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NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

Edited By Shivani Poonia | Updated on Jun 21, 2025 12:24 PM IST

Thermodynamics is an important branch of Chemistry. It deals with energy, its transformation, and the laws governing the entire process of transformation of energy. Have you ever wondered why a hot cup of tea cools down after some time or why ice melts? The only reason behind this kind of phenomenon is thermodynamics. NCERT Chapter 5 plays a very important role in Chemistry as it is the basic chapter that builds the foundations of advanced chapters, and it also helps to explain why changes occur around us.

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This Story also Contains
  1. Download PDF Of NCERT Solutions for Class 11 Chemistry Chapter 5
  2. NCERT Solutions for Class 11 Chemistry Chapter 5 (Exercise Questions with Answers)
  3. Class 11 Chemistry NCERT Chapter 5: Higher Order Thinking Skills (HOTS)
  4. Approach to Solve Questions of Class 11 Chemistry Chapter 5
  5. Topics and Subtopics Covered in the NCERT Textbook
  6. What Extra Should Students Study Beyond the NCERT for JEE/NEET?
  7. NCERT Solutions for Class 11 Chemistry
  8. NCERT solutions for class 11 subject-wise
NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics
NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

In Thermodynamics, there are 22 questions in the exercise. The solutions of thermodynamics are prepared in a very comprehensive manner. These NCERT solutions of class 11 can help you prepare for the class 11 final examination and competitive exams like JEE Mains, NEET, etc. If you are looking for an answer from any other chapter, even from any other class, then go with the NCERT Solutions, where you will get all the answers to the NCERT easily.

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Download PDF Of NCERT Solutions for Class 11 Chemistry Chapter 5

To get all the solved exercise questions, click below on the download PDF icon. In this PDF, you will get detailed solutions to all the questions that are given in the NCERT textbook.

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NCERT Solutions for Class 11 Chemistry Chapter 5 (Exercise Questions with Answers)

Question 5.1 Choose the correct answer. A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of the path
(iii) used to determine pressure-volume work
(iv) whose value depends on temperature only.

Answer :

A state function refers to the equilibrium state of a system. A thermodynamic state function is a quantity whose value is independent of the path.
Like- p,V,T depends on the state of the system, not on the path.

So, the correct option is (ii)

Question 5.2 For the process to occur under adiabatic conditions, the correct condition is:

(i) ΔT=0
(ii) Δp=0
(iii) q = 0
(iv) w = 0

Answer :

In adiabatic conditions, the heat exchange between the system and the surroundings through its boundary is not allowed.
So, the correct option is (iii) q = 0

Question 5.3 The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element

Answer :

The enthalpies of all elements in their standard states are zero.
So, the correct option is (ii)

Question 5.4 ΔU of combustion of methane is XKJmol1 . The value of ΔH is
(i) =ΔU
(ii) >ΔU
(iii) <ΔU
(iv) = 0

Answer :

Equation of combustion of methane-

CH4(g)+2O2(g)CO2(g)+2H2O(l)

Since, ΔHo=ΔUo+ΔngRT and ΔUo=XKJmol1

from the equation Δng=(npnR)=13=2

ΔHo=X2RT

ΔHo<ΔUo

So the correct option is (iii).

Question 5.5 The enthalpy of combustion of methane, graphite, and dihydrogen at 298 K are 890.KJmol1,393.5KJmol1 and 285.8KJmol1 respectively. Enthalpy of formation of CH4(g) will be

(i) –74.8 kJmol1

(ii) –52.27 kJmol1

(iii) +74.8 kJmol1

(iv) +52.26 kJmol1

Answer :

CH4+2O2CO2+2H2O ΔH ................. -890.3 kJ/mol
C+2O2CO2 ΔH ..................-393.5 kJ/mol
2H2+O22H2O ΔH ..................-258.8 kJ/mol

So, the required equation is to get the formation of CH4(g) by combining these three equations-

Thus, C+2H2CH4

ΔfHCH4=ΔcHc+2ΔcHH2
= [-393.5 + 2(-285.8) + 890.3]
= -74.8 kJ/mol

Therefore, the enthalpy of formation of methane is -74.8 kJ/mol. So, the correct option is (i)

Question 5.6 A reaction, A + B C + D + q is found to have a positive entropy change. The reaction will be
(i) Possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(v) possible at any temperature

Answer :

For the reaction to be feasible, the ΔG should be negative
ΔG=ΔHTΔS

According to the question,
ΔS = positive unit
ΔH = negative (as heat is evolved in the reaction)
Overall the ΔG is negative.

Therefore, the reaction is possible at any temperature. So, the correct option is (iv)

Question 5.7 In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

Answer :

The first law of Thermodynamics states that,
ΔU=q+W

where, ΔU = change in internal energy for the process
q = heat and W = work

Given that,
q = +701 J (heat is absorbed)
W = - 394 (work is done by the system)

Substituting the value in the equation of the first law we get,

ΔU=701+(394)=307J

So, the change in internal energy for the process is 307 J

Question 5.8 The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be 742.7KJmol1 at 298 K. Calculate the enthalpy change for the reaction at 298 K.

NH2CN(g)+32O2(g)N2+CO2(g)H2O(l)

Answer :

Given information,
ΔU=742.7kJmol1
T = 298 K
R = 8.314 ×103

Δng= ng (products) ng (reactants)
= (2 - 1.5)
= 0.5 moles

The enthalpy change for the reaction is expressed as;

ΔH=ΔU+ngRT
where, ΔU = change in internal energy and
Δng= change in no. of moles

By putting the values we get,
ΔH=(742.7)+0.5(298)(8.314×103)
=741.5kJmol1

Question 5.9 Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminum from 35C to 55C. Molar heat capacity of Al is 24Jmol1K1

Answer :

We know that
q=mcΔT
m = mass of the substance
c = heat capacity
ΔT = change in temperature

By putting all these values we get,
q=(6027mol)(24J/mol/K)(20K)
= 1066.7 J
= 1.07 kJ

Question 5.10 Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0C to ice at 10.0C.

ΔfusH=6.03KJmol1 at 0C .

Cp[H2O(l)]=75.3Jmol1K1.

Cp[H2O(s)]=36.8Jmol1K1

Answer :

Total enthalpy change is equal to the summation of all the energy required at three different stages-
(i) from 10oC to 0oC of 1 mol of water (water water )
(ii) from 0oC to 0oC of 1 mol of ice (water ice)
(iii) from 0oC to 10oC of 1 mole of ice (ice ice)

So, the total enthalpy change
ΔH=Cp[H2O(l)](ΔT)+

ΔH(freezing)+Cp[H2O(s)]ΔT

=(75.3)(10)+(6.03×103)+(36.8)(10)

=7536030368

=7151Jmol1

=7.151kJ/mol
Hence, the total enthalpy change in the transformation process is 7.151 kJ/mol

Question 5.11 Enthalpy of combustion of carbon to CO2 is 393.5KJmol1 Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.

Answer :

Formation of carbon dioxide from carbon and dioxygen reaction is-

C+O2CO2 ............. ΔHf = -393.5 kJ/mol

1 mole of CO2 = 44 g
So, the heat released in the formation of 44g of CO2 = -393.5 kJ/mol
therefore, In 35.5 g of CO2 the amount of heat released =
393.544×35.2

=314.8kJ/mol

Question 5.12 Enthalpies of formation of CO(g),CO2(g),N2O(g)andN2O4(g) are –110, – 393, 81 and 9.7 KJmol1 respectively. Find the value of ΔrH for the reaction:

N2O4(g)+3CO(g)N2O(g)+3CO2(g)

Answer :

Given,
Enthalpies of formation of CO(g),CO2(g),N2O(g)andN2O4(g) are –110, – 393, 81 and 9.7 KJmol1
N2O4(g)+3CO(g)N2O(g)+3CO2(g)
We know that the ΔrH=ΔfH (product) ΔfH (reactants)

For the above reaction,

ΔrH=[ΔfH(N2O)+3ΔfH(CO2)]

[ΔfH(N2O4)+3ΔfH(CO)]
Substituting the given values we get,
=[81+3(393)][9.7+3(110)]

=777.7kJ/mol
Thus the value of ΔrH of the reaction is -777.7 kJ/mol

Question 5.13 Given N2(g)+3H2(g)2NH3(g);

ΔrH=92.4kJmol1 .What is the standard enthalpy of the formation of NH3 gas?

Answer :

the standard enthalpy of formation of any compound is the required change in enthalpy of formation of 1 mole of a substance in its standard form from its constituent elements.

Thus we can re-write the reaction as;
12N2+32H2NH3

Therefore, standard enthalpy of formation of ammonia is = 12ΔrHΘ
= 1/2 (-92.4)
= -46.2 kJ/mol

Question 5.14 Calculate the standard enthalpy of the formation of CH3OH(l) from the following data:

CH3OH(l)+3/2O2(g)CO2(g)+2H2O(l)

ΔrH=726kJmol1.

C(graphite)+O2(g)CO2(g);

ΔcH=393kJmol1.

H2(g)+1/2O2(g)H2O(l);

ΔfH=286kJmol1.

Answer :

for the formation of CH3OH the reaction is ,

C+2H2O+12O2CH3OH
This can be obtained by the following expressions-
required eq = eq (i) + 2 (eq iii) - eq (i)

ΔfH[CH3OH]=ΔcHΘ+

2ΔfH[H2O]ΔrHΘ
=(393)+2(286)(726)
=239kJmol1

Question 5.15 Calculate the enthalpy change for the process CCl4(g)C(g)+4Cl(g) and calculate bond enthalpy of CCl in CCl4(g)
ΔvapH(CCl4)=30.5kJmol1..

ΔfH(CCl4)=135.5kJmol1..

ΔaH(C)=715.0kJmol1,

whereΔaHisenthalpyofatomisation.

ΔaH(Cl2)=242kJmol1

Answer :

We have the following chemical reaction equations-

(1) CCl4(l)CCl4(g) ................ ΔvapHo=30.5kJ/mol
(2) C(s)C(g) ........................ ΔaHo=715.0kJ/mol
(3) Cl2(g)2Cl(g) .................. ΔaHo=242kJ/mol
(4) C(g)+4ClCCl4(g) ............ ΔfH=135.5kJ/mol
The enthalpy change for the process CCl4(g)C(g)+4Cl(g) by the above reaction is calculated as;

ΔH=ΔaHo(C)+2(ΔaHo(Cl2))

ΔvapHoΔfH
=[(715)+2(242)(30.5)(135.5)]kJ/mol
=1304kJmol1

And the bond enthalpy of C-Cl bond in CCl4 (g) = 1304/4 = 326 kJ/mol

Question 5.16 For an isolated system, ΔU=0, what will be ΔS?

Answer :

Since ΔU=0
So, the ΔS>0 (positive) therefore the reaction will be feasible.

Question 5.17 For the reaction at 298 K,
2A + B C
ΔH=400kJmol1and

ΔS=0.2kJK1mol1
At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?

Answer :

From the equation,

ΔG=ΔHTΔS
Suppose the reaction is at equilibrium, So the change in temperature is given as;

T=ΔHΔGΔS
=4000.2

=2000K ( ΔG at equilibrium is zero)
To reaction should be spontaneous, ΔG should be negative. So, for the given reaction T should be greater than 2000 K

Question 5.18 For the reaction, 2Cl(g)Cl2(g) , what are the signs of ΔHandΔS?

Answer :

The given reaction represents the formation of a chlorine molecule from its atom. Bond formation taking place, therefore the energy is released during this. So, ΔH is negative.

Two moles of an atom have more randomness than one mole of chlorine. So, spontaneity is decreased. Hence ΔS is negative.

Question 5.19 For the reaction
2A(g)+B(g)2D(g)
ΔU=10.5KJandΔS=44.1JK1
Calculate ΔG for the reaction, and predict whether the reaction may occur spontaneously.

Answer :

Given reaction is 2A(g)+B(g)2D(g)

We know that, ΔGo=ΔHoTΔSo
and ΔHo=ΔUo+ΔngRT
here Δng=2(3)=1

and ΔU=10.5KJandΔS=44.1JK1

Substituting the given values in equations-
ΔG=(10.5+8.314×103×298)

298(44.1)
ΔG=(12.98+13.14)kJ

ΔG=+0.16kJ

Hence the reaction will not occur spontaneously because the ΔG value is positive for this reaction.

Question 5.20 The equilibrium constant for a reaction is 10. What will be the value of

ΔG?R=8.314JK1mol1,T=300K

Answer :

Given values are,
R=8.314JK1mol1,T=300K
and equilibrium constant = 10

It is known that,
ΔGo=2.303 RTlogKeq
=2.303(8.314)(300)log10

=5744.14Jmol1

=5.744kJmol1

Hence the value of ΔGo is -5.744 kJ/mol

Question 5.21 Comment on the thermodynamic stability of NO(g), given

1/2N2(g)+1/2O2(g)NO(g);

ΔrH=90kJmol1.

NO(g)+1/2O2(g)NO2(g):

ΔrH=74kJmol1

Answer :

The formation of NO is unstable because ΔrH is positive which means heat is absorbed during the reaction. So, NO (g) has higher energy than its reactants N2 and O2.
On the other hand, NO2 is stable because ΔrH is negative means heat is released during the formation of NO2. It is established with minimum energy. Hence unstable NO changes to stable NO2.

Question 5.22 Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ΔfH=286KJmol1

Answer :

-286 kJ/mol heat is evolved when 1 mol of the water molecule is formed. It means the same amount of energy is absorbed by the surrounding also.

qsurr=+286 kJ/mol
We know that,
Entropy changes for the surrounding is = q(surr)/ temp.
= 286000/ 298

= 959.73 J/mol/K

Class 11 Chemistry NCERT Chapter 5: Higher Order Thinking Skills (HOTS)

Question:

At standard conditions, if the change in the enthalpy for the following reaction is 109kJmol1.

H2(g)+Br2(g)2HBr(g)
Given that bond energy of H2 and Br2 is 435kJmol1 and 192kJmol1 respectively, What is the bond energy (in kJmol1 ) of HBr ?

1) 736

2) 518

3) 259

4) 368

Solution:

H22H BEH2Br22Br BEBr22H+2Br2HBr2 BEHBrH2+Br22HBr


From the Bond energy concept, we have

BEH2+BEBr22BEHBr=ΔHHBr435+1922x=1092x=736x=368 kJ mol1

Hence, the correct answer is option (4).

Question: Which of the following molecules has the highest bond dissociation energy?

(1) H2

(2) O2

(3) F2

(4) Cl2

Answer:

Bond dissociation energy is the energy required to break a bond between two atoms in a molecule. The energy level diagram for F2 shows that it has the highest bond order among the given options, making it the most stable and requiring the highest bond dissociation energy to break.

Hence, the correct answer is option (3).

Approach to Solve Questions of Class 11 Chemistry Chapter 5

Thermodynamics comes under physical chemistry. This chapter is about learning the concepts and then applying them to solve numerical problems. The approach to solving numerical problems should be simple yet time-saving.

  • First you need to understand the key concepts as a system, surroundings, internal energy, work, heat, enthalpy, etc.
  • To solve any questions, first give it a thorough reading and note down all the information given in the question.
  • Begin by identifying what is given in the question and what needs to be found. Carefully note the units and convert them if necessary to ensure consistency.
  • Use the appropriate formula based on the concept like ΔU=q+w (change in internal energy = heat + work done) or ΔH=ΔU+PΔV (enthalpy change).
  • Practice helps in mastering thermodynamics numerical. Start with simpler problems and gradually move to ones involving concepts like enthalpy change, work in isothermal or adiabatic processes, and heat capacity.

Let us understand it by taking an example. Suppose the question says "A system absorbs 300 J of heat and does 100 J of work. Calculate the change in internal energy."
Solution- Here, heat absorbed q=+300J (positive as it is absorbed) and work done by the system w= 100J (negative as energy is leaving the system).
Now use the formula:

ΔU=q+w=300+(100)=200J


So, the change in internal energy is 200 J.

Topics and Subtopics Covered in the NCERT Textbook

All the topics and subtopics covered in the NCERT textbook are listed below:

5.1 Thermodynamics Terms

5.1.1 The System and the Surroundings

5.1.2 Types of the System

5.1.3 The State of the System

5.1.4 The Internal Energy as a State Function

5.2 Applications

5.2.1 Work

5.2.2 Enthalpy, H

5.3 Measurement of ΔU and ΔH: Calorimetry

5.4 Enthalpy Change, ΔH of a Reaction – Reaction Enthalpy

5.5 Enthalpies for Different Types of Reactions

5.6 Spontaneity

5.7 Gibbs Energy Change and Equilibrium

What Extra Should Students Study Beyond the NCERT for JEE/NEET?

Beyond the NCERT, students should focus on important concepts, formulas, different types of processes, and laws of thermodynamics according to the table given below:

ConceptJEENCERT
I Thermodynamics
Thermodynamics: Properties Of System
Path, State Function, Types Of Process
Reversible, Irreversible, Polytropic Process
Thermodynamic Equilibrium
Zeroth Law Of Thermodynamics
Heat And Work
Internal Energy
Isothermal Reversible And Isothermal Irreversible
Graphical Representation Of Work Done In Thermodynamics
Heat Capacity
Relation Between Cp And Cv
Thermochemistry And Enthalpy For Chemical Reaction
Standard Enthalpy And Enthalpy Of Formation
Enthalpy Of Combustion
Enthalpy Of Dissociation, Atomisation And Phase Change
Lattice Enthalpy, Hydration Enthalpy And Enthalpy Of Solution
Enthalpy Of Neutralisation
Ionization And Electron Gain Enthalpy
Resonance Enthalpy
Kirchoff’s Equation
Born Habers Cycle
Hess’s Law
Bomb Calorimeter
Calculation Of Changes In S For Different Process
Spontaneity Criteria Through Entropy
Spontaneity Criteria Through Enthalpy (H) And Entropy (S)
Gibbs Energy And Change In Gibbs Energy
Spontaneity Criteria With Gibbs Energy (G)
Gibbs Energy At Equilibrium

NCERT Solutions for Class 11 Chemistry

Class 11 NCERT chapter-wise solutions are given below:

NCERT solutions for class 11 subject-wise

NCERT Subject-wise solutions are given below:

NCERT Books and NCERT Syllabus

The NCERT books and syllabus links for class 11 are given below:

Frequently Asked Questions (FAQs)

1. What is the difference between heat and temperature?

Temperature: It is a measure of the average kinetic energy of the particles in a substance.

Heat: The transfer of energy between two systems at different temperatures called heat. Heat flows from high temperature to low temperature.

2. What is the difference between work and energy?
  • Work: It is a form of energy transfer that occurs when a force moves an object over a distance.
  • Energy: It is the capacity to do work.

In thermodynamics, work is often associated with the mechanical work done by a system, such as the expansion or compression of a gas. Energy can exist in various forms, such as potential energy, kinetic energy, and internal energy, and it can be converted from one form to another.

3. What is the difference between an exothermic and endothermic process?

Exothermic process: In this process, energy is released from the system to its surroundings, usually in the form of heat, and the enthalpy change (ΔH) for this process is negative. Examples include combustion, condensation, and freezing. 

Endothermic process: In this process, energy is absorbed by the system from its surroundings, usually in the form of heat, and the enthalpy change (ΔH) for this process is. Examples include evaporation, melting, and photosynthesis.

4. What is the difference between an open, closed, and isolated system in thermodynamics?

Open system: An open system can exchange both energy and matter with its surroundings. Examples include a cup of hot coffee.

Closed system: A closed system can exchange energy but not matter with its surroundings. Examples include a sealed thermos.

Isolated system: An isolated system cannot exchange either energy or matter with its surroundings. Examples include a perfectly insulated container.

5. What is entropy and why is it important in thermodynamics?

Entropy is a measure of the randomness or disorderness in a system. As described by the Second Law of Thermodynamics, It is an important concept in thermodynamics because it is related to the natural tendency of energy to disperse over time. The increase in entropy in an isolated system is a measure of the irreversibility of a process, and it helps to determine the direction of spontaneous processes.

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Option 1)

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less than 3

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more than 6 but less than 9

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more than 9

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