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Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Important Questions Chapter 7 Bills of Exchange Important Questions and Answers.

Maharashtra State Board 12th Commerce BK Important Questions Chapter 7 Bills of Exchange

Objective Questions

A. Select the correct option and rewrite the sentence:

Question 1.
A bill of exchange is called a _____________ by one who is entitled to receive the amount due on it.
(a) Bills Payable
(b) Draft
(c) Bills Receivable
(d) Promissory Note
Answer:
(c) Bills Receivable

Question 2.
The person who draws a bill of exchange is called _____________
(a) Payee
(b) Drawee
(c) Endorsee
(d) Drawer
Answer:
(d) Drawer

Question 3.
A bill of exchange is required to be _____________ by drawee.
(a) drafted
(b) discounted
(c) accepted
(d) endorsed
Answer:
(c) accepted

Question 4.
A person who accepts the bill is called _____________
(a) Drawer
(b) Acceptor
(c) Payee
(d) Creditor
Answer:
(b) Acceptor

Question 5.
The person to whom the amount of the bill is made payable is called _____________
OR
_____________ is a person to whom the amount on a bill is payable.
(a) Endorsee
(b) Drawer
(c) Drawee
(d) Payee
Answer:
(d) Payee

Question 6.
When the acceptor accepts the bill with certain conditions, the acceptance is called _____________ Acceptance.
(a) Qualified
(b) General
(c) Clean
(d) Special
Answer:
(a) Qualified

Question 7.
The drawee becomes an _____________ on acceptance of a bill.
(a) acceptor
(b) owner
(c) endorser
(d) drawer
Answer:
(a) acceptor

Question 8.
Borrowing money from a bank on the security of a bill of exchange is called _____________
(a) Honouring
(b) Endorsing
(c) Discounting
(d) Retiring
Answer:
(c) Discounting

Question 9.
A bill of exchange is/can be discounted with the _____________
(a) bank
(b) payee
(c) money lenders
(d) government
Answer:
(a) bank

Question 10.
Transferring a bill of exchange before maturity to a third party is called _____________ of a bill of exchange.
(a) honouring
(b) endorsement
(c) retirement
(d) discounting
Answer:
(b) endorsement

Question 11.
The person who endorses the bill of exchange is known as _____________
(a) Drawer
(b) Endorsee
(c) Endorser
(d) Drawee
Answer:
(c) Endorser

Question 12.
A person to whom a bill of exchange is endorsed is called _____________
(a) Endorsee
(b) Drawer
(c) Endorser
(d) Payee
Answer:
(a) Endorsee

Question 13.
If a bill falls due on 15th August, payment on it must be made on _____________
(a) 14th August
(b) 16th August
(c) 13th August
(d) 17th August
Answer:
(a) 14th August

Question 14.
A bill drawn on 12th June, 2020 at two months would be payable on _____________
(a) 12th August 2020
(b) 14th August 2020
(c) 15th August 2020
(d) 16th August 2020
Answer:
(b) 14th August 2020

Question 15.
If a bill is drawn on 3rd July, 2020 for 40 days, its payment must be made on _____________
(a) 14th August 2020
(b) 15th August 2020
(c) 13th August 2020
(d) 16th August 2020
Answer:
(a) 14th August 2020

Question 16.
A bill is drawn on 23rd September, 2019 at 4 months would be payable on _____________
(a) 24th January 2020
(b) 25th January 2020
(c) 26th January 2020
(d) 25th January 2019
Answer:
(b) 25th January 2020

Question 17.
A bill is drawn on 23rd October 2016 payable after 3 months, the due date of the bill will be _____________
(a) 25th January 2017
(b) 26th January 2017
(c) 24th January 2017
(d) 25th January 2016
Answer:
(a) 25th January 2017

Question 18.
_____________ means payment of the bill before due date.
(a) Discounting of Bill
(b) Retirement of Bill
(c) Renewal of Bill
(d) Endorsement of Bill
Answer:
(b) Retirement of Bill

Question 19.
A bill of one month duration is accepted on 12th July, 2020, its due date will be _____________
(a) 12th August 2020
(b) 16th August 2020
(c) 14th August 2020
(d) 15th August 2020
Answer:
(c) 14th August 2020

Question 20.
When a bill Is dishonoured, the _____________ is held responsible for the noting charges.
(a) holder
(b) drawee
(c) drawer
(d) endorser
Answer:
(b) drawee

Question 21.
Fees charged by the Notary Public for noting facts or reasons of dishonour of bill are called _____________
(a) Discount
(b) Rebate
(c) Noting Charges
(d) Commission
Answer:
(c) Noting Charges

Question 22.
Noting charges are paid when a bill is _____________
(a) honoured
(b) dishonoured
(c) renewed
(d) retired
Answer:
(b) dishonoured

Question 23.
_____________ is done in respect of dishonour of foreign bill of exchange.
(a) Discounting
(b) Endorsement
(c) Noting
(d) Protesting
Answer:
(d) Protesting

B. Give one word/phrase/term which can substitute each of the following statements:

Question 1.
A bill of exchange is drawn and accepted for a value received.
Answer:
Trade bill

Question 2.
A person who draws a bill of exchange.
Answer:
Drawer

Question 3.
A person on whom a bill of exchange is drawn.
Answer:
Drawee

Question 4.
Payment in accordance with the apparent tenor of the bill.
Answer:
Honour

Question 5.
Non-payment in accordance with the apparent tenor of the bill.
Answer:
Dishonour

Question 6.
Acceptance without making any change in the terms of a bill.
Answer:
General acceptance

Question 7.
Acceptance with some changes as regards the terms of a bill.
Answer:
Qualified acceptance

Question 8.
A bill of which payment is to be made after the fixed period.
Answer:
After date bill

Question 9.
The bill is drawn in one country and payable in other countries.
Answer:
Foreign bill

Question 10.
Encashment of the bill before the due date.
Answer:
Discounting

Question 11.
Transfer of title of the bill from the debtor to the creditor.
Answer:
Endorsement

Question 12.
Payment of the bill before the due date.
Answer:
Retirement of bill

Question 13.
A document consists of a written order signed by the maker, directing a certain person to pay a certain sum of money on-demand or on a certain future date.
Answer:
Bill of Exchange

Question 14.
A person who accepts the bill.
Answer:
Drawee or Acceptor

Question 15.
The period for which a bill is drawn.
Answer:
Term/Tenure of a bill of exchange

Question 16.
A bill of exchange that does not contain the period for its payment.
Answer:
Demand bill/Bill at sight

Question 17.
The date on which period of the bill gets expired.
Answer:
Nominal due date

Question 18.
The date on which the payment of the bill is to be made.
Answer:
Due date or Date of maturity

Question 19.
Drawee’s signature on the face of the bill to show his consent to pay the amount of the bill.
Answer:
Acceptance of a bill of exchange

Question 20.
A bill of exchange before its acceptance.
Answer:
Draft

Question 21.
A bill is drawn, accepted, and made payable within the territory of one and the same country.
Answer:
Inland bill of exchange

Question 22.
Selling a bill to the bank before its due date for an amount slightly less than its face value.
Answer:
Discounting of a bill of exchange

Question 23.
Act of signing the bill on its back by its holder to transfer its title to a third Person.
Answer:
Endorsement of a bill of exchange

Question 24.
Discount is given by holder to acceptor on the retirement of the bill of exchange.
Answer:
Rebate

Question 25.
Non-payment of the bill on the due date.
Answer:
Dishonour of a bill of exchange

Question 26.
Recording the facts of dishonour of a bill of exchange by a Notary Public.
Answer:
Noting

Question 27.
The request by the acceptor of the bill to the drawer for issuing a new bill after canceling the old bill.
Answer:
Renewal of the bill of exchange

Question 28.
The account to which the bill is sent for collection is debited.
Answer:
Bill sent for Collection Account

Question 29.
Payment of a bill on the due date.
Answer:
Honouring of a bill of exchange

Question 30.
Drafting a new bill in cancellation of the old bill at the request of drawee.
Answer:
Renewal of the bill of exchange

Question 31.
Certificate is given by Notary Public for the fact of dishonour of the bill.
Answer:
Certificate of protest

Question 32.
The account is to be debited in case of dishonour of bill in the books of the drawer.
Answer:
Drawee’s Account

Question 33.
A foreign bill accompanied by shipping documents.
Answer:
Documentary bill

C. State True or False with reasons:

Question 1.
Bills payable are a liability.
Answer:
This statement is True.
Bill is always drawn on the debtor by the creditor. The debtor i.e. drawee has to pay the money on a future date. For the drawee or acceptor of the bill, payment of the number of Bills payable is certain and therefore, for drawee, Bills payable is a liability.

Question 2.
Drawee has no right to discount the bill with the bank.
Answer:
This statement is True.
Drawee means acceptor of a bill and for him, it is bills payable. Once he accept it, sign it, and returned it to the drawer he don’t have any bill with him to discount it with the bank. He is not the owner of the bill and hence, he has no right to discount the bill with the bank.

Question 3.
A bill of exchange needs acceptance.
Answer:
This statement is True.
A bill of exchange is drawn by the creditor on the debtor. It is signed by drawer as well as by drawee. The drawee has to give his assent to the terms and conditions of the bill by putting his signature on it. A bill without acceptance is called a draft. It becomes a valid document only when the drawee accepts it.

Question 4.
A bill can’t be deposited into the bank for collection.
Answer:
This statement is False.
When a drawer or holder of the bill needs money on the due date, he has to deposit the bill into the bank for collection purposes. So that bank can collect the amount in time. This means a bill can be deposited into the bank for collection.

Question 5.
Noting charges are payable to the Notary Public in honour of a bill.
Answer:
This statement is False.
Noting charges are payable to the Notary Public at the time of registration of a dishonoured bill, not at the time of honour of a bill.

Question 6.
A bill of exchange is a negotiable instrument.
Answer:
This statement is True.
Being a negotiable instrument, a bill of exchange is a written acknowledgment of debts and also a promise to pay the debt according to the terms of the bill and can be transferred from one person to another.

Question 7.
A bill of exchange is signed by the person on whom it is drawn.
Answer:
This statement is True.
A bill of exchange is signed by the person who draws or makes it, and the person on whom it is drawn accepts it.

Question 8.
Acceptance with some change as regards the terms of a bill is called general acceptance.
Answer:
This statement is False.
General acceptance means acceptance of a draft without any change or conditions regards the terms of a bill.

Question 9.
Drawee is a person who holds the title of the bill in due course.
Answer:
This statement is False.
A drawer is a person who holds the title of the bill in due course as drawee accepts it and returns it to the drawer.

Question 10.
A payee is an official person appointed by the Central government for noting of dishonour of the bill.
Answer:
This statement is False.
A notary public is an official person appointed by the Central government for noting of dishonour of the bill and making it legal.

D. Complete the sentences:

Question 1.
A person to whom or as per his order, amount of bill is payable is a _____________
Answer:
Payee

Question 2.
The inland bill is drawn and payable in the _____________ country.
Answer:
same

Question 3.
Discounting means encashment of the bill before its _____________
Answer:
due date

Question 4.
_____________ can transfer the ownership of the bill.
Answer:
Drawer

Question 5.
Noting charges are payable to the Notary public on _____________ of a bill.
Answer:
dishonour

Question 6.
A bill of exchange is a _____________
Answer:
negotiable instrument

Question 7.
If a discounted bill is honoured, the _____________ does not record this transaction.
Answer:
drawer

Question 8.
Days of grace are not allowed in the case of a _____________
Answer:
demand bill

Question 9.
Noting charges should be borne by _____________
Answer:
drawee

E. Answer in one sentence:

Question 1.
State the types of Bills of Exchange.
Answer:
The bills of exchange may be classified as

1. Inland bills of exchange
2. Foreign bills of exchange.

Question 2.
What is the Inland bill of exchange?
Answer:
A bill of exchange that is drafted, accepted, and made payable between the parties from one and the same country is called an Inland bill of exchange.

Question 3.
What is a Foreign bill of exchange?
Answer:
A bill of exchange that is drafted and accepted in one country and made payable in another country is called a Foreign bill of exchange.

Question 4.
Which are the parties to a bill of exchange?
Answer:
There are three parties to a bill of exchange, viz.,

1. Drawer
2. Drawee
3. Payee

Question 5.
Who is the Drawer?
Answer:
The Drawer of a bill is the person who draws or makes the bill.

Question 6.
Who is the Drawee?
Answer:
The Drawee of a bill is the person on whom the bill is drawn.

Question 7.
What is a Draft?
Answer:
A bill of exchange is called a Draft before its acceptance.

Question 8.
What is an Acceptance of the Bill of Exchange?
Answer:
The act of signing the bill of exchange by the drawee with a date to show his consent to pay the amount of the bill is called an Acceptance of the Bill of Exchange.

Question 9.
What do you mean by Clean or General Acceptance?
Answer:
A Clean or General Acceptance is an acceptance where the drawee does not make any change in the terms of the bill before accepting it.

Question 10.
What is Qualified Acceptance?
Answer:
If the drawee of a bill of exchange accepts it on condition that the time or amount of the bill is changed or adds some other conditions to the bill, his acceptance is called a Qualified Acceptance.

Question 11.
What is the term of the bill of exchange?
Answer:
The period for which the bill of exchange is drawn and accepted is called the term of the bill of exchange.

Question 12.
What is the Nominal Due Date?
Answer:
The date on which the term i.e. the period of a bill of exchange gets expired is called Nominal Due Date.

Question 13.
What is the Due Date of a Bill?
Answer:
The Due Date of a Bill of Exchange is the date on which it is falling due for payment by the drawee.

Question 14.
What is Endorsing of a Bill?
Answer:
Endorsing of a Bill is the holder’s signing on its back with the intention of transferring its title or ownership to another person.

Question 15.
Who is an Endorser?
Answer:
The drawer or the holder of a bill of the exchange who transfers or endorses the same in favour of a third party is called Endorser.

Question 16.
Who is an Endorsee?
Answer:
An Endorsee is a person to whom or in whose favour a bill is endorsed or transferred.

Question 17.
What is Retirement of a Bill?
Answer:
A bill of exchange is said to be retired if its acceptor makes payment of it before its due date, usually after deducting some discount or rebate.

Question 18.
When is a bill said to be honoured?
Answer:
A bill of exchange is said to be honoured or met when the acceptor or drawee makes payment on its due date.

Question 19.
When is the bill said to be dishonoured?
Answer:
A bill of exchange is said to be dishonoured if its acceptor or drawee fails to make payment on its due date.

Question 20.
Which account is credited in the books of the drawer when the discounted bill is dishonoured?
Answer:
Cash/Bank A/c is credited in the books of the drawer when the discounted bill is dishonoured.

Question 21.
Who is a Notary Public?
Answer:
An officer appointed by the Government to certify dishonour of bills of exchange is called Notary Public.

Question 22.
Who bears the noting charges on dishonour of a bill?
Answer:
An acceptor or drawee bears the noting charges on dishonour of a bill of exchange.

Question 23.
Who pays the noting charges?
Answer:
The holder of the bill of exchange pays the noting charges.

Question 24.
What do you mean by Renewal of a Bill?
Answer:
Renewal of a Bill of Exchange means cancellation of the original bill and drafting a new bill in exchange for that by a drawer at the request of drawee.

F. Do you agree or disagree with the following statements:

Question 1.
A bill of which payment to be made after the fixed period is after date bill.
Answer:
Agree

Question 2.
Drawee can transfer the ownership of the bill.
Answer:
Disagree

Question 3.
Endorsee is a person in whose favour the bill is transferred.
Answer:
Agree

Question 4.
The drawer and payee of a bill of exchange may be one and the same person.
Answer:
Agree

Question 5.
A bill of exchange can be endorsed only once.
Answer:
Disagree

Question 6.
Days of grace are allowed in the case of demand bills.
Answer:
Disagree

Question 7.
The noting of dishonoured bills is compulsory.
Answer:
Disagree

Question 8.
The endorser is a creditor to the endorsee.
Answer:
Disagree

Question 9.
Bills payable are a liability.
Answer:
Agree

Question 10.
A bill of exchange is a negotiable instrument.
Answer:
Agree

Solved Problems

Question 1.
On 1st April 2019 Parth draws a bill for ₹ 50,000 on Zalak for 4 months period. The bill is accepted and returned to Parth. On the same date, Parth discounted the bill with his bank @ 12% p.a.
Before the due date Zalak finds herself unable to meet the bill, hence requested Parth to renew the bill for a further period of 2 months. Parth agreed and he took the bill back from the bank and received new acceptance for ₹ 52,000 including interest. This new bill is duly honoured by Zalak on the due date.
Write Journal of Parth and Zalak for the above bill transactions.
Solution:
In the books of Parth
Journal Entries

In the books of Zalak
Journal Entries

Question 2.
Prahran owes Keyur ₹ 75,000. Keyur draws a bill for ₹ 60,000 on Prihaan for 4 months period and received the cheque for the balance. The bill is duly accepted and returned by Prahran. On the same date, Keyur endorsed Prihaan’s acceptance to Monil.
On the due date, Monil informed Keyur that Prihaan dishonored his acceptance and ₹ 1,905 paid as noting charges. Keyur then drew a new bill for 3 months on Prihaan for the amount due including noting charges and interest of ₹ 2,400. On the due date, the bill was duly honoured by Prihaan.
Write Journal Entries in the books of Keyur and prepare Keyur’s account in the books of Prihaan.
Solution:
In the books of Keyur
Journal Entries

Working Notes:
1. Amount paid by cheque = Total amount due from Prihaan – the amount of Bill accepted
= 75,000 – 60,000
= ₹ 15,000

2. Amount for which new bill is drawn = Amount of bill dishonoured + Noting charges + Interest
= 60,000 + 1,905 + 2,400
= ₹ 64,305

Note: For easy understanding, the students are advised to draft the journal entries in the journal of Prihaan first. So that they will be able to understand ledger entries of Keyur’s A/c opened in the ledger of Prihaan.

Question 3.
On 1st June, 2020 Bela draws a bill for ₹ 1,00,000 on Premila for 4 months period. The bill is duly accepted and returned to Bela. One month after the date, Bela discounted the bill with bank @ 18% p.a.
On the due date, Premila dishonoured her acceptance. Bank paid noting charges ₹ 2,250. Premila requested Bela to renew the bill for a further period of 2 months. Bela agreed and took the bill back from the bank and received new acceptance for 40% amount of the bill with the full amount of noting charges and a cheque for 60% balance plus interest @ 12% p.a.
Before the due date, Premila was declared as insolvent and 30% of the amount due could be recovered from her private estate.
Write Journal of Bela and Premila for the above bill transactions.
Solution:
In the books of Bela
Journal Entries

In the books of Premila
Journal Entries

Working Notes:
1. Discount charged by the bank on discounting 1st bill = 1,00,000 × \(\frac{3}{12} \times \frac{18}{100}\) (Period of bill is 4 months, but it is discounted 1 month later) = ₹ 4,500

2. Amount paid by Premila to Bela in Part payment = 60% of Bill amount
= 60% of 1,00,000
= ₹ 60,000

3. Balance amount still due from Premila to Bela = 40% of Bill amount
= 40% of 1,00,000
= ₹ 40,000

4. Interest is to be calculated on total amount due from Premila = Balance due + Unpaid amount noting charges
= 40,000 + 2,250
= ₹ 42,250
Interest due = Balance amount × Unexpired period × Rate of interest
= 42,250 × \(\frac{2}{12} \times \frac{12}{100}\)
= ₹ 845

5. Amount paid by Premila to Bela = 60,000 + 845 = ₹ 60,845.

6. Amount for which new bill is drafted and accepted = ₹ 42,250.

7. Amount recovered by Bela from the property of Premila = 30% of total amount due
= \(\frac{30}{100}\) × 42,250
= ₹ 12,675

8. Bad debts incurred by Bela = Total Amount due – Amount recovered
= 42,250 – 12,675
= ₹ 29,575

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 4 Chemical Thermodynamics Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 4 Chemical Thermodynamics

Question 1.
Define the term energy.
Answer:
The energy of a system is defined as its capacity to perform the work. A system with higher energy can perform more work.

Question 2.
What are different forms of energy ?
Answer:
The energy of a system has many different forms as follows :

• Kinetic energy which arises due to motion, like rotational, vibrational and translational.
• Potential energy which arises due to position and state of a matter. If depends upon the temperature of the system.
• Heat energy (or thermal energy) which is transferred from the hotter body to the colder body.
• Radiant energy which is associated with electro-magnetic or light radiation.
• Electrical energy produced in the galvanic cells.
• Chemical energy stored in chemical substances.

All these various forms of energy can be converted from one form to another without any loss.

Question 3.
Explain the concept of interconversion of different forms of energy.
Answer:
There are various forms of energy like kinetic energy, potential energy, heat or thermal energy, radiant energy, electrical energy and chemical energy.

All these forms of energy are interconvertible. For example, a body at very high level possesses higher potential energy. When it falls down, potential energy is converted into kinetic energy. Falling of water from high level is used to drive turbines converting potential energy into kinetic energy which is further converted into electrical energy.

In galvanic cells, chemical energy is converted into electrical energy.
In electrolytic cells, chemical energy is converted into electrical energy. But during interconversion, the energy can neither be created nor destroyed, and there is a conservation of energy.

Question 4.
What is thermodynamics ? What are its drawbacks?
Answer:
Thermodynamics : It is concerned with the energy changes in physical and chemical changes.

Drawbacks :

• It does not give information on the rates of physical or chemical changes.
• It does not explain mechanisms involved in physical and chemical processes.

Question 5.
Define and explain :
(1) System (2) Surroundings (3) Boundary.
Answer:
(1) System : The portion of the universe under thermo-dynamic consideration to study thermodynamic properties is called a system.
Explanation :

• As such any portion of the universe under thermodynamic consideration is a system. The thermodynamic consideration involves the study of thermodynamic parameters like pressure, volume, temperature, energy, etc.
• The system may be very small or very large.
• The system is confined by a real or an imaginary boundary.

(2) Surroundings : The remaining portion of the universe other than under thermodynamics study i.e„ the system is called the surroundings.
Explanation :

• Surroundings represent a large stock of mass and energy and can be exchanged with the system when allowed.
• For a liquid in an open vessel, the surrounding atmosphere around it represents the surroundings.

(3) Boundary : The wall or interface separating the system from its surrounding is called a boundary.
Explanation :

• This boundary may be either real or imaginary.
• Through this boundary, exchange of heat and matter between the system and surroundings can take place, e.g. when a liquid is placed in a beaker the walls of beaker represent real boundaries while open portion of the beaker is imaginary boundary.
• Everything outside the boundary represents surroundings.

Question 6.
What are the types of systems ?
Answer:
Following are the types of systems :

1. Open system
2. Closed system
3. Isolated system
4. Homogeneous system
5. Heterogeneous system.

Question 7.
Define and explain the following :
(1) Open system
(2) Closed system
(3) Isolated system.
Answer:
(1) Open system :it is defined as a system which can exchange both matter and energy with its surroundings, e.g. a beaker containing water. The water continuously absorbs energy from the surroundings and forms vapour which diffuse in the surroundings. So that this system exchanges energy and matter (or mass), with the surroundings.

Fig. 4.1 : Open system

(2) Closed system : it is defined as a system which can exchange only energy but not the matter with its surroundings, e.g. A closed vessel containing hot water so that only heat is lost to the surroundings and not the matter.

Fig. 4.2 : Closed system

(3) Isolated system : it is defined as a system which can neither exchange energy nor matter with its surroundings, e.g. hot water filled in a thermally insulated closed vessel like thermos flask.

In actual practice, perfectly isolated system is not possible.
Universe represents an isolated system.

Question 8.
‘Universe is an isolated system’. Explain.
Answer:
Universe represents an isolated system due to the following reasons :

• The total mass and energy of the universe remain constant.
• The universe has no boundary.
• The universe has no surroundings.

Question 9.
Define and explain :
(1) Homogeneous system
(2) Heterogeneous system.
Answer:
(1) Homogeneous system : A system consisting of only one uniform phase is called a homogeneous system.
Explanation :
(1) The properties of homogeneous system are uniform throughout the phase or system.
(2) The homogeneous systems are :

• Solutions of miscible liquids (water and alcohol) or soluble solids in liquids, (NaCl in water), etc.
• Mixture of gases. H₂ and N₂, NH₃ and H₂, etc.

(2) Heterogeneous system : A system consisting of two or more phases separated by interfacial boundaries is called a heterogeneous system.
Explanation : These systems are :

• Mixture of two or more immiscible liquids. E.g. Water and benzene.
• Solid in equilibrium with liquid.
  E.g. Ice ⇌ water.
• Liquid in equilibrium with vapour.
  E.g. Water ⇌ vapour.

Question 10.
Explain : (A) Extensive property (B) Intensive property of a system.
OR
What is the difference between extensive and intensive properties?
Answer:
The properties of a system are classified as (A) Extensive property and (B) Intesive property.
(A) Extensive property : It is defined as a property of a system whose magnitude depends on the amount of matter present in the system.
Explanation :

• More the quantity (or amount) of the matter of the system, more is the magnitude of extensive property, e.g., mass, volume, heat, energy, enthalpy, etc.
• The extensive properties are additive.

(B) Intensive property : It is defined as a property of a system whose magnitude is independent of the amount of matter present in the system.
Explanation :

• Intensive property is characteristic of the system, e.g., refractive index, density, viscosity, temperature, pressure, boiling point, melting point, freezing point of a pure liquid, surface tension, etc.
• The intensive properties are not additive.

Question 11.
Select extensive and intensive properties in the following :
Moles, molar heat capacity, entropy, heat capacity.
Answer:
Extensive property : Moles, entropy, heat capacity.
Intensive property : Molar heat capacity.

Question 12.
What is a state function ? Give examples.
Answer:
State function : The property which depends on the state of the system and independent of the path followed by the system to attain the final state is called a state function.
For example, pressure, volume, temperature, etc.

Question 13.
Classify the following properties as intensive or extensive :
(i) Temperature (ii) Density (iii) Enthalpy (iv) Mass (v) Energy (vi) Refractive index (vii) Pressure (viii) Viscosity (ix) Volume (x) Weight.
Answer:
(1) Intensive properties : Temperature, Density, Refractive index. Pressure, Viscosity.
(2) Extensive properties : Enthalpy, Mass, Energy, Volume, Weight.

Question 14.
What are path functions?
Answer:
Path functions : The properties which depend on the path of the process are called path functions. For example, work (W) and heat (Q).

Question 15.
Define thermodynamic equilibrium. Mention different types of thermodynamic equilibria.
Answer:
Thermodynamic equilibrium : A system is said to have attained a state of thermodynamic equilibrium if there is no change in any thermodynamic functions or state functions like energy, pressure, volume, etc. with time.

For a system to be in thermodynamic equilibrium, it has to attain following three types of equilibrium :

• Thermal equilibrium
• Chemical equilibrium
• Mechanical equilibrium

Question 16.
Distinguish between :
(1) Open system and Closed system :
Open system:

1. An open system can exchange both matter and energy with the surroundings.
2. In this, the total amount of matter does not remain constant.
3. Example : Hot water kept in an open beaker.

Closed system:

1. A closed system can exchange only energy, but not matter with the surroundings.
2. In this, the total amount of matter remains constant.
3. Example : Hot water kept in a closed glass flask.

(2) Closed system and Isolated system :
Closed system:

1. A closed system can exchange only energy, but not matter with the surroundings.
2. In this, the total amount of energy does not remain constant.
3. Example : Hot water kept in a sealed glass flask.

Isolated system:

1. An isolated system can exchange neither matter nor energy with the surroundings.
2. In this, the total amount of energy remains constant.
3. Example : Hot water kept in a thermos flask.

(3) Open system and Isolated system :
Open system:

1. An open system can exchange matter with the surroundings.
2. It can exchange energy with the surroundings.
3. In this, the total amount of energy does not remain constant.
4. Example : Hot water kept in an open beaker.

Isolated system

1. An isolated system cannot exchange matter with the surroundings.
2. It cannot exchange energy with the surroundings.
3. In this, the total amount of energy remains constant.
4. Example : Hot water kept in a thermos flask.

Question 17.
What is a thermodynamic process ? What are different types of processes ?
Answer:
(i) Thermodynamic process : It is defined as a transition by which a state of a system changes from initial equilibrium state to final equilibrium state.

The process is carried out by changing the state functions or thermodynamic variables like pressure, volume and temperature. During the process one or more properties of the system change.

(ii) Types of processes :

• Isothermal process
• Isobaric process
• Isochoric process
• Adiabatic process
• Reversible process
• Irreversible (spontaneous) process.

Question 18.
Define and explain different types of processes.
Answer:
There are following types of processes :
(1) Isothermal process : It is defined as a process in which the temperature of the system remains constant throughout the change of a state of the system.
In this, ΔT = 0.

Features :

• In this process, the temperature at initial state, final state and throughout the process remains constant.
• In this process, system exchanges heat energy with its surroundings to maintain constant temperature. E.g., in case of exothermic process liberated heat is given to the surroundings and in case of endothermic process heat is absorbed from the surroundings so that temperature of the system remains constant and ΔT = 0.
• Isothermal process is carried out with a closed system.
• Internal energy (U) of the system remains constant, hence, Δ U = 0.
• In this process, pressure and volume of a gaseous system change.

(2) Isobaric process : It is defined as a process which is carried out at constant pressure. Hence, Δ P = 0.
Features :

• In this process, the volume (of gaseous system) changes against a constant pressure.
• Since the external atmospheric pressure remains always constant, all the processes carried out in open vessels, or in the laboratory are isobaric processes.
• In this volume and temperature change.
• Internal energy of a system changes, hence, ΔU ≠ 0.

(3) Isochoric process : It is defined as a process which is carried out at constant volume of the system.
Features :

• In this process, temperature and pressure of the system change but volume remains constant.
• Since ΔV = 0, no mechanical work is performed.
• In this internal energy (U) of the system changes. The example of this process in cooking takes place in a pressure cooker.

(4) Adiabatic process : It is defined as a process in which there is no exchange of heat energy between the system and its surroundings. Hence, Q = 0.
Features :

• An adiabatic process is carried out in an isolated system.
• In this process, temperature and internal energy of a system change, ΔT ≠ 0, Δ U ≠ 0.
• During expansion, temperature and energy decrease and during compression, temperature and energy increase.
• If the process is exothermic, the temperature rises and if the process is endothermic the temperature decreases in the adiabatic process.

(5) Reversible process : A process carried out in such a manner that at every stage, the driving force is only infinitesimally greater than the opposing force and it can be reversed by an infinitesimal increase in force and the system exists in equilibrium with its surroundings throughout, is called a reversible process.
Features :

• This is a hypothetical process.
• Driving force is infinitesimally greater than the opposing force throughout the change.
• The process can be reversed at any point by making infinitesimal changes in the conditions.
• The process takes place infinitesimally slowly involving infinite number of steps.
• At the end of every step of the process, the system attains mechanical equilibrium, hence, throughout the process, the system exists in temperature-pressure equilibrium with its surroundings.
• In this process, maximum work is obtained.
• Temperature remains constant throughout the isothermal reversible process.

(6) Irreversible process : it is defined as the unidirectional process which proceeds in a definite direction and cannot be reversed at any stage and in which driving force and opposing force differ in a large magnitude. It is also called a spontaneous process.
Features :

• It takes place without the aid of external agency.
• All irreversible processes are spontaneous.
• All natural processes are irreversible processes.
• Equilibrium is attained at the end of process.
• They are real processes and are not hypothetical.

Examples :

• Flow of heat from a matter at higher temperature to a matter at lower temperature.
• Flow of a gas from higher to lower pressure.
• Flow of water from higher level to lower level.
• Flow of a solvent into a solution through a semipermeable membrane due to osmosis.
• Flow of electricity from higher potential terminal to lower potential terminal.

Question 19.
Distinguish between :
(1) Isothermal process and Adiabatic process.
(2) Reversible and irreversible processes.
Answer:
Isothermal process:

1. In an isothermal process, the temperature of the system remains constant. ΔT = 0
2. In this process, the system exchanges heat with the surroundings. Q ≠ 0 (Closed system)
3. The total internal energy of the system remains constant.
4. In this process, the system is not thermally isolated.
5. In this process, Q = -W as ΔU = 0.
6. ΔH = 0.

Adiabatic process:

1. In an adiabatic process, the temperature of the system changes. ΔT ≠ 0
2. In this process, the system does not exchange heat with the surroundings. Q = 0 (Isolated system)
3. The total internal energy of the system changes. ΔU ≠ 0
4. In this process, the system is thermally isolated.
5. In this process, W = ΔU.
6. ΔH ≠ 0.

(2) Reversible and irreversible processes.
Reversible process:

1. The process whose direction can be reversed at any stage by an infinitesimal increase in the opposing force is called a reversible process.
2. Such a process is not spontaneous and takes place infinitesimally slowly and takes infinite time for completion.
3. In this process, the thermodynamic equilibrium is always maintained between the system and the surroundings at every step.
4. The opposing force is infinitesimally less than the driving force.
5. It is an ideal or hypothetical process.
6. Maximum work can be derived from such a process.

Irreversible process:

1. The process whose direction cannot be reversed by an infinitesimal increase in the opposing force is called an irreversible process.
2. Such a process is spontaneous and takes finite time for completion.
3. The thermodynamic equilibrium is attained only at the end of the process.
4. The opposing force is significantly less than the driving force.
5. It is a practical or real and spontaneous process.
6. Work derived from such a process is always less than the maximum work.

Question 20.
Show that pressure times volume (PV) is equal to work.
Answer:
The work is defined as the energy by which a body is displaced through a distance d by applying a force f.
∴ W = f × d
If area is A = d² and volume V = d³ then,
PV = \(\frac{f}{A}\) × d³ = \(\frac{f}{d^{2}}\) × d³ = f × d = W
∴ The term PV represents the pressure-volume work.

Question 21.
Explain the process of (A) expansion and (B) compression with suitable examples.
Answer:
(A ) Expansion : Consider an ideal cylinder fitted with a piston and filled with H₂O_2(l).
2H₂O_2(l) → 2H₂O_(l) + O_2(l)

Fig. 4.5 : Decomposition of H₂O₂
The oxygen gas produced pushes the piston upwards lifting the mass. Thus, the system performs the work on the surroundings and loses energy by expansion. In this work is done by the system.

(B) Compression : Consider an ideal cylinder fitted with a piston and containing gaseous NH_3(g) and HCl_(g).

Fig. 4.6 : Reaction between NH_3(g) and HCl_(g)
NH_3(g) + HCl_(g) → NH₄Cl_(s)
As the reaction proceeds, due to consumption of gases, the volume decreases and there is work due to compression. In this work is done on the system by surroundings and the system gains energy.

Question 22.
What are the sign conventions for Q and W in (A) expansion, (B) compression?
Answer:
(A) For expansion, work is done by the system hence,
Q = -ve and W = -ve
(B) For compression, work is done on the system hence,
Q = -ve and W = +ve

Fig. 4.7 : Sign conventions

Question 23.
Explain sign convention of work during expansion and compression.
OR
Explain +W and -W.
Answer:
(A) Expansion of a gas :
(1) When a gas expands against a constant pressure, P_ex changing the volume from initial volume V₁ to final volume V₂,
Change in volume, Δ V = V₂ – V₁
The mechanical work = W = -P_ex × Δ V
= -P_ex (V₂ – V₁)

(2) During expansion V₂ > V₁. The work is said to be performed by the system on the surroundings. This results in the decrease in the (work) energy of the system. Hence the work is negative, i.e. W is -ve.

(B) Compression of a gas : During compression, V₂ < V₁. The work is said to be performed on the system by the surroundings. This results in the increase in the (work) energy of the system. Hence the work is positive, i.e. W is + ve.

Question 24.
What are different units of energy and work ?
Answer:

Question 25.
What are the characteristics of maximum work?
Answer:
(1) The process is carried out at constant temperature.
(2) During the complete process, driving force is infinitesimally greater than opposing force.
(3) Throughout the process, the system exists in equilibrium with its surroundings.
(4) The work obtained is maximum. This is given by,
W_max = -2.303 nRT log₁₀ \(\frac{V_{2}}{V_{1}}\)
OR
W_max = -2.303 nRT log₁₀ \(\frac{P_{1}}{P_{2}}\)
where n, P, V and T represent number of moles, pressure, volume and temperature respectively.
(5) ΔU = 0, ΔH = 0.
(6) The heat absorbed in reversible manner
Q_rev, is completely converted into work.
Q_rev = -W_max.
Hence work obtained is maximum.

Solved Examples 4.3

Question 26.
Solve the following :

(1) 2.5 moles of an ideal gas are expanded isothermally from 12 dm³ to 25 dm³ against a pressure of 3.0 bar. Calculate the work obtained.
Solution :
Given : n = 2.5 mol; V₁ = 12 dm³;
V₂ = 25 dm³
P_ext = 3.0 bar; W = ?
W = -P_ext × (V₂ – V₁)
= – 3 × (25 – 12)
= -39 dm³ bar
∵ V₁ dm³ = 100 J
∴ W = -39 × 100 = -3900 J = -3.9kJ
Ans. W= -3.9 kJ

(2) When 2.2 moles of an ideal gas are expanded from 3.5 dm³ to 12 dm³ against a constant pressure, the work obtained is 3910 J. Estimate the external pressure.
Solution :
Given : n = 2.2 mol; V₁ = 3.5 dm³; V₂ = 12 dm³
W= -3910 J = \(\frac{-3910}{100}\)dm³ bar
W = -P_ex (V₂ – V₁)

= 4.6 bar
Ans. P_ex = 4.6 bar

(3) Three moles of an ideal gas are expanded isothermally from a volume of 300 cm³ to 2.5 dm³ at 300 K against a pressure of 1.9 bar. Calculate the work done in L atm and joules.
Solution :
Given : Number of moles of a gas = n = 3 mol
Initial volume = V₁ = 300 cm³
= 0.3 dm³
Final volume = V₂ = 2.5 dm³
External pressure = P_ex =1.9 bar
Temperature = T = 300 K
∵ W = -P_ex (V₂ – V₁)
= -1.9 (2.5 – 0.3)
= -4.18 dm³ bar
Now, 1 dm³ bar = 100 J
∴ W = -4.18 × 100
= -4180 J
Ans. Work of expansion = W = -4180 J

(4) Calculate the constant external pressure needed to compress an ideal gas from 25 dm³ to 15 dm³. The amount of work done in the compression process is 3500 joules.
Solution :
As the compression of the gas takes place against a constant pressure, the work done is given by
W = -P_ex(V₂ – V₁)
W = Work done by the gas against the external pressure = 3500 J
∴ W = \(\frac{3500}{100}\) = 35 dm³ bar
P = Constant external pressure = ?
V₂ = Final volume = 15 dm³
V₁ = Initial volume = 25 dm³
∴ 35= -P × (15 – 25)
∴ 35 = 10 × P
∴ P = \(\frac{35}{10}\) = 3.5 bar
Ans. External pressure = 3.5 bar

(5) Three moles of an ideal gas are compressed isothermally and reversibly to a volume of 2 dm³. The work done is 2.983 kJ at 22°C. Calculate the initial volume of the gas.
Solution :
Given : Number of moles of a gas = n = 3 mol
Final volume = V₂ = 2 dm³
Initial volume = V₁ = ?
For compression,
Wmax = +2.983 kJ = 2983 J
Temperature = T = (273 + 22) K = 295 K
Wmax= -2.303 nRT log₁₀\(\frac{V_{2}}{V_{1}}\)
∴ \(\frac{2983}{2.303 \times 3 \times 8.314 \times 295}\)
= -[log₁₀2 – log₁₀V₁]
0.1760 = -log₁₀2 + log₁₀ V₁
= -0.3010 + log₁₀V₁
∴ log₁₀ V₁ = 0.1760 + 0.3010 = 0.4770
∴ V₁ = Antilog 0.4770
= 3.0 dm³
Ans. Initial volume of the gas = 3.0 dm³

(6) A chemical reaction takes place in a container of cross sectional area 100 cm². As a result of the reaction, a piston is pushed out through 10 cm against an external pressure of 1.0 bar. Calculate the work done by the system.
Solution :
Given : Cross sectional area = A = 100 cm²
Displacement of a piston = 1 = 10 cm
External pressure = P = 1.0 bar
Work = W = ?
Volume change = A × l
∴ ΔV = 100 × 10
= 1000 cm³
= 1 dm³
Work = W = -P × ΔV
= -1 × 1
= -1 dm³ bar
= – 1 × 100 J
= -100 J
Ans. Work = W = -100 J

(7) 5 moles of helium expand isothermally and reversibly from a pressure 4 atm to 0.4 at 300 K. Calculate the work done, change in internal energy and heat absorbed during the expansion. (R = 8.314 J k^-1 mol^-1)
Solution :
Given : n = 5 mol
P₁ = 4 atm
P₂ = 0.4 atm
T = 300 K
W_max = ?, ΔU = ?, Q = ?
W_max = -2.303 nRT log₁₀ (\(\frac{P_{1}}{P_{2}}\))
= -2.303 × 5 × 8.314 × 300 log₁₀ \(\frac{4}{0.4}\)
= -2.303 × 5 × 8.314 × 300 × 1
= -28720 J
= -28.72 kJ
For an isothermal process, ΔU = 0
By first law, ΔU = Q + Wmax
∴ Q = -W_max
= – (-28.72) = 28.72 kJ
Ans. W_max = – 28.72 kJ; ΔU = 0;
Q = 28.72 kJ

(8) 2.8 × 10^-2 kg of nitrogen is expanded isothermally and reversibly at 300 K from 15.15 × 10⁵ Pa when the work done is found to be -17.33 kJ. Find the final pressure.
Solution :
Given : Mass of nitrogen = m = 2.8 × 10^-2 kg
Temperature = T = 300 K
Work obtained in expansion = Wmax = -17.33 kJ
= – 17330 J
Initial pressure = P₁ = 15.15 × 10⁵ Pa
= 1.515 × 10⁶ Pa
Molar mass of nitrogen (N₂) = M_N2
= 28 × 10^-3 kg mol^-1
Final pressure = P₂ = ?
Number of moles of N₂ = n = \(\frac{m}{M_{\mathrm{N}_{2}}}\)
= \(\frac{2.8 \times 10^{-2}}{28 \times 10^{-3}}=1 \mathrm{~mol}\)
W_max = -2.303 × nRT log₁₀ \(\frac{P_{1}}{P_{2}}\)
17330 = 2.303 × 1 × 8.314 × 300 × \(\log _{10} \frac{1.515 \times 10^{6}}{P_{2}}\)
∴ \(\frac{17330}{2.303 \times 1 \times 8.314 \times 300}\)
= [log₁₀ 1.515 × 10⁶ – log₁₀P₂]
3.017 = 6.1804 – log₁₀P₂
∴ log₁₀P₂ = 6.1804 – 3.017 = 3.1634
∴ P₂ = Antilog 3.1634
= 1456.8 Pa
Ans. Final pressure = 1456.8 Pa

(9) Carbon monoxide expands isothermally and reversibly at 300 K doing 4.754 kJ of work. If the initial volume changes from 10 dm³ to 20 dm³, calculate the number of moles of carbon monoxide. (R = 8.314 JK^-1 mol^-1)
Solution :
W_max = -2.303 nRT log₁₀ \(\frac{V_{2}}{V_{1}}\)
W_max = Maximum work done = -4.754 kJ
= -4754 J, n = Number of moles = ?,
= 8.314 JK^-1 mol^-1
T = 300 K, V₁ = Initial volume of carbon monoxide = 10 dm³
V₂ = Final volume of carbon monoxide = 20 dm³
∴ -4754 = 22.303 × n × 8.314 × 300 log₁₀ \(\frac {20}{10}\)
∴ -4754 = – 2.303 × n × 8.314 × 300 × log₁₀2
∴ – 4754 = – 2.303 × n × 8.314 × 300 × 0.3010
∴ n = \(\frac{-4754}{-2.303 \times 8.314 \times 300 \times 0.3010}\)
= 2.75 mol
Ans. Number of mol = 2.75 mol

(10) Given that the work done in isothermal and reversible expansion is 6.4 kJ when 2 moles of an ideal gas expanded to double its volume. Calculate the temperature at which expansion takes place.
Solution :
Given : Work = W_max = -6.4 kJ (For expansion)
= – 6400 J
Number of moles = 2
If ₁ = x L
V₂ = 2 L
Temperature = T = ?
For isothermal reversible expansion,

(11) 300 mmol of perfect gas occupies 13 dm³ at 320 K. Calculate the work done in joules when the gas expands :
(a) Isothermally against a constant external pressure of 0.20 bar,
(b) Isothermal and reversible process,
(c) Into vacuum until the volume of gas is increased by 3 dm³. (R = 8.314 J mol^-1K^-1)
Solution :
Given : Number of moles of a gas = n
= 300 mmol = 0.3 mol
Initial volume = V₁ = 13 dm³
Increase in volume = ΔV = 3 dm³
Pressure = P_ex = 0.2 atm
Temperature = 320 K

(a) Expansion against constant pressure is an irreversible process.
∴ W = -P_ex × ΔV
= -0.2 × 3
= -0.6 dm³ bar
= -0.6 × 100 J
= -60 J

(b) For isothermal reversible process,
W_max = 2.303 nRT log₁₀ \(\frac{V_{2}}{V_{1}}\)
Now, V₂ = V₁ + ΔV = 13 + 3 = 16 dm³
W_max = – 2.303 × 0.3 × 8.314 × 320 log₁₀ \(\frac {16}{13}\)
= -165.4 J

(c) In vacuum, P_ex = 0
∴ W = -P_ex × ΔV
= -0 × 3
= 0
Ans. (a) W= -60.78 J
(b) W_max = -165.4 J
(c) W = 0.

Question 27.
Define and explain the term internal energy.
Answer:
Internal energy : It is defined as the total energy constituting potential energy and kinetic energy of the molecules present in the system.
Explanation :

• The internal energy of a system is a state function and thermodynamic function. It is denoted by U.
• Its value depends on the state of a system.
• The change in internal energy (Δ U) depends only on the initial state and the final state of the system.
  Δ U = U₂ – U₁
• It is an extensive property of the system.
• It has same unit as heat and work.
• Total internal energy U of the system is,
  Total energy = Potential energy + Kinetic energy

Question 28.
Explain the formulation of first law of thermodynamics.
OR
Deduce mathematical equation for the first law of thermodynamics. Justify its expression.
Answer:
(1) The first law of thermodynamics is based on the principle of conservation of energy.
(2) If Q is the heat absorbed by the system and if W is the work done by surroundings on the system then the internal energy of the system will increase by Δ U.
(3) From the conservation of energy we can write,
Increase in internal energy of the system = Quantity of heat absorbed by the system + Work done on the system
∴ ΔU = Q + W
(4) For an infinitesimal change,
dU = dQ + dW

Question 29.
Deduce the mathematical expression of first law of thermodynamics for the following processes :
(1) Isothermal process
(2) Isobaric process
(3) Isochoric process
(4) Adiabatic process.
Answer:
(1) Isothermal process :This is a process which is carried out at constant temperature. Since internal energy, U of the system depends on temperature there is no change in the internal energy U of the system. Hence ΔU = 0.
By first law of thermodynamics,
ΔU = Q +W
∴ 0 = Q + W
∴ Q = -W or W = -Q.

• Hence in expansion, the heat absorbed by the system is entirely converted into work on the surroundings.
• In compression, the work done on the system is converted into heat which is transferred to the surroundings by the system, keeping temperature constant.

(2) Isobaric process : In this, throughout the process pressure remains constant. Hence the system performs the work of expansion due to volume change ΔV.
W= -P_ext × ΔV
Let Q_P be the heat absorbed by the system at constant pressure.
By first law of thermodynamics,
ΔU = Q_P + W.
∴ ΔU = Q_P – P_exΔV
or Q_P = ΔU + P_exΔV
In this process, the heat absorbed Q_P is used to increase the internal energy (ΔU) of the system.

(3) Isochoric process : In this process the volume of the system remains constant. Hence ΔV = 0. Therefore, the system does not perform mechanical work.
∴ W = -PΔV = -P × (0) = 0
Let Q_V be the heat absorbed at constant volume.
By first law of thermodynamics,
ΔU = Q + W
∴ ΔU = Q_V.

(4) Adiabatic process : In this process, the system does not exchange heat, Q with its surroundings.
∴ Q = 0.
Since by first law of thermodynamics,
ΔU = Q + W
∴ ΔU = W_ad.
Hence,
(i) the increase in internal energy ΔU is due to the work done on the system by surroundings. This results in increase in energy and temperature of the system.
(ii) if the work is done by the system on surroundings, like expansion, then there is a decrease in internal energy (-ΔU) and temperature of the system decreases.

Question 30.
What are the IUPAC sign conventions of Q, U and W?
Answer:
In thermodynamics, the sign conventions are adopted according to IUPAC convention, based on acquisition of energy.
(i) Heat absorbed = +Q
Heat evolved = -Q
(ii) Internal energy change :
Increase in energy = + Δ U
Decrease in energy = – Δ U
(iii) Work done by the system = – W
Work done on the system = + W

Question 31.
Define and explain the term enthalpy.
OR
What is meant by enthalpy of a system ?
Answer:
Enthalpy (H) : It is defined as the total energy of a system consisting of internal energy (U) and pressure – volume (P × V) type of energy, i.e. enthalpy represents the sum of internal energy U and product PV energy. It is denoted by H and is represented as

Explanation :

• Enthalpy represents total heat content of the system.
• Enthalpy is a thermodynamic state function.
• Enthalpy is an extensive property.
• The absorption of heat by a system increases its enthalpy. Hence enthalpy is called heat content of the system.

Question 32.
Derive the expression, ΔH = ΔU + PΔV.
Answer:
Enthalpy (H) of a system is defined as
H = U + PV
where U is internal energy
P is pressure and V is volume.
Consider a process in which a state of a system changes from an initial state A to a final state B. Let H₁, U₁, P₁, V₁ and H₂, U₂, P₂, V₂ be the state functions of the system in initial and final states.

Then,
H₁ = U₁ + P₂V₂ and H₂ = U₂ + P₂V₂
The enthalpy change ΔH is given by,
ΔH = H₂ – H₁
= (U₂ + P₂V₂) – (U₁ + P₁V₁)
= (U₂ – U1) + (P₂V₂ – P₁V₁)
= ΔU + ΔPV
where ΔU = U₂ – U₁
At constant pressure, P₁ = P₂ = P
∴ P₂V₂ – P₁V₁ = PV₂ – PV₁
= P(V₂ – V₁)
= P × ΔV
Hence, ΔH = ΔU + PΔV
This is a relation for enthalpy change.

Question 33.
Show that the heat absorbed at constant pressure is equal to the change in enthalpy of the system.
OR
Why is enthalpy called heat content of the system?
Answer:
By the first law of thermodynamics,
ΔU = Q + W
where ΔU is the change in internal energy
Q is heat supplied to the system
W is the work obtained.
∴ Q = ΔU – W
If Q_P is the heat absorbed at constant pressure by the system, so that the volume changes by Δ V against constant pressure P then,
W = -PΔV
∴ Q_P = ΔU – (-PΔV)
∴ Q_P = ΔU + PΔV ……… (1)
If ΔH is the enthalpy change for the system, then
ΔH = ΔU + PΔV ……….. (2)
By comparing above equations, (1) and (2), we can write, Q_P = ΔH
Hence heat absorbed at constant pressure is equal to the enthalpy change for the system.
Since by increase in enthalpy heat content of the system increases, enthalpy is also called as the heat content of the system.

Question 34.
What are the conditions under which ΔH = ΔU?
Answer:

1. For any thermodynamic process or a chemical reaction at constant volume, Δ V = 0.
2. Since ΔH = ΔU + PΔV, at constant volume ΔH = ΔU.
3. In the reactions, involving only solids and liquids, Δ V is negligibly small, hence ΔH = ΔH.
4. In a chemical reaction, in which number of moles of gaseous reactants and gaseous products are equal, then change in number of moles, Δ n = n₂ – n₁ = 0. Since ΔH = ΔU + ΔnRT, as Δn = 0, ΔH = ΔU.

Solved Examples 4.6 – 4.8

Question 35.
Solve the following :

(1) Calculate the change in internal energy when a gas is expanded by supplying 1500 J of heat energy. Work done in expansion is 850 J.
Solution :
Given : Q = 1500 J
W = -850 J (For expansion work is negative)
ΔU = ?
By first law of thermodynamics,
ΔU = Q + W
= 1500 + (- 850)
= 650 J
Ans. Change in internal energy = Δ U = 650 J

(2) A system absorbs 520 J of heat and performs work of 210 J. Calculate the change in internal energy.
Solution :
Given : Since the heat is absorbed by the system, the work is of expansion.
Q = 520 J
W= -210 J
ΔV = ?
ΔU = Q + W
= 520 + (- 210)
= 310 J
Ans. Internal energy change = Δ U = 310 J

(3) A gas expands from 6 litres to 20 dm³ at constant pressure 2.5 atmosphere. If the system is supplied with 5000 J of heat, calculate W and ΔU.
Solution :
Given : V₁ = 6 dm³
V₂ = 20 dm³
P = 2.5 atm
Q = 5000 J
W = ?; ΔU = ?
For expansion,
W = -P_ex(V₂ – V₁)
= -2.5 (20 – 6)
= – 35
= – 35 × 100 J
= – 3500 J
ΔU = Q + W
= 5000 + (-3500)
= -1500 J
Ans. W= -3500 J; ΔU = – 1500 J

(4) An ideal gas expands against a constant pressure of 2.026 × 10⁵ Pa from 5 dm³ to 15 dm³. If the change in the internal energy is 418 J, calculate the change in enthalpy.
Solution :
As the expansion takes place at a constant pressure, the change in enthalpy is given by
ΔH = ΔU + P(V₂ – V₁)
ΔH = Change in enthalpy = ?
ΔU = Change in internal energy = 418 J
P = Constant pressure = 2.026 × 10⁵ Pa
V₂ = 15 dm³ = 15 × 10^-3 m³
V₁ = 5 dm³ = 5 × 10^-3 m³
∴ ΔH = 418 + 2.026 × 10⁵ × (15 × 10^-3 – 5 × 10^-3)
= 418 + 2026
∴ ΔH = 2.444 × 10³ J = 2.444 kJ
Ans. Change in enthalpy = 2.444 × 10³ J
= 2.444 kJ

(5) In a reaction, 2.5 kJ of heat is released from the system and 5.5 kJ of work is done on the system. Find ΔU.
Solution :
Given : Q = -2.5 kJ (since heat is released)
W= + 5.5 kJ (since the work will be of compression)
ΔU = ?
ΔU = Q + W
= -2.5 + 5.5
= +3 kJ
Internal energy of the system will increase by 3 kJ.
Ans. Δ U = 3 kJ

(6) A chemical reaction is carried out by supplying 8 kJ of heat. The system performs the work of 2.7 kJ. Calculate ΔH and ΔU.
Solution :
Given : Q = + 8 kJ (since heat is absorbed by the system)
W = -2.7 kJ (It will be a work of expansion)
ΔH = ?, ΔU = ?
ΔU = Q + W = 8 + (-2.7) = 5.3 kJ
Internal energy of the system will increase by 5.3 kJ.
Due to expansion, Δ V > 0,
∴ PΔV = +2.7 kJ
ΔH = ΔU + PΔV = 5.3 + 2.7 = 8 kJ
Enthalpy of the system will increase by 8 kJ
Ans. ΔU = 5.3 kJ, ΔH = 8 kJ

(7) A sample of gas absorbs 4000 kJ of heat, (a) if volume remains constant. What is ΔU? (b) Suppose that in addition to absorption of heat by the sample, the surroundings does 2000 kJ of work on the sample. What is Δ U ? (c) Suppose that as the original sample absorbs heat, it expands against atmospheric pressure and does 600 kJ of work on its surroundings. What is ΔU ?
Solution :
Given : Q = + 4000 kJ (since heat is absorbed)
(a) Since volume remains constant, Δ V = 0.
W = -P_ex (V₂ – V₁)
= -P_exΔV = -P_ex(0) = 0
∴ ΔU = Q + W = 4000 + 0 = 4000 kJ

(b) Q = + 4000 kJ
W = + 2000 kJ (Work done on the system)
ΔU = Q + W = 4000 + 2000 = 6000 kJ

(c) W = -600 kJ (Work of expansion)
ΔU = Q + W
ΔU = 4000 + (-600) = 3400 kJ
Ans. (a) Δ U = 4000 kJ
(b) Δ U = 6000 kJ
(c) Δ U = 3400 kJ

(8) Calculate the internal energy change at 298 K for the following reaction :
\(\frac{1}{2} \mathbf{N}_{2(\mathrm{~g})}+\frac{3}{2} \mathbf{H}_{2(\mathrm{~g})} \rightarrow \mathbf{N H}_{3(\mathrm{~g})}\)
The enthalpy change at constant pressure is -46.0 kJ mol^-1. (R = 8.314 JK^-1 mol^-1)
Solution :
Given : \(\frac{1}{2} \mathbf{N}_{2(\mathrm{~g})}+\frac{3}{2} \mathbf{H}_{2(\mathrm{~g})} \rightarrow \mathbf{N H}_{3(\mathrm{~g})}\)
ΔH= -46.0 kJ mol^-1
ΔH = Heat of formation of NH₃ at constant pressure
= -46.0 kJ mol^-1 = -4600 J mol^-1
Δ U = Change in internal energy = ?
Δn = (Number of moles of ammonia) – (Number of moles of hydrogen + Number of moles of nitrogen)
= [1 – (\(\frac {1}{2}\) + \(\frac {3}{2}\))]= -1 mol
R= 8.314 JK^-1 mol^-1
T = Temperature in kelvin = 298 K
ΔH = Δ U + ΔnRT
∴ -46000 = ΔU + (-1 × 8.314 × 298)
∴ -46000 = ΔU – 2477.0
∴ ΔU = -46000 + 2477.0
= -43523 J
= -43.523 kJ
Ans. Change in internal energy = -43.523 kJ

(9) 5 moles of helium expand isothermally and reversibly from a pressure 40 × 10^-5 Nm^-2 to 4 × 10^-5 Nm^-2 at 300 K. Calculate the work done, change in internal energy and heat absorbed during the expansion. (R = 8.314 JK^-1 mol^-1)
Solution :
As the expansion takes place isothermally and reversibly, the work done is given by
Wmax = -2.203 nRT log\(\frac{P_{1}}{P_{2}}\)
Wmax = Maximum work done
n = Number of moles of helium = 5 moles
R = Gas constant = 8.314 JK^-1 mol^-1
T = 300 K
P₁ = Initial pressure = 40 × 10^-5 Nm^-2
P₂ = Final pressure = 4 × 10^-5 Nm^-2
∴ W_max = -2.303 × 5 × 8.314 × 300 × log \(\frac {40}{4}\)
= – 2.303 × 5 × 8.314 × 300 × log 10
= -2.303 × 5 × 8.314 × 300 × 1
= – 28720J
As the expansion takes place isothermally, i.e., at the same temperature, there is no change in the internal energy of the system.
Q = ΔU + W
∴ Q = – W= + 28720 J as ΔU = 0
Ans. Work done = -28720 J, Heat absorbed = 28720 J, Change in internal energy = 0

(10) Calculate the work done in each of the following reactions. State whether work is done on or by the system.
(a) The oxidation of one mole of SO₂ at 50°C.
2SO_2(g) + O_2(g) → 2SO_3(g)
(b) Decomposition of 2 moles of NH₄NO₃ at 100°C
NH₄NO_3(s) → N₂O_(g) + 2H₂O_(g)
Solution :
(a) Given reaction :
2SO_2(g) + O_2(g) → 2SO_3(g)
For 1 mole of SO₂,
SO_2(g) + \(\frac {1}{2}\)O₂(g) → SO_2(g)
∴ Δn = (n₂)_{gaseous products} – (n₁)_{gaseous reactants}
= 1 – (1 + \(\frac {1}{2}\))
= -0.5 mol
Since there is decrease in number of moles of gases, there will be compression, hence, the work will be done on the system by the surroundings.
Work is given by,
∴ W = – ΔnRT
= – (- 0.5) × 8.314 × (273 + 50)
= + 1342.7 J

(b) Given reaction :
NH₄NO_3(s) → N₂O_(g) + 2H₂O_(g)
For 2 moles of NH₄NO₃,
2NH₄NO_3(s) → 2N₂O_(g) + 4H₂O_(g)
∴ Change in number of moles,
Δn = (n₂)_{gaseous products} – (n₁)_{gaseous reactants}
= (2 + 4) – 0
= 6 mol
Since there is an increase in number of moles of gases, work of expansion is done by the system on the surroundings.
∴ W = -ΔnRT
= – 6 × 8.314 × (273+ 100)
= – 18606 J
= – 18.606 kJ
Ans. (a) W = 1342.7 J (b) W= -18.606 kJ

(11) The amount of heat evolved for the combustion of ethane is -900kJ mol^-1 at 300K and 1 atm.
C₂H_6(g) + \(\frac {7}{2}\)O_2(g) → 2CO_2(g) + 3H₂O_(l)
Calculate W, ΔH and ΔU for the combustion of 12 × 10^-3 kg ethane.
Solution :
Given : ΔH = -900 kJ mol^-1
Temperature = T = 300 K
Pressure = P = 1 atm
Mass of ethane = m = 12 × 10^-3 kg
Molar mass of ethane (C₂H₆) = 30 × 10^-3 kg mol^-1
ΔH = ? Δ U = ? for given ethane.
Number of moles of C₂H₆ = n = \(\frac{m \mathrm{~kg}}{M \mathrm{~kg} \mathrm{~mol}^{-1}}\)
= \(\frac{12 \times 10^{-3}}{30 \times 10^{-3}}\)
= 0.4 mol
For the given reaction,
C₂H_6(g) + \(\frac {7}{2}\)O_2(g) → 2CO_2(g) + 3H₂O_(l)
Δn = (n₂)_{gaseous products} – (n₁)_{gaseous reactants}
= 2 – (1 + \(\frac {7}{2}\)) = -2.5 mol
For 1 mol of C₂H₆, Δn = -2.5 mol
∴ For 0.4 mol of C₂H₆, Δn = -2.5 × 0.4
= -1 mol
Since there is a decrease in number of moles, the work is of compression on the system.
W = -ΔnRT
= – (-1) × 8.314 × 300
= + 2494 J
= + 2.494 kJ
For 1 mol of C₂H₆ ΔH = -900 kJ
∴ For 0.4 mol of C₂H₆, ΔH= – 900 × 0.4
= – 360 kJ
ΔH = ΔU + ΔnRT
ΔU = ΔH – ΔnRT
= -360 – (-1) × 8.314 × 300× 10^-3
= – 360 + 2.494
= – 357.506 kJ
Ans. W = + 2.494 kJ, ΔH = -360 kJ;
ΔU= – 357.506 kJ

(12) The latent heat of evaporation of water is 80 kJ mol^-1. If 100 g water are evaporated at 100 °C and 1 atm, calculate W, ΔH, ΔU and Q.
Solution :
Given : Latent heat of evaporation = ΔH
= 80 kJ mol^-1 of water
Temperature = T = 273 + 100 = 373 K
Pressure = P = 1 atm
Mass of water = m = 100 g
Molar mass of water = 18 g mol^-1
W = ?, ΔH = 1, U = ?, Q = ?
Number of moles of water = \(\frac{m}{M}=\frac{100}{18}\) = 5.556 mol
H₂O_(l) → H₂O_(g)
5.556 mol 5.556 mol
Change in number of moles = Δn = 5.556 – 0
= 5.556 mol
For evaporation of 1 mol H₂O, ΔH = 80 kJ
For 5.556 mol H₂O, ΔH= 80 × 5.556 = 444.5 kJ
In this reaction, the work will be of expansion.
W= -ΔnRT
= -5.556 × 8.314 × 373
= – 17230 J
= -17.23 kJ
Now,
ΔH = ΔU + ΔnRT
ΔU = ΔH – ΔnRT
= 444.5 – 5.556 × 8.314 × 373 × 10^-3
= 444.5 – 17.23
= 427.27 kJ
In this, Q = Q_P = ΔH = 444.5 kJ
Ans. W= -17.23 kJ; ΔH = 444.5 kJ
ΔU= 427.21 kJ, Q = 444.5 kJ

(13) Oxidation of propane is represented as
C₃H_8(g) + 5O_2(g) → 3CO_2(g) + 4H₂O_(g), ΔH⁰ = -2043 kJ
How much pressure volume work is done and what is the value of AU at constant pressure of 1 bar when the volume change is + 22.4 dm³.
Solution :
Given :
ΔH⁰ = – 2043 kJ
Change in volume = ΔV = +22.4 L
ΔU = ?
C₃H_8(g) + 5O_2(g) → 3CO_2(g) + 4H₂O_(g)
Δn = (n₂)_{gaseous products} – (n₁)_{gaseous reactants}
= (3 + 4) – (1 + 5)
= 1 mol
Since there is an increase in number of moles, the work will be of expansion.
W = -P × ΔV dm³ bar
= – 1 × 22.4
= – 1 × 22.4 × 100 J
= – 2240 J
= -2.240
ΔH = ΔU + PΔV
ΔU = ΔH – PΔV
= – 2043 – (2.24)
= – 2040.7 kJ
Ans. W = -2.27 kJ, ΔU = -2040.7 kJ

(14) How much heat is evolved when 12 g of CO reacts with NO₂ ? The reaction is :
4CO_(g) + 2NO_2(g) → 4CO_2(g) + N_2(g),
Δ_rH⁰ = -1200 kJ
Solution :
Given : 4CO_(g) + 2NO_2(g) → 4CO_2(g) + N_2(g)
Molar mass of CO = 28 g mol^-1
Δ_rH⁰ = – 1200 kJ;
Molar mass of CO = 28 g mol^-1
m_co = 12 g, ΔH = ?
From the reaction,
∵ For 4 × 28 g CO, ΔH⁰ = – 1200 kJ
∵ For 12g CO ΔH⁰ = \(\frac{(-1200) \times 12}{4 \times 28}\)
= -128.6 kJ
Ans. Heat evolved = 128.6 kJ

Question 36.
What is phase transformation?
Answer:
Phase transformation (or transition) :

• The transition of one phase (physical state) of a matter to another phase at constant temperature and pressure without change in chemical composition is called phase transformation.
• During phase transformation, both the phases exist at equilibrium.
  Solids ⇌ Liquid; Liquid ⇌ Vapour.

Question 37.
Mention different types of phase transitions.
Answer:
The following are the types of phase changes :
(1) Fusion : This involves the change of a matter from solid state to liquid state. In this heat is absorbed, hence it is endothermic (ΔH > 0).
H₂O_(s) → H₂O_(l)
(2) Vaporisation or evaporation : This involves the change of a matter from liquid state to gaseous state. In this heat is absorbed, hence it is endothermic (ΔH > 0).
H₂O_(l) → H₂O_(g)
(3) Sublimation : This involves the change of matter from solid state directly into gaseous state. In this heat is absorbed, hence it is endothermic (ΔH > 0).
Camphor_(s) → Camphor_(g)

Question 38.
Define and explain enthalpy of freezing.
Answer:
Enthalpy of freezing (Δ_freezH) : The enthalpy change that accompanies the solidification of one mole of a liquid into solid at constant temperature and pressure is called enthalpy of freezing.
For example,
\(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \stackrel{1 \mathrm{~atm}, 273 \mathrm{~K}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})}\)
Δ_freezH = -6.01 kJ mol^-1
This equation describes that when one mole of water freezes (solidifies) at 0 °C (273 K) and 1 atmosphere, 6.01 kJ of heat will be released to the surroundings.

Question 39.
Define and explain the following :
(A) Enthalpy of vaporisation.
(B) Enthalpy of sublimation.
Answer:
(A) Enthalpy of vaporisation (Δ_vapH) : The enthalpy change that accompanies the vaporisation of one mole of a liquid at constant temperature and pressure is called heat of vaporisation or evaporation. For example,
\(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \stackrel{1 \mathrm{~atm}, 373 \mathrm{~K}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\)
Δ_vapH = +40.7 kJ mol^-1
This equation describes that when one mole of water is evaporated at 100 °C (373 K) and 1 atmosphere, 40.7 kJ of heat will be absorbed.

(B) Enthalpy of sublimation (Δ_subH) : The enthalpy change or the amount of heat absorbed that accompanies the sublimation of one mole of a solid directly into its vapour at constant temperature and pressure is called enthalpy of sublimation.
For example,
\(\mathrm{CO}_{2(\mathrm{~s})} \stackrel{1 \mathrm{~atm}, 195 \mathrm{~K}}{\longrightarrow} \mathrm{CO}_{2(\mathrm{~g})}\)
Δ_subH = 25.2 kJ mol^-1
This equation describes that when 1 mole of dry solid carbon dioxide, CO_2(s) sublimes forming gaseous CO_2(g), 25.2 kJ of heat will be absorbed.

Question 40.
Explain process of sublimation and enthalpy of sublimation ?
OR
How is enthalpy of sublimation related to enthalpy of fusion and enthalpy of vaporisation ?
Answer:
(1) The sublimation involves the conversion of a solid into vapour at constant temperature and pressure. For example,
H₂O_(s) → H₂O_(g), Δ_subH = 51.08 kJ mol^-1 at 0°C.
(2) This conversion takes place in two steps, first melting of solid into liquid and second vaporisation of the liquid.

Hence we can write,
Δ_subH = Δ_fusH + Δ_vapH

Question 41.
Arrange the following in order of increasing enthalpy :
H₂O_(s), H₂O_(g), H₂O_(l)
Answer:
The increasing order of enthalpy of the given substance will be,
H_H2O(g), < H_H2O(l), < H_H2O(s)
This is because the conversion of H₂O_(s) to H₂O_(l) and further to H₂O_(g) involves absorption of heat.

Question 42.
Define and explain :
(A) Enthalpy of atomisation
(B) Enthalpy of ionisation.
Answer:
(A) Enthalpy of atomisation (Δ_atoH) : The enthalpy change or amount of heat absorbed accompanying the dissociation of the molecules in one mole of a gaseous substance into free gaseous atoms at constant temperature and pressure is called enthalpy of atomisation.
For example,
Cl_2(g) → 2Cl_(g), Δ_atoH = 242 kJ mol^-1
CH_4(g) → C_(g) + 4H_(g), Δ_atoH = 1660 kJ mol^-1.

(B) Enthalpy of ionisation (Δ_ionH) : The enthalpy change or amount of heat absorbed accompanying the removal of one electron from each atom or ion in one mole of gaseous atoms or ions is called enthalpy of ionisation.
For example,
Na_(g) → Na⁺_(g) + e^– Δ_ionH = 494 kJ mol^-1
This equation describes that when one mole of gaseous sodium atoms, Na_(g) are ionised forming gaseous ions, Na⁺_(g), the energy required is 494 kJ.

Question 43.
Define and explain electron gain enthalpy.
Answer:
Electron gain enthalpy (Δ_egH) : It is defined as the enthalpy change, when mole of gaseous atoms of an element accept electrons to form gaseous ion.
E.g. Cl_(g) + e^– → Cl^– _(g) Δ_egH = – 349 kJ mol^-1.
It is the reverse of ionisation process.

Question 44.
Define enthalpy of solution.
Answer:
Enthalpy of solution : It is defined as the enthalpy change in a process when one mole of a substance is dissolved in specified amount of a solvent.
NaCl_(s) + aq ⇌ NaCl_(aq) Δ_solnH = 4 kJ mol^-1

Question 45.
Define enthalpy of solution at infinite dilution.
Answer:
Enthalpy of solution (Δ_solnH) : It is defined as the enthalpy change when one mole of a substance is dissolved in a large excess of a solvent, so that further dilution will not change the enthalpy at constant temperature and pressure.
For example,
HCl_(g) + aq → HCl_(aq) Δ_solnH
= -75.14 kJ mol^-1

Question 46.
Explain the enthalpy of solution of an ionic compound.
Answer:
An ionic compound dissolves in a polar solvent like water in two steps as follows :
Step-I : The ions are separated from the molecule involving crystal lattice enthalpy Δ_LH.
\(\mathrm{MX}_{(\mathrm{s})} \rightarrow \mathrm{M}_{(\mathrm{g})}^{+}+\mathrm{X}_{(\mathrm{g})}^{-} \quad \Delta_{\mathrm{L}} H\)
Δ_LH is always positive.

Step-II : The gaseous ions are hydrated with water molecules involving hydration energy, Δ_hydH.
\(\mathbf{M}_{(\mathrm{g})}^{+}\) + xH₂O_(l) → [M(H₂O)_x]⁺
\(\mathrm{X}_{(\mathrm{g})}^{-}\) + yH₂O_(l) → [X(H₂O)_y]^–
Δ_hydH is always negative.
The enthalpy change ΔsolnH of solution is given by,
Δ_solnH = Δ_LH + Δ_hydH
For example, consider enthalpy of solution of NaCl(s).
Δ_LH_NaCl = 790 kJ mol^-1
Δ_hydH_NaCl = -786 kJ mol^-1
Hence enthalpy change for solution of NaCl(s) is,
Δ_solnH = Δ_LH_NaCl + Δ_hydH_NaCl
= 790 + (-786)
= + 4 kJ mol^-1
Therefore dissolution of NaCl in water is an endothermic process.

Solved Examples 4.9

Question 47.
Solve the following :

(1) Heat of fusion of ice at 0 °C and 1 atmosphere is 6.1 kJ mol^-1 and heat of evaporation of water at 100 °C is 40.7 kJ mol^-1. Calculate the enthalpy change for the conversion of 1 mole of ice at 0 °C into vapour at 100 °C. Heat capacity of water is 4.184 JK^-1 mol^-1.
Solution :
Given : Heat of fusion of ice = Δ_fusH = 6.01 kJ mol^-1
Heat of evaporation of water = Δ_vapH = 40.7 kJ mol^-1
Temperature of ice = 273 K
Temperature of vapour = (273 + 100) K = 373 K
Heat capacity of water = 4.184 JK^-1 g^-1
Heat capacity of 1 mole of water
= C_H2O = 4.184 × 18
= 75.312 JK^-1 mol^-1
The conversion of ice at 0 °C to water at 100°C takes place in three steps as follows:

ΔH₁ = n_ice × Δ_fusH = 1_mol × 6.01_{kJ mol^-1} = 6.01 kJ
ΔH₂ is the enthalpy change for raising the temperature from 273 K to 373 K.
ΔH₂ = n_water × C_H2O × (T₂ – T₁)
= 1_mol × 75.312_{JK^-1 mol^-1} × (373 – 273)K
= 7531 J
= 7.531 kJ
ΔH₃ = n_water × Δ_vapH
= 1 × 40.7
= 40.7 kJ
Hence total enthalpy change will be ΔH = ΔH₁ + ΔH₂ + ΔH₃
= 6.01 + 7.531 + 40.7
= 54.241 kJ
Ans. ΔH = 54.241 kJ

(2) Heat of sublimation of ice at 0°C and 1 atmosphere is 6.01 kJ mol^-1 and heat of evaporation of water at 0 °C and 1 atmosphere is 45.07 kJ mol^-1. Calculate the heat of sublimation of one mole of ice at 0 °C and 1 atmosphere. Write the equation for the same.
Solution :
The sublimation of ice can be represented by following equation,
\(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})} \stackrel{1 \mathrm{~atm}, 0^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\)
This is a process of two steps.

ΔH₁ = n_H2O × Δ_fusH
= 1 × 6.01 =6.01 kJ
ΔH₂ = n_H2O × Δ_vapH
= 1 × 45.07 = 45.07
Heat of sublimation = ΔH = ΔH₁ + ΔH₂
= 6.01 + 45.07
= 51.08 kJ
Ans. Heat of sublimation of ice = 51.08 kJ

(3) Heat of evaporation of ethyl alcohol at 78.5 °C and 1 atmosphere is 38.6 kJ mol^-1. If 100 g ethyl alcohol vapour is condensed, what will be ΔH ?
Solution :
Given : Δ_vapH_C2H5OH = 38.6 kJ mol-1
Mass of C₂H₅OH = m = 100 g
Molar mass of C₂H₅OH = M = 46 g mol-1
Δ_condH_C2H5OH = ?
Number of moles of C₂H₅OH = \(\frac{m}{M}=\frac{100}{46}\)
= 2.174 mol
Heat of condensation = Δ_condH = -38.6 kJ mol^-1
∴ Δ_condH = n × Δ_condH
= 2.174_mol × (-38.6)_{kJ mol^-1} kJ = – 83.9 kJ
Ans. Heat of condensation = Δ_condH = -83.9 kJ

(4) The hydration enthalpies of Li⁺_(g), and Br^–_(g) are -500 kJ mol^-1 and -350 kJ mol^-1 respectively and the lattice energy of LiBr_(s) is 807 kJmol^-1. Write the thermochemical equations for enthalpy of solution of LiBr(s) and calculate its value.
Solution :
Given : Enthalpy of hydration of Li⁺_(g)
= Δ_hydH1
= -500 kJmol^-1
Enthalpy of hydration of Br^–_(g) = Δ_hydH₂
= -350 kJ mol^-1
Lattice energy of LiBr_(s) = Δ_LH₃ = 807 kJ mol^-1
Enthalpy of solution of LiBr_(s) = Δ_solnΔH = ?
The thermochemical equation for the dissolution of LiBr_(s) forming a solution is,
LiBr_(s) + aq → Li⁺_(aq) + Br^–_(aq) (I) Δ_solH = ?
This takes place in two steps as follows :
(i) LiBr_(s) → Li⁺_(g) + Br^–_(g) Δ_LH₃
(ii) (a) Li⁺_(g) + aq → Li⁺_(aq) Δ_hydH₁
(b) Br^–_(g) + aq → Br^–_(aq) Δ_hydH₂
Hence by adding equations (i) and (ii) (a) and (b) we get equation I.
∴ Δ_solH = Δ_LH₃ + Δ_hydH₁ + Δ_hydH₂
= 807 + (-500) + (-350)
= -43 kJ mol^-1
Ans. Heat of solution of LiBr_(s) = Δ_solH = -43 kJ mol^-1

(5) Heat of solution of NaCl is 3.9 kJ mol^-1. If the lattice energy of NaCl is 787 kJ mol^-1, calculate the hydration energies of ions of the salt.
Solution :
Given : Heat of solution of NaCl
= Δ_solnH⁰
= ΔH₁ = 3.9 kJmol^-1
Lattice energy of NaCl = ΔLH
= ΔH₂ = 787 kJ mol^-1
Hydration energy of Na⁺_(g) and Cl^–_(g)
= Δ_hydH(Na⁺ + Cl^–)
= ΔH₃ = ?
Thermochemical equation for dissolution of NaCl(s) is;
NaCl_(s) + aq → Na⁺_(aq) + Cl^–_(aq)…ΔH₁
NaCl_(s) → Na⁺_(g) + Cl^–_(g)… ΔH₂
Na⁺_(g) + Cl^–_(g) + aq → Na⁺_(aq) + Cl^–_(aq) ΔH₃
∴ ΔH₁ = ΔH₂ + AH₃
3.9 = 787 + AH3
∴ ΔH₃ = -787 + 3.9= -783.1 kJmol^-1
Ans. Hydration energy of Na⁺_(g) and Cl^–_(g)
= -783.1 kJmol^-1

(6) Enthalpies of solution are given as follows :
CuSO_4(s) + 10H₂O → CUSO₄(10H₂O)
ΔH₁ = -54.5 kJ mol^-1
CuSO₄(s) + 100 H₂O → CUSO₄(100H₂O)
ΔH₂ = -68.4 kJ mol^-1
A solution contains 1 mol of CuSO₄ in 180 g water at 25 °C. If it is diluted by adding 1620 g water, calculate the enthalpy of dilution.
Solution :
Given : Enthalpy of solution of CuSO₄ in 10 mol H₂O
= Δ_solnH = ΔH₁ = -54.5 kJ mol^-1
Enthalpy of solution of CuSO₄ in 100 mol
H₂O = ΔH₂
= -68.4 kJmol^-1
Mass of water = 1620 g
For dilution, Δ_dilH = ?
Now 180 g H₂O = \(\frac {180}{18}\) = 10 mol H₂O
And, 1620 g H₂O = \(\frac {1620}{18}\) = 90 mol H₂O
Hence for heat of dilution,
CUSO₄(10H₂O) + 90H₂O_(l) → CUSO₄(100H₂O) Δ_dilH = ?
∴ Δ_dilH = ΔH₂ -ΔH₁
= -68.4 – (54.5)
= -13.9 kJmol^-1
Ans. Heat of dilution = Δ_dilH = -13.9 kJ mol^-1

(7) Heat of solution and heat of hydration of AgF are -20.5kJmol^-1 and -930kJmol^-1 respectively. Calculate lattice energy of AgF.
Solution :
Given : Heat of solution of AgF = Δ_solnH
= -ΔH₁ = -20.5 kJmol^-1
Heat of hydration of AgF = Δ_hydH = ΔH₂
= -930 kJ mol^-1
Lattice energy of AgF = Δ_LH = ΔH₃ = ?
For heat of solution, AgF_(s) + aq → AgF_(aq) ΔH₁
For heat of hydration,
Ag⁺_(g) + F^–_(g) + aq → Ag⁺_(aq) + F^–_(aq) ΔH₂
For Lattice energy, Ag⁺_(g) + F^–_(g) → AgF_(s)
Δ_LH = ?
From above equations,
∴ ΔH₃ = ΔH₂ – ΔH₁
= -930 – (-20.5)
= -909.5 kJmol^-1
Ans. Lattice energy of AgF_(s) = -909.5 kJ mol^-1

(8) Bond enthalpy of H₂ is 436kJmol^-1 while hydration energy of hydrogen ion is -1075 kJ mol^-1. Calculate the enthalpy of formation of H⁺_(aq). (Ionisation energy of hydrogen is 1312 kJ mol^-1
Solution :
Given : Bond enthalpy of H_2(g) = ΔH⁰H_2(g)
= 436 kJ mol^-1
Hydration energy of H⁺_(g) = ΔH₂ = -1075 kJmol^-1
Ionisation energy of H_(g) = ΔH₃ = 1312 kJ mol^-1
Enthalpy of formation of H⁺_(aq) = Δ_fH = ?
Thermochemical equation for the formation of H⁺_(aq)
\(\frac {1}{2}\)H_2(g) + aq → H⁺_(aq)Δ_fH
This takes place in three steps as follows :

Hence heat of formation of H⁺_(aq) is
Δ_fH = ΔH₁ + ΔH₂ + ΔH₃
= \(\frac {1}{2}\) × 436 + (-1075) + 1312
= 218 – 1075 + 1312.
= 455 kJ mol^-1
Ans. Enthalpy of formation of H⁺_(aq) = 455 kJmol^-1

(9) Calculate lattice energy of crystalline sodium chloride from the following data :

Solution :
Given : Bond enthalpy of Cl₂ = ΔH⁰ = 244 kJmol^-1
Thermochemical equation for the formation of 1 mole of NaCl_(s),
\(\mathrm{Na}_{(\mathrm{s})}+\frac{1}{2} \mathrm{Cl}_{2} \longrightarrow \mathrm{NaCl}_{(s)} \quad \Delta_{\mathrm{f}} H_{\mathrm{NaCl}}^{0}=-411 \mathrm{~kJ}\)
Lattice energy, Δ_LH = ?
Since enthalpy is a state function, this reaction can be written in various steps as follows :

By Hess’s law,
Δ_LH⁰ = Δ_subH⁰+ Δ_ionH⁰ + \(\frac {1}{2}\)ΔH⁰cl₂ + ΔegH⁰ + Δ_LH⁰
-411 = 109 + 496 + \(\frac {1}{2}\) × 244 + (-348) + Δ_LH
= 109 + 496 + 122 – 348 + Δ_LH
∴ Δ_LH= -790 kJ mol^-1.
Ans. Lattice energy of NaCl_(s) = -790 kJ mol^–

(10) Calculate the internal energy at 298 K for the formation of one mole of ammonia, if the enthalpy change at constant pressure is -42.0 kJ mol^-1.
(Given : R = 8.314 J K^-1 mol^-1)
Solution :
Given : ΔH = -42.0 kJ mol^-1, T = 298 K, ΔU = ?
\(\frac{1}{2} \mathrm{~N}_{2(\mathrm{~g})}+\frac{3}{2} \mathrm{H}_{2(\mathrm{~g})} \longrightarrow \mathrm{NH}_{3(\mathrm{~g})}\)
Δn = 1 – (\(\frac {1}{2}\) + \(\frac {3}{2}\)) = -1 mol
ΔH = ΔU + ΔnRT
∴ ΔU = ΔH – ΔnRT
= -42 – (- 1) × 8.314 × 298 × 10^-3
= -42 + 2.477
= -39.523 kJ
Ans. ΔU = -39.523 kJ

Question 48.
What is thermochemistry ? Explain.
Answer:
Thermochemistry : Thermodynamic study of heat changes or enthalpy changes during the chemical reactions is called thermochemistry.
Consider a reaction, Reactants → Products
The heat changes ΔH for the reaction may be represented as,
ΔH_reaction = Σ H_products – Σ H_reactants
where H represents enthalpy.
The energy released or absorbed during a chemical change appears in the form of heat energy.

Question 49.
Define and explain the term, enthalpy or heat of reaction.
Answer:
Enthalpy or heat of reaction : The enthalpy of a chemical reaction is the difference between the sum of the enthalpies of products and that of the reactants with every substance in a definite physical state and in the amounts (moles) represented by the coefficients in the balanced equation.
Explanation : Consider the following general reaction,
aA + bB → cC + dD
The heat of the reaction ΔH is the difference in sum of enthalpies of products and sum of enthalpies of reactants.
∴ ΔH = Σ H_products – Σ H_reactants
= [cH_C + dH_D] – [aH_A + bH_B]
= Σ_PH – Σ_RZ
where H represents enthalpy of the substance.
For endothermic reaction, ΔH is positive, (ΔH > 0).
For exothermic reaction, ΔH is negative, (ΔH < 0).

Question 50.
Explain the sign convention used for ΔH.
Answer:
The change in enthalpy or heat of reaction ΔH is given by,

(i) If the sum of enthalpies of products, ΣH and reactants, Σ_RH are equal then ΔH for the reaction is zero, (ΔH = 0).
i. e. Σ_PH = Σ_RH
∴ ΔH = Σ_PH – Σ_RH = 0
(ii) If the sum of enthalpies of products Σ_PH is greater than the sum of enthalpies of reactants Σ_RH, then ΔH is positive, (ΔH > 0). Since such reactions take place with the absorption of heat from surroundings, they are called endothermic reactions.
∴ ΣH_products > ΣH_reactants
∴ ΔH > 0
(iii) If the sum of enthalpies of products Σ_PH is less than the sum of enthalpies of reactants, Σ_RH then ΔH is negative, (ΔH < 0). Since in such reactions heat is given out to the surroundings, they are called exothermic reactions.
∴ Σ_PH < Σ_RH
∴ ΔH < 0

Question 51.
Define : (i) Exothermic process (ii) Endothermic process.
Answer:
(i) Exothermic process : A process taking place with the evolution of heat is called exothermic process.
For this process, Q is -ve, ΔH is -ve.
(ii) Endothermic process : A process taking place with the absorption of heat (from the surroundings) is called endothermic process.
For this process, Q is +ve, ΔH is +ve.

Question 52.
Distinguish between Endothermic reaction and Exothermic reaction.
Answer:
Endothermic reaction:

• In endothermic reaction heat is absorbed from suroundings.
• Sum of enthalpies of products is greater than sum of enthalpies of reactants i.e. Σ_PH > Σ_RH
• Heat of reaction, ΔH is positive.
• Products are less stable than reactants.
• C_(s) + O_2(g) → CO_2(g)
  ΔH = -394 kJ
• This reaction requires supply of thermal energy.

Exothermic reaction:

• In exothermic reaction heat is given out to surroundings.
• Sum of enthalpies of products is less than sum of enthalpies of reactants.
  i.e. Σ_PH < Σ_RH
• Heat of reaction, ΔH is negative.
• Products are more stable than reactants.
• N_2(g) + O_2(g) → 2NO
  ΔH = + 180 kJ
• This reaction does not require supply of thermal energy.

Question 53.
Explain the standard state of an element.
Answer:
Standard state of an element : It is defined as the most stable state of an element at 298 K and 1 atmosphere (or 1 bar).
In this state, the enthalpy of the element is assumed to be zero.
∴ H⁰_element or in general H_element = 0
∴ H⁰_graphite = H_H2(g) = 0; H_Na(s) = 0; H_Hg(l) = 0

Question 54.
What is a thermochemical equation? Explain with an example.
Answer:
Thermochemical equation : It is defined as a balanced chemical equation along with the corresponding heat of reaction (ΔH) and physical states and number of moles of all reactants and all products appropriately mentioned.
E.g. C₆H₁₂O_6(s) + 6O_2(g) = 6CO_2(g) + 6H₂O_(l)
ΔH = -2808 kJ mol^-1

Question 55.
What are the guidelines followed for writing thermochemical equations?
Answer:
According to IUPAC conventions, while writing thermochemical equations, following rules must be followed :
(1) Reaction is represented by balanced chemical equation for the number of moles of the reactants and the products. E.g.
CH_4(g) + 2O_2(g) = CO_2(g) + 2H₂O_(l)
ΔrH°= -890 kJ mol^-1
(2) The physical states of all the substances in the reaction must be mentioned. E.g. (s) for solid, (l) for liquid and (g) for gas.
(3) Heat or enthalpy changes are measured at 298 K and 1 atmosphere (or 1 bar).
(4) ΔH⁰ is written at right hand side of thermochemical equation.
(5) Proper sign must be indicated for ΔH⁰. For endothermic reaction ΔH⁰ is positive, (+ΔH⁰) and for exothermic reaction ΔH is negative, (-ΔH⁰).
(6) The enthalpy of the elements in their standard states is taken as zero. (H⁰_Element = 0; H⁰_C(s) = 0, H⁰_H2(g) = 0)
(7) When all the substances taking part in the reaction are in their standard states, the enthalpy change is written as ΔH⁰.
(8) The enthalpy of any compound is equal to its heat of formation.
(9) In case of elements, the allotropic form must be mentioned. E.g. C_(graphite), S_(rhombic), Sn_(white)
(10) For the reverse reaction, ΔH⁰ value has equal magnitude but opposite sign.

Question 56.
Define the following terms giving examples :
(1) Standard enthalpy of reaction.
(2) Standard enthalpy of formation or standard heat of formation
(3) Standard enthalpy of combustion or standard heat of combustion.
Answer:
(1) Standard enthalpy of reaction : it is defined as the difference between the sum of enthalpies of products and that of the reactants with every substance in its standard state at constant temperature (298 K) and pressure (1 atm).
Reactants → Products
ΔH⁰_reaction = ΣH⁰_products – ΣH⁰_reactants

(2) Standard enthalpy of formation or standard heat of formation (Δ_fH⁰) : It is defined as the enthalpy change ΔH⁰ when one mole of a pure compound is formed in its standard state from its constituent elements in their standard states at constant temperature (298 K) and pressure (1 atmosphere or 1 bar). It is denoted by Δ_fH⁰. E.g.
C_(s) + O_2(g) = CO_2(g) Δ_fH⁰= -394 kJ mol^-1
(Δ_fH⁰ may be positive or negative.)

(3) Standard enthalpy of combustion or standard heat of combustion : it is defined as the enthalpy change when one mole of a substance in the standard state undergoes complete combustion in a sufficient amount of oxygen at constant temperature (298 K) and pressure (1 atmosphere or 1 bar). It is denoted by Δ_cH⁰.
E.g. CH₃OH_(l) + \(\frac {3}{2}\)O_2(g) = CO_2(g) + 2H₂O
ΔCH⁰ = -726 kJ mol^-1
(Δ_cH⁰ is always negative.)
[Note : Calorific value : It is the enthalpy change or amount of heat liberated when one gram of a substance undergoes combustion.

Question 57.
Show that the standard heat of formation of a compound is equal to its enthalpy.
Answer:
Consider the formation of one mole of gaseous CO₂ in the standard state at 298 K and 1 atmosphere. The thermochemical equation for formation can be represented as,
C_(s) + O_2(g) = CO_2(g) Δ_fH⁰ = -394 kJ mol^-1
Now heat of this reaction, ΔH⁰ is,
ΔH⁰_reaction = Σ_PH⁰ – Σ_RH⁰
∴ Δ_fH⁰_co2(g) = H⁰_co2(g) – [H⁰c_(s) + H⁰O_2(g)]
Since the enthalpies of elements in their standard states are zero,
i.e.
H⁰c_(s) = o, H⁰O_2(g) = 0
∴ Δ_fH⁰co_2(g) = H⁰co_2(g) – [0 + 0]
∴ Δ_fH⁰co2 = Hco_2(g)
Therefore standard heat of formation of a compound is equal to its enthalpy.

Question 58.
Standard enthalpy of formation of various following compounds are given. Write thermochemical equation for each :

Compound	ΔfH0 KJ mol-1
Cao(s)	-635.1
Al2Cl6(s)	-1300
C2H6(g)	-84.7
CH3COOH(l)	-484.7
C2H5OH(l)	-277.7
NaNO3(s)	-950.8

Answer:
(Hint: Write one mole of the compound on right hand side and corresponding constituent elements along with their standard physical states on left hand side.)

Question 59.
Standard enthalpy of combustion of different substances are given. Write thermochemical equation for each.

Substance	ΔCH0 KJ mol-1
C(graphite)	-393.5
C6H6(l)	-3268
C2H5OH(l)	-1409
CH3CHO	-1166

Answer:
In the combustion reaction, C forms CO_2(g),
H forms H₂O_(l), etc.
(1) C_(graphite) + O_2(g) → CO_2(g)
Δ_CH⁰ = -393.5 kJ mol^-1

Question 60.
Write the thermochemical equations for enthalpy of solution of :
(1) Glucose (C₆H₁₂O₆)
(2) NaCl_(s)
(3) CaBr_2(s)
Answer:

Question 61.
How is standard enthalpy of formation useful to calculate standard enthalpy of reaction ?
Answer:
(1) The standard enthalpies of formation, Δ_fH⁰ of the compounds can be used to determine the standard enthalpy of reaction (Δ_rH⁰).
(2) Δ_rH⁰ of a reaction can be obtained by subtracting the sum of Δ_fH⁰ values of all the reactants from the sum of Δ_fH⁰ values of all the products with each Δ_fH⁰ value multiplied by the appropriate coefficient of that substance in the balanced thermochemical equation.
(3) Consider following reaction :
aA + bB → cC + dD
The standard enthalpy of the reaction is given by,

where a, b, c and d are the coefficients (moles) of the substances A, B, C and D respectively.

Question 62.
Write the balanced chemical equation that have ΔH⁰ value equal to Δ_fH⁰ for each of the following substances :
(1) C₂H_2(g)
(2) KCIO_3(s)
(3) C₁₂H₂₂O_11(s) (4) CH₃-CH₂-OH₍₁₎
Answer:
Δ_fH⁰ represents the standard enthalpy of formation of each given substance. Hence it is necessary to write thermochemical equation for the formation of each substance. ΔH⁰ of this formation reaction is equal to standard heat of formation, Δ_fH⁰.

Question 63.
Consider the chemical reaction,
OF_2(g) + H₂O_(g) → O_2(g) + 2HF_(g) ΔH⁰ = -323 kJ
What is ΔH⁰ of the reaction if (a) the equation is multiplied by 3, (b) direction of reaction is reversed?
Answer:
(a) If the given thermochemical equation is multiplied by 3 then,
Δ_rH⁰ = 3ΔH⁰ = 3 × (-323) = -969 kJ
(b) If the direction of equation is reversed, then the reaction will be,
O_2(g) + 2HF_(g) → OF_2(g) + H₂O_(g)
∴ Δ_rH⁰ = – ΔH⁰ = – (- 323) = + 323 kJ

Question 64.
Define bond enthalpy (or bond energy).
Answer:
Bond enthalpy (or Bond energy) : The enthalpy change or amount of heat energy required to break one mole of particular covalent bonds of gaseous molecules forming free gaseous atoms or radicals at constant temperature (298 K) and pressure (1 atmosphere) is called bond enthalpy or bond energy. For example, bond enthalpy of H₂ is 436.4 kJ mol^-1.

Question 65.
Explain bond enthalpy of diatomic molecules.
Answer:
In case of diatomic molecules, since there is only one bond, the bond enthalpy is equal to heat of atomisation. For example, heat of atomisation of
HCl_(g) is 431.9 kJ mol^-1.

(Bond enthalpy is generally denoted by D).

Question 66.
Explain bond enthalpy in polyatomic molecules.
Answer:
Consider bond enthalpy in H₂O. The thermochemical equation for dissociation of H₂O_(g) is,
H₂O_(g) → 2H_(g) + O_(g), Δ_rH⁰ = 927 kJ mol^-1
In this, two O – H bonds are broken. It can be represented in stepwise as follows :

In above, even if two identical O – H bonds are borken, the energies required to break each bond are different.
The average bond enthalpy of O – H bond is,
Δ_rH⁰ = \(\frac {927}{2}\) = 463.5 kJ mol^-1

Solved Examples 4.10

Question 67.
Solve the following :

(1) Standard enthalpy of formation of ethane, C₂H_6(g) is -84.7 kJ mol^-1. Calculate the enthalpy change for the formation of 0.1 kg ethane.
Solution :
Given : Enthalpy of formation of C₂H_6(g)
= Δ_fH⁰ = ΔH₁ = -84.7 kJ mol^-1
Mass of C₂H_6(g) = 0.1 kg = 100 g
Molar mass ofC₂H₆ = 30 g mol^-1
ΔH⁰ for the formation of 0.1 kg C₂H₆ = 100 g
C₂H₆ = ΔH₂ = ?

Ans. Heat of formation = -282.3 kJ

(2) When 10 g C₂H₅OH_(l) are formed, 51 kJ heat is liberated. Calculate standard enthalpy of formation of C₂H₅OH_(l).
Solution :
Given : Mass of C₂H₅OH_(l) = m = 10 g
Heat liberated = ΔH1 = -51 kJ
Molar mass of C₂H₅OH = 46 gmol^-1
Standard enthalpy of formation of C₂H₅OH_(l)
= Δ_fH = ?
Standard enthalpy of formation is the enthalpy change for the formation of 1 mole C₂H₅OH_(l) i.e., 46 g C₂H₅OH_(l).
Now,
∵ For the formation of 10 g C₂H₅OH_(l)
ΔH₁ = -51 kJ
∴ For the formation of 46 g C₂H₅OH,
Δ_fH⁰ = \(\frac{-51 \times 46}{10}\) = – 234.6 kJ mol^-1
Ans. Standard enthalpy of formation of C₂H₅OH = Δ_fH⁰ = – 234.6 kJ mol^-1

(3) Standard enthalpy of combustion of CH₃OH is -726kJ mol^-1. Calculate enthalpy change for the combustion of 0.5 kg CH₃OH.
Solution :
Given : Standard enthalpy of combustion of
CH₃OH = Δ_CH⁰ = ΔH₁ = -726 kJ mol^-1
Mass of CH₃OH = m = 0.5 kg = 500 g
Molar mass of CH₃OH = 32 g mol^-1
Enthalpy of combustion = Δ_CH = ΔH₂ = ?
Now,
Enthalpy of combustion is ΔH for the combustion of 1 mole CH₃OH = 32 g CH₃OH.
∵ For 1 mole CH₃OH = 32g CH₃OH
ΔH₁ = – 726 kJ
∴ For 500 g CH₃OH, ΔH₂ = \(\frac{-726 \times 500}{32}\)
= -11344 kJ
Ans. Enthalpy change for combustion of 0.5 kg CH₃OH = – 11344 kJ

(4) The heat evolved in a reaction of 7.5 g of Fe₂O₃ with enough CO is 1.164 kJ.
Calculate ΔH⁰ for the reaction,
Fe₂O_3(s) + 3CO_(g) → 2Fe_(s) + 3CO_2(g)
Solution :
Fe₂O_3(s) + 3CO_(g) → 2Fe_(s) + 3CO_2(g)
ΔH = -1.164 kJ
Atomic mass of Fe = 56 g mol^-1
Atomic mass of O = 16 g mol^-1
Mass of Fe₂O₃ = 7.5 g
ΔH = -1.164 kJ
ΔH⁰ for reaction = ?
Molar mass of Fe₂O₃ = 2 × 56 + 3 × 16
= 160 g mol^-1
∵ For 7.5 g Fe₂O₃ ΔH= – 1.164 kJ
∴ For 160 g Fe₂O₃
ΔH⁰ = \(\frac{-1.164 \times 160}{7.5}\) = -24.86 kJ mol^-1
Ans. ΔH⁰ for the reaction = -24.83 kJ mol^-1

(5) Calculate the standard enthalpy of the reaction,
2C (graphite) + 3H_2(g) → C₂H_6(g), ΔH⁰ = ? from the following ΔH⁰ values :

Solution :

Ans. Standard enthalpy of formation of C₂H₆ = -84.4 kJ mol^-1

(6) The enthalpy of combustion of ethane is -1300 kJ. How much heat will be evolved by combustion of 1.3 × 10^-3 kg of ethane?
Solution :
Given : Δ_CH_C2H6(g) = -1300 kJ mol^-1
ΔH = ?
Amount of C₂H_6(g) = 1.3 × 10^-3 kg
Molar mass of C₂H₆ = 30 × 10^-3 kg mol^-1
Number of moles of C₂H₆
= nC₂H₆ = \(\frac{1.3 \times 10^{-3}}{30 \times 10^{-3}}\) = 4,333 × 10^-2 mol
For, combustion of 1 mol C₂H₆ ΔH = -1300 kJ
∴ For combustion of 4.333 × 10^-2 mol C₂H₆,
ΔH = 4.333 × 10^-2 × ( -1300) = – 56.33 kJ
Ans. Heat evolved is -56.33 kJ

(7) Calculate heat of formation of pentane from the following data :
(i) C_(s) + O_2(g) = CO_2(g) ΔH⁰ = -393.51 kJ
(ii) H_2(g) + \(\frac {1}{2}\)O_2(g) = H₂O_(l) ΔH⁰ = -285.80 kJ
(iii) C₅H₁₂ + 😯_2(g) = 5CO_2(g) + 6H₂O₁ ΔH⁰ = -3537 kJ
Solution :
Given :
(i) CO_(s) + O_2(g) = CO_2(g) ….. (1)
\(\Delta H_{1}^{0}\) = -393.51 kJ mol^-1
(ii) H_2(g) + \(\frac {1}{2}\)O_2(g) = H₂O_(l) … (2)
\(\Delta H_{2}^{0}\) = – 285.80 kJ mol^-1
(iii) C₅H_12(g) + 😯_2(g) = 5CO_2(g) + 6H₂O_(l) ….. (3)
\(\Delta H_{3}^{0}\) = -3537 kJ mol^-1
Required thermochemical equation :
5C_(s) + 6H_2(g) → C₅H_12(g) – ΔH = ?
Add 5 × equation (1) and 6 × equation (2) and subtract equation (3), then we get the required equation.
∴ ΔH⁰ = 5 \(\Delta H_{1}^{0}\) + 6 \(\Delta H_{2}^{0}\) – \(\Delta H_{3}^{0}\)
= 5( -393.52) + 6( -285.8) – (-3537)
= -1967.6 – 1714.8 + 3537
= -145.4 kJ mol^-1
Ans. Δ_fH⁰C₅H₁₂ = -145.4 kJ mol^-1

(8) How much heat is evolved when 12 g of CO react with NO₂ according to the following reaction,
4CO_(g) + 2NO_2(g) → 4CO_2(g) + N_2(g), ΔH⁰ = -1198 kJ ?
Solution :
Given : Mass of CO(g) = m = 12 g
Molar mass of CO = 28 g mol^-1
4CO_(g) + 2NO_2(g) → 4CO_2(g) + N_2(g)
Δ_rH⁰ = -1198 kJ
Mass of 4 moles of CO = 4 × 28 g CO = 112 g CO

Ans. Heat evolved during combustion of 12 g CO = 128.4 kJ, or Δ_CH = -128.4 kJ

(9) The heats of formation of C₁₂H₂₂O_11(S), CO_2(g) and H₂O_(l) are -2271.82, – 393.50 and 285.76 kJ respectively. Calculate the amount of cane sugar (C₁₂H₂₂O_11(S)) which will supply 11296.8 kJ of energy.
Solution :
Given : Δ_fH_C12H22O11(S) = -2271.82 J mol^-1
Δ_fH_CO2(g) = – 393.5 kJ mol^-1
Δ_fH_H2O(l) = – 285.76 kJ mol^-1
Energy required = 11296.8 kJ
Thermochemical equation for combustion of C₁₂H₂₂O_11(S) is represented as,

= [ 12(-393.5) + 11 (-285.76] – [-2271.82 + 12(0)]
= [ -4722 – 3143.36] + 2271.82
= -5593.54 kJ mol^-1
Molar mass of C₁₂H₂₂O_11(S) = 342
To obtain 5593.5 kJ energy, C₁₂H₂₂O_11(S) required is 342 gram.
Hence for 11296.8 energy, the amount of C₁₂H₂₂O_11(S) required as = \(\frac{11296.8 \times 342}{5593.54}\)
= 690.7 g
Ans. Amount of sugar required = 690.7 g

(10) 6.24 g of ethanol are vaporized by supplying 5.89 kJ of heat energy. What is enthalpy of vaporization of ethanol?
Solution :
Given : Mass of ethanol (C₂H₅OH) = m = 6.24 g
Heat energy supplied = ΔH = 5.89 kJ
Heat of vaporisation of ethanol = Δ_vapH = ?
Molar mass of ethanol, C₂H₅OH = 46 g mol^-1
∵ For 6.24 g C₂H₅OH ΔH = 5.89kJ
∴ For 1 mole C₂H₅OH = 46 g C₂H₅OH
ΔH = \(\frac{5.89 \times 46}{6.24}\)
= 43.42 kJ mol^-1
∴ Enthalpy of vaporisation of C₂H₅OH_(l)
= 43.42 kJ
Ans. Enthalpy of vaporisation of C₂H₅OH_(l)
= 43.42 kJ

(11) Given the following equations calculate the standard enthalpy of the reaction :

Solution :

By subtracting eq. (ii) from eq. (iii), we get eq. (i)
∴ eq. (i) = eq. (iii) – eq. (ii)
\(\Delta H_{1}^{0}=\Delta H_{3}^{0}-\Delta H_{2}^{0}\)
= -1670 – (-847.6)
= – 822.4 kJ
∴ Δ_rH⁰ = ΔH⁰₁ = -822.4 kJ
Ans. Standrad enthalpy of the reaction = Δ_rH⁰ = -822.4 kJ

(12) Calculate the standard enthalpy of combustion of CH₂COOH_(l) from the following data : Δ_fH⁰(CO₂) = -393.3 kJ mol^-1
Δ_fH⁰(H₂O) = -285.8 kJ mol^-1
Δ_fH⁰(CH₃COOH) = -483.2 kJ mol^-1
Solution :


∴ ΔH₁ = 2ΔH₂ + 2ΔH₃ – ΔH₄
= 2(-393.3) + 2(-285.8) – (-483.2)
= -786.6 – 571.6 + 483.2
= -875 kJ mol^-1
Ans. Standard enthalpy of combustion of CH₃COOH = -875 kJ mol^-1.

(13) The bond enthalpies of H_2(g), Br_2(g) and HBr_(g) are 436 kJ mol^-1, 193 kJ mol^-1 and 366 kJ mol^-1 respectively. Calculate the enthalpy change for the following reaction,
H_2(g) + Br_2(g) → 2HBr_(g).
Solution :
Given : Bond enthalpy of H_2(g) = ΔH⁰H_2(g)
= 436 kJ mol-1
Bond enthalpy of Br_2(g) = ΔH⁰Br_2(g) = 193 kJ mol^-1
Bond enthalpy of HBr_(g) = ΔH⁰HBr_(g) = 366 kJ mol^-1
Given reaction,
H_2(g) + Br_2(g) → 2HBr_(g)
OR
H-H_(g) + Br-Br_(g) → 2H-Br_(g)
The enthalpy change of the reaction is,

= [436 + 193] – 2[366]
= 629 – 732
= -103 kJ
Ans. Enthalpy change for the reaction = Δ_rH⁰
= -103 kJ

(14) Calculate Δ_rH⁰ of the reaction
CH_4(g) + O_2(g) → CH₂O_(g) + H₂O_(g)
From the following data:

Solution:
Given:

Standard enthalpy change for the reaction = Δ_rH⁰ = ?

= [ 2ΔH⁰_C-H + ΔH⁰_o=o ] – [ΔH⁰_C=o + 2ΔH⁰_o-H]
= [2 × 414 + 499] – [745 + 2 × 464]
= [828 + 499] – [745 + 928]
= -346 kJ
Ans. Standard enthalpy change for the reaction = ΔrH⁰ = -346 kJ

(15) Calculate C-Cl bond enthalpy from the following data :
CH₃Cl_(g) + Cl_2(g) → CH₂Cl_2(g) + HCl_(g) ΔH⁰ = – 104 kJ

Bond	C–H	Cl–Cl	H–Cl
ΔH0/KJ mol-1	414	243	431

(330 kJ mol^-1)
Solution :
Given :

Bond	C–H	Cl–Cl	H–Cl
ΔH0/KJ mol-1	414	243	431

For the given reaction, Δ_rH⁰ = -104 kJ
Bond enthalpy of C-Cl = ΔH⁰C–Cl] = ?

In this reaction, 1 C–H, 1 Cl–Cl bonds of the reactants are broken while 1C–Cl and 1H–Cl bonds of the products are formed.
Sum of bond enthalpies of bonds formed of the products

Ans. Bond enthalpy of C–Cl = ΔH⁰C–Cl
= 330 kJ mol^-1

(16) The enthalpy change for the atomisation of 10¹⁰ molecules of ammonia is 1.94 × 10^-11 kJ. Calculate the bond enthalpy of N-H bond.
Solution :
Given : Enthalpy change for atomisation of 10¹⁰ molecules = 1.94 × 10^-11 kJ
Number of NH₃ molecules dissociate = 10¹⁰
Bond enthalpy of N-H = ΔH = ?
1 mole of NH₃ contains 6.022 × 10²³ NH₃ molecules.
∵ For atomisation of 1010 molecules of NH₃
ΔH = 1.94 × 10^-11 kJ
∴ For atomisation of 6.022 × 10²³ molecules of NH₃,
ΔH = \(\frac{1.94 \times 10^{-11} \times 6.022 \times 10^{23}}{10^{10}}\)
= 1168 kJ mol^-1
In NH₃ three N-H bonds are broken on atomisation.
NH_3(g) → N_(g) + 3H_(g) ΔH = 1168 kJ mol^-1
∴ Average bond enthalpy of N-H bond is,
ΔH = \(\frac{1168}{3}\) = 389.3 kJ mol^-1
Ans. Bond enthalpy of N-H bond
= 389.3 kJ mol^-1

(17) Calculate the enthalpy of atomisation (or dissociation) of CH₂Br_2(g) at 25°C from the following data :

Bond enthalpies	C-H	C-Br
ΔH0 kJ mol-1	414	352

Solution :
Given : Bond enthalpies : ΔH⁰_C-H
= 414 kJ mol-1;
ΔH⁰_C-Br = 352 kJ mol^-1
Enthalpy of atomisation of CH₂Br_2(g) = ?
Thermochemical equation for atomisation (or dissociation) of CH₂Br₂ is,

ΔatomH° = sum of bond enthalpies of all bonds broken
= 2ΔH⁰_C-H + 2ΔH⁰_C-Br
= 2 × 414 + 2 × 352
= 828 + 704
= 1532 kJ mol^-1
Ans. Enthalpy of atomisation of CH₂Br_2(g)
= 1532 kJ mol^-1

(18) Enthalpy of sublimation of graphite is 716 kJ mol^-1.

Bond enthalpy	H-H	C-H
ΔH0 kJ mol-1	436.4	414

Calculate standard enthalpy of formation of CH₄.
Solution :
Given : Δ_subH⁰_graphite = 716 kJ mol^-1

Bond enthalpy	H-H	C-H
ΔH0 kJ mol-1	436.4	414

Thermochemical equation for the formation of CH₄,

= [716 + 2 × 436.4] – [4 × 414]
= [716+ 872.8] – [1656]
= 1588.8 – 1656
= -67.2 kJ mol^-1
Ans. Standard enthalpy of formation of CH4 = Δ_fH⁰_CH4(g) = -67.2 kJ mol^-1

(19) Calculate enthalpy of formation of propane from the following data :
Heat of sublimation of graphite is 716 kJ mo^-1.

Bond enthalpy	H-H	C-H	C-C
ΔH0 kJ mol-1	436.4	414	350

Solution :
Given: Enthalpy of sublimation of graphite = Δ_subH⁰_C
= 716 kJmol^-1

Bond enthalpy	H-H	C-H	C-C
ΔH0 kJ mol-1	436.4	414	350

Enthalpy of formation of propane = Δ_fH⁰ = ?
Thermochemical equation of the formation of propane, CH₃-CH₂-CH₃,

= [3 × 716 + 4 × 436.4] – [2 × 350 + 8 × 414]
= [2148 + 1745.6] – [700 + 3312]
= -118.4 kJmol^-1
Ans. Enthalpy of formation of propane (C₃H₈)
= -118.4 kJmol^-1

(20) The standard enthalpy of formation of propene, CH₃-CH = CH₂ is -13.2 kJ mol^-1. Enthalpy of sublimation (atomisation) of graphite is 716 kJmol^-1.

Bond enthalpy	H-H	C-H	C-C
ΔH0 kJ mol-1	436.4	414	350

Calculate bond enthalpy of C = C
Solution :

Bond enthalpy of C = C = ΔH⁰_C=C = ?
For the formation of propene, (CH₃ – CH = CH₂),

13.2 = [3 × 716 + 3 × 436.4] – [6 × 414 + 350 + \(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\) ]
= [2148 + 1309.2] – [2484 + 350 + \(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\) ]
= [3457.2] – [2834 + \(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\) ]
= 3457.2 – 2834 – \(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\)
\(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\) = 3457.2 – 2834 – 13.2
= 610 kJmol^-1
Ans. Bond enthalpy of C = C = \(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\)
= 610 kJ mol^-1

(21) Calculate the enthalpy of the reaction,
CH₃COOH_(g) + CH₃CH₂OH_(g) → CH₃COOCH₂CH_3(g) + H₂O_(g)
Bond enthalpies of O-H, C-O, in kJmol^-1 are 464, 351 respectively.
Solution :

In this reaction 1O-H and 1C-0 bond of the reactants are broken while 1C-0 and 1O-H bonds of the products are formed. Enthalpy of reaction,

Ans. Hence Enthalpy change for the reaction = Δ_rH⁰ = 0.

Question 68.
(1) What is a spontaneous process?
(2) What are its characteristics?
Answer:
(1) Spontaneous process : It is defined as a process that takes place on its own or without the intervention of the external agency or influence. For example, expansion of a gas or flow of a gas from higher pressure to low pressure or a flow of heat from higher temperature to lower temperature.

(2) Characteristics :

• It occurs on its own and doesn’t require external agency.
• It proceeds in one direction and can’t be completely reversed by external stimulant.
• These processes may be fast or slow.
• These processes proceed until an equilibrium is reached.

Question 69.
Give the examples of spontaneous processes.
Answer:
The examples of the spontaneous processes are as follows :

1. All natural processes are spontaneous.
2. A flow of gas from higher pressure to lower pressure.
3. Flow of water on its own from higher level to lower level.
4. Flow of heat from hotter body to colder body.
5. Acid-base neutralisation is a spontaneous reaction.

Question 70.
What is relation between spontaneity and energy of a system ?
Answer:
(1) The spontaneous process takes place in a direction in which energy of the system decreases. For example, neutralisation reaction between NaOH and HCl solution is exothermic with release of energy.

(2) The spontaneous process also takes place with the increase in energy by absorbing heat. For example,
(a) Melting of ice at 0 °C by absorption of heat
(b) Dissolution of NaCl,
NaCl_(s) + aq → NaCl_(aq) → Na⁺_(aq) + Cl^–_(aq)
ΔH⁰ = + 3.9 kJ mol^-1

Question 71.
Which of the following are spontaneous ?
(a) Dissolving sugar in hot coffee.
(b) Separation of Ar and Kr from their mixture.
(c) Spreading of fragrance when a bottle of perfume is opened.
(d) Flow of heat from cold object to hot object.
(e) Heat transfer from ice to room temperature at 25 °C.
Answer:
The spontaneous processes are :
(a) Dissolving sugar in hot coffee.
(b) Separation of Ar and Kr from their mixture.
(c) Spreading of fragrance when a bottle of perfume is opened.

The non-spontaneous processes are :
(d) Flow of heat from cold object to hot object.
(e) Heat transfer from ice to room temperature at 25 °C.

Question 72.
Explain : (a) Order in a system.
(b) Disorder in a system.
Answer:
(a) (i) When the atoms, molecules or ions constituting the system are arranged in a perfect order then the system is said to be in order. For example, in the solid state, the constituent atoms, molecules or ions are tightly placed at lattice points in the crystal lattice.
(ii) When solid melts forming a liquid or when a liquid vaporises, the constituents are separated and are in random motion imparting maximum disorder.
(iii) As energy of the system decreases order increases.

(b) Increase in entropy is a measure of disorder in the system. Consider following process :

Fig. 4.11 : Order decreases and disorder increases, Entropy increases

Question 73.
What is the change in order and entropy in the following :
(i) Dissolution of solid I₂ in water.
(ii) Dissociation of H_2(g) into atoms ?
Answer:
(i) For dissolution of solid I₂,

In the solid I2, there is ordered arrangement which collapses in solution increasing disorder and entropy, hence ΔS is positive.

(ii) In the dissociation of H_2(g)
H_2(g) → 2H_(g) (ΔS > 0)
In the molecular state, two H atoms in every molecule are together but in atomic state the disorder is increased with the increase in entropy and hence ΔS > 0.

Question 74.
How does addition of heat to a system at different temperatures changes disorder or ΔS ?
Answer:

• The amount of heat added to a system at higher temperature causes less disorder than when the heat is added at lower temperature.
• Since disorder depends on the temperature at which heat is added, ΔS relates reciprocally to temperature.
• This can also be explained from equation,
  ΔS = \(\frac{Q_{\text {rev }}}{T}\)

Question 75.
Explain the change in entropy for the following processes :
(i) 2H₂O_2(l) → 2H₂O_(l) + O_2(g)
(ii) 2H_2(g) + O_2(g) → 2H₂O_(l)
(iii) When ice melts at 0 °C and water vaporises at 100 °C.
Answer:
(i) In the following reaction,
2H₂O_2(l) → 2H₂O_(l) + O_2(g) ΔS = + 126 JK
Due to formation of O₂ gas from liquid, entropy increases.
(ii) In the reaction, entropy decreases due to formation of liquid H₂O from gaseous H₂ and O₂.
(iii) \(\text { Ice } \stackrel{0^{\circ} \mathrm{C}}{\longrightarrow} \text { water } \stackrel{100^{\circ} \mathrm{C}}{\longrightarrow} \text { vapour }\)
In these two steps, entropy increases due to increase in disorder from solid ice to liquid water and further to gaseous state.

Question 76.
How does entropy change in the following processes ? Explain.
(a) freezing of a liquid
(b) sublimation of a solid
(c) dissolving sugar in water
(d) condensation of vapour.
Answer:
(a) Freezing of a liquid results in decrease in randomness and disorder, hence entropy decreases, ΔS < 0.
(b) Sublimation of a solid converts it into vapour where the molecules or atoms are free to move randomly. Hence disorder increases accompanying increase in entropy, ΔS > 0.
(c) Dissolving sugar in water separates the molecules of sugar in the solution increasing disorder and entropy, ΔS > 0.
(d) Condensation of vapour decreases the disorder and randomness, hence entropy, ΔS < 0.

Question 77.
Predict the sign of ΔS in the following processes. Give reasons for your answer :
OR
Explain with reason sign conventions of ΔS in the following reactions.

Answer:
(a) N₂O_4(g) → 2NO_2(g)
Since 1 mole N₂O₄ on dissociation gives two moles of NO₂, the number of molecules increase, disorder increases hence entropy increases, ΔS > 0.

(b) Fe₂O_3(s) + 3H_2(g) → 2Fe_(s) + 3H₂O_(g)
In the reaction number of moles of gaseous reactants and products are same, hence ΔS = 0.

(c) N_2(g) + 3H_2(g) → 2NH_3(g)
In the reaction, 4 moles of gaseous reactants form 2 moles of gaseous products (Δn < 0). Therefore disorder decreases and hence entropy decreases, ΔS < 0.

(d) MgCO_3(s) → MgO_(s) + CO_2(g)
In this 1 mole of orderly solid MgCO₃ gives 1 mole of solid MgO and 1 mole of gaseous CO₂ (Δn > 0) with more disorder. Hence entropy increases, ΔS > 0.

(e) CO_2(g) → CO_2(s)
In this system from higher disorder in gaseous state changes to less disorder in the solid state, hence entropy decreases, ΔS < 0.

(f) Cl_2(g) → 2Cl_(g)
Since the dissociation of Cl₂ gas gives double Cl atoms, the number of atoms increases (Δn >0) increasing the disorder of the system. Hence ΔS > 0.

Question 78.
Identify which of the following pairs has larger entropy ? Why ?
(a) He_(g) in a volume of 1 L or He_(g) in a volume of 5 L both at 25 °C.
(b) O_2(g) at 1 atm or O_2(g) at 10 atm both at the same temperature.
(c) C₂H₅OH_(l) or C₂H₅OH_(g)
(d) 5 mol of Ne or 2 mol of Ne.
Answer:
(a) Atoms of He in 5 L at 25 °C occupy more volume than in 1 L. Hence, the randomness and disorder is more in 5 L. Expansion of a gas always increases its entropy. Therefore He_(g) in 5L will have larger entropy.

(b) O_2(g) at 1 atm will occupy more volume than O_2(g) at 10 atm at the same temperature. Hence at 1 atm O_2(g) will have higher disorder and hence higher entropy.

(c) The molecules of gaseous C₂H₅OH_(g) will have more disorder and randomness due to free motion of molecules than C₂H₅OH_(l). Hence entropy of C₂H₅OH_(g) will be larger.

(d) 5 mol Ne will contain more Ne atoms than 2 mol Ne. Hence disorder in 5 mol will be more. Therefore 5 mol Ne will have larger entropy.

Question 79.
Mention entropy change (ΔS) for :
(i) spontaneous process
(ii) nonspontaneous process
(iii) at equilibrium.
Answer:
(a) ΔS_total > 0, the process is spontaneous
(b) ΔS_total < 0, the process is non-spontaneous
(c) ΔS_total = 0, the process is at equilibrium.

Question 80.
Define Gibbs free energy and change in free energy. What are the units of Gibbs free energy ?
OR
Derive the relation between ΔG and ΔS Total.
Answer:
(i) Gibbs free energy, G is defined as,
G = H – TS
where H is the enthalpy, S is the entropy of the system at absolute temperature T.
Since H, T and S are state functions, G is a state function and a thermodynamic function.

(ii) At constant temperature and pressure, change in free energy ΔG for the system is represented as, ΔG = ΔH – TΔS

This is called Gibbs free energy equation for ΔG. In this ΔS is total entropy change, i.e., ΔS_Total.

(iii) The SI units of ΔG are J or kJ (or Jmol^-1 or kJmol^-1).
The c.g.s. units of ΔG are cal or kcal (or cal mol^-1 or kcal mol^-1.)

Question 81.
Explain Gibbs free energy and spontaneity of the process.
Answer:
The total entropy change for a system and its surroundings accompanying a process is given by,
ΔS_Total = ΔS_system + ΔS_surr
By second law, for a spontaneous process,
ΔS_Total > 0. If + ΔH is the enthalpy change (or enthalpy increase) for the process, or a reaction at constant temperature (T) and pressure, then enthalpy change (or enthalpy decrease) for the surroundings will be -ΔH.

By Gibbs equation,
ΔG = ΔH – TΔS
By comparing above two equations,
∴ ΔG = -TΔS_Total
As ΔS_Total increases, ΔG decreases.
For a spontaneous process, ΔS_Total > 0
which is according to second law of thermodynamics.
∴ ΔG < 0.
Hence in a spontaneous process, Gibbs free energy decreases (ΔG < 0) while entropy increases (ΔS > 0).
Therefore for a non-spontaneous process Gibbs free energy increases (Δ G > 0).
It can be concluded that for a process at equilibrium, ΔG=0.
Hence,

• For the spontaneous process, Δ G < 0
• For the non-spontaneous process, Δ G > 0
• For the process at equilibrium, Δ G = 0.

Question 82.
How does second law of thermodynamics explain the conditions of spontaneity ?
Answer:
The second law explains the conditions of spontaneity as below :
(i) ΔS_total  > 0 and ΔG < 0, the process is spontaneous.
(ii) ΔS_total  < 0 and ΔG > 0, the process is nonspontaneous.
(iii) ΔS_total = 0 and ΔG = 0, the process is at equilibrium.

Question 83.
Discuss the factors, ΔH, ΔS and ΔG for spontaneous and non-spontaneous processes.
OR
What can be said about the spontaneity of reactions when (1) ΔH and ΔS are both positive (2) ΔH and ΔS are both negative (3) ΔH is positive and ΔS is negative (4) ΔH is negative and ΔS is positive.
Answer:
For a spontaneous or a non-spontaneous process, ΔH and ΔS may be positive or negative (ΔH < 0 or ΔH > 0; ΔS < 0 or ΔS > 0). But ΔG must decrease, i.e., ΔG < 0. If ΔG > 0, the process or a reaction will definitely be non-spontaneous. This can be explained by Gibbs equation, ΔG = ΔH – TΔS.
(1) If ΔH and ΔS are both negative, then ΔG will be negative only when TΔS < ΔH or when temperature T is low. Such reactions must be carried out at low temperatures.
(2) If ΔH and ΔS are both positive then ΔG will be negative if, TΔS > ΔH; such reactions must be carried out at high temperature.
(3) If ΔH is negative (ΔH < 0) and ΔS is positive (ΔS > 0) then for all temperatures ΔG will be negative and the reaction will be spontaneous. But as temperature increase, ΔG will be more negative, hence the reaction will be more spontaneous at higher temperature.
(4) If ΔH is positive, (ΔH > 0) and ΔS is negative (ΔS < 0), ΔG will be always positive (ΔG > 0) and hence the reaction will be non-spontaneous at all temperatures.

This can be summarised in the following table :

Question 84.
Obtain a temperature condition for equilibrium.
Answer:
For a system at equilibrium, free energy change ΔG is,
ΔG = ΔH – TΔS
where ΔH is enthalpy change, ΔS is entropy change at temperature, T. Since ΔG = 0 at equilibrium,
O = ΔH – TΔS
∴ TΔS = ΔH
OR T = \(\frac{\Delta H}{\Delta S}\)
Hence at temperature T, changeover between forward spontaneous step and backward non-spontaneous step occurs and the system attains an equilibrium.
Here ΔH and ΔS are assumed to be independent of temperature.

Question 85.
Predict the signs of ΔH, ΔS and ΔG of the system when a solid melts at 1 atmosphere and at (a) -55 °C (b) -95 °C (c) -77 °C, if the normal melting point of the solid is -77 °C.
Answer:
Since the normal melting point of the solid is -77°C, to melt the solid at any temperature other than at -77 °C, the pressure is required to be changed. During the phase change, the system will be at equilibrium, hence Δ G = 0.
(a) In case a solid at -55 °C, the temperature should be decreased (ΔH < 0, ΔS < 0) to -77 °C and then it will melt, so ΔH > 0, ΔS > 0, ΔG = 0.
(b) In case of a solid at -95 °C, it represents supercooled system and the temperature should be raised to -77 °C (ΔH > 0, ΔS > 0) and then it will melt so ΔH > 0, ΔS > 0, ΔG = 0.
(c) At -77 °C, solid will melt, solid and liquid will be at equilibrium. Melting involves absorption of heat,

ΔH >0, ΔS > 0, ΔG = 0.

Solved Examples 4.11

Question 86.
Solve the following :

(1) In an isothermal reversible process, 6 kJ heat is absorbed at 27 °C. Calculate the entropy change.
Solution :
Given : Temperature = T = 273 + 21 = 300 K
Heat absorbed = Q_rev = 6 kJ = 6000 J
Entropy change = ΔS = ?
ΔS = \(\frac{Q_{\text {rev }}}{T}=\frac{6000}{300}\) = 20 JK^-1
Ans. Entropy change = ΔS = 20 JK^-1

(2) The latent heat of evaporation of water is 2.26 kJ g^-1 at 1 atm and 100 °C. Calculate the entropy change for evaporation of 1 mole of water at 100 °C.
Solution :
Given : Latent heat of evaporation = Δ_vapH⁰
= 2.26 kJ g^-1
Temperature = T = 273 + 100 = 373 K
Molar mass of water = 18 g mol^-1
ΔS = ?
For 1g H₂O_(l) Δ_vapH⁰ = 2.26 KJ
∴ For 1 mol H₂O_(l) = 18 g H₂O_(l)
Δ_vapH⁰ = 2.26 × 18
= 40.68 kJ
= 40680 J
Entropy change, ΔS is given by,
ΔS = \(\frac{\Delta_{\mathrm{vap}} H^{0}}{T}=\frac{40680}{373}\) = 109.06 JK^-1 mol^-1
Ans. Entropy change = ΔS = 109.06 JK^-1 mol^-1

(3) Calculate the standard (absolute) entropy change for the formation of CO_2(g).

Substance	C(graphite)	O2(g)	CO2(g)
Standard molar enthalpy JK-1 mol-1	5.74	205	213.7

Solution:
Given:

Substance	C(graphite)	O2(g)	CO2(g)
Standard molar enthalpy S0 JK-1 mol-1	5.74	205	213.7

For the formation of CO_2(g),

(4) The standard entropies of H_2(g), O_2(g) and H₂O_(g) in JK^-1 mol^-1 are 130, 205 and 189 respectively. The heat of formation of H₂O_(g) is -242 kJ mol^-1. Calculate ΔS for formation of H₂O_(g), for the surroundings and the universe at 298 K. Mention whether the reaction is spontaneous or non-spontaneous.
Solution :
Given :

Substance	H2(g)	O2(g)	H2O(g)
Standard entropy S0 JK-1 mol-1	130	205	189

Δ_fH⁰ = -242 kJ mol^-1
ΔS_universe = ?, ΔS_surr = ?
Thermochemical equation for the formation of H₂O_(g)
H_2(g) + \(\frac {1}{2}\)O_2(g) → H₂O_(g)
ΔS⁰ = [S⁰_H2O] – [H⁰_H2 + \(\frac {1}{2}\) H⁰_O2]
= 189 – [130 + \(\frac {1}{2}\)(205)]
= 189 – [232.5]
= -43.5 JK^-1 mol^-1
Hence, ΔSsystem = -43.5 JK-1 mol^-1
Since for the formation of H₂O_(g)
Δ_fH⁰ = -242 kJmol^-1 = -242 × 10³ Jmol^-1, the reaction is exothermic. Hence the surroundings gains heat energy 242 × 10³J. Therefore entropy of the system decreases while entropy of surroundings increases.

Hence, ΔS_sys < 0 but ΔS_universe > 0, hence the reaction is spontaneous.
Ans. ΔS_H2O(g) = -43.5 JK^-1 mol^-1
ΔS_surr = 813 JK^-1 mol^-1
ΔS_universe = 769.5 JK^-1.

(5) Calculate ΔS_Total and hence show whether the following reaction is spontaneous at 25 °C.
Hg_(s) + O_2(g) → Hg_(l) + SO_2(g) ΔH⁰ = – 238.6 kJ
ΔS⁰ = +36.7 JK^-1
Solution :
Given : Hg_(s) + O_2(g) → Hg_(l) + SO_2(g)
Δ_rH⁰ = – 238.6 kJ
ΔS⁰ = +36.7 JK^-1
T = 273 + 25 = 298 K
ΔS_Total = ?
ΔS_Total = ΔS_sys + Δ S_surr
Now, ΔS_sys = 36.7 JK^-1
Since the reaction is exothermic, system loses heat to surroundings. Hence the entropy of the surroundings increases.
ΔH_surr = + 238.6 kJ = 238600 J

∵ ΔS_Total > 0, the reaction is spontaneous.
Ans. ΔS_Total = 837.4 JK^-1
The reaction is spontaneous.

(6) What is the value of AS_surr for the following reaction at 298 K ?
6CO_2(g) + 6H₂O_(l) → C₆H₁₂O_6(s) + 6O_2(g),
ΔG⁰ = 2879 kJ mol^-1, ΔS⁰ = -210 JK^-1 mol^-1.
Solution :
Given :
6CO_2(g) + 6H₂O_(l) → C₆H₁₂O_6(s) + 6O_2(g),
ΔG⁰ = 2879 kJmol^-1;
ΔS⁰ = -210 JK^-1mol^-1 = -0.210 kJ K^-1 mol^-1
T = 298 K ΔH⁰ = ?
ΔG⁰ = ΔH⁰ – TΔS⁰
∴ ΔH⁰ = ΔG⁰ + TΔS⁰
= 2879 + 298(-0.210)
= 2879 – 62.58
= 2816.42 kJ mol^-1
Since ΔH⁰ > 0, the reaction is endothermic, and system absorbs heat from surroundings. Hence surroundings loses heat,

Ans. ΔS⁰surr = -9.45 kJ K^-1

(7) Calculate ΔS_surr when on mole of methanol (CH₃OH) is formed from its elements under standard conditions if Δ_fH⁰(CH₃OH) = -238.9 J mol^-1.
Solution :
Given : Number of moles of ethanol,
(C₂H₅OH) = n = 1 mol
Δ_fH⁰(CH₃OH) = -238.9 kJ mol^-1
= -238.9 × 10³J mol^-1
Temperature = T = 298 K
ΔS = ?
ΔS_surr = ?
Since Δ_fH⁰ is negative, the reaction for the formation of one mole of C₂H₅OH is exothermic.
As heat is released to the surroundings,
ΔH⁰_surr = + 238.9 kJ mol^-1
∴ ΔS_surr = \(\frac{\Delta H_{\text {surr }}^{0}}{T}=\frac{+238.9 \times 10^{3}}{298}\)
= +801.7 JK^-1
Thus entropy of the surroundings increases.
Ans. ΔS_surr = +801.7 JK^-1

(8) What is the value of ΔSsurr for the following reaction at 298 K –
6CO_2(g) + 6H₂O_(l) → C₆H₁₂O_6(s) + 6O_2(g)
Given that: ΔG° = 2879 KJ mol^-1
ΔS⁰ = -210 J K^-1 mol^-1
Solution :
Given : ΔG⁰ = 2879 KJ mol^-1 = 2879 × 10³ J mol^-1
ΔS⁰ = -210 JK^-1 mol^-1
ΔS_surr = ?
ΔG⁰ = ΔH⁰ – TΔS⁰
∴ ΔH⁰ = ΔG⁰ + TΔS⁰
= 2879 × 10³ + 298 × (- 210)
= 2879 × 10³ – 62580
= 2816420 J
Since, for a system, ΔH⁰ is +2816420 J, the surrounding loses heat to system,
∴ ΔH⁰_surr = – 2816420 J
∴ ΔS⁰_surr = \(\frac{\Delta H_{\text {surr }}^{0}}{T}\)
= \(\frac{-2816420}{298}\)
= -9451 JK^-1
= -9.451 kJ K^-1
Ans. ΔS_surr = -9.451 kJ K^-1

(9) Determine whether the reactions with the following ΔH and ΔS values are spontaneous or non-spontaneous. State whether they are exothermic or endothermic.
(a) ΔH = -110 kJ and ΔS = +40 JK^-1 at 400 K
(b) ΔH = +50 kJ and ΔS = -130 JK^-1 at 250 K.
Solution :
(a) Given : ΔH= -110 kJ ΔS = 40 JK^-1 = 0.04 kJK^-1
Temperature = T = 400 K ΔG = ?
Since ΔH is negative, the reaction is exothermic
ΔG = ΔH – TΔS
= -110 – 400 × 0.04
= -110 – 16
= -126 kJ
Since ΔG is negative, the reaction is spontaneous.

(b) Given : ΔH=50 kJ,
ΔS= -130 JK^-1 = -0.13 kJ K^-1
Temperature = T = 250 K
ΔG = ?
Since ΔH is positive, the reaction is endothermic.
ΔG = ΔH – TΔS
= 50 – 250 × (-0.13)
= 50 + 32.5
= 82.5 kJ
Since ΔG > 0, the reaction is non-spontaneous.
Ans. (a) ΔG = -126 kJ; The reaction is exothermic and spontaneous.
(b) ΔG = 82.5 kJ; The reaction is endothermic and non-spontaneous.

(10) For a certain reaction, ΔH⁰ = -224 kJ and ΔS⁰ = -153 JK^-1. At what temperature will it change from spontaneous to non-spontaneous ?
Solution :
Given : ΔH⁰ = – 224 kJ = – 224000 J
ΔS⁰ = – 153 JK^-1
Temperature (T) at which, reaction changes from spontaneous to non-spontaneous = ?
Find the temperature at equilibrium, where ΔG⁰ = 0
ΔG⁰ = ΔH⁰ – TΔS⁰
0 = ΔH⁰ – TΔS⁰
∴ TΔS⁰ = ΔH⁰
∴ T = \(\frac{\Delta H^{0}}{\Delta S^{0}}\)
= \(\frac{224000}{153}\)
= 1464 K.
Hence reaction will be spontaneous below 1464 K. It will be at equilibrium at 1464 K and non-spontaneous above 1464 K.
Ans. Change over temperature from spontaneous to non-spontaneous = 1464 K.

(11) Determine whether the reactions with the following ΔH and ΔS values are spontaneous or non-spontaneous. State whether the reactions are exothermic or endothermic.
(a) ΔH = -110 kJ, ΔS = +40 JK^-1 at 400 K
(b) ΔH = + 40 kJ, ΔS = – 120 JK^-1 at 250 K
Solution :
(a) Given : ΔH = -110 kJ, ΔS = +40 JK^-1 at T = 400K
ΔG = ΔH – TΔS

= -110 – 16
= -126 kJ
Since ΔG is negative, the reaction is spontaneous.
Since ΔH is negative, the reaction is exothermic.

(b) Given : ΔH = + 40 kJ, ΔS = -120 JK^-1 at T = 250 K
ΔG = ΔH – TΔS

= 40 + 30
= 70 kJ
Since ΔG is negative, the reaction is spontaneous.
Since ΔH is negative, the reaction is exothermic.

(12) Determine whether the following reaction will be spontaneous or non-spontaneous under standard conditions.
Zn_(s) + Cu²⁺ → Zn²⁺ +Cu_(s) ΔH⁰ = -219 kJ, ΔS⁰ = -21 JK^-1
Solution :
Given : ΔH⁰ = -219 kJ;
ΔS⁰ = -21 JK^-1= 0.021 kJ K^-1
ΔG⁰ = ?
For standard conditions : Pressure = 1 atm
Temperature = T = 298 K
ΔG⁰ = ΔH⁰ – TΔS⁰
= -219 – 298 × (-0.021)
= -219 + 6.258
= -212.742 kJ
Since ΔG < 0, the reaction is spontaneous.
Ans. The reaction is spontaneous.

(13) The equilibrium constant for a gaseous reversible reaction at 200 °C is 1.64 × 10³ atm². Calculate ΔG° for the reaction.
Solution :
Given : Equilibrium constant = KP = 1.64 × 10³ atm²
Temperature = T = 273 + 200 = 473 K
ΔG⁰ = ?
ΔG⁰ = -2.303 RTlog₁₀ K_p
= – 2.303 × 8.314 × 473 × log₁₀ 1.64 × 10³
= – 2.303 × 8.314 × 473 × (3.2148)
= -29115 J
= -29.115 kJ
Ans. ΔG⁰ = -29.115 kJ

(14) Calculate ΔG for the reaction at 25°C
CO_(g) + 2H_2(g) ⇌ CH₃OH_(g)
ΔG⁰ = -24.8 kJ mol^-1.
if P_co = 4 atm, P_H2 = 2 atm, P_CH3OH = 2 atm.
Solution :
Given : Partial pressures : pco = 4 atm,
P_H2 = 2 atm,
P_CH3OH = 2 atm
Temperature = T = 273 + 25 = 298 K
ΔG⁰ = -24.8 kJ mol^-1
CO_(g) + 2H_2(g) ⇌ CH₃OH_(g)
The reaction quotient, Q is,

= – 24.8 + 2.303 × 8.314 × 298 × log₁₀ 0.125
= – 24.8 + 2.303 × 8.314 × 298 × (\(\overline{1} \cdot 09691\))
= – 24.8 + 2.303 × 8.314 × 298 × (- 0.90709)
= – 24.8 – 2.303 × 8.314 × 298 × 0.90709 × 10^-3
= -24.8 – 5.176
= -29.976 kJ mol^-1.
Ans. ΔG = – 29.976 kJ mol^-1

(15) Calculate K_P for the reaction,
C₂H_4(g) + H_2(g) ⇌ C₂H_6(g),
ΔG⁰ = -100 kJ mol^-1, at 25°C.
Solution :
Given : ΔG⁰ = – 100 kJ mol^-1 = – 100 × 10³ J mol^-1
= -1 × 10⁵ Jmol^-1
Temperature = T = 273 + 25 = 298 K
Equilibrium constant = K_P = ?

(16) K_P for the reaction,
MgCO_3(s) → MgO_(s) + CO_2(g) is 9 × 10^-10.
Calculate ΔG⁰ for the reaction at 25 °C.
Solution :
Given : K_P = 9 Δ 10^-10
Temperature = T = 273 + 25 = 298 K
ΔG⁰ = ?
ΔG⁰ = -2.303 RTlog₁₀K_P
= -2.303 × 8.314 × 298 × log109 × 10^-10
= -2.303 × 8.314 × 298 × \([\overline{10} \cdot 9542]\)
= – 2.303 × 8.314 × 298 × [ – 9.0458]
= 51683 Jmol^-1
= 51.683 kJmol^-1
Ans. ΔG⁰ = 51.653 kJ mol^-1

(17) Calculate ΔH, ΔS and ΔG for melting of 10 g ice at 0 °C and 1 atm. (Δ_fusH⁰ = 6.02 kJ mol^-1 for ice)
Solution :
Given : Δ_fusH⁰ = 6.02 kJ mol^-1 = 6.02 × 10³ Jmol^-1
Temperature = T = 273 + 0 = 273 K
Mass of ice = 10 g
Molar mass of H₂O = 18 g mol^-1
ΔH= ?, ΔS = ?, ΔG = ?
For melting of ice,
H₂O_(s) ⇌ H₂O_(l)
For 1 mol ice = 18 g ice Δ_fusionH = 6.05 kJ
∴ For 10 g ice
ΔH = \(\frac{6.02 \times 10}{18}\)
= 3.344 kJ
ΔH = 3.344 kJ = 3.344 × 10³ J
∴ ΔS = \(\frac{\Delta H}{T}=\frac{3.344 \times 10^{3}}{273}\) = 12.25 JK^-1
ΔG = ΔH – TΔS
= 3.344 – 273 × 12.25 ×10^-3 kJ
= 3.344 – 3.344
= 0
Since ΔG = 0, the system is at equilibrium.
Ans. ΔH = 3.344 kJ; ΔS = 12.25 JK^-1; ΔG = 0

(18) Calculate Kp, ΔG⁰ for the reaction,
C_(s) + H₂O_(g) ⇌ CO_(g) + H_2(g)
at 990 K if the equilibrium concentrations are as follows :
[H₂O] = 1.10 mol dm-^-3,
[CO] = [H₂] = 0.2 mol dm^-3,
R = 0.08206 L atm K^-1 mol^-1.
Solution :
Given : [H₂O] = 1.1 mol dm^-3,
[CO] = 0.2 mol dm^-3,
[H₂] = 0.2 mol dm^-3, T = 990 K,
R = 0.08206 L atm K^-1 mol^-1
K_P = ? ΔG⁰ = ?
C_(s) + H₂O_(g) ⇌ CO_(g) + H_2(g)

K_P = K_C × (RT)^Δn
= 0.03636 × (0.08206 × 990)
= 2.954 atm
ΔG⁰ = -2.303 RTlog₁₀K_P
= -2.303 × 8.314 × 990 × log₁₀ 2.954
= -2.303 × 8.314 × 990 × 0.4704
= -8917 J
= -8.917 kJ
Ans. K_P = 2.954 atm; ΔG⁰ = -8.917 kJ

Multiple Choice Questions

Question 87.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. For an isochoric process, the change in
(a) pressure is zero
(b) volume is negative
(c) volume is zero
(d) temperature is zero
Answer:
(c) volume is zero

2. Which of the following is an extensive property ?
(a) Surface tension
(b) Refractive index
(c) Energy
(d) Temperature
Answer:
(c) Energy

3. Which of the following is an intensive property ?
(a) Enthalpy
(b) Weight
(c) Refractive index
(d) Volume
Answer:
(c) Refractive index

4. Which of the following pairs is an intensive property ?
(a) Density, viscosity
(b) Surface tension, mass
(c) Viscosity, internal energy
(d) Heat capacity, volume
Answer:
(a) Density, viscosity

5. The property which is not intensive is
(a) freezing point
(b) viscosity
(c) temperature
(d) free energy
Answer:
(d) free energy

6. Which of the following is not an extensive property ?
(a) molarity
(b) molar heat capacity
(c) mass
(d) volume
Answer:
(b) molar heat capacity

7. Which of the following is NOT a state function ?
(a) Work
(b) Enthalpy
(c) Temperature
(d) Pressure
Answer:
(a) Work

8. In an adiabatic process
(a) ΔT ≠ 0
(b) ΔU ≠ 0
(c) Q = 0
(d) All of these
Answer:
(d) All of these

9. For an isothermal and reversible process
(a) P₁V₁ = P₂V₂
(b) P₁V₁ ≠ P₂V₂
(c) ΔV ≠ 0
(d) ΔH ≠ 0
Answer:
(a) P₁V₁ = P₂V₂

10. For the process to occur under adiabatic conditions, the correct condition is :
(a) ΔT = 0
(b) Δp = 0
(c) Q = 0
(d) W = 0
Answer:
(c) Q = 0

11. What is true for an adiabatic process ?
(a) ΔT = 0
(b) ΔU
(c) ΔH = ΔU
(d) Q = 0
Answer:
(d) Q = 0

12. ΔU = 0 is true for
(a) Adiabatic process
(b) Isothermal process
(c) Isobaric process
(d) Isochoric process
Answer:
(b) Isothermal process

13. When a gas expands in vacuum, the work done by the gas is
(a) maximum
(b) zero
(c) less than zero
(d) greater than zero
Answer:
(b) zero

14. When a sample of an ideal gas is allowed to expand at constant temperature against an atomospheric pressure,
(a) surroundings does work on the system
(b) ΔU = 0
(c) no heat exchange takes place between the system and surroundings
(d) internal energy of the system increases
Answer:
(b) ΔU = 0

15. In what reaction of the following work is done by the system on the surroundings ?
(a) Hg_(l) → Hg_(g)
(b) 3O_2(g) → 2O_3(g)
(c) H_2(g) + Cl_2(g) → 2HCl_(g)
(d) N_2(g) + 3H_2(g) → 2NH_3(g)
Answer:
(a) Hg_(l) → Hg_(g)

16. A gas does 0.320 kJ of work on its surroundings and absorbs 120 J of heat from the surroundings. Hence, ΔU is
(a) 440 kJ
(b) 200 J
(c) 120.32 J
(d) -200J
Answer:
(d) -200J

17. For an isothermal and reversible expansion of 0.5 mol of an ideal gas, Wmax is -3.918 kJ. The value of ΔU is
(a) 3.918 kJ
(b) zero
(c) 1.959 kJ
(d) 3918 J
Answer:
(b) zero

18. The mathematical expression of the first law of thermodynamics for an adiabatic process is
(a) W = Q
(b) W = -ΔU
(c) W = +ΔU
(d ) W = -Q
Answer:
(c) W = +ΔU

19. A gaseous system absorbs 600 kJ of heat and performs the work of expansion equal to 130 kJ. The internal energy change is
(a) 730 kJ
(b) -470 kJ
(c) -730 kJ
(d) 470 kJ
Answer:
(d) 470 kJ

20. When a gas is compressed, the work obtained is 360 J while the heat transferred is 190 J. Hence the change in internal energy is
(a) -170 J
(b) 170 J
(c) 550 J
(d) -550 J
Answer:
(b) 170 J

21. For the reaction N_2(g) + 3H_2(g) = 2NH_3(g); Which of the following is valid ?
(a) ΔH = ΔU
(b) ΔH < ΔU
(c) ΔH > ΔU
(d) ΔH = 2ΔH
Answer:
(b) ΔH < ΔU

22. For which reaction ΔH = ΔU ?

Answer:
(b) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6(\mathrm{~s})}+6 \mathrm{O}_{2(\mathrm{~g})} \rightarrow 6 \mathrm{CO}_{2(\mathrm{~g})}+6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)

23. For the following reaction at 298 K
H_2(g) + \(\frac {1}{8}\)O_2(g) = H₂O_(l)
Which of the following alternative is correct ?
(a) ΔH = ΔU
(b) ΔH > ΔU
(c) ΔH < ΔU
(d) ΔH = 1.5 ΔU
Answer:
(c) ΔH < ΔU

24. The heat of combustion of carbon is 394 kJ mol^-1. The heat evolved in combustion of 6.023 × 10²¹ atoms of carbon is
(a) 3940 kJ
(b) 3940.0 kJ
(c) 3.94 kJ
(d) 0.394 kJ
Answer:
(c) 3.94 kJ

25. Which of the reactions defines the heat of formation ?

Answer:
(d) \(\frac{1}{2} \mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{HCl}_{(\mathrm{g})}\)

26. ΔU^o of combustion of methane is -X kJ mol^-1. The value of ΔH⁰ is
(a) = ΔU^o
(b) > ΔU^o
(c) < ΔU^o
(d) =0
Answer:
(c) < ΔU^o

27. The enthalpy of combustion of 5(rhomic)is -297.4 kJ mol^-1. The amount of sulphur required to produce 29.74 kJ of heat is
(a) 32 × 10^-2 kg
(b) 3.2 × 10^-3 kg
(c) 3.2 × 10^-2 kg
(d) 6.4 × 10^-3 kg
Answer:
(b) 3.2 × 10^-3 kg

28. The heat of formation of SO_2(g) and SO_3(g) are -269 kJ mol^-1 and -395 kJ mol^-1 respectively the value of ΔH for the reaction
SO_2(g) + \(\frac {1}{2}\)O_2(g) → SO_3(g) is
(a) -664 kJ mol^-1
(b) -126 kJ mol^-1
(c) 63 kJ mol^-1
(d) 126 kJ mol^-1
Answer:
(b) -126 kJ mol^-1

29. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.3 kJ mol^-1, -393.5 kJ mol^-1 and -285.8 kJ mol^-1 respectively. Enthalpy of formation of CH_4(g) will be
(a) – 74.8 kJ mol^-1
(b) – 52.27 kJ mol^-1
(c) + 74.8 kJ mol^-1
(d) + 52.26 kJ mol^-1
Answer:
(a) – 74.8 kJ mol^-1

30. The enthalpies of formation of N₂O_(g) and NO_(g) are 82 kJ mol^-1 and 90 kJ mol^-1 respectively. Then enthalpy of a reaction 2N₂O_(g) + O_2(g) → 4NO_(g) is …………
(a) 8 kJ
(b) -16 kJ
(c) 88 kJ
(d) 196 kJ
Answer:
(d) 196 kJ

31. The heat of combustion of naphthalene (C₁₀H₈) to CO₂ gas and water vapour at 298 K and at constant pressure is -5.1567 × 10⁶ J. The heat of combustion at constant volume at 298 K is (R = 8.314 JK^-1 mol^-1)
(a) -5.1567 × 10⁶ J
(b) -5.6161 × 10⁶ J
(c) -5.1616 × 10⁶ J
(d) -5.7161 × 10⁶ J
Answer:
(c) -5.1616 × 10⁶ J

32. Given the reaction,
2NH_3(g) → N_2(g) + 3H_2(g) ΔH = 92.6 kJ
The enthalpy of formation of NH₃ is
(a) -92.6 kJ
(b) 92.6 kJ mol^-1
(c) -46.3 kJmol^-1
(d) -185.2 kJmol^-1
Answer:
(c) -46.3 kJmol^-1

33. Calculate the heat of reaction at 298 K for the reaction C₂H_4(g) + H_2(g) = C₂H_6(g)
Given that the heats of combustion of ethylene, hydrogen and ethane are 337.0, 68.4 and 373.0 kcal respectively.
(a) 23.4 kcal
(b) 62.2 kcal
(c) 32.4 kcal
(d) 34.2 kcal
Answer:
(c) 32.4 kcal

34. Entropy change for a process is given by,
(a) Q_rev × T
(b) Q_rev/T
(c) \(\frac{T}{Q_{\text {rev }}}\)
(d) ΔH_rev × T
Answer:
(b) Q_rev/T

35. For a spontaneous process, total entropy change for a system and its surroundings is
(a) ΔS_total < 0
(b) ΔS_total = 0
(c) ΔS_total > 0
(d) ΔS_total ≤ 0
Answer:
(c) ΔS_total > 0

36. For a system at equilibrium,
(a) ΔS_total = 0
(b) ΔS_total > 0
(c) ΔS_total < 0
(d) ΔS_total ≥ 0
Answer:
(a) ΔS_total = 0

37. If the enthalpy of vaporisation of water at 100°C is 186.5 J·mol^-1, the entropy of vaporization will be-
(a) 4.0 J·K^-1 mol^-1
(b) 3.0 J·K^-1 mol^-1
(c) 1.5 J·K^-1 mol^-1
(d) 0.5 J·K^-1 mol^-1
Answer:
(d) 0.5 J·K^-1 mol^-1

38. Heat of fusion of ice is 6.02kJmol-1 at 0 °C. If 100 g water is frozen at 0 °C, entropy change will be
(a) -0.1225 JK^-1
(b) 310.6 JK^-1
(c) -122.6 JK^-1
(d) 92.8 JK^-1
Answer:
(c) -122.6 JK^-1

39. If for a reaction ΔH is negative and ΔS is positive then the reaction is
(a) spontaneous at all temperatures
(b) non-spontaneous at all temperatures
(c) spontaneous only at high temperatures
(d) spontaneous only at low temperature
Answer:
(a) spontaneous at all temperatures

40. The relationship between ΔG^o of a reaction and its equilibrium constant is

Answer:
(c) \(\frac{R T \ln K}{\Delta G^{0}}=-1\)

41. Which of the following has highest entropy?
(a) Al_(s)
(b) CaCO_3(s)
(c) H₂O_(l)
(d) CO_2(g)
Answer:
(d) CO_2(g)

42. The entropy change for the formation of 3.5 mol NO_(g) from the following data will be,

Answer:
(b) 42.875 JK^-1

43. Gibbs free energy change at equilibrium is
(a) ΔG = 0
(b) ΔG > 0
(c) ΔG < 0
(d) ΔG ≤ 0
Answer:
(a) ΔG = 0

44. For spontaneous process,
(a) ΔG = 0
(b) ΔG > 0
(c) ΔG < 0
(d) ΔG ≤ 0
Answer:
(c) ΔG < 0

45. A substance which shows highest entropy is
(a) SrCO_3(S)
(b) Cu_(S)
(c) NaC_(aq)
(d) Cl_2(g)
Answer:
(d) Cl_2(g)

46. For which of the following reactions ΔS is negative ?

Answer:
(a) \(\mathrm{Mg}_{(s)}+\mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{MgCl}_{2(s)}\)

47. For a reaction, at 300K enthalpy is 138 kJ and entropy change is 115 JK^-1. Hence the free energy change of the reaction is
(a) 130.5 kJ
(b) 103.5 kJ
(c) 82.8 kJ
(d) – 60.5 kJ
Answer:
(b) 103.5 kJ

48. Bond enthalpies of H_2^-1, I_2(g) and HI are 436, 151 and 298 kJ mol^-1 respectively. Hence enthalpy of formation of HI_(g) is
(a) -9 kJmol^-1
(b) -4.5kJmol^-1
(c) 4.5 kJ mol^-1
(d) 9 kJ mol^-1
Answer:
(b) -4.5kJmol^-1

49. The average bond energy of C-H bond is 410 kJmol^-1. The enthalpy change of atomisation of 3.2 g CH_4(g) is
(a) 1312 kJ
(b) 29.8 kJ
(c) 328 kJ
(d) 120 kJ
Answer:
(c) 328 kJ

50. For a chemical reaction ΔS = -0.035 kJ/K and ΔH = -20 kJ. At what temperature does the reaction turn non-spontaneous ?
(a) 5.14 K
(b) 57.14 K
(c) 571.4 K
(d) 5714.0 K
Answer:
(c) 571.4 K

51. For a certain reaction, ΔH = -50 kJ and ΔS = -80 JK^-1, at what temperature does the reaction turn from spontaneous to non-spontaneous.
(a) 6.25 K
(b) 62.5 K
(c) 625 K
(d) 6250 K
Answer:
(c) 625 K

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Important Questions Chapter 6 Dissolution of Partnership Firm Important Questions and Answers.

Maharashtra State Board 12th Commerce BK Important Questions Chapter 6 Dissolution of Partnership Firm

1. Objective Questions:

A. Select the most appropriate answer from the alternatives given below and rewrite the sentences.

Question 1.
_____________ means winding up of partnership firm.
(a) Dissolution
(b) Formation
(c) Retirement
(d) Death
Answer:
(a) Dissolution

Question 2.
When a partner takes over a liability, his Capital Account is _____________
(a) debited
(b) credited
(c) deducted
(d) none of these
Answer:
(b) credited

Question 3.
Dissolution expenses are debited to the _____________ Account.
(a) Profit and Loss
(b) Trading
(c) Capital
(d) Realisation
Answer:
(d) Realisation

Question 4.
The debit balance on Realisation A/c indicates _____________
(a) profit
(b) loss
(c) gain
(d) deficiency
Answer:
(b) loss

Question 5.
The partner who is unable to pay his liabilities is called an _____________ partner.
(a) solvent
(b) working
(c) insolvent
(d) sleeping
Answer:
(c) insolvent

Question 6.
Debit balance of insolvent Partner’s Capital Account is known as _____________
(a) capital deficiency
(b) capital surplus
(c) profit
(d) loss
Answer:
(a) capital deficiency

B. Give the word/term/phrase which can substitute each of the following statements.

Question 1.
The account records all realisable assets and external liabilities of the firm on dissolution.
Answer:
Realisation Account

Question 2.
The partner who bears capital deficiency of an insolvent partner.
OR
The person who bears insolvency loss of an insolvent partner.
Answer:
Solvent Partner

Question 3.
Account to which the ultimate unpaid balances on the outside liability accounts are transferred on dissolution.
Answer:
Deficiency Account

C. State whether the following statements are True or False with reasons.

Question 1.
The cash and bank balances are not transferred to the Realisation A/c.
Answer:
This statement is True.
Cash or bank balance are liquid assets. They cannot be sold or realised. The cash and/or bank balances are recorded in the Cash and/or Bank Account. All cash and/or bank transactions, at the time of Realisation, are recorded in the Cash and Bank Accounts. Therefore, Cash and Bank balances are not transferred to Realisation Account.

Question 2.
On dissolution, sundry debtors are transferred to Realisation A/c at their net figure.
Answer:
This statement is False.
On dissolution, sundry debtors are transferred to Realisation A/c at their Gross value book value and not at their net figure. R.D.D. which is deducted from debtors is not an asset and therefore R.D.D. is transferred to the credit side of Realisation A/c and the remaining debtors are transferred to the debit side of Realisation A/c.

Question 3.
On dissolution of the firm, the partner’s wife loan is transferred to Realisation A/c.
Answer:
This statement is True.
A loan taken from the partner’s wife is an external liability and it is a third party’s liability. So, the partner’s wife’s loan is transferred to Realisation A/c at the time of dissolution of the firm.

Question 4.
A liability that is not shown in the Balance Sheet on the date of dissolution cannot be repaid.
Answer:
This statement is False.
Liability of the firm which is not yet recorded in the book of accounts is called unrecorded liability. At the time of dissolution unrecorded liability is supposed to be paid though it is not shown in the Balance Sheet.

Question 5.
A debit balance of Realisation A/c indicates profit on realisation.
Answer:
This statement is False.
A debit balance of Realisation A/c means payment is more than the receipt. When payments exceed receipts there is a loss. Hence, debit balance on realisation account indicates loss on realisation and not profit.

D. Answer in one sentence only.

Question 1.
What is Realisation Account?
Answer:
An account that is opened by the firm at the time of its dissolution to determine profit or loss on realisation of assets and payment of liabilities is known as Realisation Account.

Question 2.
Why is the Realisation Account opened?
Answer:
Realisation Account is opened to find out profit or loss made on the sale of assets and discharge of liabilities of the partnership firm.

Question 3.
What are realisation or dissolution expenses?
Answer:
The expenses incurred by the firm to realise the assets and to liquidate the liabilities of the firm on its dissolution are called realisation or dissolution expenses.

Solved Problem

Question 1.
Asha, Bela, and Nisha were partners sharing profits and losses in the ratio of 3 : 2 : 1. On 31st March 2020 their Balance Sheet was as follows:
Balance Sheet as of 31st March 2020

The firm was dissolved on 31st March 2020 and the assets realised as follows:
1. Joint Life Policy was taken over by Asha at ₹ 10,000.
2. Stock realised: ₹ 36,000, Debtors realised: ₹ 29,000, Machinery was sold for ₹ 72,000.
3. Liabilities were paid in full. In addition, one bill for ₹ 700 under discount was dishonoured and had to be taken up by the firm.
4. There were no realisation expenses.
Give the Journal entries and prepare necessary Ledger Accounts to close the books of the firm.
Solution:
In the Journal of Partnership Firm



Ledger Accounts:
In the books of Partnership Firm

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 5 Cell Structure and Organization Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 5 Cell Structure and Organization

Question 1.
Define cell.
Answer:
Cell is defined as a structural and functional unit of life of all living organisms capable of independent existence and can perform all functions of life.

Question 2.
Write information about the instrument which is used for observing smaller organisms or cells.
Answer:

1. To observe cells or organisms of smaller size we use a microscope.
2. Larger cells can be seen through simple microscope but to observe smaller cells we require compound microscope.
3. Simple microscope can magnify image 50 to 100 times but a compound microscope can do so 1000 times or more.
4. In the microscope a beam of light is used to make things visible hence it is light microscope.
5. To observe interior of cell we need electron microscope which can magnify image 500000 times.

Question 3.
Write the shapes of the cells that can be observed.
Answer:
There is no typical shape of a cell. Cells may be spherical, rectangular, flattened, polygonal, oval, triangular, conical, columnar, etc.

Question 4.
1. Smallest cell
2. Longest cell in animals
3. Largest cell
Answer:
1. Mycoplasma (0.3 µm)
2. Nerve cell
3. Ostrich egg

Question 5.
Explain the term totipotency.
Answer:

1. Totipotency (totus – entire, potential – power) is the capacity or the potential of living nucleated cell, to differentiate into any other type of cell and thus, can form a complete new organism.
2. A cell is totipotent as it has the entire genetic information of the organism stored in its nucleus.
3. Embryonic animal cells are totipotent and are termed as stem cells.
4. Stem cells are used in curing many diseases. Therefore, they have great potential for medical applications.

Question 6.
Who proposed the cell theory?
Answer:
Schwann and Schleiden proposed the cell theory.

Question 7.
Give the postulates of modern cell theory.
Answer:
Postulates of modern cell theory:

1. All living organisms are made up of cells.
2. Cell is the basic structural and functional unit of life.
3. All cells arise from pre-existing cells. (Rudolf Virchow 1858 – “Omnis cellula-e-cellula”).
4. Total activities of cells are responsible for activity of an organism.
5. Cells show transformation of energy.
6. Cells contain nucleic acids; DNA and RNA in the nucleus and cytoplasm.

Question 8.
State the two general categories on which living organisms are grouped.
Answer:
Living organisms are grouped into two main categories the Prokaryotes and Eukaryotes.

Question 9.
State the general characteristics of prokaryotic cell.
Answer:
General characteristics of prokaryotic cell:
1. Prokaryotic cells are primitive type of cells.
2. It does not have membrane bound cell organelles (like endoplasmic reticulum, Golgi complex, mitochondria, etc.) and well-defined nucleus (nuclear membrane is absent).
3. Genetic material is in the form of nucleoid.

4. Cell envelope:
a. Prokaryotic cell has chemically complex protective cell envelope having glycocalyx, cell wall and plasma membrane.
b. In some bacteria, glycocalyx occurs in the form of a slime layer (loose sheath). Other bacteria may have a thick and tough covering called capsule. It helps in protection of bacterial cell.

5. Cell wall:
The Gram-positive bacteria show presence of peptidoglycan layer in the cell wall and Gram-negative bacteria show presence of murein in the cell wall. It gives mechanical strength to the cell.
[Note: In Gram-negative bacteria, cell wall is made up of two layers; inner layer of Murein or peptidoglycan and outer layer of Lipopolysaccharides.]

6. Cell membrane:
a. It is the innermost covering of the cell envelope, chemically composed of lipids and proteins.
b. It helps in intercellular communication.
c. Cell membrane shows infoldings called mesosomes which help in cell wall formation, cellular respiration and DNA replication.
d. The cyanobacteria show longer extensions called as chromatophores which carry photosynthetic pigments.

7. In motile bacteria either cilia or flagella are found. Both are driven by rotatory movement produced by basal body (which works as motor) of flagellum. Other parts of flagellum are filament and hook.

8. Some other surface projections are the tubular pili (which help in inter-cellular communication) and fimbriae (for clinging to support).

9. Ribosomes:
Bacterial cell cytoplasm contains dense particles called ribosomes which help in protein synthesis. Ribosomes are 70S type (composed of a larger sub-unit 50S and + smaller sub-unit 30S).

Question 10.
What is the difference between Gram-positive and Gram-negative bacterial cells? Name the technique used for differentiating such bacterial cells.
Answer:
The Gram-positive bacteria show presence of peptidoglycan layer in the cell wall and Gram-negative bacteria show presence of murein in the cell wall. The technique used for differentiating bacterial cells is Gram staining.
[Note: Murein is similar to peptidoglycan in structure and function. It is present in the cell walls of archaebacteria.

Question 11.
Write the constituents of prokaryotic cytoplasm.
Answer:
1. Cytoplasm of prokaryotes is a pool of all necessary materials like water, enzymes, elements, amino acids, etc.
2. Some inclusion bodies in form of organic (cyanophycean starch and glycogen) and inorganic granules (phosphate and sulphur) are also found.

Question 12.
Fill in the blanks.

1. Genetic material in bacterium is a single chromosome made up of circular and coiled _______.
2. The bacterial chromosome remains attached to _________.
3. The _________ model of replication is observed in bacterial cells.
4. _________ present in the bacterial cells are known as extrachromosomal self-replicating DNA.

Answer:

1. DNA
2. Mesosomes
3. Theta
4. Plasmids

Question 13.
What are eukaryotic cells?
Answer:
1. Eukaryotic cells are the cells possessing well-defined nucleus and membrane bound organelles (like mitochondria, endoplasmic reticulum, ribosomes, Golgi complex etc.).
2. Eukaryotes include protists, plants, animals and fungi.

Question 14.
Write a note on cell wall in Eukaryotic cells.
Answer:

• The rigid, protective and supportive covering, outside the cell membrane is called cell wall. It is present in plant cells, fungi and some protists.
• Algae show presence of cellulose, galactans, mannans and minerals like calcium carbonate in cell wall.
• In other plants, it is made up of hemicelluloses, pectin, lipids and protein.
• Microfibrils of plant cell wall show presence of cellulose which is responsible for rigidity.
• Some of the depositions of cell wall are silica (grass stem), cutin (epidermal walls of land plants), suberin (endodermal cells of root), wax, lignin.
• Function:
  • Provides support, rigidity and shape to the cell.
  • Protects the protoplasm against mechanical injury and infections.

Question 15.
Explain the structure of plant cell wall.
Answer:
In plants, cell wall shows middle lamella, primary wall and secondary wall

1. Middle lamella:
It is thin and present between two adjacent r cells. It is the first structure formed from cell plate during cytokinesis. It is mainly made up of pectin, calcium and magnesium pectate. Softening of ripe fruit is due to solubilization of pectin.
2. Primary wall:
In young plant cell, it is capable of growth. It is laid inside to middle lamella.
It is the only wall seen in meristematic tissue, mesophyll, pith, etc.
3. Secondary wall:
It is present inner to primary wall. Once the growth of primary wall stops, secondary wall is laid. At some places thickening is absent which leads to formation of pits.

Question 16.
Draw a well labelled diagram of a plant cell.
Answer:

Question 17.
Give an account of eukaryotic plasma membrane.
Answer:
Eukaryotic plasma membrane/ Cell membrane/ Biomembrane:

1. It is thin, quasi-fluid structure present both extracellularly and intracellularly.
2. Extracellularly, it is present around protoplast and intracellularly, it is present around most of the cell organelles in eukaryotic cell. It separates cell organelles from cytosol.
3. Thickness of bio-membrane is about 75A.
4. Cell membrane appears trilaminar (made up of three layers) when observed under electron microscope. It shows presence of lipids (mostly phospholipids) arranged in bilayer.
5. Lipids possess one hydrophilic polar head and two hydrophobic non-polar tails. Therefore, phospholipids are amphipathic.
6. Lipid molecules are arranged in two layers (bilayer) in such a way that their tails are sandwiched in between heads. Due to this, tails never come in direct contact with aqueous surrounding.
7. Cell membrane also shows presence of proteins and carbohydrates.
8. Ratio of proteins and lipids varies in different cells. For example, in human beings, RBCs show approximately 52% protein and 40% lipids.

Question 18.
Explain the structure of plasma membrane on the basis of Fluid mosaic model.
Answer:
Fluid mosaic model:

1. Fluid mosaic model was proposed by Singer and Nicholson (1972).
2. This model states that plasma membrane is made up of phospholipid bilayer and proteins.
3. Proteins are embedded in the lipid membrane like icebergs in the sea of lipids.
4. Phospholipid bilayer is fluid in nature.
5. Quasi-fluid nature of lipid enables lateral movement of proteins. This ability to move within the membrane is measured as fluidity.
6. Based on organization of membrane proteins they are of two types, as:

a. The intrinsic proteins occur at different depths of bilayer i.e. they are tightly bound to the phospholipid bilayer and are embedded in it. They span the entire thickness of the membrane. Therefore, they are known as transmembrane proteins. They form channels for passage of water.
b. The extrinsic or peripheral proteins are found on two surfaces of the membrane i.e. are loosely held to the phospholipid layer and can be easily removed.

Question 19.
Draw neat and labelled diagram of structure of plasma membrane proposed by Singer and Nicolson.
Answer:
Fluid mosaic model:

1. Fluid mosaic model was proposed by Singer and Nicholson (1972).
2. This model states that plasma membrane is made up of phospholipid bilayer and proteins.
3. Proteins are embedded in the lipid membrane like icebergs in the sea of lipids.
4. Phospholipid bilayer is fluid in nature.
5. Quasi-fluid nature of lipid enables lateral movement of proteins. This ability to move within the membrane is measured as fluidity.
6. Based on organization of membrane proteins they are of two types, as:

a. The intrinsic proteins occur at different depths of bilayer i.e. they are tightly bound to the phospholipid bilayer and are embedded in it. They span the entire thickness of the membrane. Therefore, they are known as transmembrane proteins. They form channels for passage of water.
b. The extrinsic or peripheral proteins are found on two surfaces of the membrane i.e. are loosely held to the phospholipid layer and can be easily removed.

Question 20.
Give the functions of plasma membrane.
Answer:
1. The significant function of plasma membrane is transport of molecules across it. Plasma membrane is selectively permeable.

2. Passive transport:
a. Many molecules move across the membrane without spending energy.
b. Some molecules move by simple diffusion along the concentration gradient i.e. from higher to lower concentration.
c. Neutral molecules may move across the membrane by the process of simple diffusion.
d. Water may also move by osmosis.

3. Active transport:
a. Few ions or molecules are transported against concentration gradient i.e. from lower to higher concentration.
b. This requires energy, hence ATP is utilized. As such a transport is an energy dependent process in which ATP is utilized, it is called Active transport e.g. Na⁺ /K⁺ pump.
c. Polar molecules cannot pass through non-polar lipid bilayer. Therefore, they require carrier proteins to facilitate their transport across the membrane.

Question 21.
Write a note on cytoplasm in Eukaryotic cell.
Answer:
Cytoplasm in Eukaryotic cell:

1. The cell contains ground substance called cytoplasmic matrix or cytosol.
2. This colloidal jelly like material shows streaming movements called cyclosis.
3. The cytoplasm contains water as major component along with organic and inorganic molecules like sugars, amino acids, vitamins, enzymes, nucleotides, minerals and waste products.
4. It also contains various membrane-bound cell organelles like endoplasmic reticulum, Golgi complex, mitochondria, plastids, nucleus, microbodies and cytoskeletal elements like microtubules.
5. Cytoplasm acts as a source of raw materials as well as seat for various metabolic activities taking place in the cell.
6. It helps in distribution and exchange of materials between various cell organelles.

Question 22.
Explain the endomembrane system of the cell.
Answer:

1. Cell organelles are compartments in the cell that carry out specific functions.
2. Some of these organelles coordinate with each other and complete the specific function of the cell.
3. Nuclear membrane, endoplasmic reticulum, Golgi complex, lysosomes and various types of vesicles and vacuoles form such a group and are together considered as endomembrane system of the cell.

Question 23.
Why mitochondria and chloroplasts are not considered as a part of endomembrane system?
Answer:
1. Organelles having distinct functions are not included in endomembrane system.
2. Mitochondria or chloroplast carry out specific type of energy conversions in the cell. Therefore, mitochondria and chloroplasts are not considered as a part of endomembrane system.

Question 24.
Describe the structure of Endoplasmic Reticulum.
Answer:

1. Endoplasmic reticulum is a network present within the cytosol.
2. It is present in all a cells except ova and mature red blood corpuscles.
3. Under the electron microscope, it appears like network of membranous tubules and sacs called cisternae.
4. This network of ER divides the cytoplasm in two parts viz. one within the lumen of ER called laminal cytoplasm and non-laminal cytoplasm that lies outside ER.
5. Membrane of ER is continuous with nuclear envelope at one end and extends till cell membrane. It thus acts as intracellular supporting framework and helps in maintaining position of various cell organelles in the cytoplasm.
6. Depending upon the presence or absence of ribosomes, endoplasmic reticulum is called rough endoplasmic reticulum (RER) or smooth endoplasmic reticulum (SER) respectively.

Question 25.
Label the diagram
Answer:

1. Endoplasmic reticulum is a network present within the cytosol.
2. It is present in all a cells except ova and mature red blood corpuscles.
3. Under the electron microscope, it appears like network of membranous tubules and sacs called cisternae.
4. This network of ER divides the cytoplasm in two parts viz. one within the lumen of ER called laminal cytoplasm and non-laminal cytoplasm that lies outside ER.
5. embrane of ER is continuous with nuclear envelope at one end and extends till cell membrane. It thus acts as intracellular supporting framework and helps in maintaining position of various cell organelles in the cytoplasm.
6. Depending upon the presence or absence of ribosomes, endoplasmic reticulum is called rough endoplasmic reticulum (RER) or smooth endoplasmic reticulum (SER) respectively.

Question 26.
Explain the structure, location and functions of Golgi complex.
Answer:
Golgi complex or Golgi apparatus or Golgi body act as a assembly, manufacturing cum packaging and transport unit of cell.
1. Structure of Golgi complex:
a. Golgi complex consists of stacks of membranous sacs called cistemae.
b. Diameter of cistemae varies from 0.5 to 1pm.
c. A Golgi complex may have few to several cistemae depending on its function.
d. The thickness and molecular composition of membranes at one end of the stack of a Golgi sac differ from those at the other end.
e. The Golgi sacs show specific orientation in the cell.
f. Each cistema has a forming or ‘cis’ face (cis: on the same side) and maturing or ‘trans’ face (trAnswer:the opposite side).
g. Transport vesicles that pinch off from transitional ER merge with cis face of Golgi cistema and add its contents into the lumen.

2. Location of Golgi complex:
Golgi bodies are usually located near endoplasmic reticulum.

3. Functions of Golgi complex:
a. Golgi body carries out two types of functions, modification of secretions of ER and production of its own secretions.
b. Cistemae contain specific enzymes for specific functions.
c. Refining (modification) of product takes place in a sequential manner.
d. For example, certain sugar component is added or removed from glycolipids and glycoproteins that are brought from ER, thus forming a variety of products.
e. Golgi bodies also manufacture their own products. Golgi bodies in many plant cells produce non-cellulose polysaccharides like pectin.
f. Manufactured or modified, all products of Golgi complex leave cistemae from trans face as transport vesicles.

Question 27.
How transport vesicles identify their target cell or cell membrane?
Answer:
While transport vesicles are leaving from the trans face of the Golgi, certain markers get impregnated on their membrane. These markers help them to identify their specific target cell or cell organelle.

Question 28.
Label the diagrams and write down the details of concept in your words.
Answer:
Golgi complex or Golgi apparatus or Golgi body act as a assembly, manufacturing cum packaging and transport unit of cell.
1. Structure of Golgi complex:
a. Golgi complex consists of stacks of membranous sacs called cistemae.
b. Diameter of cistemae varies from 0.5 to 1pm.
c. A Golgi complex may have few to several cistemae depending on its function.
d. The thickness and molecular composition of membranes at one end of the stack of a Golgi sac differ from those at the other end.
e. The Golgi sacs show specific orientation in the cell.
f. Each cistema has a forming or ‘cis’ face (cis: on the same side) and maturing or ‘trans’ face (trAnswer:the opposite side).
g. Transport vesicles that pinch off from transitional ER merge with cis face of Golgi cistema and add its contents into the lumen.

2. Location of Golgi complex:
Golgi bodies are usually located near endoplasmic reticulum.

3. Functions of Golgi complex:
a. Golgi body carries out two types of functions, modification of secretions of ER and production of its own secretions.
b. Cistemae contain specific enzymes for specific functions.
c. Refining (modification) of product takes place in a sequential manner.
d. For example, certain sugar component is added or removed from glycolipids and glycoproteins that are brought from ER, thus forming a variety of products.
e. Golgi bodies also manufacture their own products. Golgi bodies in many plant cells produce non-cellulose polysaccharides like pectin.
f. Manufactured or modified, all products of Golgi complex leave cistemae from trans face as transport vesicles.

Question 29.
Write a note on lysosomes and make a list of lysosomal enzymes.
Answer:
Lysosomes:

1. Lysosomes are considered as dismantling and restructuring units of a cell.
2. These are membrane bound vesicles containing hydrolytic enzymes. The enzymes in lysosomes are used by most eukaryotic cells to digest (hydrolyse) macromolecules.
3. The lysosomal enzymes show optimal activity in acidic pH.
4. Lysosomes arise from Golgi associated endoplasmic reticulum.
5. Lysosomes are polymorphic in nature and are classified as primary lysosomes, secondary or hybrid lysosomes, residual body and autophagic vesicle.
6. The list of lysosomal enzymes includes:
   All types of hydrolases viz, amylases, proteases and lipases.

Question 30.
“Lysosomes are polymorphic in nature.” Justify the statement.
Answer:

1. Lysosomes are classified as, Primary lysosomes; which are nothing but membrane bound vesicles in which enzymes are in inactive state.
2. Secondary lysosomes or hybrid lysosomes, which are formed by fusion of lysosome with endocytic vesicle containing materials to be digested, represented as heterophagic vesicle. This is larger in size than primary lysosome.
3. When organic molecules or membrane bound old cell organelle to be recycled fuses with primary lysosome, autophagic vesicles are formed.
4. Residual body is the vesicle containing undigested remains left over in the heterophagic vesicle after releasing the products of digestion in the cytosol. Hence, lysosomes are polymorphic in nature.

Question 31.
“Lysosomes are called suicide bags of the cells”. Why?
Answer:

1. Lysosomes which bring about digestion of cell’s own organic material like a damaged cell organelle are called autophagic vesicle (suicide bags).
2. An autophagic vesicle essentially consists of lysosome fused with membrane bound old cell organelle or organic molecules to be recycled.
3. Thus, lysosomes are capable of destructing all kinds of material in the cell. Therefore, can digest its own cell organelles due to presence of lysosome. Hence, lysosomes are also called as suicide bags.

Question 32.
Write a note on vacuoles.
Answer:
The organelle which helps in maintaining turgidity of the cell and a proper internal balance of cellular contents is known as vacuole.

1. The vacuoles are bound by semipermeable membrane, called tonoplast membrane. This membrane helps in maintaining the composition of vacuolar fluid (cell sap), different from that of the cytosol.
2. Composition of cell sap differs in different types of cells.
3. In vacuoles along with excretory products other compounds are stored that are harmful or unpalatable to herbivores, thereby protecting the plants.
4. Attractive colours of the petals are due to storage of such pigments in vacuoles.
5. Generally, there are two or three permanent vacuoles in a plant cell.
6. In some large plant cells, a single large vacuole occupies the central part of the cell. It is called central vacuole. In such cells, vacuole can occupy about 90% of the total volume of the cell.
7. The cell sap of central vacuole is a store house of various ions and thus is hypertonic to cytosol.
8. Small vacuoles in seeds of certain plants store organic materials like proteins.
9. In animal cells, they are few in number and smaller in size.
10. Intake of food or foreign particle by phagocytosis involves formation of food vacuole.

Question 33.
What is the function of contractile vacuole in Paramoecium?
Answer:
Contractile vacuole performs excretion and osmoregulation in fresh water unicellular forms like Paramoecium.

Question 34.
What are microbodies? Mention their types and functions.
Answer:
Microbodies are minute membrane bound sacs found in both plant and animal cells. Microbodies contain various types of enzymes based on which they are classified into following types:
1. Sphaerosomes:
a. These are found mainly in cells involved in synthesis and storage of fats. For e. g. endosperm of oil seeds.
b. The membrane of sphaerosome is half unit membrane i.e. this membrane has only one phospholipid layer.

2. Peroxisomes:
a. Peroxisomes contain enzymes that remove hydrogen atoms from substrate and produce toxic hydrogen peroxide by utilisation of oxygen.
b. At the same time peroxisome also contains enzymes that convert toxic H₂0₂ to water. Conversion of toxic substances like alcohol takes place in liver cells by peroxisomes.

Question 35.
Draw a neat and labelled diagram and explain the functions of glyoxysomes.
Answer:

Glyoxysomes are membrane bound organelles containing enzymes that convert fatty acids to sugar. They are observed in cells of germinating seeds where the cells utilize sugar (formed by conversion of stored fatty acids) till it starts photosynthesising on its own.

Question 36.
Describe the structure of mitochondria.
Answer:
Mitochondrion is known as the power house of the cell. It plays significant role in aerobic respiration. Mitochondria are absent in prokaryotic cells and red blood corpuscles (RBCs).

The structure of mitochondrion:

1. Shape of the mitochondria may be oval or spherical or like spiral strip.
2. It is a double membrane bound organelle.
3. Outer membrane is permeable to various metabolites due to presence of a protein-Porin or Parson’s particles.
4. Inner membrane is selectively permeable to few substances only.
5. Both membranes are separated by intermembrane space.
6. Inner membrane shows several finger like or plate like folds called as cristae which bears numerous particles oxysomes and cytochromes / electron carriers.
7. Inner membrane encloses a cavity called inner chamber, containing a fluid-matrix.
8. Matrix contains few coils of circular DNA, RNA, 70S types of ribosomes, lipids and various enzymes of Krebs’ cycle and other pathways.

Question 37.
Label the diagram and write down the details of concept in your words.

Answer:
Mitochondrion is known as the power house of the cell. It plays significant role in aerobic respiration. Mitochondria are absent in prokaryotic cells and red blood corpuscles (RBCs).

The structure of mitochondrion:

1. Shape of the mitochondria may be oval or spherical or like spiral strip.
2. It is a double membrane bound organelle.
3. Outer membrane is permeable to various metabolites due to presence of a protein-Porin or Parson’s particles.
4. Inner membrane is selectively permeable to few substances only.
5. Both membranes are separated by intermembrane space.
6. Inner membrane shows several finger like or plate like folds called as cristae which bears numerous particles oxysomes and cytochromes / electron carriers.
7. Inner membrane encloses a cavity called inner chamber, containing a fluid-matrix.
8. Matrix contains few coils of circular DNA, RNA, 70S types of ribosomes, lipids and various enzymes of Krebs’ cycle and other pathways.

Question 38.
Identify and label the following structure. Write a note on it.

Answer:
1. The given structure is of Oxysome/ F₁ Particle.
2. A: Head (F₁)
B: Pedicel
C: Foot (Base / F₀)
3. Structure of Oxysome:
a. Inner membrane of mitochondria bears numerous particles called as Oxysomes (F₁ – F_o / Fernandez – Moran Elementary particles / Mitochondrial particles).
b. Each particle consists of head, stalk (Pedicel) and base.
c. Head (F₁) / lollipop head faces towards matrix and foot (F0) is embedded in inner membrane.
d. Head acts as an enzyme ATP synthase and foot (base) as proton channel. Oxysomes are involved in proton pumping and ATP synthesis.

Question 39.
What are plastids?
Answer:
Plastids are double membraned organelles containing DNA, RNA and 70S ribosomes.

Question 40.
Draw a labelled diagram of the organelle which plays a significant role in synthesis of starch in plants. Write a note on its structure.
Answer:
Chloroplast plays a significant role in synthesis of starch in plants.
Structure of chloroplast:

1. In plants, chloroplast is found mainly in mesophyll of leaf.
2. Chloroplast is lens shaped but it can also be oval, spherical, discoid or ribbon like.
3. A cell may contain single large chloroplast as in Chlamydomonas or there can be 20 to 40 chloroplasts per cell as seen in mesophyll cells.
4. Chloroplasts contain green pigment called chlorophyll along with other enzymes that help in production of sugar by photosynthesis.
5. Inner membrane of double membraned chloroplast is comparatively less permeable.
6. Inside the cavity of inner membrane, there is another set of membranous sacs called thylakoids.
7. Thylakoids are arranged in the form of stacks called grana (singular: granum).
8. The grana are connected to each other by means of membranous tubules called stroma lamellae.
9. Space outside thylakoids is filled with stroma.
10. The stroma and the space inside thylakoids contain various enzymes essential for photosynthesis.
11. Stroma of chloroplast contains DNA and ribosomes (70S).

Question 41.
Insulin is the protein hormone synthesized by pancreatic cells. Name the component that performs the role of protein factory and draw their labelled structure as seen in prokaryotes and eukaryotes.
Answer:
Ribosomes are the protein factories that synthesize proteins using genetic information.

Question 42.
Give the detailed information on ribosomes found in eukaryotic cell.
Answer:

1. Ribosomes are protein factories of cell and were first observed as dense particles in electron micrograph of a cell by scientist Palade in 1953.
2. Ribosomes lack membranous covering around them and are made up of Ribosomal RNA and proteins.
3. In a eukaryotic cell, ribosomes are present in mitochondria, plastids (in plant cells) and in cytosol.
4. Ribosomes are either found attached to outer surface of Rough Endoplasmic Reticulum and nuclear membrane or freely suspended in cytoplasm.
5. Both are of 80S type. Each ribosome is made up of two subunits- a large (60S) and a small (40S) subunit.
6. Bound ribosomes generally produce proteins that are transported outside the cell after processing in ER and Golgi body. e.g. Bound ribosomes of acinar cells of pancreas produce pancreatic digestive enzymes.
7. Free ribosomes come together and form chains called polyribosomes for protein synthesis.
8. Free ribosomes generally produce enzymatic proteins that are used up in cytoplasm, like enzymes required for breakdown of sugar.
9. Both types of ribosomes (bound and free) can interchange position and function.
10. Number of ribosomes is high in cells actively engaged in protein synthesis.

Question 43.
What is Svedberg unit?
Answer:
The particle size of ribosomes is measured in terms of Svedberg unit (S). It is a measure of sedimentation rate of a particle in ultracentrifuge. It is thus a measure of density and size of a particle. 1S = 10^-13 sec.

Question 44.
Describe the structure of nucleus.
Answer:
Nucleus is known as the master cell organelle as it regulates various metabolic activities through synthesis of various proteins and enzymes.
The nucleus in eukaryotic cell is made up of nuclear envelope, nucleoplasm, nucleolus and chromatin network.
1. Nuclear envelope:
a. Nuclear envelope is a double layered delimiting membrane of nucleus.
b. Two membranes are separated from each other by perinuclear space (10 to 50nm).
c. Outer membrane is connected with endoplasmic reticulum at places and harbours ribosomes on it.
d. The inner membrane is lined by nuclear lamina- a network of protein fibres that helps in maintaining shape of the nucleus.
e. The two membranes along with perinuclear space help in separating nucleoplasm from cytoplasm. However, nuclear membrane is not continuous.
f. There are small openings called nucleopores on the nuclear membrane.
g. The nucleopores are guarded by pore complexes which regulate flow of substances from nucleus to cytoplasm and in reverse direction.

2. Nucleoplasm or karyolymph:
a. The nucleoplasm or karyolymph contains various substances like nucleic acids, protein molecules, minerals and salts.
b. It contains chromatin network and nucleolus.

3. Nucleolus:
a. Nucleolus is made up of rRNA and ribosomal proteins and it is known as the site of ribosome biogenesis.
b. The rRNA and ribosomal proteins are transported to cytoplasm and are assembled together to form ribosomes.
c. Depending on synthetic activity of a cell, there are one or more nucleoli present in the nucleoplasm. For e.g. cells of oocyte contain large nucleolus whereas sperm cells contain small inconspicuous one.
d. Nucleolus appear as dense spherical body present near chromatin network.

Question 45.
Write the functions of the controlling unit of the cell.
Answer:
Nucleus is known as the controlling unit of the cell.
Functions of the nucleus:
1. The nucleus contains entire genetic information; hence play important role in heredity and variation.
2. It is the site for synthesis of DNA, RNA and ribosomes.
3. It plays important role in protein synthesis.

Question 46.
Write a note on chromatin material.
Answer:

1. Nucleus contains genetic information in the form of chromosomes which are DNA molecules associated with proteins.
2. In a non-dividing cell, the chromosomes appear as thread like network and cannot be identified individually. This network is called chromatin material.
3. The chromatin material contains DNA, histone and non-histone proteins and RNA.
4. In some regions of chromatin, DNA is more and is genetically active called euchromatin.
5. Some regions that contain more of proteins and less DNA and are genetically inert, are called
   heterochromatin.

[Note: Heterochromatin is a region in chromatin that is highly compacted during interphase and is generally not accessible for transcription of genes.]

Question 47.
What is the significance of having constant chromosome number in a species?
Answer:
Constant chromosome number in a species is important in phylogenetic studies.

Question 48.
Explain the cytoskeletal system of a cell.
Answer:

1. The cytoskeleton is a supportive structure built from microtubules, intermediate filaments, and
   microfilaments.
2. Microtubules are made up of protein- tubulin.
3. Microfilaments are made up of actin.
4. Intermediate filaments are composed of fibrous proteins.

Question 49.
Compete the following concept map representing the functions of cytoskeleton.
Answer:

Question 50.
Explain in detail the structures of components that help in locomotion of unicellular organisms.
Answer:

1. Cilium or flagellum helps in locomotion of unicellular organisms.
2. They consist of basal body, basal plate and shaft.
3. Basal body is placed in outer part of cytoplasm. It is derived from centriole. It has nine peripheral triplets of fibrils.
4. Shaft is exposed part of cilia or flagella. It consists of two parts- sheath and axoneme.
5. Sheath is covering membrane of cilium or flagellum.
6. Core called axoneme possesses 11 fibrils (microtubules) running parallel to long axis.
7. It shows 9 peripheral doublet microtubules and two single central microtubules (9+2).
8. The central tubules are enclosed by central sheath.
9. This sheath is connected to one of the tubules of peripheral doublets by a radial spoke.
10. Central tubules are connected to each other by bridges.
11. The peripheral doublet microtubules are connected to each other through linkers or inter-doublet bridge.

Question 51.
Draw a labelled diagram of the structure of cilia.
Answer:
Cilia act as oars causing movement of cell.

Question 52.
Spindle apparatus is formed during cell division. Write the information on the components of cell which help in formation of this.
Answer:

1. Centrioles and centrosomes play significant role in formation of spindle apparatus during cell division.
2. Centrosome is usually found near the nucleus of an animal cell.
3. It contains a pair of cylindrical structures called centrioles.
4. The cylinder (centriole) are perpendicular to each other and are surrounded by amorphous substance called pericentriolar material.
5. Each cylinder of centriole is made up of nine sets of triplet microtubules made up of tubulin.
6. Evenly spaced triplets are connected to each other by means of non-tubulin proteins.
7. At the proximal end of centriole, there is a set of tubules called hub.
8. The peripheral triplets are connected to hub by means of radial spokes. Due to this proximal end of centriole looks like a cartwheel.
9. Centriole forms basal body of cilia and flagella.

Question 53.
Draw a labelled diagram of the structure of centriole.
Answer:

Question 54.
Match the column I with column II.

Column I	Column II
1. Mitochondria	(a) Synthesis of protein
2. Nucleus	(b) Photosynthesis
3. Chloroplast	(c) Respiration
4. Ribosomes	(d) Nucleoplasm

Answer:

Column I	Column II
1. Mitochondria	(c) Respiration
2. Nucleus	(d) Nucleoplasm
3. Chloroplast	(b) Photosynthesis
4. Ribosomes	(a) Synthesis of protein

Question 55.
Distinguish between Plant cell and Animal cell.
Answer:

Plant cell	Animal cell
(a) Cell wall is present.	Cell wall is absent.
(b) Plastids present.	Plastids absent.
(c) Chloroplast present.	Chloroplast absent.
(d) Centrioles are present only in lower plant forms.	Centrioles are present in all animal cells.
(e) Lysosomes absent.	Lysosomes present in all animal cells.
(f) Two or three large and permanent vacuoles.	Small and temporary vacuoles are present.
(g) Carbohydrates stored as starch.	Carbohydrates stored as glycogen.

Question 56.
Label the A, B, C, and D in above diagram and write the functions of organelles A and B.
Answer:
1. A: Mitochondria B: Endoplasmic Reticulum
C: Golgi complex D: Amyloplast
2. Functions of Mitochondria: Mitochondrion is known as the power house of the cell. It plays significant role in aerobic respiration. Mitochondria are absent in prokaryotic cells and red blood corpuscles (RBCs).
3. Functions of Endoplasmic Reticulum: Refer Q.33.

Question 57.
Draw a labelled diagram of an animal cell.
Answer:

Question 58.
Classify the following organelles / cellular components on the basis of presence or absence in prokaryotic and eukaryotic cells.
(Ribosomes, Nucleus, Plasma membrane, Mitochondria, mRNA, Endoplasmic Reticulum, Golgi complex, Centrioles, Nucleoid)
Answer:

Prokaryotic cell	Ribosomes, Plasma membrane, mRNA, Nucleoid
Eukaryotic cell	Ribosomes, Plasma membrane, mRNA, Nucleus, Mitochondria, Endoplasmic Reticulum, Golgi complex, Centrioles

Question 59.
Apply Your Knowledge

Question 1.
After learning organization of cell, to test one of the postulates of cell theory, Ananya requested her teacher to guide and allow her to perform a small experiment. The aim of the experiment Avas to form new cells in the laboratory using isolated cellular organelles from other cells. Though Ananya did not succeed to form new cells, teacher-guided and motivated her explaining why experiment performed by them failed.
1. Which postulate Ananya was willing to test and why new cells failed to form from the isolated organelles from other cells?
2. From the above mentioned data could you guess which type of cells they were trying to form whether eukaryote or prokaryote?
Answer:
1. The postulate Ananya was willing to test was, ‘all cells arise from pre-existing cells’. According to this postulate, to form new cells, pre-existing cells are must; therefore, cellular organelles did not form new cells.
2. The cells which Ananya and her teacher were trying to form were eukaryotic cells, as cellular organelles are present in eukaryotes.

Question 2.
A mix bacterial culture was given to different teams of students and was asked to write their observation regarding the shapes of bacterial cells they observed under microscope. Students discussed the characteristics among their respective teams and mentioned major types of shapes they observed.
1. Which types of bacterial shapes were observed by the students?
2. Mention why they were named in a specific manner with respect to their shapes?
Answer:
1. The bacterial shapes observed by the students are cocci, bacilli, vibrios, spirilla.
2. Under microscope, cocci appear spherical shape, bacilli appear rod shape, vibrios appear comma shape and spirilla appear twisted, therefore they are named accordingly.

Question 60.
Quick Review:

Question 61.
Exercise

Question 1.
Define cell.
Answer:
The first microscope was made by two Dutch spectacle makers Hans and Zacharias Janssen.
[Note: The Dutch scientist Anton van Leeuwenhoek made microscopes capable of magnifying single-celled organisms in a drop of pond water.]

Question 2.
Write a note on microscope.
Answer:
Cell is defined as a structural and functional unit of life of all living organisms capable of independent existence and can perform all functions of life.

Question 3.
Write a short note on totipotency.
Answer:

1. Totipotency (totus – entire, potential – power) is the capacity or the potential of living nucleated cell, to differentiate into any other type of cell and thus, can form a complete new organism.
2. A cell is totipotent as it has the entire genetic information of the organism stored in its nucleus.
3. Embryonic animal cells are totipotent and are termed as stem cells.
4. Stem cells are used in curing many diseases. Therefore, they have great potential for medical applications.

Question 4.
What are the characteristics of cells in which genetic material is known as nucleoid?
Answer:
General characteristics of prokaryotic cell:
1. Prokaryotic cells are primitive type of cells.
2. It does not have membrane bound cell organelles (like endoplasmic reticulum, Golgi complex, mitochondria, etc.) and well-defined nucleus (nuclear membrane is absent).
3. Genetic material is in the form of nucleoid.

4. Cell envelope:
a. Prokaryotic cell has chemically complex protective cell envelope having glycocalyx, cell wall and plasma membrane.
b. In some bacteria, glycocalyx occurs in the form of a slime layer (loose sheath). Other bacteria may have a thick and tough covering called capsule. It helps in protection of bacterial cell.

5. Cell wall:
The Gram-positive bacteria show presence of peptidoglycan layer in the cell wall and Gram-negative bacteria show presence of murein in the cell wall. It gives mechanical strength to the cell.
[Note: In Gram-negative bacteria, cell wall is made up of two layers; inner layer of Murein or peptidoglycan and outer layer of Lipopolysaccharides.]

6. Cell membrane:
a. It is the innermost covering of the cell envelope, chemically composed of lipids and proteins.
b. It helps in intercellular communication.
c. Cell membrane shows infoldings called mesosomes which help in cell wall formation, cellular respiration and DNA replication.
d. The cyanobacteria show longer extensions called as chromatophores which carry photosynthetic pigments.

7. In motile bacteria either cilia or flagella are found. Both are driven by rotatory movement produced by basal body (which works as motor) of flagellum. Other parts of flagellum are filament and hook.

8. Some other surface projections are the tubular pili (which help in inter-cellular communication) and fimbriae (for clinging to support).

9. Ribosomes:
Bacterial cell cytoplasm contains dense particles called ribosomes which help in protein synthesis. Ribosomes are 70S type (composed of a larger sub-unit 50S and + smaller sub-unit 30S).

Question 5.
Which technique is used to differentiate between Gram positive and Gram negative bacteria?
Answer:
The Gram-positive bacteria show presence of peptidoglycan layer in the cell wall and Gram-negative bacteria show presence of murein in the cell wall. The technique used for differentiating bacterial cells is Gram staining.
[Note: Murein is similar to peptidoglycan in structure and function. It is present in the cell walls of archaebacteria.

Question 6.
What are mesosomes?
Answer:
Cytoplasm does not show streaming movement. Cytoplasm shows streaming movement.

Question 7.
What are the functions of pili and fimbriae?
Answer:
Respiratory enzymes are present on the infoldings of the plasma membrane called mesosomes. Respiratory enzymes are present within mitochondria.
e-g- Cyanobacteria (Blue green algae) and bacteria. Algae, fungi, plants and animals.

Question 8.
Enlist the organelles present in eukaryotic cells.
Answer:
It also contains various membrane bound cell organelles like endoplasmic reticulum, Golgi complex, mitochondria, plastids, nucleus, microbodies and cytoskeletal elements like microtubules.

Question 9.
Who proposed the fluid-mosaic model?
Answer:
Fluid mosaic model was proposed by Singer and Nicholson (1972).

Question 10.
What are nuclear membrane?
Answer:
Nucleus is known as the master cell organelle as it regulates various metabolic activities through synthesis of various proteins and enzymes.
The nucleus in eukaryotic cell is made up of nuclear envelope, nucleoplasm, nucleolus and chromatin network.
1. Nuclear envelope:
a. Nuclear envelope is a double layered delimiting membrane of nucleus.
b. Two membranes are separated from each other by perinuclear space (10 to 50nm).
c. Outer membrane is connected with endoplasmic reticulum at places and harbours ribosomes on it.
d. The inner membrane is lined by nuclear lamina- a network of protein fibres that helps in maintaining shape of the nucleus.
e. The two membranes along with perinuclear space help in separating nucleoplasm from cytoplasm. However, nuclear membrane is not continuous.
f. There are small openings called nucleopores on the nuclear membrane.
g. The nucleopores are guarded by pore complexes which regulate flow of substances from nucleus to cytoplasm and in reverse direction.

Question 11.
Name two types of chromatin.
Answer:
1. In some regions of chromatin, DNA is more and is genetically active called euchromatin.
2. Some regions that contain more of proteins and less DNA and are genetically inert, are called heterochromatin.

Question 12.
What are lysosomes commonly known as?
Answer:
1. Lysosomes which bring about digestion of cell’s own organic material like a damaged cell organelle are called autophagic vesicle (suicide bags).
2. An autophagic vesicle essentially consists of lysosome fused with membrane bound old cell organelle or organic molecules to be recycled.
3. Thus, lysosomes are capable of destructing all kinds of material in the cell. Therefore, can digest its own cell organelles due to presence of lysosome. Hence, lysosomes are also called as suicide bags.

Question 13.
What are ribosomes?
Answer:
Ribosomes are the protein factories that synthesize proteins using genetic information.

Question 14.
What are glyoxysomes? Where do they occur?
Answer:
Glyoxysomes are membrane bound organelles containing enzymes that convert fatty acids to sugar. They are observed in cells of germinating seeds where the cells utilize sugar (formed by conversion of stored fatty acids) till it starts photosynthesising on its own.

Question 15.
Sketch and label the fluid mosaic model of cell membrane.
Answer:
Fluid mosaic model:

1. Fluid mosaic model was proposed by Singer and Nicholson (1972).
2. This model states that plasma membrane is made up of phospholipid bilayer and proteins.
3. Proteins are embedded in the lipid membrane like icebergs in the sea of lipids.
4. Phospholipid bilayer is fluid in nature.
5. Quasi-fluid nature of lipid enables lateral movement of proteins. This ability to move within the membrane is measured as fluidity.
6. Based on organization of membrane proteins they are of two types, as:

a. The intrinsic proteins occur at different depths of bilayer i.e. they are tightly bound to the phospholipid bilayer and are embedded in it. They span the entire thickness of the membrane. Therefore, they are known as transmembrane proteins. They form channels for passage of water.
b. The extrinsic or peripheral proteins are found on two surfaces of the membrane i.e. are loosely held to the phospholipid layer and can be easily removed.

Question 16.
State the functions of Endoplasmic reticulum.
Answer:
Smooth endoplasmic reticulum (SER):
1. Depending on cell type, it helps in synthesis of lipids for e.g. Steroid secreting cells of cortical region of adrenal gland, testes and ovaries.
2. Smooth endoplasmic reticulum plays a role in detoxification in the liver and storage of calcium ions (muscle cells).

Rough Endoplasmic Reticulum (RER):

1. Rough ER is primarily involved in protein synthesis. For e.g. Pancreatic cells synthesize the protein insulin in the ER.
2. These proteins are secreted by ribosomes attached to rough ER and are called secretory proteins. These proteins get wrapped in membrane that buds off from transitional region of ER. Such membrane bound proteins depart from ER as transport vesicles.
3. Rough ER is also involved in formation of membrane for the cell. The ER membrane grows in place by addition of membrane proteins and phospholipids to its own membrane. Portions of this expanded membrane are transferred to other components of endomembrane system.

Question 17.
Write short note on lysosomes.
Answer:
Lysosomes:

1. Lysosomes are considered as dismantling and restructuring units of a cell.
2. These are membrane bound vesicles containing hydrolytic enzymes. The enzymes in lysosomes are used by most eukaryotic cells to digest (hydrolyse) macromolecules.
3. The lysosomal enzymes show optimal activity in acidic pH.
4. Lysosomes arise from Golgi associated endoplasmic reticulum.
5. Lysosomes are polymorphic in nature and are classified as primary lysosomes, secondary or hybrid lysosomes, residual body and autophagic vesicle.
6. The list of lysosomal enzymes includes:
   All types of hydrolases viz, amylases, proteases and lipases.

Question 18.
Lysosomes are known as suicide bags of the cell. Give reason.
Answer:

1. Lysosomes which bring about digestion of cell’s own organic material like a damaged cell organelle are called autophagic vesicle (suicide bags).
2. An autophagic vesicle essentially consists of lysosome fused with membrane-bound old cell organelle or organic molecules to be recycled.
3. Thus, lysosomes are capable of destructing all kinds of material in the cell. Therefore, can digest its own cell organelles due to presence of lysosome. Hence, lysosomes are also called as suicide bags.

Question 19.
Describe the structure of plant cell wall.
Answer:
In plants, cell wall shows middle lamella, primary wall and secondary wall

1. Middle lamella:
   It is thin and present between two adjacent r cells. It is the first structure formed from cell plate during cytokinesis. It is mainly made up of pectin, calcium and magnesium pectate. Softening of ripe fruit is due to solubilization of pectin.
2. Primary wall:
   In young plant cell, it is capable of growth. It is laid inside to middle lamella.
   It is the only wall seen in meristematic tissue, mesophyll, pith, etc.
3. Secondary wall:
   It is present inner to primary wall. Once the growth of primary wall stops, secondary wall is laid. At some places thickening is absent which leads to formation of pits.

Question 20.
Describe the cell wall of eukaryotic cells and state their function.
Answer:

1. The rigid, protective and supportive covering, outside the cell membrane is called cell wall. It is present in plant cells, fungi and some protists.
2. Algae show presence of cellulose, galactans, mannans and minerals like calcium carbonate in cell wall.
3. In other plants, it is made up of hemicelluloses, pectin, lipids and protein.
4. Microfibrils of plant cell wall show presence of cellulose which is responsible for rigidity.
5. Some of the depositions of cell wall are silica (grass stem), cutin (epidermal walls of land plants), suberin (endodermal cells of root), wax, lignin.
6. Function:
   1. Provides support, rigidity and shape to the cell.
   2. Protects the protoplasm against mechanical injury and infections.

Question 21.
1. Draw neat and labelled diagram of ultrastructure of mitochondria,
2. Explain the structure of mitochondria.
Answer:
Mitochondrion is known as the power house of the cell. It plays significant role in aerobic respiration. Mitochondria are absent in prokaryotic cells and red blood corpuscles (RBCs).

The structure of mitochondrion:

1. Shape of the mitochondria may be oval or spherical or like spiral strip.
2. It is a double membrane bound organelle.
3. Outer membrane is permeable to various metabolites due to presence of a protein-Porin or Parson’s particles.
4. Inner membrane is selectively permeable to few substances only.
5. Both membranes are separated by intermembrane space.
6. Inner membrane shows several finger like or plate like folds called as cristae which bears numerous particles oxysomes and cytochromes / electron carriers.
7. Inner membrane encloses a cavity called inner chamber, containing a fluid-matrix.
8. Matrix contains few coils of circular DNA, RNA, 70S types of ribosomes, lipids and various enzymes of Krebs’ cycle and other pathways.

Question 22.
1. Draw neat and labelled diagram of structure of plasma membrane proposed by Singer and Nicholson,
2. Write any two functions of plasma membrane.
Answer:
1. Fluid mosaic model:

1. Fluid mosaic model was proposed by Singer and Nicholson (1972).
2. This model states that plasma membrane is made up of phospholipid bilayer and proteins.
3. Proteins are embedded in the lipid membrane like icebergs in the sea of lipids.
4. Phospholipid bilayer is fluid in nature.
5. Quasi-fluid nature of lipid enables lateral movement of proteins. This ability to move within the membrane is measured as fluidity.
vi. Based on organization of membrane proteins they are of two types, as:
a. The intrinsic proteins occur at different depths of bilayer i.e. they are tightly bound to the phospholipid bilayer and are embedded in it. They span the entire thickness of the membrane. Therefore, they are known as transmembrane proteins. They form channels for passage of water.
b. The extrinsic or peripheral proteins are found on two surfaces of the membrane i.e. are loosely held to the phospholipid layer and can be easily removed.

(ii)

1. The significant function of plasma membrane is transport of molecules across it. Plasma membrane is selectively permeable.

2. Passive transport:
a. Many molecules move across the membrane without spending energy.
b. Some molecules move by simple diffusion along the concentration gradient i.e. from higher to lower concentration.
c. Neutral molecules may move across the membrane by the process of simple diffusion.
d. Water may also move by osmosis.

3. Active transport:
a. Few ions or molecules are transported against concentration gradient i.e. from lower to higher concentration.
b. This requires energy, hence ATP is utilized. As such a transport is an energy dependent process in which ATP is utilized, it is called Active transport e.g. Na⁺ /K⁺ pump.
c. Polar molecules cannot pass through non-polar lipid bilayer. Therefore, they require carrier proteins to facilitate their transport across the membrane.

Question 23.
1. Draw neat and labelled diagram of nucleus,
2. Write a short note on nuclear envelope.
Answer:
Nucleus is known as the master cell organelle as it regulates various metabolic activities through synthesis of various proteins and enzymes.
The nucleus in eukaryotic cell is made up of nuclear envelope, nucleoplasm, nucleolus and chromatin network.
1. Nuclear envelope:
a. Nuclear envelope is a double layered delimiting membrane of nucleus.
b. Two membranes are separated from each other by perinuclear space (10 to 50nm).
c. Outer membrane is connected with endoplasmic reticulum at places and harbours ribosomes on it.
d. The inner membrane is lined by nuclear lamina- a network of protein fibres that helps in maintaining shape of the nucleus.
e. The two membranes along with perinuclear space help in separating nucleoplasm from cytoplasm. However, nuclear membrane is not continuous.
f. There are small openings called nucleopores on the nuclear membrane.
g. The nucleopores are guarded by pore complexes which regulate flow of substances from nucleus to cytoplasm and in reverse direction.

2. Nucleoplasm or karyolymph:
a. The nucleoplasm or karyolymph contains various substances like nucleic acids, protein molecules, minerals and salts.
b. It contains chromatin network and nucleolus.

3. Nucleolus:
a. Nucleolus is made up of rRNA and ribosomal proteins and it is known as the site of ribosome biogenesis.
b. The rRNA and ribosomal proteins are transported to cytoplasm and are assembled together to form ribosomes.
c. Depending on synthetic activity of a cell, there are one or more nucleoli present in the nucleoplasm. For e.g. cells of oocyte contain large nucleolus whereas sperm cells contain small inconspicuous one.
d. Nucleolus appear as dense spherical body present near chromatin network.

Question 24.
Which components of a cell help in formation of spindle apparatus formed during cell division?
Answer:
Centrioles and centrosomes play significant role in formation of spindle apparatus during cell division.

Question 25.
Write a note on control unit of a cell.
Answer:
a. Nucleus contains the genetic material of an organism.
b. This genetic material is present in the form of Deoxyribonucleic Acid (DNA) which is responsible for synthesis of various proteins and enzymes.
c. These proteins and enzymes in turn regulate metabolic activities of the cells.
Therefore, nucleus is considered as control unit of a cell.

Question 26.
What are the various types of plastids? ii. Describe the chemical composition and functions of eukaryotic cell wall.
Answer:
(i)

1. Plastids are classified according to the pigments present in it. Three main types of plastids are – leucoplasts, chromoplasts and chloroplasts.
2. Leucoplasts do not contain any photosynthetic pigments they are of various shapes and sizes. These are meant for storage of nutrients:
a. Amyloplasts store starch. b. Elaioplasts store oils. c. Aleuroplasts store proteins.
3. Chromoplasts contain pigments like carotene and xanthophyll etc.
a. They impart yellow, orange or red colour to flowers and fruits.
b. These plastids are found in the coloured parts of flowers and fruits.
iv. Chloroplasts are plastids containing green pigment chlorophyll along with other enzymes that help in production of sugar by photosynthesis. They are present in plants, algae and few protists like Euglena.

(ii)

1. The rigid, protective and supportive covering, outside the cell membrane is called cell wall. It is present in plant cells, fungi and some protists.
2. Algae show presence of cellulose, galactans, mannans and minerals like calcium carbonate in cell wall.
3. In other plants, it is made up of hemicelluloses, pectin, lipids and protein.
4. Microfibrils of plant cell wall show presence of cellulose which is responsible for rigidity.
5. Some of the depositions of cell wall are silica (grass stem), cutin (epidermal walls of land plants), suberin (endodermal cells of root), wax, lignin.
6. Function:
   Provides support, rigidity and shape to the cell.
   Protects the protoplasm against mechanical injury and infections.

Question 27.
1. Explain the structure of ribosomes in detail.
2. What are sphaerosomes?
3. What is totipotency?
Answer:
(i) Ribosomes are the protein factories that synthesize proteins using genetic information.

1. Ribosomes are protein factories of cell and were first observed as dense particles in electron micrograph of a cell by scientist Palade in 1953.
2. Ribosomes lack membranous covering around them and are made up of Ribosomal RNA and proteins.
3. In a eukaryotic cell, ribosomes are present in mitochondria, plastids (in plant cells) and in cytosol.
4. Ribosomes are either found attached to outer surface of Rough Endoplasmic Reticulum and nuclear membrane or freely suspended in cytoplasm.
5. Both are of 80S type. Each ribosome is made up of two subunits- a large (60S) and a small (40S) subunit.
6. Bound ribosomes generally produce proteins that are transported outside the cell after processing in ER and Golgi body. e.g. Bound ribosomes of acinar cells of pancreas produce pancreatic digestive enzymes.
7. Free ribosomes come together and form chains called polyribosomes for protein synthesis.
8. Free ribosomes generally produce enzymatic proteins that are used up in cytoplasm, like enzymes required for breakdown of sugar.
9. Both types of ribosomes (bound and free) can interchange position and function.
10. Number of ribosomes is high in cells actively engaged in protein synthesis.

(ii) Sphaerosomes:
a. These are found mainly in cells involved in synthesis and storage of fats. For e. g. endosperm of oil seeds.
b. The membrane of sphaerosome is half unit membrane i.e. this membrane has only one phospholipid layer.

(iii) 1. Totipotency (totus – entire, potential – power) is the capacity or the potential of living nucleated cell, to differentiate into any other type of cell and thus, can form a complete new organism.
2. A cell is totipotent as it has the entire genetic information of the organism stored in its nucleus.
3. Embryonic animal cells are totipotent and are termed as stem cells.
4. Stem cells are used in curing many diseases. Therefore, they have great potential for medical applications.

Question 28.
1. Give any two functions of each of the following:
a. Golgi complex
b. Lysosomes
2. What are the major differences between eukaryotic and prokaryotic cells? Write any two points.
3. Explain the structure of cilia and flagella.
Answer:
1. a. Functions of Golgi complex:
a. Golgi body carries out two types of functions, modification of secretions of ER and production of its own secretions.
b. Cistemae contain specific enzymes for specific functions.
c. Refining (modification) of product takes place in a sequential manner.
d. For example, certain sugar component is added or removed from glycolipids and glycoproteins that are brought from ER, thus forming a variety of products.
e. Golgi bodies also manufacture their own products. Golgi bodies in many plant cells produce non-cellulose polysaccharides like pectin.
f. Manufactured or modified, all products of Golgi complex leave cistemae from trans face as transport vesicles.
b. i. Lysosomes which bring about digestion of cell’s own organic material like a damaged cell organelle are called autophagic vesicle (suicide bags).
2. An autophagic vesicle essentially consists of lysosome fused with membrane bound old cell organelle or organic molecules to be recycled.
3. Thus, lysosomes are capable of destructing all kinds of material in the cell. Therefore, can digest its own cell organelles due to presence of lysosome. Hence, lysosomes are also called as suicide bags.

(ii)

1. Cilium or flagellum helps in locomotion of unicellular organisms.
2. They consist of basal body, basal plate and shaft.
3. Basal body is placed in outer part of cytoplasm. It is derived from centriole. It has nine peripheral triplets of fibrils.
4. Shaft is exposed part of cilia or flagella. It consists of two parts- sheath and axoneme.
5. Sheath is covering membrane of cilium or flagellum.
6. Core called axoneme possesses 11 fibrils (microtubules) running parallel to long axis.
7. It shows 9 peripheral doublet microtubules and two single central microtubules (9+2).
8. The central tubules are enclosed by central sheath.
9. This sheath is connected to one of the tubules of peripheral doublets by a radial spoke.
10. Central tubules are connected to each other by bridges.
11. The peripheral doublet microtubules are connected to each other through linkers or inter-doublet bridge.
    Cilia act as oars causing movement of cell.

Question 29.
Write a note on glycoprotein molecules found on membranes of RBC.
Answer:
Glycoproteins are protein molecules modified within the Golgi complex by having a short sugar chain (polysaccharide) attached to them.
The polysaccharide part of glycoproteins located on the surfaces of red blood cells acts as the antigen responsible for determining the blood group of an individual.
Different polysaccharide part of glycoproteins act as different type of antigens that determine the blood groups.
Four types of blood groups A, B, AB, and O are recognized on the basis of presence or absence of these antigens.

Question 30.
Describe in detail the structure of nucleus.
Answer:
Nucleus is known as the master cell organelle as it regulates various metabolic activities through synthesis of various proteins and enzymes.
The nucleus in eukaryotic cell is made up of nuclear envelope, nucleoplasm, nucleolus and chromatin network.
1. Nuclear envelope:
a. Nuclear envelope is a double layered delimiting membrane of nucleus.
b. Two membranes are separated from each other by perinuclear space (10 to 50nm).
c. Outer membrane is connected with endoplasmic reticulum at places and harbours ribosomes on it.
d. The inner membrane is lined by nuclear lamina- a network of protein fibres that helps in maintaining shape of the nucleus.
e. The two membranes along with perinuclear space help in separating nucleoplasm from cytoplasm. However, nuclear membrane is not continuous.
f. There are small openings called nucleopores on the nuclear membrane.
g. The nucleopores are guarded by pore complexes which regulate flow of substances from nucleus to cytoplasm and in reverse direction.

2. Nucleoplasm or karyolymph:
a. The nucleoplasm or karyolymph contains various substances like nucleic acids, protein molecules, minerals and salts.
b. It contains chromatin network and nucleolus.

3. Nucleolus:
a. Nucleolus is made up of rRNA and ribosomal proteins and it is known as the site of ribosome biogenesis.
b. The rRNA and ribosomal proteins are transported to cytoplasm and are assembled together to form ribosomes.
c. Depending on synthetic activity of a cell, there are one or more nucleoli present in the nucleoplasm. For e.g. cells of oocyte contain large nucleolus whereas sperm cells contain small inconspicuous one.
d. Nucleolus appear as dense spherical body present near chromatin network.

Question 31.
Observe the diagram given below and answer the questions based on it.
1. Identify the structure labelled as ‘A’.
2. Mention the two types of the given cell organelle.
3. Which type of ribosomes would be seen on the membrane of the given structure.
Answer:
1. Smooth endoplasmic reticulum (SER):

1. Depending on cell type, it helps in synthesis of lipids for e.g. Steroid secreting cells of cortical region of adrenal gland, testes and ovaries.
2. Smooth endoplasmic reticulum plays a role in detoxification in the liver and storage of calcium ions (muscle cells).

Rough Endoplasmic Reticulum (RER):
1. Rough ER is primarily involved in protein synthesis. For e.g. Pancreatic cells synthesize the protein insulin in the ER.
2. These proteins are secreted by ribosomes attached to rough ER and are called secretory proteins. These proteins get wrapped in membrane that buds off from transitional region of ER. Such membrane bound proteins depart from ER as transport vesicles.
3. Rough ER is also involved in formation of membrane for the cell. The ER membrane grows in place by addition of membrane proteins and phospholipids to its own membrane. Portions of this expanded membrane are transferred to other components of endomembrane system.

3. Both are of 80S type. Each ribosome is made up of two subunits- a large (60S) and a small (40S) subunit.

Multiple Choice Questions:

Question 1.
Which of the following is the smallest cell?
(A) Red Blood Cell
(B) Plant cell
(C) Mycoplasma
(D) Euglena
Answer:
(C) Mycoplasma

Question 2.
From the following identify the CORRECT range of size of the bacteria.
(A) 0.3 pm to 1 mm
(B) 1 pm to 1mm
(C) 1 nm to 1 pm
(D) 3 pm to 5 pm
[Note: Prokaryotic cells generally range between 1 to 10 pm in size.]
Answer:
(D) 3 pm to 5 pm

Question 3.
Identify the CORRECT statements.
1. Nerve cells are the longest cells.
2. The concept ‘Omnis cellula-e-cellulla’ was explained by Rudolf Virchow.
3. The cell theory was proposed by Nicolson and Singer.
(A) Statements i and ii are correct.
(B) Statements ii and iii are correct.
(C) Statements i and iii are correct.
(D) Statements i, ii and iii are correct.
Answer:
(A) Statements i and ii are correct.

Question 4.
New cells generate from
(A) bacterial fermentation
(B) regeneration of old cells
(C) pre-existing cells
(D) abiotic materials
Answer:
(C) pre-existing cells

Question 5.
Mesosonle is produced by the infoldings of
(A) mitochondria
(B) chloroplast
(C) golgi complex
(D) plasma membrane
Answer:
(D) plasma membrane

Question 6.
The ribosomes present in prokaryotic cells is of type.
(A) 30S
(B) 80S
(C) 70S
(D) 50S
Answer:
(C) 70S

Question 7.
Complete the analogy.
F-plasmid: Reproduction :: R-plasmid: ________
(A) Respiration
(B) Resistance against antibiotics
(C) Packaging and transportation
(D) Apposition
Answer:
(B) Resistance against antibiotics

Question 8.
A rigid, supportive and protective outer covering of plasma membrane of fungi is called
(A) cell wall
(B) lamella
(C) plasmodesmata
(D) cell membrane
Answer:
(A) cell wall

Question 9.
The cytoplasmic connections from cell to cell are known as
(A) middle lamella
(B) plasmodesmata
(C) cell membrane system
(D) endoplasmic reticulum
Answer:
(B) plasmodesmata

Question 10.
Due to presence of ________, endoplasmic reticulum is termed as rough endoplasmic reticulum.
(A) cistemae
(B) RNA
(C) ribosomes
(D) tubules
Answer:
(C) ribosomes

Question 11.
Golgi body is absent in
(A) Prokaryotes
(B) Mature mammalian RBC
(C) Akaryotes
(D) All of the above
Answer:
(D) All of the above

Question 12.
Lysosomes are not helpful in
(A) Osteogenesis
(B) Cellular digestion
(C) Metamorphosis
(D) Lipogenesis
Answer:
(D) Lipogenesis

Question 13.
Identify the INCORRECT statements from the following.
1. Lysosomal enzymes do not digest their own membrane proteins.
2. Accidental release of lysosomal enzymes in limited amount does not harm the cell because pH of cytosol is near neutral.
3. Any insufficiency in secretion of lysosomal enzymes leads to disorders e.g. in genetic disorder- Klinefelter syndrome.
iv. Due to insufficiency of protease brain gets impaired resulting from accumulation of fats.
(A) Statements i and ii are incorrect.
(B) Statements i, ii and iii are incorrect.
(C) Statements iii and iv are incorrect.
(D) Statements i, ii and iv are incorrect.
Answer:
(C) Statements iii and iv are incorrect.

Question 14.
Tonoplast is a differentially permeable membrane surrounding the
(A) cytoplasm
(B) vacuole
(C) nucleus
(D) mitochondria
Answer:
(B) vacuole

Question 15.
Which organelle is surrounded by two membranes?
(A) Ribosomes
(B) Peroxisomes
(C) Vacuoles
(D) Mitochondria
Answer:
(D) Mitochondria

Question 16.
F₁ particles are present in
(A) plasmids
(B) mitochondria
(C) chloroplast
(D) ribosomes
Answer:
(B) mitochondria

Question 17.
_______ are green plastids containing green pigment chlorophyll.
(A) Chloroplasts
(B) Leucoplast
(C) Chromoplasts
(D) Xanthophyll
Answer:
(A) Chloroplasts

Question 18.
Select the INCORRECT statement about ribosome.
(A) Each ribosome consists of two sub- units-large and small subunit.
(B) Ribosomes are double membrane bound cell organelles.
(C) Ribosomes are made up of ribosomal RNA and protein.
(D) Ribosomes are involved in protein synthesis.
Answer:
(B) Ribosomes are double membrane bound cell organelles.

Question 19.
The space between the two nuclear membranes is known as
(A) peritonial space
(B) periplasmic space
(C) perinuclear space
(D) none of the above
Answer:
(C) perinuclear space

Question 20.
In eukaryotic cells, the chromosomes are located in
(A) nucleus
(B) nucleolus
(C) golgi complex
(D) lysosomes
Answer:
(A) nucleus

Question 21.
What is the normal chromosome number in humans?
(A) 23
(B) 46
(C) 48
(D) 16
Answer:
(B) 46

Question 22.
During which stage of cell division chromosomes become distinct and can be clearly identified?
(A) Interphase
(B) Prophase
(C) Pachytene
(D) Metaphase
Answer:
(D) Metaphase

Question 23.
Microtubules are made up of protein.
(A) tubulin
(B) fibrion
(C) collagen
(D) myosin
Answer:
(A) tubulin

Competitive Corner:

Question 1.
Match the column I with column II.

Column I	Column II
(a) Golgi apparatus	(i) Synthesis of protein
(b) Lysosomes	(ii) Trap waste and excretory products
(c) Vacuoles	(iii) Formation of glycoproteins and glycolipids
(d) Ribosomes	(iv) Digesting  biomolecules

Choose the right match from options given below:
(A) a-i, b-ii, c-iv, d-iii
(B) a-iii, b-iv, c-ii, d-i
(C) a-iv, b-iii, c-i, d-ii
(D) a-iii, b-ii, c-iv, d-i
Answer:
(B) a-iii, b-iv, c-ii, d-i

Question 2.
The concept of “Omnis cellula – e- cellula” regarding cell division was first proposed by:
(A) Schleiden
(B) Aristotle
(C) Rudolf Virchow
(D) Theodore Schwann
Answer:
(C) Rudolf Virchow

Question 3.
The Golgi complex participates in
(A) respiration in bacteria
(B) formation of secretory vesicles
(C) fatty acid breakdown
(D) activation of amino acid
Answer:
(B) formation of secretory vesicles

Question 4.
Which of the following is true for nucleolus?
(A) It takes part in spindle formation.
(B) It is a membrane-bound structure.
(C) Larger nucleoli are present in dividing cells.
(D) It is a site for active ribosomal RNA synthesis.
Hint: Large nucleoli are found in cells that are actively engaged in protein synthesis. Nucleolus is non-membranous structure.
Answer:
(D) It is a site for active ribosomal RNA synthesis.

Question 5.
Given below are cell organelles and their functions. Select the INCORRECT match.
(A) Lysosome – Phagocytosis
(B) Centriole – Spindle formation
(C) Sphaerosomes – Storage and synthesis of fats
(D) Leucoplast – Photosynthesis
Hint: Leucoplasts store food material.
Answer:
(D) Leucoplast – Photosynthesis

Question 6.
Which of the following cell organelles is responsible for extracting energy from carbohydrates to form ATP?
(A) Lysosome
(B) Ribosome
(C) Chloroplast
(D) Mitochondria
Hint: Glucose (carbohydrate) on complete oxidation from ATP during respiration. The ATP synthesis during carbohydrate oxidation takes place in the Mitochondria (site of aerobic respiration). Mitochondria produce cellular energy in the form of ATP.
Answer:
(D) Mitochondria

Question 7.
Which of the following components provides sticky character to the bacterial cell?
(A) Cell wall
(B) Nuclear membrane
(C) Plasma membrane
(D) Glycocalyx
Hint: In some bacteria, glycocalyx is rich in glycoproteins and could be a loose sheath called as slime layer. This slime layer or glycocalyx imparts sticky character to bacterial cell wall or bacteria.
Answer:
(D) Glycocalyx