NCERT Solutions for Class 11 Chemistry Chapter 6

NCERT Solutions for Class 11 Chemistry Chapter 6 – Thermodynamics 

Chapter 6 Chemistry Class 11 is about Thermodynamics, which is the study of relations between heat, temperature, energy, and work. The chapter discusses the laws of thermodynamics that describe how the energy in a system changes, and other concepts related to the laws.

NCERT Solutions for Class 11 Chemistry Chapter 6- Thermodynamics

Click here to download NCERT Solutions for Class 11 Chemistry Chapter 6- Thermodynamics 

Access NCERT Solutions for Class 11 Chemistry Chapter 6- Thermodynamics

NCERT Solutions Class 11 Chemistry Thermodynamics

Chapter 6 Thermodynamics involves topics like open, closed, and isolated systems, enthalpy, isothermal and free expansion of gas, internal energy as a state function, etc. The Laws of Thermodynamics can be understood well only after reading all these topics in detail. 

Class 11 Chemistry Ch 6 NCERT Solutions can be downloaded from the website, and can even be accessed on Extramarks app.

NCERT Class 11 Chemistry Thermodynamics 

NCERT Solutions Class 11 Chemistry cover all the questions that are listed at the end of the Thermodynamics chapter in the textbook. As the solutions are prepared by our subject-matter experts, students can be assured that they are referring to authentic study material.

Why should one refer to CBSE Class 11 Chemistry Thermodynamics NCERT Solutions by Extramarks? 

Questions and answers in Class 11 Chemistry Chapter 6 are curated in a way that eases a student’s learning process. Following the answer pattern in the NCERT Solutions will assure that students secure good grades in their exams.

Here are the benefits of following NCERT Solutions Class 11:

  • Subject matter experts have prepared the solutions, which ensures their high accuracy level.
  • The answers in the solutions are delivered concisely without any filler to aid students in getting better marks.
  • The solutions have images and tables to explain the concept in a better way.

The Class 11 Chemistry Chapter 6 solutions are offered by Extramarks. Students can download them from the website or the Extramarks app.

Related Questions 

Q.1 A thermodynamic state function is a quantity

(i) Used to determine heat changes

(ii) Whose value is independent of path

(iii) Used to determine pressure volume work

(iv) Whose value depends on temperature only.


Correct answer is (ii)


A thermodynamic state function is a quantity whose value is independent of a path.

Functions like p, V, T etc. depend only on the state of a system and not on the path.

Q.2 For the process to occur under adiabatic conditions, the correct condition is:

(i) ∆T = 0

(ii) ∆p = 0

(iii) q = 0

(iv) w = 0


Correct answer is (iii)


When there is no exchange of heat between the system and its surroundings then the system is said to be under adiabatic conditions. Hence, under adiabatic conditions, q = 0.

Q.3 The enthalpies of all elements in their standard states are:

(i) unity

(ii) zero

(iii) < 0

(iv) Different for each element


Correct answer is (ii)


The enthalpy of all elements in their standard state is zero.

Q.4 ∆Uθof combustion of methane is – X kJ mol–1. The value of ∆Hθ is

(i) = ∆Uθ

(ii) > ∆Uθ

(iii) < ∆Uθ

(iv) = 0


Correct answer is (iii)


Since ∆Hθ = ∆Uθ + ∆ngRT and ∆Uθ = –X kJ mol–1,

∆Hθ = (–X) + ∆ngRT.

⇒ ∆Hθ < ∆Uθ

Q.5 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 CH4 (g) will be

(i) –74.8 kJ mol–1 (ii) –52.27 kJ mol–1

(iii) +74.8 kJ mol–1 (iv) +52.26 kJ mol–1.


Correct answer is (i)


Given: The enthalpy of combustion of,

  1. CH4(g) + 2O2(g) → CO2(g) +2H2O(g) ∆H=-890.3 kJ mol -1
  2. C(s)+O2(g) → CO2(g) ∆H=-393.5kJ mol -1
  3. 2H2(g)+ O2(g) → 2H2O(g) ∆H=-285.8 kJ mol -1

Thus, the desired equation is the one that represents the formation of CH4

C(s)+2H2(g) → CH4(g)

fHCH4 =∆cHc+2∆cHH2 -∆CHCO2

=[-393.5+2(-285.8)-(-890.3)]kJ mol-1

=-74.8 kJ mol-1

Enthalpy of formation of CH4(g) = –74.8 kJ mol–1

Hence, alternative (i) is correct.

Q.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

(iv) possible at any temperature


For a reaction to be spontaneous, ∆G should be negative.

∆G = ∆H – T∆S

According to the question, for the given reaction,

∆S = positive

∆H = negative (since heat is evolved)

⇒ ∆G = negative

Therefore, the reaction is spontaneous at any temperature.

Hence, alternative (iv) is correct.

Q.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?


According to the first law of thermodynamics,

∆U = q + W (i)


∆U = change in internal energy for a process

q = heat

W = work


q = + 701 J (Since heat is absorbed)

W = –394 J (Since work is done by the system)

Substituting the values in expression (i), we get

∆U = 701 J + (–394 J)

∆U = 307 J

Hence, the change in internal energy for the given process is 307 J.

Q.8 The reaction of cyanamide, NH2CN(s), with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7 kJ mol–1 at298 K. Calculate enthalpy change for the reaction at 298 K.

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


Enthalpy change for a reaction (∆H) is given by the expression,

∆H = ∆U + ∆ngRT


∆U = change in internal energy

∆ng = change in number of moles

For the given reaction,

∆ng = ∑ng (products) – ∑ng (reactants)

= (2 – 1.5) moles

∆ng = 0.5 moles


∆U = –742.7 kJ mol–1

T = 298 K

R = 8.314 × 10–3 kJ mol–1 K–1

Substituting the values in the expression of ∆H:

∆H = (–742.7 kJ mol–1) + (0.5 mol) (298 K) (8.314 × 10–3 kJ mol–1 K–1)

= 742.7 – 1.2

∆H = 741.5 kJ mol–1

Q.9 Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol–1 K–1.


From the expression of heat (q),

q = m. c. ∆T


c = molar heat capacity

m = mass of substance

∆T = change in temperature

Substituting the values in the expression of q:

q=[60/27 mole](24 j mole-1 K-1)(20K)

q = 1066.7 J

q = 1.07 kJ

Q.10 Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C.

fusH = 6.03 kJ mol–1 at 0°C.

Cp[H 2O(l)] = 75.3 J mol–1 K–1

Cp[H 2O(s)] = 36.8 J mol–1 K–1


Total enthalpy change involved in the transformation is the sum of the following changes:

(a) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.

(b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.

(c) Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at –10°C.

Total ∆H=Cp[H2OCl]∆T+∆H freezing +Cp [H2O(S)]∆T

= (75.3 J mol–1 K–1) (0 – 10)K + (–6.03 × 103 J mol–1) + (36.8 J mol–1 K–1) (–10 – 0)K

= –753 J mol–1 – 6030 J mol–1 – 368 J mol–1

= –7151 J mol–1

= –7.151 kJ mol–1

Hence, the enthalpy change involved in the transformation is –7.151 kJ mol–1.

Q.11 Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol–1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.


Formation of CO2 from carbon and dioxygen gas can be represented as:

C(s )+O2(g) →CO2(g) H=-393.5 kJ mol-1

(1 mole = 44 g)

Heat released on formation of 44 g CO2 = 393.5 kJ mol–1

Heat released on formation of 35.2 g CO2

= 314.8 kJ mol–1

Q.12 Enthalpies of formation of CO(g), CO2(g), N 2 O(g) and N 2 O4(g) are –110 kJ mol–1, – 393 kJ

mol–1, 81 kJ mol–1 and 9.7 kJ mol–1 respectively. Find the value of ∆rH for the reaction:

N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)


rH for a reaction is defined as the difference between ∆fH value of products and ∆fH

value of reactants.

Δ rH = ∑ ΔH (Products) – ∑ ΔH (reactants)

For the given reaction,

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

ΔrH =[{ ΔH(N2O)+3 ΔH (CO2)}-{ ΔH(N2O4)+3 ΔH(CO)}]

Substituting the values of ∆fH for N2O, CO2, N2O4, and CO from the question, we get:

ΔrH =[{81 kJ mol-1 +3(-393)kJ mol-1}-{9.7 kJ mol-1 +3(-110)kJ mol-1}]

ΔrH =-777.7 kJ mol-1

Hence, the value of ∆rH for the reaction is -777.7 kJ mol-1

Q.13 Given

N2(g)+3H2(g) → 2NH3(g) ∆rHθ = –92.4 kJ mol–1

What is the standard enthalpy of formation of NH3 gas?


Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state.Re-writing the given equation for 1 mole of NH3(g),

1 2 N 2( g ) + 3 2 H 2( g ) N H 3( g ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacuaaqlaaWBaaa2aayibay0YaaSaaaeaacaaIXaaabaGaaGOmaaaacaWGobWaaSbaaSqaaiaaikdadaqadaqaaiaadEgaaiaawIcacaGLPaaaaeqaaOGaey4kaSYaaSaaaeaacaaIZaaabaGaaGOmaaaacaWGibWaaSbaaSqaaiaaikdadaqadaqaaiaadEgaaiaawIcacaGLPaaaaeqaaOGaeyOKH4QaamOtaiaadIeadaWgaaWcbaGaaG4mamaabmaabaGaam4zaaGaayjkaiaawMcaaaqabaaaaa@53AB@

...Standard enthalpy of formation of NH3(g)

= ½ ∆rHθ

= ½ (–92.4 kJ mol–1)

= –46.2 kJ mol–1

Q.14 Calculate the standard enthalpy of formation of CH3OH(l) from the following data:

C H 3 O H ( I ) + 2 3 O 2( g ) C O 2 O ( I ) ; ? r H θ =726 kJ mol 1 C ( g ) + O 2( g ) C O 2( g ) ;


c H θ =393 kJ mol 1 H 2( g ) + 1 2 O 2( g ) H 2 O ( I ) ;


f H θ =286 kJ mol 1


The reaction that takes place during the formation of CH3OH(l) can be written as:

C( s )+2 H 2 O( g )+ 1 2 O 2( g ) C H 3 O H ( I ) ( 1 ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacGaaqlaaWBaaa2aayibaOzbayJbaa+bay0Iaam4qamaabmaabaGaam4CaaGaayjkaiaawMcaaiabgUcaRiaaikdacaWGibWaaSbaaSqaaiaaikdaaeqaaOGaam4tamaabmaabaGaam4zaaGaayjkaiaawMcaaiabgUcaRmaalaaabaGaaGymaaqaaiaaikdaaaGaam4tamaaBaaaleaacaaIYaWaaeWaaeaacaWGNbaacaGLOaGaayzkaaaabeaakiabgkziUkaadoeacaWGibWaaSbaaSqaaiaaiodaaeqaaOGaam4taiaadIeadaWgaaWcbaWaaeWaaeaacaWGjbaacaGLOaGaayzkaaaabeaakiaabccacaqGGaWaaeWaaeaacaaIXaaacaGLOaGaayzkaaaaaa@5F30@

The reaction (1) can be obtained from the given reactions by following the algebraic

calculations as:

Equation (ii) + 2 × equation (iii) – equation (i)

fHθ [CH3OH(l)] = ∆cHθ + 2∆fHθ [H2O(l)] – ∆rHθ

= (–393 kJ mol–1) + 2(–286 kJ mol–1) – (–726 kJ mol–1)

= (–393 – 572 + 726) kJ mol–1

fHθ [CH3OH(l)] = –239 kJ mol–1

Q.15 Calculate the enthalpy change for the process

CCl4(g) → C(g) + 4Cl(g)

and calculate bond enthalpy of C–Cl in CCl4(g).

vapHθ (CCl4) = 30.5 kJ mol–1.

fHθ (CCl4) = –135.5 kJ mol–1.

aHθ (C) = 715.0 kJ mol–1, where ∆aHθ is enthalpy of atomisation

aHθ (Cl2) = 242 kJ mol–1


The chemical equations implying to the given values of enthalpies are:

CCI4(l) → CCI4(g)vapHθ = 30.5 kJ mol–1

C(s) → C(g)aHθ = 715.0 kJ mol–1

Cl2(g) → 2Cl(g)aHθ = 242 kJ mol–1

C(g)+Cl(g) → CCI4(g)fH = –135.5 kJ mol–1

Enthalpy change for the given processCCI4(g) → C(g)+4CI(g)’ can be calculated using the following algebraic calculations as:

Equation (ii) + 2 × Equation (iii) – Equation (i) – Equation (iv)

∆H = ∆aHθ(C) + 2∆aHθ (Cl2) – ∆vapHθ – ∆fH

= (715.0 kJ mol–1) + 2(242 kJ mol–1) – (30.5 kJ mol–1) – (–135.5 kJ mol–1)

∆H = 1304 kJ mol–1

Bond enthalpy of C–Cl bond in CCl4(g)

=326 kJ mol-1

Q.16 For an isolated system, ∆U = 0, what will be ∆S?


∆S will be positive i.e., greater than zero

Since ∆U = 0, ∆S will be positive and the reaction will be spontaneous.

Q.17 For the reaction at 298 K,

2A + B → C

∆H = 400 kJ mol–1 and ∆S = 0.2 kJ K–1 mol–1

At what temperature will the reaction become spontaneous considering ∆H and ∆S to be

constant over the temperature range?


From the expression,

∆G = ∆H – T∆S

Assuming, the reaction at equilibrium, ∆T for the reaction would be:

T= ΔH ΔS ( ΔG=0 at equilibrium ) 400 kJ mol 1 0.2 kJ K 1 mo l 1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@7231@

T = 2000 K

For the reaction to be spontaneous, ∆G must be negative. Hence, for the given reaction

to be spontaneous, T should be greater than 2000 K.

Q.18 For the reaction,

2Cl(g) → Cl2(g), what are the signs of ∆H and ∆S ?


∆H and ∆S are negative

The given reaction represents the formation of chlorine molecule from chlorine atoms.

Here, bond formation is taking place. Therefore, energy is being released. Hence, ∆H is negative.

Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, ∆S is negative for the given reaction.

Q.19 For the reaction

2A(g) + B(g) → 2D(g)

∆Uθ = –10.5 kJ and ∆Sθ= –44.1 JK–1.

Calculate ∆Gθ for the reaction, and predict whether the reaction may occur spontaneously.


For the given reaction,

2 A(g) + B(g) → 2D(g)

∆ng = 2 – (3)

= –1 mole

Substituting the value of ∆Uθ in the expression of ∆H:

∆Hθ = ∆Uθ + ∆ngRT

= (–10.5 kJ) – (–1) (8.314 × 10–3 kJ K–1 mol–1) (298 K)

= –10.5 kJ – 2.48 kJ

∆Hθ = –12.98 kJ

Substituting the values of ∆Hθ and ∆Sθ in the expression of ∆Gθ:

∆Gθ = ∆Hθ – T∆Sθ

= –12.98 kJ – (298 K) (–44.1 J K–1)

= –12.98 kJ + 13.14 kJ

∆Gθ = + 0.16 kJ

Since ∆Gθ for the reaction is positive, the reaction will not occur spontaneously.

Q.20 The equilibrium constant for a reaction is 10. What will be the value of ∆Gθ? R = 8.314

JK–1 mol–1, T = 300 K.


From the expression,

∆Gθ = –2.303 RT logKeq

∆Gθ for the reaction,

= (2.303) × (8.314 JK–1 mol–1)× (300 K) × log10

= –5744.14 Jmol–1

= –5.744 kJ mol–1


Comment on the thermodynamic stability of NO( g ), given 1 2 N 2( g ) + 1 2 O 2( g ) N O ( g ) ; Δ r H θ =90 kJ mo l 1 1 2 N ( g ) + 1 2 O 2( g ) N O 2( g ) ; Δ r H θ =74 kJ mo l 1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=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@B75C@


The positive value of ∆rH indicates that heat is absorbed during the formation of NO(g).

This means that NO(g) has higher energy than the reactants (N2 and O2). Hence, NO(g) is unstable.

The negative value of ∆rH indicates that heat is evolved during the formation of NO2(g) from NO(g) and O2(g). The product, NO2(g) is stabilized with minimum energy. Hence, unstable NO(g) changes to unstable NO2(g).

Q.22 Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ∆fHθ = –286 kJ mol–1.


It is given that 286 kJ mol–1 of heat is evolved on the formation of 1 mol of H2O(l). Thus, an equal amount of heat will be absorbed by the surroundings.

qsurr = +286 kJ mol–1

Entropy change (∆Ssurr) for the surroundings =

q surr T MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacGaaqlaaWBaaa2aayibaOzbayJbaa+bay0YaaSaaaeaacaWGXbWaaSbaaSqaaiaadohacaWG1bGaamOCaiaadkhaaeqaaaGcbaGaamivaaaaaaa@4865@

286 kJ mol 1 298 k MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbi9G8qqLqFD0xd9wqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqacGaaqlaaWBaaa2aayibaOzbayJbaa+bay0YaaSaaaeaacaaIYaGaaGioaiaaiAdacaqGGaGaae4AaiaabQeacaqGGaGaaeyBaiaab+gacaqGSbWaaWbaaSqabeaacqGHsislcaaIXaaaaaGcbaGaaGOmaiaaiMdacaaI4aGaaeiiaiaabUgaaaaaaa@5041@

∆Ssurr =959.73 J mol -1 K-1

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FAQs (Frequently Asked Questions)

1. What are the three types of systems?

A thermodynamic system is a system that is delimited from the surroundings by hypothetical or real boundaries. The part of the universe in which observations are made is a thermodynamic system. The system has a boundary and all that is beyond the system, i.e. the remaining universe, is called surroundings. The system and surroundings can have an exchange of mass and energy. 

There are three types of systems discussed in this chapter;

Open System: A system that is able to exchange both energy and matter with the area beyond the boundary (its surroundings) is said to be an open system.

Closed System: A system that can exchange energy but cannot exchange matter with its surroundings is called a closed system.

Isolated System: A system that cannot exchange energy or matter with its surroundings is called an isolated system. 

2. What are the different branches of Thermodynamics?

The branches of thermodynamics are;

Classical Thermodynamics

In this branch of Thermodynamics, a macroscopic approach is used to analyse the behaviour of matter. Pressure and temperature units being taken into consideration help in calculating other properties and in predicting the characteristics of matter (which is undergoing the process).

Statistical Thermodynamics

Every molecule is under the spotlight in statistical thermodynamics. This means that every molecule’s properties and how they interact is taken into consideration to determine the behaviour of a collection of molecules.

Chemical Thermodynamics

Chemical thermodynamics is the study of the interrelation of work and heat during chemical reactions as well as changes in states.

Equilibrium Thermodynamics

The study of energy and matter transformations as they approach the state of equilibrium is called equilibrium thermodynamics.

3. What is Enthalpy or Heat Reaction?

The heat of a reaction caused because of a chemical reaction that takes place when pressure is constant is called Enthalpy. The amount of energy per mole in a reaction is calculated using thermodynamics measurement. Volume, internal energy, and pressure are used to derive Enthalpy. 

H= E+PV 

Where H= Enthalpy, P= Pressure, V= Volume, and E= Internal Energy.

The heat of reaction in a chemical process is calculated using heat reaction. The flow of heat in a calorimeter can be calculated using a change in enthalpy. This is utilised to evaluate Joule Thomson Expansion or the throttling process. Additionally, the minimum power used in a compressor is utilised to measure enthalpy. 

4. What is the Thermodynamic Process?

The thermodynamic process includes the movement of heat energy within a system. The four types of thermodynamic cycles are Isobaric, Isothermal, Isochoric, and Adiabatic.

Isobaric: This is a process that sees no change in pressure. 

Isothermal: This is a process that does not see any change in temperature.

Isochoric: In this process, there is no change in volume and the system remains constant.

Adiabatic: Here the heat does not go in or out of the system. 

5. What is Thermodynamic equilibrium according to Chapter 6 of Class 11 Chemistry?

If all the thermodynamic processes stay constant with passing time, a system is said to be in thermodynamic equilibrium (as per thermodynamic concepts). All properties of a system have fixed values at a given state, hence, a change in even one property can change the system’s state into a different one altogether. A system in equilibrium sees no change occur in the value of properties when it is isolated from its surroundings. 

When it is observed that at any point in time, there is no change in pressure, it is called mechanical equilibrium. 

Chemical equilibrium occurs when a system’s chemical composition does not vary with time. 

When the entire system observes temperature being constant, it is called thermal equilibrium. 

When the mass of each phase in a two-phase system reaches an equilibrium level, it is called phase equilibrium. 

When a thermodynamic system is in mechanical, chemical, and thermal equilibrium, and the relevant parameters cease changing with time, the system is believed to be in thermodynamic equilibrium.

6. How is studying Thermodynamics with the help of NCERT Solutions for Chapter 6 of Class 11 Chemistry helpful for exam preparation?

NCERT Solutions For Class 11 Chemistry Chapter 6 is the go-to guide for students to prepare better for their examinations. Thermodynamics is a topic often asked in exams, which is why students must refer to Class 11 Chemistry Ch 6 NCERT Solutions to get clarity of the concept through accurate and detailed answers to the textbook questions.