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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 Jul 16, 2025 11:54 AM IST

Do you know why heat is absorbed in some chemical reactions while others release heat, what makes hot tea cool down when left on the table and why ice melts faster in cold water as compared to hot water? The answer to all these questions lies in Thermodynamics, a branch of Chemistry that deals with energy changes during Physical and Chemical processes. NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics helps students to develop strong understanding of concepts like internal energy, enthalpy, entropy, spontaneity, and the laws of Thermodynamics.

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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

NCERT Class 11 Chemistry Solutions of Chapter 5 Thermodynamics are designed by our subject experts in a systematic way to help you understand complex topics. These NCERT solutions include step by step answers , HOTS questions and approaches to solve questions to strengthen conceptual clarity and prepare you for exams. NCERT Solutions for Class 11 serves as an important resource for mastering Chapter 5 Thermodynamics and helps you to enhance performance in board exams as well as in the competitive exams like JEE Advanced, NEET, JEE Mains, etc.

Also Read :

Download PDF Of NCERT Solutions for Class 11 Chemistry Chapter 5

To get NCERT Solutions Class 11 Chemistry Thermodynamics PDF, 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.

Download PDF

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

Get detailed and accurate NCERT solutions for Class 11 Chemistry Thermodynamics covering all exercise questions with step-by-step answers. These solutions help you understand important concepts like enthalpy, internal energy, entropy, and Gibbs free energy.

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 :

A system is said to be under adiabatic conditions if there is zero exchange of heat between the system and its surroundings. Hence, under adiabatic conditions, q = 0.
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. The standard state of an element refers to its most stable form under standard conditions, typically at a pressure of 1 bar and a temperature of 298 K.
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 5.10Calculate 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

314.8 KJ/mol heat is released when 35.2 g of CO2 is formed from carbon and dioxygen.

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 (ii) + 2 (eq iii) - eq (i)

ΔfH[CH3OH]=ΔcHΘ+2ΔfH[H2O]ΔrHΘ
=(393)+2(286)(726)
=239kJmol1

Therefore, ΔfHθ[CH3OH()]=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
Means change in internal energy ΔU for an isolated system is zero, it does not exchange any energy with the surroundings. But entropy tends to increase in case of spontaneous reaction. Therefore, ΔS > 0 or positive.

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)

Enhance your conceptual understanding with HOTS questions from Class 11 Chemistry Chapter 5 Thermodynamics. These questions help you think deeper and prepare better for CBSE exams and entrance tests like NEET and JEE.

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

Answer:

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: The correct statement amongst the following is :

(1) The term 'standard state' implies that the temperature is 0C

(2) The standard state of pure gas is the pure gas at a pressure of 1 bar and temperature 273 K

(3) ΔfH298θ is zero for O(g)

(4) ΔfH298θ is zero for O2( g)

Answer:

Option (1)

It asserts that the term "standard state" implies that the temperature is 0C(273 K).
In fact, "standard state" typically specifies a standard pressure (1 bar) and a chosen temperature-commonly 298 K for tabulated thermodynamic data-not necessarily 0C.

Thus, Option 1 is incorrect.

Option (2)

It claims that the standard state for a pure gas is defined at 1 bar and 273 K .
In modern thermochemistry, the standard state generally means the pure substance at 1 bar, but the standard temperature most commonly used is 298 K(25C), not 273 K (0C).

Thus, Option 2 is incorrect.

Option (3)

It claims that ΔfH298 is zero for O(g) (atomic oxygen).
However, the standard state of oxygen is diatomic O2(g), not atomic oxygen. Since O(g) is not the most stable form of oxygen under standard conditions, its formation enthalpy is not zero.

Hence, Option 3 is incorrect.

Option (4)

ΔfH2980 for O2 is zero and not for O.

The standard enthalpy of formation of any element in its most stable form is zero, regardless of temperature (though typically reported at 298 K ).

Thus, option (4) is correct

Hence, the correct answer is option (4).

Question: The hydration energies of K+and Clare -x and -y kJ/mol respectively. If lattice energy of KCl is -z kJ/mol, then the heat of solution of KCl is :

(1) +xyz

(2) x+y+z

(3) z(x+y)

(4) z(x+y)

Answer:

ΔHsolution =(ΔHL.E. )+ΔHhydration =(z)+(xy)=z(x+y)

Hence, the correct answer is option (3).

Approach to Solve Questions of Class 11 Chemistry Chapter 5

The basic understanding of how to solve questions from the Thermodynamics Class 11 NCERT is essential for building a strong foundation in physical chemistry. The approaches given below help students in developing the right techniques to solve questions.

1. Study the key concepts:

Learning about the basic concepts like system, surrounding, internal energy, work, heat, enthalpy and Laws of Thermodynamics is very important for solving questions of NCERT Class 11 Chemistry Chapter 5 Thermodynamics.

2). Identify the data and Read questions carefully

Before solving any question, first give it a thorough reading and note down all the information given in the question. Identify given data like pressure, volume, temperature, etc. and what we need to find.

3). Use the appropriate formula based on the concept like:

Using the correct formula is very important for Solving NCERT Class 11 Chemistry Chapter 5 Thermodynamics questions

(i). Internal Energy Change

ΔU=Ufinal Uinitial 

(ii). First Law of Thermodynamics

ΔU=q+W

(iii). Work Done in Expansion/Compression

W=PΔV

(iv). Enthalpy and Enthalpy Change

H=U+PV

ΔH=ΔU+PΔV

(v). tandard Enthalpy Change of Reaction

ΔHreaction =ΔHproducts ΔHreactants 

(vi). Gibbs Free Energy

ΔG=ΔHTΔS

4). Convert units into standard form

Be careful while dealing with units and convert them if necessary to ensure consistency like temperature in Kelvin or volume in litres, etc.

5). Practice questions

Practice again and again, as it will help in mastering thermodynamics numerically.

  • Start with simpler problems
  • Then gradually move to ones involving concepts like enthalpy change, work in isothermal or adiabatic processes, and heat capacity.

6). Solve the NCERT textbook and NCERT exemplar questions.

Chapter 5 Thermodynamics Questions from the NCERT books are asked directly in the boards and other competitive exams. Do previous year questions from NEET and JEE to get used to question patterns.

Topics and Subtopics Covered in the NCERT Textbook

All the topics and subtopics included in the NCERT Class 11 Chemistry Chapter 5 Thermodynamics 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:

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 Thermodynamics?

Thermodynamics is the study of energy transfer, particularly the relationships and conversions between heat, work, and chemical reactions.

2. How NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics helpful for annual exam preparation?

NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics help students prepare well for the annual examinations. NCERT Solutions for Class 11 will help students to solve the problems given in the textbook. They give detailed and stepwise explanations of the problems given in the exercises of NCERT Solutions for Class 11. 

3. What is enthalpy, and how is it related to thermodynamics?

Enthalpy is a thermodynamic quantity that represents the total heat content of a system. It is defined as the internal energy plus the product of pressure and volume. In thermodynamics, enthalpy changes indicate whether a reaction is exothermic or endothermic.

4. What is the significance of the Gibbs free energy in Class 11 Chemistry Chapter 5 Thermodynamics?

Gibbs free energy is a crucial thermodynamic potential that indicates the spontaneity of a process. It combines enthalpy and entropy into one value (G = H - TS), where T is temperature.

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. 

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