.
(a) Realisation
(b) Accounting period
(c) Accrual
(d) Matching cost
Answer:
(c) Accrual

Question 19.
According to ____________ convention, while preparing planning anticipate losses.
(a) Materiality
(b) Consistency
(c) Conservatism
(d) disclosure
Answer:
(c) Conservatism

Question 20.
Customs and traditions which guide the accountants to prepare accounting statements are called ____________
(a) Conventions
(b) Principles
(c) Concepts
(d) Procedure
Answer:
(a) Conventions

Question 21.
According to ____________ concept, assets are recorded at a price paid to acquire them.
(a) Cost
(b) Money measurement
(c) Entity
(d) Dual aspect
Answer:
(a) Cost

4. State whether the following statements are true or false with reasons:

Question 1.
The trading concern is established for rendering services to society.
Answer:
This statement is False.
The main objective of trading concern is to earn profit. It is for traders’ livelihood, so the trading concern is not established for rendering services to the society.

Question 2.
Book-keeping is an art as well as a science.
Answer:
This statement is True.
Book-keeping is also considered as an art of recording business transactions because the writing of accounts in a specific style and format requires education, knowledge, training, skill, and experience. From another point of view, Book-keeping is a continuous process of collecting, analyzing, classifying, summarising, and recording the different types of business transactions. In brief, bookkeeping may be defined as “A science as well as an art of collecting, analyzing, classifying, summarising and recording all types of business transactions in a significant manner and in terms of money in a separate set of books.”

Question 3.
In Book-keeping & Accountancy non-monetary transactions are also recorded.
Answer:
This statement is False.
Only monetary transactions are recorded in Book-keeping & Accountancy. It is done to find financial results for the purpose of analyzing and interpreting.

Question 4.
Perpetual succession is explained by the concept of the entity.
Answer:
This statement is False.
The concept of entity is different from the perpetual succession. Entity means separate existence of business from the owner whereas perpetual succession means long life and continuation.

Question 5.
Accounting is useful only to the owner.
Answer:
This statement is False.
Accounting is not only useful for the owner but also to the suppliers, government, customers, lenders, etc.

Question 6.
Writing of account does not require specific skill and knowledge.
Answer:
This statement is False.
Writing of account requires specific skill and knowledge as book-keeping is an art and science of correctly recording business transactions.

Question 7.
The main objective of Bookkeeping is to keep permanent records of business transactions.
Answer:
This statement is True.
The record of business transactions is prepared for a specific period of time i.e. one year and it is preserved for a long period of time.

Question 8.
In credit transactions, goods or services are purchased for cash only.
Answer:
This statement is False.
In credit transactions, goods and services are purchased and sold for credit only. Credit means post-pone payments for a future date.

Question 9.
In barter transactions, goods or services are purchased for other goods or services.
Answer:
This statement is True.
Barter means exchange of goods or services for goods or services. No money is involved in barter transactions.

Question 10.
In cash transactions, goods or services are purchased for a certain value to be paid in the future.
Answer:
This statement is False.
In a cash transaction, goods or services are purchase or sold for spot payments or receipts. No credit is allowed.

Question 11.
Good is a commodity that is purchased for self-use.
Answer:
This statement is False.
Good is a commodity that is purchased for resale purposes. It is an article or a commodity in which businessman deals.

Question 12.
Amount due from other person is known as debt.
Answer:
This statement is True.
Debt is a total sum of money due from a person with whom the business has dealings. Accordingly, a person from whom such debt is due to a business is called the debtor.

Question 13.
Liabilities represent the debts or obligations that a business must receive in terms of money.
Answer:
This statement is False.
Liabilities refer to the total amount of obligations that a business has to pay in the future. Liability means the total amount owed by the business to other persons.

Question 14.
Capital = Liabilities – Assets.
Answer:
This statement is False.
Excess of business assets over business liabilities is called capital.
Capital = Assets – liabilities.

Question 15.
Drawings made by the businessman increase his capital.
Answer:
This statement is False.
Drawing means cash or goods withdrawn by the proprietor for his personal or family use. Drawings made by the businessman decreases his capital.

Question 16.
A person whose assets are equal to or greater than liabilities is known as insolvent.
Answer:
This statement is False.
Insolvent means a person whose liabilities are greater than assets is known as ‘Insolvent’.

Question 17.
A person whose assets are less than business liabilities is known as insolvent.
Answer:
This statement is True.
When a person is not able to pay liabilities is called insolvent. Liabilities are greater than assets.

Question 18.
The cost or sacrifice which is incurred to run the business is known as an income.
Answer:
This statement is False.
The cost or sacrifice which is incurred to run the business is known as an expenditure.

Question 19.
Benefits which are given by giver to the receiver is known as discount.
Answer:
This statement is False.
Discount is a concession given by the seller to the buyer. There are two types of discount,

Cash discount
Trade discount

Question 20.
Posting implies recording a transaction in a journal.
Answer:
This statement is False.
Posting means transferring or recording journal entries from the journal to the respective ledger accounts. Ledger is the main book of accounting.

Question 21.
Contingent liabilities can also be called doubtful liabilities.
Answer:
This statement is True.
The contingent liabilities are the liabilities whose occurrence depends upon the happening of a certain event that may or may not take place. So a contingent liabilities can also be called doubtful liabilities.

Question 22.
A sports club is a trading concern.
Answer:
This statement is False.
The sports club is a nontrading concern as they do not work for profit-making. They work for the promotion of sports.

Question 23.
A cash discount is an incentive allowed for the speedy recovery of income.
Answer:
This statement is True.
A cash discount is given by the creditor or seller to the debtor or buyer to induce him to make prompt payment. It is given at the time of cash payment.

Question 24.
Excess of revenue over expenses is called loss.
Answer:
This statement is False.
Excess of revenue over expenses is called profit. Your income is less and expenditures are less so you get profit.
Income ₹ 50,000 – Expenses ₹ 35,000 = Profit is ₹ 15,000.

Question 25.
Excess of liabilities over assets represents the solvency of a business.
Answer:
This statement is False.
Excess of liabilities over assets represents insolvency of business. A trader cannot pay his debts as liabilities are greater than his assets.
Liabilities ₹ 1,50,000 Assets ₹ 80,000.
₹ 1,50,000 – ₹ 80,000 = ₹ 70,000 deficiency.

Question 26.
A business can not run without bookkeeping.
Answer:
This statement is False.
Book-keeping is a recording of business financial transactions which is useful to prepare ledger, trial balance, financial statements. All the above are useful to take decisions, planning, liabilities, and assets so a businessman cannot run his business without book-keeping.

Question 27.
Bookkeeping is necessary only for organizing with profit objectives.
Answer:
This statement is False.
Books of accounts are maintained by trading and non-trading concerns. Non-trading concerns do not earn any profits so book-keeping is not necessary only with the objective of profit.

Question 28.
The figures of profit and net worth are disclosed by books of accounts.
Answer:
This statement is True.
The purpose of preparing books of accounts is to know financial details of organization from journal ledger, trial balance, final accounts proprietor to get an idea of his financial status. The purpose of financial books of accounts is to get profit and net worth etc.

Question 29.
Financial statements are an effective weapon in the hands of management.
Answer:
This statement is True.
With help of financial statements, management does planning, decision-making for the future. The past record gives ideas to the management to improve financial decisions like more profit better financial conditions etc. so financial statements are effective weapons in the hands of management.

Question 30.
Irregularities, frauds, and misappropriations can be detected only because of books accounts.
Answer:
This statement is False.
Irregularities frauds and misappropriations can be detected by auditing. Continuous and periodical auditing will detect the irregularities frauds and misappropriations which is done by a chartered accountant.

5. Do you agree or disagree with the following statements:

Question 1.
Goodwill is a tangible asset.
Answer:
Disagree

Question 2.
For income tax purposes an accounting year starts on 1st January and ends on 31st March.
Answer:
Disagree

Question 3.
Earning profits is an aim of Not for Profit Concern.
Answer:
Disagree

Question 4.
Trade discount is recorded in the books of accounts.
Answer:
Disagree

Question 5.
Wages paid on installation of machinery are revenue expenditures.
Answer:
Disagree

Question 6.
An amount that is not recoverable from debtors is called bad debt.
Answer:
Agree

Question 7.
The double-entry book-keeping system is invented in India.
Answer:
Disagree

Question 8.
Book-keeping does not have any rules and regulations for recording business transactions.
Answer:
Disagree

Question 9.
The owner of the business does not have any utility of book-keeping.
Answer:
Disagree

Question 10.
Bookkeeping cannot be used as financial evidence in a court of law.
Answer:
Disagree

Question 11.
It is not possible for management to plan and take decisions in business with help of book-keeping and accountancy.
Answer:
Disagree

Question 12.
Cash-withdrawn by the proprietor from his business for personal use is called capital.
Answer:
Disagree

Question 13.
A person who has to pay the business for getting goods and services on credit is known as the debtor.
Answer:
Agree

Question 14.
Trade discount is not recorded in the books of accounts.
Answer:
Agree

Question 15.
Purchase of goods is revenue expenditure.
Answer:
Agree

Question 16.
Heavy expenditure on advertising is deferred revenue expenditure.
Answer:
Agree

Question 17.
A solvent person’s assets are more than his liabilities.
Answer:
Agree

Question 18.
Net Worth = Total Assets – Outsiders Liabilities
Answer:
Agree

Question 19.
Conservatism’s concept means to play safe.
Answer:
Agree

Question 20.
The dual aspect means every transaction has two effects i.e. debit and credit.
Answer:
Agree

6. Complete the following sentences:

Question 1.
In barter transaction goods are exchange for ____________
Answer:
Goods

Question 2.
Asset is ____________ of the business.
Answer:
property

Question 3.
Capital expenditures are ____________ in nature.
Answer:
Non Recurring

Question 4.
A ____________ liability is an uncertain liability.
Answer:
Contingent

Question 5.
A person whose liabilities are more than his assets is known as ____________
Answer:
Insolvent

Question 6.
An article or a commodity which is purchase for resale in business is called ____________
Answer:
Goods

Question 7.
Customs and traditions which guide the accountants to prepare accounting systems are called ____________
Answer:
Conventions

Question 8.
Discount given by the creditor to the debtor on payment of cash is called ____________
Answer:
Cash discount

Question 9.
For tax purpose an year starts on 1st April and ends on 31st March is called ____________
Answer:
Accounting year

Question 10.
An institution which provides standard of accounting in India called ____________
Answer:
Institute of Chartered Accountants of India

Question 11.
Which accounting standard is followed for depreciation on fixed assets ____________
Answer:
AS-6

Question 12.
Brief explanation of an entry is called as ____________
Answer:
Narration

Question 13.
Businessman open ____________ type of bank account.
Answer:
Current

Question 14.
The main objective of business concern ____________
Answer:
making profits

Question 15.
Assets which are held in the business for a period of less than one year ____________
Answer:
Current Assets

Question 16.
10 years loan taken from bank by organisation is ____________ term loan.
Answer:
long

Question 17.
Heavy expenses paid on formation of an organisation is known as ____________ expenditure.
Answer:
Deferred Revenue

Question 18.
Dividend received on shares is ____________ income.
Answer:
Revenue

Question 19.
Loss on sale of old machine is ____________ loss.
Answer:
Capital

Maharashtra Board Class 12 Chemistry Important Questions Chapter 13 Amines

July 19, 2024July 18, 2024 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 13 Amines Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 13 Amines

Question 1.
What are amines?
Answer:
Amines : The alkyl or aryl derivatives of ammonia in which one, two or all the three hydrogen atoms attached to nitrogen are replaced by same or different alkyl or aryl groups are called amines. OR Amines are nitrogen-containing organic compounds having basic character.

Example : methyl amine : CH₃ – NH₂ Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 1

Question 2.
Classify the following amines as primary, secondary and tertiary.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 7

Question 3.
Mention the functional group in :
(1) Primary amine
(2) Secondary amine
(3) Tertiary amine.
Answer:
(1) A primary amine has a functional group – NH₂ (amino group).
Example : ethylamine, C₂H₅ – NH₂
(2) A secondary amine has a functional group – NH – (imino group).
Example : Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 8
(3) A tertiary amine has a functional group (tertiary nitrogen atom)

Example :
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 10

Question 4.
Write common and IUPAC names of following compounds :
Answer:
(A) Primary amines :
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 14

(B) Secondary amine :

(C) Tertiary Aimines :
.

Question 5.
Give the structures of the following :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 22

Question 6.
Give the IUPAC names of the following amines :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 23

Question 7.
Write the IUPAC names of the following amines :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 24

Question 8.
Give the structures and IUPAC names of the following amines :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 25

Question 9.
Classify the following amines as primary, secondary and tertiary and write the IUPAC names.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 26

Question 10.
Write the structures and classify the following amines as primary, secondary, tertiary amines.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 27

Question 11.
Write the common and IUPAC name of a tertiary amine in which one methyl, one ethyl and one w-propyl group is attached to nitrogen.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 28

Question 12.
How will you prepare ethanamine from ethyl iodide?
Answer:
When ethyl iodide is heated with excess of alcoholic ammonia, under pressure at 373 K ethanamine is obtained as a major product.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 29

Question 13.
How is a nitroalkane converted to a primary amine?
OR
What is the action of LiAlH4/ether on (i) 1-Nitropropane (ii) 2-MethyI-l-nitropropane?
Answer:
When a nitroalkane is refluxed with tin (or iron) and concentrated HCl it gives corresponding primary amine.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 33
For example, (1) nitromethane on reduction by refluxing with Sn and concentrated HCl gives methylamine.

(2) 1-Nitropropane on reduction with Sn and concentrated HCl gives propan-1-amine.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 35

(3) Niirobenzcnc on reducion with tin and concentrated HCI or by using H₂/Pd in ethanol gives anilinc.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 36

(4) When nitropropane is reduced in the presence of LiAlH₄ in ether, n-propyl amine is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 37

(5) When 2-methyl-1-nitropropane is reduced in the presence of LiAlH₄ in ether, 2-methyl propan-1-amine is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 38

Question 14.
How will you prepare aniline from nitrobenzene?
OR
How is aniline prepared from nitro compounds?
Answer:
Nitrobenzene is reduced to aniline by passing hydrogen gas in the presence of finely divided nickel, palladium or platinum.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 39

Question 15.
Identify the compounds A and B in the following reactions

Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 41

Question 16.
How will you obtain a primary amine from an alkyl cyanide (nitrile)?
OR
Write a short note on Mendius reduction.
Answer:
Alkyl cyanides (nitriles) on reduction by sodium and ethyl alcohol form corresponding primary amines. This reaction is called Mendius reduction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 42
For example; propionitrile on reduction by sodium and ethanol gives n-propyl amine (Propan-1-amine).

Methyl cyanide or acetonitrile on reduction by sodium and ethanol gives ethanaminc.

Question 17.
How will you prepare ethylamine from acetonitrile?
OR
How is ethanamine prepared from methyl cyanide?
OR
What is the action of a mixture of sodium and alcohol on acetonitrile?
Answer:
Methyl cyanide or acetonitrile on reduction by sodium and ethyl alcohol forms ethanamine. The reaction is called Mendius reduction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 45

Question 18.
How will convert phenyl acetonitrile to β-phenylethylamine?
Answer:
When phenyl acetonitrile is reduced in the presence of sodium and ethanol, β-phenyl ethylamine is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 46

Question 19.
How will you obtain primary amine from an acid amide?
Answer:
Acid amides on reduction with lithium aluminium hydride or sodium, ethanol form corresponding primary amines.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 50
For example : Acetamide on reduction with lithium aluminium hydride or sodium, ethanol gives ethylamines.

Question 20.
Explain Hoffmann degradation of amides.
Write a note on Hoffmann bromamide degradation.
Answer:
The conversion of amides into amines in the presence of bromine and alkali is known as Hoffmann degradation of amides. An important characteristic of this reaction is that an amine with one carbon less than those in the amide is formed. Thus, decreasing the length of carbon chain. This reaction is an example of molecular rearrangement and involves the migration of an alkyl or aryl group from the carbonyl carbon to the adjacent nitrogen atom. For example,

(1) When propanamide is treated with bromine and aqueous or alcoholic sodium hydroxide, ethanamine is obtained which has one carbon atom less.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 53
(2) When benzamide is treated with bromine and aqueous or alcoholic sodium hydroxide, aniline is obtained.

Question 21.
How will you obtain methyl amine from acetamide?
Answer:
When acetamide is treated with bromine and aq or alcoholic solution of KOH, methyl amine is obtained, which has one cabon atom less.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 55

Question 22.
How will you convert the following?

(1) Ethyl bromide to ethylamine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 60

(2) Propionitrile to n-propyl amine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 61

(3) Acetonitrile to ethylamine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 62

(4) Phenyl acetonitrile to β-phenylethyl amine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 63

(5) Acetamide to ethylamine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 64

(6) Nitropropane to propan-l-amine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 65

(7) Nitrobenzene to Aniline.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 66

(8) Benzamide to aniline.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 67

Question 23.
How will you prepare propan-l-amine from (1) butane nitrile (2) 1-nitropropane (3) propanamide (4) butanamide?
Answer:
(1) From butane nitrile :
When butane nitrile is reduced by sodium and ethanol, it gives propan-l-amine.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 68

(2) From 1-nitropropane :
When 1-nitropropane is reduced in the presence of tin and cone, hydrochloric acid, propan-l-amine is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 69

(3) From propanamide :
When propanamide is reduced in the presence of lithium aluminium hydride, propan-l-amine is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 70

(4) From butanamide :
When butanamide is treated with bromine and aq. KOH, propan-l-amine is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 71

Question 24.
Write a reaction to, convert acetic acid into methyl amine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 78

Question 25.
Primary and secondary amines have boiling points higher than the tertiary amines. Explain why?
Answer:
(1) The N – H bond in amines is polar in nature because of electronegativities of nitrogen (3.0) and hydrogen (2.1) are different.
(2) Due to the polar nature of N – H bond, primary and secondary have strong intermolecular hydrogen bonding. Tertiary amines do not have intermolecular hydrogen bonding as there is no hydrogen atom on nitrogen of tertiary amine. Thus, intermolecular forces of attraction are strongest in primary and secondary amines and weakest in to tertiary amines. Hence, primary and secondary amines have boiling points higher than the tertiary amines.

Question 26.
Amines have boiling points higher than the hydrocarbon but lower than the alcohols of comparable masses. Explain, why?
Answer:
Amines are polar than alkanes but less polar than alcohols. Primary and secondary amines form intermolecular hydrogen bonds. This hydrogen bonding leads to an associated structure. The association is more in primary amines than that in secondary amines as there are two hydrogen atoms attached to the nitrogen atom. However, tertiary amines do not form intermolecular hydrogen bonds because they do not contain any hydrogen atoms attached to the nitrogen atom. Hence, amines have higher boiling points than the hydrocarbons but lower boiling points than the alcohols of comparable masses.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 79

Compound	Molar mass	Boiling points (K)
nC₂H₅CH(CH₃)₂	72	300
nC₄H₉NH₂	73	350.8
nC₄H₉OH	74	391

Question 27.
Arrange the following compounds in the decreasing order of their solubility in water.
(a) Ethyl amine, diethyl amine and triethyl amine.
Answer:
Diethyl amine > triethyl amine > ethyl amine
(The reason that ethyl group has greater +1 effect than methyl group)

(b) Ethyl amine, n-propyl amine and n-butyl amine.
Answer: n-butyl amine < n-propyl amine < ethyl amine

Question 28.
Arrange the following compounds in the decreasing order of their boiling points.
(a) Ethane, ethyl amine and ethyl alcohol.
Answer:
Ethyl alcohol < ethyl amine < ethane

(b) Ethyl amine, n-propyl amine and n-butyl amine.
Answer:
n-butyl amine < n-propyl amine < ethyl amine

(c) n-propyl amine, ethyl methyl amine and trimethyl amine.
Answer:
n-propyl amine < ethyl methyl amine < trimethyl amine.

(d) Ethyl alcohol, dimethyl amine and ethyl amine.
Answer:
Ethyl alcohol < ethyl amine < dimethyl amine.

Question 29.
Explain the basic nature of amines with a suitable examples.
OR
Explain why amines are basic.

Question 38.
Tertiary amine (R₃N) or 3° amine is weaker base than secondary amine R₂NH or 2° amine. Explain.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 81

The increase in basic strength from 1° amine to 2° amine is explained on the basis of increased stabilization of conjugate acids by +1 effect of the increased number of the alkyl group. However, decreased basic strength of 3° implies that the conjugate acid of 3° amine is less stabilized and is weak base though the +1 effect of three alkyl groups in R₃NH is large.

R₂NH is best stabilized by solvation while the stabilization by solvation is very poor in R₃NH. Hence (R₃N) or tertiary amine or 3° amine is weaker base than secondary amine (R₂NH) or 2° amine.

Question 30.
Primary or aliphatic amine is a stronger base than ammonia. Explain.
Answer:
(1) The alkyl group in primary amines has +I effect i.e. (electron releasing).

The alkyl group tends to increase the electron density on the nitrogen atom. As a result, amines can donate the lone pair of electrons on nitrogen more easily than ammonia.

(2) The amine being a base, can donate a pair of electrons to an acid. The alkyl group with +I effect will disperse the positive charge on the cation more than ammonia.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 84

Due to +I effect of alkyl group cation formed by primary amine is more stable compared to cation formed from ammonia. Also it is seen that observed increasing basic strength from ammonia to primary amine is explained on the basis of increased stabilization of conjugate acids by +I effect for the presence of alkyl (R) groups. Hence, primary or aliphatic amine is a stronger base than ammonia.

Question 31.
Aniline is less basic than ammonia. Explain.
Answer:
The less basic character of aniline can be explained on the basis of resonance shown by aniline.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 85

Due to resonance, the nitrogen atom of amino group in aniline acquires a positive charge, hence, lone pair of electrons is less available for protonation as compared to that of ammonia. Aniline is resonance stabilized by five resonance structures. On the other hand, aniline in aqueous medium, accepts a proton does not have lone pair of electrons on nitrogen to produce a very low concentration of anilium ion and anilium ion shows only two resonance structures and therefore less stabilized than anline.

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 86

Thus, aniline is more stable than anilium ion. Hence aniline accepts proton less readily or less basic in nature than ammonia.

Question 32.
Explain the order of basicity in ammonia and aliphatic amines.
Answer:
Since nitrogen atom in ammonia molecule has a lone pair of electrons, it is a Lewis base.
Greater the availability of an electron pair, more is the basic character.

Since alkyl group (R -) is an electron releasing group with (+I) inductive effect, alkyl amines act as a stronger base than ammonia.

The decreasing order of basicity is –
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 87

The availability of a lone pair of electrons on a nitrogen atom in amines is influenced by steric factor due to crowding of alkyl groups which affects solvation along with inductive effect of alkyl groups.

Due to high energy of solvation of \(\mathrm{NH}_{4}^{+}\) ions, they acquire higher stability in aqueous solutions.

The presence of alkyl groups in secondary and tertiary amines, due to steric hindrance decrease the solvation energy.

This effect is more in tertiary amines making the tertiary ammonium ions (R₃NH⁺) unstable as compared to secondary ammonium ion (R₂N⁺H₂).
Hence the cumulative effect on the order of basicity of amines is, secondary amine > primary amine > tertiary amine > ammonia (NH₃).

Question 33.
Arrange the following amines in the decreasing order of their basic nature.
(a) Aniline, propan-l-amine and N-methylethanamine.
Answer:
N-methylethanamine < propan-l-amine < aniline

(b) Benzene-1, 4-diamine, ammonia and 4-aminobenzoic acid.
Answer:
Ammonia < benzene-1, 4-diamine < 4-aminobenzoic acid

(c) N-Methylaniline, phenylmethylamine and N-phenylaniline.
Answer:
N-Methylaniline < N-phenylaniline < phenylmethylamine

Question 34.
Arrange the following amines in the increasing order of their pKb values.
(a) Aniline, N-methylaniline and cyclohexalamine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 88

(b) Phenylmethylamine, 2-aminotoluene and 2-fluoroaniline.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 89

(c) Aniline, 4-methoxyaniline and 4-nitroaniline.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 90

Question 35.
Arrange the following compounds in the decreasing order of their basic nature in the gaseous phase.
Ammonia, N-methylhexanamine, propan-1-amine and N, N-dimethylethanolamine.
Answer:
Propan-1-amine < N-methylethanamine < N,N-dimethylmethanamine < ammonia

Question 36.
Explain laboratory test for amines.
Answer:
(1) All amines are basic compounds. Aqueous solution of water soluble amines turns red litmus blue.

(2) When water insoluble amine is dissolved in aqueous HCl, forms water soluble substituted ammonium chloride, further a substituted ammonium chloride on reaction with excess aqueous NaOH regenerates the original insoluble amine.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 91

(3) Diazotization reaction/ Orange dye test: In a sample of aromatic primary amine, 1-2 mL of cone. HCl is added. The aqueous solution of NaNO₂ is added with cooling. This solution is transfered to a test tube containing solution of β naphthol in NaOH. Formation of orange dye indicates presence of aromatic primary amino group. (It may be noted that temperature of all the solutions and reaction mixtures is maintained near 0 °C throughout the reaction).

Question 37.
Explain Hofmann’s exhaustive alkylation.
OR
Explain Hofmann’s exhaustive methylation of amines.
Answer:
Hofmann’s Exhaustive alkylation : When a primary amine is heated with excess of primary alkyl halide it gives a mixture of secondary amine, tertiary amine along with tetraalkylammonium halide
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 92

If excess of alkyl halide is used, tetraalkyl ammonium halide is obtained as major product. The reaction is known as exhaustive alkylation of amines.

Hofmann’s Exhaustive Methylation : The process of converting a primary, secondary or tertiary amine into quaternary ammonium halide by heating them with excess of methyl iodide, is called exhaustive methylation or Hoffmann’s exhaustive methylation.

Thus when methyl amine is heated with excess of methyl iodide it forms dimethylamine (secondary amine), then trimethylamine (a tertiary amine) and finally of quaternary ammonium iodide. The reaction is carried out in the presence of mild base NaHCO₃, to neutralize the large quantity of HI formed.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 93

Question 38.
Predict the products of exhaustive methylation of following compounds.
(1) Ethylamine.
Answer:
A primary amine, ethylamine (CH₃ – CH₂ – NH₂) on exhaustive methylation, i.e., on heating with excess methyl iodide, forms secondary amine, tertiary amine and finally a quaternary ammonium salt, ethyl-trimethyl ammonium iodide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 97

(2) Benzylamine.
Answer:
Benzylamine C6H5CH2NH2 on exhaustive methylation i.e., on heating with excess methyl iodide forms benzylmethyl amine, benzyldimethyl ammonium chloride and finally benzyltrimethyl ammonium iodide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 98

Question 39.
Explain Hofmann elimination.
OR
Write a note on Hoffmann elimination.
Answer:
When tetra alkyl ammonium halide is heated with moist silver hydroxide, a quaternary ammonium hydroxide is obtained. Quaternary ammonium hydroxides are deliquescent crystalline solids and are basic in nature. Quaternary ammonium hydroxides on strong heating undergo ^-elimination to give tertiaryamine, alkenes and water, the reaction is called Hofmann elimination. The major product is least substituted alkene.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 99

Question 40.
Write the bond line formula of the alkene which is obtained as major product from the following amines, on heating with excess of methyl iodide followed by strong heating with moist silver oxide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 102
Answer:

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 104
Answer:

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 106
Answer:

Question 41.
Compound X with a molecular formula C₅H₁₃N did not react with nitrous acid, but reacted with one mole of CH3I to form a salt. What is the structure of X?
Answer:
The structure of compound X is Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 108 ethyl-N-methylethanamine since compound X is tertiary amine. It reacts with one mole of CH₃I to give a quaternary ammonium salt.

Question 42.
What is the action of acetyl chloride on :
(1) ethyl amine (ethanamine)
(2) diethyl amine (N-Ethylethanamine)
(3) triethyl amine?
OR
Write a short note on acylation of amines.
Answer:
The reaction of amines with acetyl chloride is called acetylation of amines.

(1) Acetyl chloride on reaction with ethylamine forms monoacetyl derivative, N-ethylacetamide (or N-acetyl ethylamine).
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 109
(2) Diethyl amine on reaction with acetyl chloride forms N-acetyl dimethylamine.

(3) Triethyl amine, being a tertiary amine does not have H atom attached to nitrogen of amine, hence it does not react with acetyl chloride.

Question 43.
What is the action of acetic anhydride on aniline?
Answer:
Aniline on reaction with acetic anhydride forms N-phenyl acetamide.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 116

Question 44.
What is the action of benzoyl chloride on ethanamine?
Answer:
When benzoyl chloride is treated with ethanamine, N-ethyl benzamide is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 117

Question 45.
What is the action of nitrous acid on ethylamine?
Answer:
Ethyl amine on reaction with nitrous acid in cold forms aliphatic diazonium salt, (unstable intermediate), which decomposes immediately by reaction with solvent water to produce ethyl alcohol and nitrogen gas.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 123

Question 46.
What is the action of nitrous acid on aniline?
Answer:
Aniline reacts with nitrous acid in cold to form diazonium salt which has reasonable solubility at 273 K
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 124

Question 47.
How is benzenediazon|um chloride prepared?
Answer:
Benzenediazonium chloride is prepared by the action of nitrous acid on aniline at 273-278 K. Nitrous acid being unstable, is prepared in situ by the reaction between sodium nitrite and dilute hydrochloric acid.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 125

Question 48.
Write resonance stabilized structures of aryl diazonium salt.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 126

Question 49.
Write a note on Sandmeyer’s reaction.
OR
How is aryl chloride or aryl bromide or aryl cyanide prepared from diazonium salt?
Answer:
[Replacement by Cl^–, Br^– and -CN : Sandmeyer reaction.] Freshly prepared aromatic diazonium salt on reaction with cuprous chloride gives aryl chloride, on reaction with cuprous bromide gives aryl bromide and on reaction with cuprous cyanide give aryl cyanide. The reaction in which copper (I) salts are used to replace nitrogen in diazonium salt is called Sandmeyer reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 128

Question 50.
How is aryl chloride or aryl bromide prepared by Gattermann reactions?
Answer:
The aryl chloride or bromides can also be prepared by Gattermann reactions in which diazonium salt reacts with
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 129

Question 51.
How is aryl iodide obtained from diazonium salt?
Answer:
When diazonium salt is warmed with potassium iodide, aryl iodide is obtained.

Question 52.
Explain the reduction of arene diazonium salt?
OR
How is arene obtained from arene diazonium salt?
OR
What is the action of benzene diazonium chloride on ethanol?
Answer:
Arene diazonium salt on treatment with mild reducing agents like phosphinic acid (hypophosphoric acid) or ethanol, arene is obtained.

Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 131

Question 53.
How is phenol obtained from arene diazonium salt?
Answer:
When arene diazonium salt is slowly added to a large volume of boiling dilute sulphuric acid, phenol is obtained,
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 133

Question 54.
How is aryl fluoride obtained from diazonium salt?
Answer:
When fluoroboric acid is treated with the solution of diazonium salt, a precipitate of diazonium fluoroborate is obtained, which is filtered and dried. When dry diazonium fluoroborate is heated, it decomposes to give aryl fluoride. This reaction is called Balz-Schiemann reaction.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 134

Question 55.
How is nitrobenzene obtained from benzene diazonium fluoroborate?
Answer:
When benzene diazonium fluoroborate is heated with aqueous solution of sodium nitrite in the presence of copper powder, nitrobenzene is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 135
Benzene diazonium fluorobate can be obtained by reaction of benzene diazonium chloride with HBF₄.

Question 56.
What is meant by a coupling reaction? Explain with suitable examples.
OR
What is the action of benzene diazonium chloride on (a) phenol in alkaline medium (b) aniline?
OR
Write a note on the coupling reaction.
Answer:
Diazonium salts react with certain aromatic compounds having an electron-rich group (e.g.-OH, – NH₂, etc.) to form azo compounds. This reaction is an electrophilic substitution and is called coupling reaction. Azo compounds are brightly coloured and are used as dyes and indicators. Coupling reaction is an electrophilic substitution reaction. Benzene diazonium chloride reacts with alkaline solution of phenol to give p-hydroxy azo benzene (orange dye).
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 139
Benzene diazonium chloride reacts with aniline in mild alkaline medium to give p-aminobenzene (yellow dye).

Question 57.
What is the action of p-toluene sulphonyl chloride on ethyl amine and diethyl amine?
Answer:
(1) When ethyl amine is treated with p-toluene sulphonyl chloride, N-ethyl p-toluene sulphonamide is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 144
(2) When diethyl amine is treated with p-toluene suiphonyl chloride. N.N-dicthyl p-toluene suiphonyl amide is formed.

Question 58.
How will you distinguish between :
(1) Ethylamine, diethyl amine and triethyl amine by using (i) nitrous acid (ii) Hinsberg’s reagent.
(2) Diethyl amine and triethyl amine by using acetic anhydride.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 150

Question 59.
Give a chemical test to distinguish between following pairs of compounds.
(i) Ethylamine and diethyl amine :
Answer:
Ethylamine (C₂H₅NH₂) is a primary amine while diethyl amine ( (C₂H₅)₂NH) is a secondary amine. So the two can be distinguished by the following test.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 152

(ii) Ethyl amine and aniline :
Answer:
Ethylamine is an aliphatic amine, while aniline is an aromatic amine. So the two can be distinguished by the following test :
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 153

(iii) Aniline and benzyl amine :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 154

(iv) Aniline and N-ethylaniline :
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 155

Question 60.
Compound ‘X’ with a molecular formula C₄H₁₁N did not react with Hinsberg’s reagent, but reacted with one mole of CH₃I to form a salt. What is the structure of ‘X’?
Answer:
The structure of compound ‘X’ is :
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 156
Since the compound ‘X’ does not react with NaN02 and HC1 i.e. nitrous acid (HO – N = O), it must be a tertiary amine.

The tertiary amine reacts with one mole of CH3I to give a quaternary ammonium salt.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 157

Question 74.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 158
p-(dimethylamino) azobenzene is yellow dye which was formerly used as a colouring agent in margarine. Write the structures of the reactants used in the preparation of this dye.
Answer:

Question 61.
Convert 3-Methyl aniline into 3-nitrotoluene.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 160

Question 62.
How will you bring about following conversions?
(1) N.Methyl aniline into N-methyl benzanilide.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 161

(2) 1.4-Dichlorobutane Into hexane-1,6-diamlne.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 162

(3) Benzene into 3-bromo aniline.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 163

(4) Chlorobenzene into 4-chioroanilinc.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 164

(5) 11enaniide into toluene.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 165

Question 63.
What is the action of aqueous bromine on aniline?
Answer:
Action of aqueous bromine on aniline : When aniline is treated with bromine water at room temperature, a white precipitate of 2, 4, 6-tri bromoaniline is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 166

Question 64.
Explain the action of cone, nitric acid (nitrating mixture) on aniline.
Answer:
When aniline is warmed with a mixture of cone, nitric acid and cone, sulphuric acid (a nitrating mixture), a mixture of ortho, meta and para nitroaniline is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 169

Question 65.
What is the action of acetic anhydride on aniline?
Answer:
When aniline is heated with acetic anhydride, an acetanilide is obtained.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 170

Question 66.
How will you convert aniline to p-nltroanhline? (major product)
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 171

Question 67.
What is the action of cone, sulphuric acid on aniline?
Answer:
Aniline on treatment with cold sulphuric acid forms anilium hydrogen sulphate which on heating with sulphuric acid at 453 K-475 K gives sulphanilic acid, (p-aminobenzene sulphonic acid) as major product. Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 172
Sulphanilic acid exists as a salt; called dipolar ion or zwitter ion. It is produced by the reaction between an acidic group and a basic group present in the same molecule.

Question 68.
How will you convert the following?
(1) Ethylamine to ethyl alcohol.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 173

(2) N-Methyl aniline to N-Nitroso-N-methyl aniline.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 174

(3) Diethylamine to N-nitrosodiethylamine
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 175

(4) Triethylamine to triethyl ammonium nitrite.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 176

(5) Ethyl amine to N-ethylacetamide.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 177

(6) Diethyl amine to N-acetyl diethylamine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 178

(7) Aniline to acetanilide.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 179

(8) Aniline to N-ethyl henzamide.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 180

(9) Ethylamine to ethyl isocyanide.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 181

(10) Aniline to phenyl isocyanide.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 182

(11) Aniline to 2,4,6-tribromoaniline.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 183

Question 69.
Give a plausible explanation for each of the following statements :
(1) Ethylamine is soluble in water whereas aniline is not.
Answer:
Ethylamine is soluble in water due to intermolecular hydrogen bonding resulting in the formation of C₂H₅NH₃ ion. Whereas in anline the hydrogen bonding with water is negligible due to the phenyl group (C₆H₅) is bulky and has -I effect. Therefore, aniline is nearly insoluble in water.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 184

(2) Butan-1-ol is more soluble in water than butani-amine.
Answer:
Rutan- l-al is more soluble in watcr duc to intermoiccular hydrogen bonding. In alcohols, hydrogen bonding is through oxygen atoms. WIereas hutani-amine is less soluble in water due to the larger hydrocarbon part is hydrophobic in nature. Hence, butan-l-ol is more soluble in water than butani-amine.

(3) Butan-1-amlne has higher boiling point than N-ethylethanamine.
Answer:
Due to the presence of two H-atoms on N-atom in butait- I -amine, they undergo extensive intermolecular H-bonding while in N-cthylethanamine due to the presence of one-H atom on the N-atom, they undergo least intermolecular H-bonding. Hence, butan- l-amine has higher boiling point than-N-ethyl ethanamine.

(4) AnIline Is less basic than ethyl afine.
Answer:
Aniline (K_b4-2 x 10^-10) is less basic than ethyl amine (K_b5.1 x 10^-4). This is because -I effect of phenyl group in aniline as compared to + 1 effect of ethyl group in ethyl amine.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 185
Due to resonance, the lone pair of electrons on the nitrogen atom gets delocalized over the benzene ring and thus less available for protonation. On the other hand, in ethyl anine, delocalization of the lone pair of electrons on the nitrogen atom by resonance is not possible. Further more, the electron density on the nitrogen atom is increased by +1 effect of the ethyl group. Hence, aniline is less basic than ethyl amine.

(5) pK_b value of diethyl amine is less than that of ethyl amine.
Answer:
The basic strength of amines is expressed in terms of pK_b values. Smaller is the value of pK_b more basic is the amine. The pK_b value of ethyl amine is 3.29 and that of diethyl amine is 3.00. Therefore, diethyl amine is more basic than ethyl amine.

(6) Aniline cannot be prepared by Gabriel phthalimide synthesis.
Answer:
In Gabriel-phthalimide synthesis of aniline, potassium phthalimide requires the treatment with chlorobenzene or bromobenzene. Since aryl halides do not undergo nucleophilic substitution reaction. Therefore, chlorobenzene or bromobenzene does not react with potassium phthalimide to give N-phenylphthalimide and hence aniline cannot be prepared by Gabriel phthalimide synthesis.

(7) Gabriel phthalimide synthesis is preferred for the preparation of aliphatic primary amines.
Answer:
In aromatic amines, the lone pair of electrons on the N-atom is delocalized over the benzene ring. As a result electron density on the nitrogen atom decreases. Whereas in aliphatic primary amines, due to +1 effect of alkyl group, electron density on nitrogen atom increases. As the pKh value of aliphatic amines is more than that of aromatic amines, aromatic amines are less basic than primary aliphatic amines. Hence, Gabriel phthalimide synthesis is preferred for the preparation of aliphatic amines.

(8) Arere diazonium salts are relatively more stable than alkyl diazonium salts.
Answer:
Arene diazonium salts are stable due to the dispersal of the positive charge over the benzene ring as shown below :
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 186
Alkane diazonium salts are unstable due to their tendency to eliminate a stable molecule of nitrogen to form carbocation. Aromatic diazonium salts have much lower tendency to remove nitrogen than aliphatic diazonium salts. Hence, arene diazonium salts are relatively more stable than alkyl diazonium salts.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 187

(9) Tertiary amines cannot be acylated.
Answer:
Tertiary amines do not react with acetic anhydride or acetyl chloride i.e. they can be acylated because they do not contain a H-atom on the N-atom.

(10) Besides the ortho and para derivatives, considerable amount of meta derivatives is also formed during nitration of aniline.
OR
Although amino group is o- and p-directing in electrophilic substitution reactions, aniline on nitration gives substantial amount of m-nitroaniline.
Answer:
In aromatic amines, -NH₂ is an electron releasing or activating group. It activates the ortho and para positions in the benzene ring towards electrophilic substitution. When aniline is treated with nitrating mixture (cone. HNO₃+ cone. H₂SO₄), a mixture of ortho and para nitroaniline is obtained. However, a substantial amount of m-nitroaniline is also formed. Aniline being a base gets protonated in acidic medium to form anilium cation, which deactivates the ring and the substitution takes place at the meta position.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 188

Question 70.
How will you convert :
(1) Aniline into benzyl alcohol.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 191

(2) Aniline into 4-bromoaniline.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 192

(3) Aniline into 1,3,5-tribromo benzene.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 193

(4) Aniline into 2,4,6-tribromo fluoro benzene.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 194

Question 71.
How will you convert :
(1) Propanoic acid into ethanoic acid.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 197

(2) Propanoic acid into ethanol
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 198

(3) Ethanamine into propan-l-amine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 199

(4) Propan-l-amine into ethanamine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 200

(5) Propanoic acid into ethanamine.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 201

(6) Ethanamine into propanoic acid.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 202

(7) Benzene to aniline.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 203

(8) Aniline to Benzene.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 204

(9) Aniline into benzoic acid.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 205

(10) Benzoic acid into aniline.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 206

(11) Aniline into benzamide.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 207

(12) 3-Nitrotoluene into 3-methyl aniline.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 208

(13) 3-Methyl aniline into 3-Nitrotoluene.
Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 209

Question 72.
An organic compound ‘A’ having molecular formula C₂H₆O evolves hydrogen gas on treatment with sodium metal and on treatment with red phosphorous and iodine gives compound ‘B’. The compound ‘B’ on treatment with alcoholic KCN and on subsequent reduction gives compound ‘C’. The compound ‘C’ on treatment with nitrous acid evolves nitrogen gas. Write the balanced chemical equations for all the reactions involved and identify the compounds ‘A’, ‘B’ and ‘C;.
Answer:
A = C₂H₅OH ethanol
B = C₂H₅I ethyl iodide
C = C₂H₅CH₂NH₂ n-propyl amine
Compound C₂H₆O = C₂H₅OH
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 215

Question 73
Identify B, C and D write complete reactions :

Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 217

Question 74.
Identify the compounds B, C and D in the following series of reactions and rewrite the complete equations :

Answer:
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 220

Question 75.
Identify the compounds ‘A’ and ‘B’ in the following equation :
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 221
Answer:

Question 76.
Answer in one sentence :

(1) Arrange the following compounds in decreasing order of basic strength in their aqueous solutions. NH₃, C₂H₅NH₂, (CH₃)₂NH, (CH₃)₃N
Answer:
The decreasing order of basic strength is – (C₂H₅)₂NH > (C₂H₅)₃N > (C₂H₅)₂NH > NH₃
(The reason that ethyl group has greater +1 effect than methyl group).

(2) Arrange the following compounds in an increasing order of their solubility in water.
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 226
Answer:
The solubility increases in order in which molecular mass decreases.

(3) What is Hinsberg’s reagent?
Answer:
Benzenesulphonyl chloride (C₆H₅SO₂Cl) is known as Hinsberg’s reagent.

(4) Name the reaction in which a primary amine is formed from amide.
Answer:
Hoffmann bromamide degradation.

(5) NH₃ is a Lewis base.
Answer:
Since nitrogen in ammonia molecule has a lone pair of electrons, it is a Lewis base.

(6) How many primary amines are possible for the compound C₃H₉N?
Answer:
For the compound C₃H₉N, two primary amines are possible.

(7) State the hybridization of the nitrogen atom in amines.
Answer:
The hybridization of nitrogen atom in amines is sp³.

(8) Arrange the following compounds in an increasing order of basic strength. Aniline, p-nitroaniline, p-toluidine.
Answer:
p-nitroaniline < aniline < p-toluidine.

(9) Which of the two is more basic and why? CH₃NH₂ or NH₃
Answer:
Due to +1 effect of -CH₃ group, electron density on N-atom increases, hence methyl amine is a stronger base than ammonia.

(10) Which of the two is more basic and why? p-toluidine or aniline.
Answer:
p-toluidine is more basic due to the presence of -CH₃ group at para position. Due to +1 effect of -CH₃ group, electron density on nitrogen increases, hence the tendency to donate pair of electrons increases.

Multiple Choice Questions

Question 77.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. Which of the following is an amine?
(a) C₂H₅N(COCH₃)₂
(b) (C₂H₅)₂N – N = 0
(c) (C₂H₅)₃N
(d) All of these
Answer:
(d) All of these

2. N-methyl-N-ethyl-n-propyl amine is
(a) a primary amine
(b) a secondary amine
(c) a tertiary amine
(d) an alkyl nitrile
Answer:
(c) a tertiary amine

3. Which of the following is a tertiary amine?
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 229
Answer:
(d)

4. Tertiary butyl amine is a
(a) primary amine
(b) secondary amine
(c) tertiary amine
(d) quaternary ammonium salt
Answer:
(a) primary amine

5. The IUPAC name of
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 230
(a) ethyl propanamine
(b) ethyl butylamine
(c) 2-pentanamine
(d) 3-hexanamine
Answer:
(d) 3-hexanamine

6. The IUPAC name of ethyl dimethyl amine is ……………..
(a) 2-amino propane
(b) N,N-dimethyl ethanolamine
(c) ethyl methanamine
(d) propanamine
Answer:
(b) N,N-dimethyl ethanolamine

7. Isopropyl amine and trimethyl amine are ……………..
(a) acidic in nature
(b) electrophilic compounds
(c) structural isomers
(d) optically active compounds
Answer:
(c) structural isomers

8. N, N-dimethylethanolamine is ……………
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 231
Answer:
(b)

9. IUPAC name of diethylmethyl amine is ………………
(a) methyl amino propane
(b) N-Ethyl-N-methylhexanamine
(c) methyl diethanamine
(d) amino pentane
Answer:
(b) N-Ethyl-N-methylhexanamine

10. Ethyl bromide reacts with excess of alcoholic ammonia, the major product is …………..
(a) ethyl amine
(b) diethylamine
(c) triethylamine
(d) tetraethyl ammonium bromide
Answer:
(a) ethyl amine

11. Isopropylamine is obtained by the reduction of
(a) acetoxime
(b) acetaldoxime
(c) formaldoxime
(d) aldoxime
Answer:
(a) acetoxime

12. Which of the following compounds can be converted into amines in the presence of Na and alcohol?
(a) Alkyl nitriles
(b) Aldoxime
(c) Ketoxime
(d) All of these
Answer:
(d) All of these

13. Chloroethane when boiled with excess of aqueous-alcoholic ammonia gives hydrochloric acid and
(a) triethyl amine
(b) trimethyl amine
(c) diethyl amine
(d) ethyl amine
Answer:
(d) ethyl amine

14. How many hydrogen atoms are required for the reduction of 1-nitropropane to n-propyl amine?
(a) Four
(b) Three
(c) Six
(d) Two
Answer:
(c) Six

15. A secondary alkyl halide is heated with excess of ammonia, the major product obtained is
(a) primary amine
(b) secondary amine
(c) tertiary amine
(d) quaternary ammonium salt
Answer:
(a) primary amine

16. The true statement about ethylamine is
(a) it is weaker base than ammonia
(b) it is stronger base than diethyl amine
(c) it is stronger base than triethyl amine
(d) it is stronger base than alkali
Answer:
(c) it is stronger base than triethyl amine

17. The reaction which is given only by primary amines is
(a) acetylation
(b) alkylation
(c) reaction with HNO₂
(d) carbyl amine test
Answer:
(d) carbyl amine test

18. The amine which reacts with NaNO₂ and dil. HCl to give yellow oily compound is
(a) ethylamine
(b) isopropylamine
(c) sec-butylamine
(d) dimethylamine
Answer:
(d) dimethylamine

19. The name of the compound ‘C’ in the following series of reactions, is
(a) propan-l-ol
(b) propan-2-ol
(c) butan-l-ol
(d) butan-2-ol
Answer:
(b) propan-2-ol

20. Triethylamine when treated with nitrous acid gives
(a) an alcohol
(b) a nitrosamine
(c) a monoacetyl derivative
(d) a soluble nitrite salt
Answer:
(d) a soluble nitrite salt

21. Ammes are basic in nature because
(a) of the nitrogen atom contain or lone pair of electrons
(b) they give H+ ions in aqueous medium
(c) they form quaternary ammonium salts when heated with acids
(d) both (a) and (c)
Answer:
(a) of the nitrogen atom contain or lone pair of electrons

22. An aqueous solution of primary amine contains
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 233
Answer:
(d)

23. The basic nature of amines in an aqueous solution is in the order of
(a) tert. > sec. > pri.
(b) sec. > pri. > tert.
(b) pri. > sec. > tert.
(d) pri. > tert. > sec.
Answer:
(b) pri. > sec. > tert.

24. In trimethyl ammonium ion, the number of sigma bonds attached to nitrogen are
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

25. The number of coordinate bond/bonds in a trialkyl ammonium ion is
(a) one
(b) two
(c) three
(d) four
Answer:
(a) one

26. The number of electrons in the valence shell of nitrogen in methyl amine is
(a) 5
(b) 3
(c) 8
(d) 7
Answer:
(c) 8

27. Ethanamine reacts with excess of acetyl chloride to form
(a) C₂H₅NHCOCH₃
(b) C₂H₅N(CH₃)₂
(c) C₂H₅N(COCH₃)₂
(d) C₂H₅N+H₃Cl
Answer:
(c) C₂H₅N(COCH₃)₂

28. The compound used for acylation of amine is
(a) (CH₃CO)₂O
(b) CH₃COOH
(c) CH₃COCl
(d) both (a) and (c)
Answer:
(d) both (a) and (c)

29. Dimethyl amine reacts with acetyl chloride to give
(a) N-acetyl methyl amine
(b) N-acetyl ethyl amine
(c) N-acetyl dimethyl amine
(d) N-acetyl diethyl amine
Answer:
(c) N-acetyl dimethyl amine

30. Identify ‘A’ in the following reaction :
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 234
Answer:
(c)

31. n-propyl alcohol is obtained when HNO₂ is treated with
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 235
Answer:
(c)

32. A mixture of CH₃NH₂, (CH₃)₂NH, (CH₃)₃N can be distinguished by using
(a) HCI
(b) HNO₂
(c) HNO₃
(d) H₂SO₄
Answer:
(b) HNO₂

33. In the acetylation reaction the H-atom of an amine is replaced by
(a) a carbonyl group
(b) an alkyl group
(c) an acetyl group
(d) an imino group
Answer:
(c) an acetyl group

34. Amines are basic in nature
(a) as they have a fishy odour
(b) as they form quaternary ammonium salts with alkyl halides
(c) due to the presence of an unshared pair of electrons on the nitrogen atom
(d) all of these
Answer:
(c) due to the presence of an unshared pair of electrons on the nitrogen atom

35. The correct order of increasing basic strength is
(a) NH₃ < CH₃NH₂ < (CH₃)₂NH
(b) CH₃NH₂ < (CH₃)₂NH < NH₃
(c) CH₃NH₂ < NH₃ < (CH₃)₂NH
(d) (CH₃)₂NH < NH₃ < CH₃NH₂
Answer:
(a) NH₃ < CH₃NH₂ < (CH₃)₂NH

36. Which of the following is the strongest base?
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 236
Answer:
(d)

37. Identify the weakest base amongst the following :
(a) p-methoxyaniline
(b) o-toluidine
(c) benzene-1, 4-diamine
(d) 4-aminobenzoic acid
Answer:
(d) 4-aminobenzoic acid

38. Amine that cannot be prepared by Gabriel phthalimide synthesis is
(a) aniline
(b) benzyl amine
(c) methyl amine
(d) iso-butyl amine
Answer:
(a) aniline

39. Which of the following exist as Zwitter ion?
(a) Salicylic acid
(b) Suphanilic acid
(c) p-Aminophenol
(d) p-Amino acetophenone
Answer:
(b) Suphanilic acid

40. Reduction of benzene diazonium chloride with Zn/HCl gives
(a) phenyl hydrazine
(b) hydrazine hydrate
(c) aniline
(d) ozo benzene
Answer:
(c) aniline

41. When primary amine reacts with CHCl₃ in alcoholic KOH, the product is
(a) aldehyde
(b) alcohol
(c) cyanide
(d) an isocyanide
Answer:
(d) an isocyanide

42. Which of the following amines cannot be prepared by Gabriel phthalimide synthesis?
(a) sec-Propylamine
(b) tert-Butylamine
(c) 2-Phenylethylamine
(d) N-Methyl benzyl amine
Answer:
(d) N-Methyl benzyl amine

43. Which of the following compounds has highest boiling point?
(a) Ethane
(b) Ethanoic acid
(c) Ethanol
(d) Ethanamine
Answer:
(b) Ethanoic acid

44. Identify the statement about the basic nature of amines.
(a) Alkylamines are weaker bases than ammonia.
(b) Arylamines are stronger bases than alkylamines.
(c) Secondary aliphatic amines are stronger bases than primary aliphatic amines.
(d) Tertiary aliphatic amines are weaker bases than arylamines.
Answer:
(c) Secondary aliphatic amines are stronger bases than primary aliphatic amines.

45. The compounds ‘A’, ‘B’ and ‘C’ react with methyl iodide to give finally quaternary ammonium iodides. Only ‘C’ gives carbylamines test while only ‘A’ forms yellow oily compound on reaction with nitrous acid. The compounds ‘A’, ‘B’ and ‘C’ are respectively.
(a) butan-1-amine, N-ethylethanamine and
N, N-dimethylethanamine.
(b) N-ethylethanamine, N, N-dimethylethanamine and butan-1 – amine.
(c) N, N-dimethylethanamine, N-ethylethanamine and butan-1-amine.
(d) N-ethylethanamine, butan-1-amine and N-ethylethanamine.
Answer:
(b) N-ethylethanamine, N, N-dimethylethanamine and butan-1 – amine.

46. Which of the following amines is most basic in nature?
(a) 2, 4-Dichloroaniline
(b) 2, 4-Dimethylaniline
(c) 2, 4-Dinitroaniline
(d) 2, 4-Dibromoaniline
Answer:
(b) 2, 4-Dimethylaniline

47. How many moles of methyl iodide are required to convert ethylamine, diethylamine and triethylamine into quaternary ammonium salt, respectively?
(a) 1, 2 and 3
(b) 2, 3 and 1
(c) 3, 2 and 1
(d) 3, 1 and 2
Answer:
(c) 3, 2 and 1

48. Which of the following amines does not undergo acetylation?
(a) t-Butylamine
(b) Ethylamine
(c) Diethylamine
(d) Triethylamine
Answer:
(d) Triethylamine

49. n-Propylamine can be prepared by catalytic reduction of
(a) n-propyl cyanide
(b) propionaldoxime
(c) acetoxime
(d) nitroethane
Answer:
(b) propionaldoxime

50. Identify the compound ‘B’ in the following series of reactions :
Maharashtra Board Class 12 Chemistry Solutions Chapter 13 Amines 237
Answer:
(c)

51. Chloropicrin is used as
(a) antiseptic
(b) antibiotic
(c) insecticide
(d) anaesthetic
Answer:
(c) insecticide

52. Identify the compound B in the following series of reactions.
(a) n-propyl chloride
(b) propanamine
(c) n-propyl alcohol
(d) Isopropyl alcohol
Answer:
(c) n-propyl alcohol

53. Which of the following amines yields foul smelling product with haloform and alcoholic KOH?
(a) Ethyl amine
(b) Diethyl amine
(c) Triethyl amine
(d) Ethyl methyl amine
Answer:
(a) Ethyl amine

54. Which of the following compounds is NOT prepared by the action of alcoholic NH₃ on alkyl halide?
(a) CH₃NH₂
(b) CH₃-CH₂-NH₂
(c) CH₃ – CH₂ – CH₂ – NH₂
(d) (CH₃)₃CNH₂
Answer:
(d) (CH₃)₃CNH₂

Maharashtra Board 12th BK Important Questions Chapter 2 Accounts of ‘Not for Profit’ Concerns

July 19, 2024July 18, 2024 by Bhagya

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Important Questions Chapter 2 Accounts of ‘Not for Profit’ Concerns Important Questions and Answers.

Maharashtra State Board 12th Commerce BK Important Questions Chapter 2 Accounts of ‘Not for Profit’ Concerns

1. Objective Type Questions:

A. Select the most appropriate alternatives from those given below.

Question 1.
Purchases of stationery is a ______________ expenditure.
(a) capital
(b) revenue
(c) long-term
(d) deferred revenue
Answer:
(b) revenue

Question 2.
Usually ______________ is a major source of revenue income for ‘Not for Profit’ concerns.
(a) subscription
(b) donations
(c) legacies
(d) entrance fees
Answer:
(a) subscription

Question 3.
An Income and Expenditure Account and a Balance Sheet are prepared as final accounts by a ______________
(a) ‘Not for Profit’ concern
(b) Trading concern
(c) commercial organisation
(d) Public Limited Company
Answer:
(a) ‘Not for Profit’ concern

Question 4.
Non-cash items are not recorded in ______________
(a) Income and Expenditure Account
(b) Receipts and Payments Account
(c) Balance Sheet
(d) Profit and Loss Account
Answer:
(b) Receipts and Payments Account

Question 5.
The excess of assets over liabilities is termed as ______________
(a) surplus
(b) deficit
(c) capital fund
(d) loan
Answer:
(c) capital fund

Question 6.
For a library, expenditure on the purchase of books is a ______________ Expenditure.
(a) Capital
(b) Revenue
(c) General
(d) Recurring
Answer:
(a) Capital

Question 7.
______________ Account starts with the opening Cash Balance.
(a) Income and Expenditure
(b) Receipts and Payments
(c) Capital Fund
(d) Subscriptions
Answer:
(b) Receipts and Payments

Question 8.
Only ______________ incomes and expenses are shown in the Income and Expenditure Account.
(a) revenue
(b) capital
(c) business
(d) non-recurring
Answer:
(a) revenue

Question 9.
A debit balance on the Income and Expenditure Account denotes ______________
(a) deficit
(b) surplus
(c) profit
(d) excess of income over expenditure
Answer:
(a) deficit

Question 10.
Non-trading organisation writes summary of all cash transactions in the ______________ Account.
(a) Cash
(b) Receipts and Payments
(c) Income and Expenditure
(d) Profit and Loss
Answer:
(b) Receipts and Payments

Question 11.
Both capitalised receipts and capital expenditure are shown in the ______________
(a) Profit and Loss A/c
(b) Balance Sheet
(c) Trading A/c
(d) Income and Expenditure A/c
Answer:
(b) Balance Sheet

Question 12.
‘Not for Profit’ organisation prepares ______________ to find out its financial position.
(a) Receipts and Payments A/c
(b) Balance Sheet
(c) Trading A/c
(d) Income and Expenditure A/c
Answer:
(b) Balance Sheet

Question 13.
Subscriptions received from the members is considered as ______________ receipts.
(a) capital
(b) revenue
(c) non-recurring
(d) commercial
Answer:
(b) revenue

Question 14.
Fund which provides permanent source of income to non-trading organisation is called ______________ fund.
(a) endowment
(b) general
(c) specific
(d) capital
Answer:
(a) endowment

Question 15.
For a public hospital, expenditure on the purchase of medicines is a ______________ Expenditure.
(a) General
(b) Non-recurring
(c) Capital
(d) Revenue
Answer:
(d) Revenue

Question 16.
Which of the following items will not appear in the Balance Sheet of a club?
(a) Subscriptions received in advance
(b) Special donation received during the year
(c) Subscriptions due for the year
(d) Entrance fees paid by new members
Answer:
(d) Entrance fees paid by new members

Question 17.
The main purpose of incurring a ______________ expenditure is to earn an income or to increase the earning capacity of the business.
(a) recurring
(b) capital
(c) revenue
(d) business
Answer:
(b) capital

Question 18.
Excess of Expenditure over Income is termed as ______________
(a) Surplus
(b) Deficit
(c) Capital Fund
(d) Profit
Answer:
(b) Deficit

Question 19.
Receipts and Payments Account is a ______________
(a) Personal Account
(b) Real Account
(c) Nominal Account
(d) Profit and Loss Account
Answer:
(b) Real Account

B. Write the Word/Term/Phrase which can substitute each of the following statements.

Question 1.
Such concerns, which are formed for rendering some useful services to its members without having profit motive.
Answer:
‘Not for Profit’ concern

Question 2.
Excess of expenditure over income of ‘Not for Profit’ concerns.
Answer:
Deficit

Question 3.
A statement showing the financial position of a concern on a particular date.
Answer:
Balance Sheet

Question 4.
The debit balance of an Income and Expenditure Account.
Answer:
Deficit

Question 5.
Fees received from the member only once at the time of his entry into the ‘Not for Profit’ concern.
Answer:
Life membership fees

Question 6.
Specific amount paid by the members annually to non-trading organisation to get certain services or benefits.
Answer:
Subscription

Question 7.
Donation or gift received from the members or outsiders for specific purpose.
Answer:
Specific donation

Question 8.
Donation received for general purpose like welfare of the members or society.
Answer:
General donation

Question 9.
The gifts received from legal representatives as per the will of a deceased person.
Answer:
Legacies

Question 10.
An expenditure which is incurred for carrying the day-to-day business activities.
Answer:
Revenue Expenditure

Question 11.
Capital resources which are owned and possessed by the ‘Not for Profit’ concern.
Answer:
Capital Fund

Question 12.
An account which is prepared by the ‘ Not for Profit’ concern to record summary of all types of receipts and payments.
Answer:
Receipts and Payments Account

Question 13.
Closing debit balance of Receipts and Payments Account.
Answer:
Cash in Hand and or Cash at Bank

Question 14.
A payment made by the non-trading organisation periodically for consecutive issue of magazines, newspapers, etc.
Answer:
Subscriptions paid

Question 15.
The major source of revenue to a ‘Not for Profit’ concern, from its members.
Answer:
Subscriptions

Question 16.
Fees paid by persons to become members of a ‘Not for Profit’ concern.
Answer:
Entrance Fees or Admission Fees

Question 17.
The concerns which prepare Income and Expenditure Account instead of Profit and Loss Account.
Answer:
‘Not for Profit’ concern

Question 18.
The expenditure which is recurring in nature.
Answer:
Revenue Expenditure

Question 19.
The expenditure which is non-recurring in nature.
Answer:
Capital Expenditure

Question 20.
An account opened by non-trading concerns, to find out surplus/deficit during the particular financial year.
Answer:
Income and Expenditure Account

C. State whether the following statements are True or False with reasons.

Question 1.
All receipts are the items of revenue income.
Answer:
This statement is False.
In ‘Not for Profit’ concern receipts includes revenue receipts as well as capital receipts of current year or of previous year or of next year, so we can say that all receipts are not the items of revenue income.

Question 2.
In the Income and Expenditure Account, all incomes received during the year irrespective of the year for which they are received, are to be recorded.
Answer:
This statement is False.
In the Income and Expenditure Account, all revenue incomes, received during the current year, are to be recorded only.

Question 3.
Receipts and Payments Account shows the amount of profit earned or loss suffered during a year.
Answer:
This statement is False.
Receipts and Payments Account shows the amount of receipts and payments (in cash or through bank) of any year in the current year and do not shows any profit earned or loss suffered during a year.

Question 4.
Every year ‘Not for Profit’ concerns, prepares Income and Expenditure Account.
Answer:
This statement is True.
To get the idea of sufficient income, other expenditure, for smooth run of concern, every year ‘Not for Profit’ concern prepares Income and Expenditure Account.

Question 5.
‘Deficit’ means excess of income over expenditure in the Income and Expenditure Account.
Answer:
This statement is False.
In the Income and Expenditure Account ‘Deficit’ means excess of expenditure over incomes.

Question 6.
‘Revenue receipts’ means receipts which are not recurring in nature.
Answer:
This statement is False.
Revenue Receipts means receipts which frequently takes place and which are recurring in nature.

D. Fill in the blanks.

Question 1.
An Income and Expenditure Account and a Balance Sheet are prepared as final account by a ______________
Answer:
‘Not for Profit’ concern

Question 2.
______________ is a major source of revenue income for ‘Not for Profit’ concern.
Answer:
Subscription

Question 3.
Non-cash items are not recorded in ______________
Answer:
Receipts and Payments Account

Question 4.
______________ concerns have profit motive.
Answer:
Trading

Question 5.
In a Trading concerns ______________ is prepared to know the financial position of the business.
Answer:
Balance Sheet

Question 6.
Excess of Receipts over Payments means ______________
Answer:
Cash or Bank balance

Question 7.
Legacy is received by ‘Not for Profit’ concerns on a ______________ basis.
Answer:
permanent

Question 8.
Incomes which are to be capitalised are to be added to ______________
Answer:
Capital Fund

Question 9.
Subscription received in advance, in current year, is to be ______________ from subscription amount.
Answer:
subtracted

Question 10.
Subscription received in advance, in previous year is to be ______________ to subscription amount.
Answer:
added

Question 11.
Locker’s rent is ______________ for ‘Not for Profit’ concern.
Answer:
revenue income

Question 12.
Life membership fees, Legacy, Surplus, etc. are to be ______________ to Capital Fund.
Answer:
added

Question 13.
All revenue incomes and revenue expenses are to be recorded in ______________
Answer:
Income and Expenditure Account

Question 14.
All Receipts and Payments are recorded in the Receipts and Payment Account ______________ of the year.
Answer:
irrespective

Question 15.
Incomes or Expenses having recurring nature are known as ______________ incomes or expenses.
Answer:
revenue

E. Answer in one sentence only.

Question 1.
What is deferred revenue expenditure?
Answer:
Expenditure which is basically revenue expenditure but benefits of which accured to the organisation for more than one year is called deferred revenue expenditure.

Question 2.
What is Entrance Fees?
Answer:
The fees which is paid by the persons who wish to become a member of the organisation are called Entrance Fees.

Question 3.
What is Deficit?
Answer:
Excess of expenditure over income shown by Income and Expenditure Account is called Deficit for the financial year.

Question 4.
State the meaning of Revenue Expenditure.
Answer:
An expenditure which is incurred for carrying day-to-day business activities and maintaining fixed assets in working condition is called Revenue Expenditure.

Question 5.
What do you mean by Capital Expenditure?
Answer:
An expenditure which is non recurring is nature and incurred to purchase new fixed assets, to increase earning capacity, efficiency and working life of the existing fixed assets and to achieve economy of operation an existing fixed assets is called Capital Expenditure.

Question 6.
Which final accounts do the ‘Not for Profit’ concern prepare?
Answer:
‘Not for Profit’ concern prepare Income and Expenditure Account and Balance Sheet as their final accounts.

Question 7.
Give the examples for ‘Not for Profit’ concerns.
Answer:
Examples of ‘Not for Profit’ concerns are: sports club, charitable hospitals, schools, colleges, universities, welfare association, chamber of commerce, etc.

Question 8.
What do you mean by Recurring Expenses?
Answer:
Recurring expenses are those expenses, benefit of which lasts for a maximum period of one year and is increased on purchase of goods or services, in order to carry out the main activity of the business.

Question 9.
Why Receipts and Payments Account is prepared?
Answer:
To record all cash and Bank transactions taken place in the organisation, Receipt and Payment Account is prepared.

Question 10.
In Income and Expenditure Account, which kind of incomes and expenses are to be recorded?
Answer:
In Income and Expenditure Account, ‘Revenue’ incomes and expenses are to be recorded.

Solved Problems

Question 1.
With the information given below regarding ‘Subscription’ give accounting effects of it in the Final Accounts of a ‘Not for Profit’ concern:
Maharashtra Board 12th BK Important Questions Chapter 2 Accounts of 'Not for Profit' Concerns Q1
Additional Information:
1. Subscriptions received during the year includes:
Subscriptions received for 2018-2019 ₹ 8,750 and for 2020-21 ₹ 7,500.
2. There are 500 members of the concern and each member pays ₹ 500 as annual subscription.
3. During the year 2018-19 subscription received for the year 2019-20 was ₹ 12,500.
Solution:
In the books of _____________________
Maharashtra Board 12th BK Important Questions Chapter 2 Accounts of 'Not for Profit' Concerns Q1.1
Balance Sheet as on 31st March, 2020

Working Notes:
Amount of subscriptions outstanding for current year 2019-20 is calculated as follows:
Subscriptions outstanding (receivable) = (Subscriptions due from all members) – (Subscriptions received)
= (500 members × ₹ 500 per member) – (Subscriptions received during current year + Subscription received during previous year)
= (500 × 500) – (2,27,500 + 12,500)
= 2,50,000 – 2,40,000
= ₹ 10,000.

Question 2.
Show the following items in the Income and Expenditure Account for the year ended 31st March, 2018:
Maharashtra Board 12th BK Important Questions Chapter 2 Accounts of 'Not for Profit' Concerns Q2
Adjustments:
1. Outstanding salaries for 2016-17 is ₹ 11,250 and of 2017 – 18 is ₹ 8,125.
2. Opening stock of stationery is ₹ 5,000 and Closing stock of stationery is ₹ 2,500.
3. There are 70 members paying annual subscription of ₹ 250/- each.
4. Insurance is paid for year ended 30th June, 2018.
Solution:
In the books of _____________________
Maharashtra Board 12th BK Important Questions Chapter 2 Accounts of 'Not for Profit' Concerns Q2.1
Working Notes:
1. Outstanding subscriptions for the current year 2017-18 are calculated as follows:
Outstanding subscriptions = Subscriptions due or receivable – Subscriptions received
= 70 × 250 – 15,250
= 17,500 – 15,250
= ₹ 2,250.
Subscriptions for 2016 – 17 and 2018 – 19 will not appear in the Income and Expenditure Account prepared for 2017 – 18.

2. Prepaid insurance is calculated as follows:
Insurance is paid in advance for 3 months i.e. from 1st April, 2018 to 30th June, 2018.
Prepaid insurance = \(\frac{3}{12}\) × Insurance premium paid
= \(\frac{3}{12}\) × 10,000
= ₹ 2,500.
Prepaid insurance deducted from Insurance on the Debit side of Income and Expenditure Account.

Question 3.
The following is the Receipts and Payments Account for the year ended on 31st March, 2019:
Maharashtra Board 12th BK Important Questions Chapter 2 Accounts of 'Not for Profit' Concerns Q3
Adjustments:
1. Outstanding picnic receipts ₹ 2,550.
2. Furniture was purchased on 01 – 10 – 2018 and it is to be depreciated @ 10% p.a.
3. Outstanding subscriptions for current year ₹ 4,920.
4. Stock of Stationery on 1st April 2018 was ₹ 390 and on 31st March 2019 was ₹ 690.
5. Entire amount of legacies and 50 % of donations are to be capitalized.
With the above information, you are required to prepare Income and Expenditure Account for the year ending on 31st March 2019.
Solution:
In the books of _____________________
Maharashtra Board 12th BK Important Questions Chapter 2 Accounts of 'Not for Profit' Concerns Q3.1

Question 4.
From the information given below of Sarthi Education, you are required to prepare Income and Expenditure Account and Balance Sheet for the year ended on 31st March, 2019:
Balance Sheet as on 1st April, 2018
Maharashtra Board 12th BK Important Questions Chapter 2 Accounts of 'Not for Profit' Concerns Q4

Adjustments:
1. Tuition fees outstanding ₹ 6,750.
2. Outstanding interest on loan ₹ 30,000.
3. Entire admission Fees are to be capitalized.
4. Depreciation is to be written off as under:
Library Books ₹ 25,000, Furniture ₹ 15,000, Laboratory Equipment ₹ 10,000, Building ₹ 15,000.
Solution:
In the books of Sarthi Education
Maharashtra Board 12th BK Important Questions Chapter 2 Accounts of 'Not for Profit' Concerns Q4.2
Balance Sheet as on 31st March, 2019

Working Notes:
1. Tuition fees outstanding ₹ 6,750 are first added to Tuition Fees on the credit side of Income and Expenditure A/c and then such outstanding tuition fees are shown on the Assets side of the Balance Sheet.
2. Outstanding interest on Loan ₹ 30,000 is first debited to Income and Expenditure A/c and it is added to Loan on the Liabilities side of Balance Sheet.
3. Government grant (Revenue income) ₹ 1,75,000 is recorded on the credit side of the Income and Expenditure Account.
4. Debit balance of Income and Expenditure Account ₹ 55,850 indicates a deficit. It is deducted from the Capital fund on the Liabilities side of the Balance Sheet.

Maharashtra Board Class 11 Political Science Important Questions Chapter 5 Concept of Representation

July 19, 2024July 18, 2024 by Bhagya

Balbharti Maharashtra State Board Class 11 Political Science Important Questions Chapter 5 Concept of Representation Important Questions and Answers.

Maharashtra State Board 11th Political Science Important Questions Chapter 5 Concept of Representation

1A. Choose the correct alternative and complete the following statements.

Question 1.
Today, in most countries, the form of government is ___________ democracy. (direct, indirect, concurrent, national)
Answer:
indirect

Question 2.
After the Civil War (1640’s) UK become a ___________ (Republic, Constitutional Monarchy, Absolute Monarchy, Federation)
Answer:
Constitutional Monarchy

Question 3.
The first general elections was held in the year ___________ in India. (1947-48, 1950, 1951-52, 1935)
Answer:
1951-52

Question 4.
In the plurality method, ___________ constituency is required. (single member, multi member, transferable, non-official)
Answer:
single member

Question 5.
___________ System is used in India for presidential elections. (List, FPTP, Majority, Single Transferable vote)
Answer:
Majority

Question 6.
In India ___________ classifies parties as ‘State’ or ‘National’ and allots symbols to them. (President, Parliament, Election Commission, Judiciary)
Answer:
Election Commission

1B. Identify the incorrect pair in every set, correct it and rewrite.

Question 1.
BMS – Trade Union.
ABVP – Peasant’s Union.
FICCI – Business Group
BKU – Agricultural Unions
Answer:
ABVP – Student Union

Question 2.
Government of India Act – 1935
Queen’s Proclamation – 1858
French Revolution – 776
Morley – Minto Reforms – 1909
Answer:
French Revolution – 1789

1C. State the appropriate concept for the given statements.

Question 1.
The idea in the middle ages in Europe was that the king was God’s representative on earth.
Answer:
Divine Rights of Kings

Question 2.
Distinct geographical areas from which representatives are elected.
Answer:
Constituencies

Question 3.
The electoral system in which the candidate who secures the maximum number of votes is declared as elected.
Answer:
Plurality System

Question 4.
Views, objectives of a political party taken together.
Answer:
Ideology

Question 5.
All Indian parties have at least 11 seats in the Lok Sabha from at least three states.
Answer:
National Party

1D. Answer in one sentence.

Question 1.
What is a single-member constituency?
Answer:
A single-member constituency is one from which only a single member can be declared elected.

Question 2.
What is a multi-member constituency?
Answer:
A constituency from which several members can be elected is called a multi-member constituency.

Question 3.
What is the First Part of the Post System?
Answer:
It is a system wherein the candidate with the maximum number of votes is declared elected.

Question 4.
What is Single Transferable Vote System?
Answer:
It is a type of Proportional Representation where voters rank candidates in order of preference.

Question 5.
What is meant by the ideology of a political party?
Answer:
The reflection of the overall views, objectives, and policies of a political party is called its ideology.

Question 6.
Name some state parties in Maharashtra.
Answer:
Shiv Sena, Maharashtra Navnirman Sena, Vanchit Bahujan Aghadi, Rashtriya Samaj Paksha.

1E. Complete the following sentence using the appropriate reason.

Question 1.
Representative democracy is referred to as responsible Government because
(a) representatives are ultimately responsible to the people.
(b) people are responsible for electing the government.
(c) direct democracy is not possible today.
Answer:
(a) representatives are ultimately responsible to the people.

Question 2.
In many European countries, a struggle between the representative assemblies and monarchs arose because
(a) monarchs believed in the Divine Rights of Kings.
(b) it was a period of national awakening.
(c) representative assemblies started insisting on a share in the decision-making process.
Answer:
(c) representative assemblies started insisting on a share in the decision-making process.

Question 3.
Elections to the Lok Sabha is called the ‘First Past the Post’ method because
(a) it is a single-member constituency.
(b) candidates are ranked in order of preference.
(c) the candidate who gets a maximum number of votes is declared as elected.
Answer:
(c) the candidate who gets a maximum number of votes is declared as elected.

1F. Find the odd word in the given set.

Question 1.
Indian National Congress, Bharatiya Janata Party, Shiv Sena, Communist Party of India (Marxist).
Answer:
Shiv Sena (it is a regional party)

Question 2.
Nationalist Congress Party, All India Anna Dravida Munnetra Kazhagam, Telugu Desam Party, Akali Dal.
Answer:
Nationalist Congress Party (it is a national party)

Question 3.
National Students Union of India, Hind Mazdoor Sangh, Akhil Bharatiya Vidyarthi Parishad, Student Federation of India.
Answer:
Hind Mazdoor Sangh (it is a labour pressure group)

2A. State whether the following statements are true or false with reasons.

Question 1.
Today, most countries have an indirect or representative democracy.
Answer:
This statement is True.

Today, most countries have large territories and populations. Hence, direct democracy is not possible. The form of democracy today is indirect democracy or representative democracy.
People elect representatives from among themselves to govern the country for e.g., in India, members of Parliament (MP’s), Members of State Legislative Assemblies/Councils (MLA’s, MLC’s), of corporations, etc. are all our representatives.

Question 2.
First Past the Post system is followed for Lok Sabha elections.
Answer:
This statement is True.

Lok Sabha (general) elections are held all over the country.
The candidate who gets the maximum number of votes is declared as elected from that constituency. He / She does not need a majority of votes.

Question 3.
Proportional Representation has limited scope.
Answer:
This statement is True.

In Proportional Representation the number of candidates of a political party to be elected depends on the proportion of votes that it receives.
It is a lengthy process and needs a multi-member constituency. Hence it is unsuitable for large-scale elections such as elections to the Lok Sabha.

Question 4.
Political parties are important channels for political representation.
Answer:
This statement is True.

Political parties serve as the primary channels of political representation. In democracies, parties seek to obtain power through elections. Members of various parties contest elections as candidates of their respective parties.
During the election, the parties present before the voters a programme based on their ideology and promise them that this programme would be implemented if elected to power. Thus, the aspirations and wishes of the voters are represented in the decision-making process through the channel or the medium of a given political party.

Question 5.
Telugu Desam (TDP) is a national party.
Answer:
This statement is False.

A national party must have a political presence in at least four states.
TDP is significant only in Andhra Pradesh and to some extent in Telangana.

Question 6.
The Communist Party of India (CPI) can be described as the first political party in the country.
Answer:
This statement is False.

CPI was formed in 1925 by people who were influenced by the Bolshevik Revolution in Russia (1917) and the communist ideology.
In 1885, the Indian National Congress was formed as a united front against British Rule. It is considered the first political party in India.

2B. Complete the concept maps.

Question 1.
Maharashtra Board Class 11 Political Science Important Questions Chapter 5 Concept of Representation 2B Q1

Answer:
Maharashtra Board Class 11 Political Science Important Questions Chapter 5 Concept of Representation 2B Q1.1

3. Explain the correlation between the following.

Question 1.
Political Parties and Pressure Groups.
Answer:
Political parties are the most important channels for political representation. They are organized groups comprising of persons who hold similar views on a variety of issues or have similar objectives. They seek to obtain political power, generally, through the process of elections. The views of a party taken together are called the party’s ideology. At election times, political parties issue ‘Manifestos’ i.e., what policies/programmes they would implement if voted to power. Every party puts up its candidates who contest election.

Interest and Pressure groups are informal channels that seek to represent the people. A pressure group is an interest group that is organized to influence public opinion and government policy towards the fulfillment of its objectives and without active participation in the electoral process. This includes interest groups in the fields of business such as the Federation of Indian Chambers of Commerce and Industry (FICCI), for labour e.g., Indian National Trade Union Congress (INTUC), Bharatiya Kamgar Sena (BKS), for peasants such as Shetkari Sanghatana, for students such as Akhil Bharatiya Vidyarthi Parishad (ABVP), National Students Union of India (NSUI). In the USA, pressure groups are also called Lobby Groups.

Many pressure groups today are closely affiliated with political parties e.g., ABVP is the student wing of BJP and BKS is the Shiv Sena’s Trade Union. Both the pressure group and the political party will then support each other in times of elections and decision-making in aspects like finance, manpower, and publicity.

4. Answer the following questions.

Question 1.
Explain the Divine Rights Theory.
Answer:
The Divine Rights of Kings Theory was propagated in Europe by Kings like James I (England). It explained that the Monarchs were God’s representatives on earth to whom, the people had to render habitual obedience. The King was infallible and unquestionable as he derived his power from God. Disobedience to the King was akin to sinning against God. This theory was used to strengthen the Absolute Monarchy in Europe.

Question 2.
Explain Representative Assemblies in Europe.
Answer:
Representative Democracy has its origins in medieval Europe. Till that time, Absolute Monarchies existed in most countries. The Divine Rights of Kings Theory was in application. As time went by, monarchs in many countries like England started having Representative Assemblies that represented the population. Soon, these assemblies asked for a share in the decision-making process of the country leading to conflicts between the monarchs and the Assemblies for e.g. French Revolution. Most conflicts ended with reduced/power to the monarchs. The Representative assemblies, now become Political Representatives as they dealt with all government activities.

Question 3.
What are the three methods of representation?
Answer:
The three methods of representation are

Electoral Method: Persons are directly or indirectly elected by the citizens to govern them as members of representative assemblies e.g., General elections to Lok Sabha and Legislative Assemblies.
Non-electoral Method: Representatives occupy their position through nomination or appointment for e.g., the President of India appoints 12 members to the Rajya Sabha.
Non-official Method: Civil society represents the people through various pressure groups like trade unions, student groups, etc.

Question 4.
Explain the Proportional System of Representation.
Answer:
Proportional Representation is generally used in multi-member constituencies. In this system, the number of candidates of a given political party to be elected depends upon the proportion of votes that it receives. For instance, if a political party receives 40% of the votes in a five-member constituency, then two of its candidates will be elected from that constituency. This system is not used in India. There is a sub-type of the proportional system which is known as the Single-Transferable Vote (STV) system. Here the voters have to rank the candidates in order of preferences. This system is used in elections to the Rajya Sabha and to the State Legislative Councils in India.

Question 5.
What is a National Party?
Answer:
Political parties can be classified as National or State parties. The Election commission has decided that a political party shall be eligible to be recognized as a National party if-

It secures at least six percent (6%) of the valid votes polled in any four or more states, at a general election to the
House of the People or, to the State Legislative Assembly; and
In addition, it wins at least four seats in the House of the People from any State or state.

It wins at least two percent (2%) seats in the House of the People (i.e., 11 seats in the existing House having 543 members), and these members are elected from at least three different States.
In India, some National Parties include I.N.C., B.J.P., G.P.M., N.C.P., etc.

Question 6.
Name six regional (state) parties.
Answer:

Telugu Desam Party (TDP) – Andhra Pradesh,
Telangana Rashtriya Samiti – Telangana,
Dravida Munnetra Kazhagam (DMK) and All India Dravida Munnetra Kazhagam (AIADMK) – Tamil Nadu,
National Conference – Jammu & Kashmir
Assam Gana Parishad (AGP) – Assam.
Shiv Sena – Maharashtra.

Question 7.
Name 4 trade unions and the political parties they are affiliated to.
Answer:

Trade Unions	Political Parties affiliated to
Indian National Trade Union Congress (INTUC)	Indian National Congress
All India Trade Union Congress (AITU)	Communist Party of India
Bharatiya Kamgar Sena (BKS)	Shiv Sena
The Bharatiya Mazdoor Sangh (BMS)	Bharatiya Janata Party (B.J.P)
Centre for Indian Trade Unions (CITU)	Communist Party of India (Marxist) (CPM)

Question 8.
How do pressure groups differ from social movements?
Answer:
Pressure groups are different from social movements. The pressure groups usually have a more formalized structure. This is why sometimes interest groups are described as representing ‘organized interest’. Social movements usually do not have a formal structure or organisation. They take up a cause and pursue it. (Example: Chipko Movement)

Question 9.
Write two examples of NGOs in India in the following field.

Child Welfare
Animal Welfare
Aged Persons
Disabled Persons
Environment
Women’s Welfare

Answer:

Child welfare – Child Rights and You (CRY), Akansha
Animal Welfare – PETA, People for Animals (PFA)
Aged Persons – Help Age, Dignity Foundation.
(iv) Disabled Persons – National Association for the Blind (NAB), GCCI, ADAPT.
Environment – BNHS, BEAG.
Women’s Welfare – SEWA, SNEHA, WIT

5. Observe the given image and writes in brief about it.
Maharashtra Board Class 11 Political Science Important Questions Chapter 5 Concept of Representation 5
Answer:
This is a heart-warming and motivating photograph.
We can observe the following about it.

In India, women have participated in the election process since 1950 when they were given voting rights.
It shows political awareness and participation of women, dressed in traditional attire, braving the hot sun standing in the queue to vote.
They are proudly holding up their identity proof which shows how motivated and proud they are to have the political right to vote.