Class 11 Chemistry NCERT Solutions

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Chemical Thermodynamics Class 11 Chemistry NCERT Solutions

The complete NCERT exercise solutions for Chapter 5, Chemical Thermodynamics — 22 questions from 5.1 to 5.22, each worked through step by step in the CBSE marking pattern. Enthalpy and entropy changes, Hess's law, heat capacity and calorimetry, the standard enthalpy of formation, and Gibbs free energy.

Class:11Subject:ChemistryChapter:5

3 Key Formulas25 Practice MCQs

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

How many questions are in NCERT Class 11 Chemistry Chapter 5?

Chapter 5 carries 22 exercise questions, numbered 5.1 to 5.22. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

Table of Contents23 sections

Revision & Practice

Chapter Overview

Chemical thermodynamics connects microscopic composition with measurable heat changes, work, entropy and spontaneous change. These exercises develop state functions, the first law, enthalpy and heat capacity, Hess’s law, bond and formation enthalpies, entropy, the Gibbs energy criterion and the equilibrium-constant relation. Each question is repaired from OCR, renumbered to the official Chapter 5 sequence and worked with explicit laws, substitutions, unit conversions and conclusions.

Board pattern

Begin every calculation by naming the system and writing physical states. Use the chemistry sign convention: heat absorbed by the system and work done on the system are positive, so q = mcΔT follows the sign of ΔT, while work done by the system makes w negative. For Hess’s law, write each thermochemical equation separately, reverse an equation when required, multiply by the matching factor and sign, and cancel species. Use ΔG = ΔH − TΔS for spontaneity, with ΔG < 0, and use ΔG° = −RT ln K for equilibrium.

$Ex 5.1 — Identifying a thermodynamic state function$
$Ex 5.2 — Condition for an adiabatic process$
$Ex 5.3 — Standard enthalpies of elements$
$Ex 5.4 — Comparing combustion enthalpy and internal energy$
$Ex 5.5 — Standard enthalpy of formation of methane$
$Ex 5.6 — Temperature dependence of spontaneity$
$Ex 5.7 — Internal energy change from the first law$
$Ex 5.8 — Enthalpy change from a bomb-calorimeter value$
$Ex 5.9 — Heat required to warm aluminium$
$Ex 5.10 — Enthalpy change from water to ice$
$Ex 5.11 — Heat released in forming carbon dioxide$
$Ex 5.12 — Reaction enthalpy from formation enthalpies$
$Ex 5.13 — Standard enthalpy of formation of ammonia$
$Ex 5.14 — Standard enthalpy of formation of methanol$
$Ex 5.15 — Atomisation enthalpy and C-Cl bond enthalpy$
$Ex 5.16 — Entropy change of an isolated system$
$Ex 5.17 — Temperature for a spontaneous reaction$
$Ex 5.18 — Signs of ΔH and ΔS during bond formation$
$Ex 5.19 — Gibbs energy from ΔU and ΔS$
$Ex 5.20 — Gibbs energy from the equilibrium constant$
$Ex 5.21 — Thermodynamic stability of nitric oxide$
$Ex 5.22 — Entropy change of the surroundings$

Work in two natural groups: (5.1–5.10) establishes signs, the first law, enthalpy and heat-capacity calculations, while (5.11–5.22) applies Hess’s law, formation and bond enthalpies, entropy, Gibbs energy and equilibrium constants. All exercise references use the official Chapter 5 numbering.

NCERT Exercise 5.1 — Identifying a Thermodynamic State Function

1Exercise question

Step-by-step solution

1Apply the definition: a state function depends only on the current state of the system and not on the path used to reach that state.
2Pressure, volume, temperature, internal energy, enthalpy and entropy are examples of state functions.
3Heat and work are path functions, while a state function need not depend on temperature alone.
4Only option (ii) matches the definition.

Final answer

A state function has a path-independent value, so the correct option is (ii).

NCERT Exercise 5.2 — Condition for an Adiabatic Process

1Exercise question

Step-by-step solution

1An adiabatic process allows no heat transfer across the system boundary.
2
$q = 0$
3The temperature, pressure and work may still change, so the other zero quantities are not required.
4Therefore option (iii) is correct.

Final answer

An adiabatic process satisfies q = 0, so the correct option is (iii).

NCERT Exercise 5.3 — Standard Enthalpies of Elements

1Exercise question

Step-by-step solution

1By convention, the standard molar enthalpy of formation of every element in its most stable reference state is assigned a value of zero.
2
$Δ_{f} H^{\circ} [element in its standard state] = 0 kJ mo l^{- 1}$
3This convention applies to each element separately, not to all compounds or all states.
4Thus the common assigned value is zero, not unity.

Final answer

The standard enthalpy of every element in its standard state is 0 kJ mol⁻¹, so option (ii) is correct.

NCERT Exercise 5.4 — Comparing Combustion Enthalpy and Internal Energy

1Exercise question

Step-by-step solution

1Write the balanced combustion reaction:
2
$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$
3Apply the gas-mole relation and count only gaseous species:
4
$Δ n_{g} = 2 - (1 + 2) = - 1 mol$
5Use ΔH° = ΔU° + Δn(gas)RT:
6
$Δ_{r} H^{\circ} = - X kJ mo l^{- 1} - R T$
7Because R, T and X are positive, subtracting RT makes ΔH° more negative than ΔU°.

Final answer

ΔH° = ΔU° − RT, so ΔH° < ΔU° and option (iii) is correct.

NCERT Exercise 5.5 — Standard Enthalpy of Formation of Methane

1Exercise question

Step-by-step solution

1Write the three combustion equations with their enthalpy changes:
2
$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (g), Δ H = - 890.3 kJ mo l^{- 1}$
3
$C (s) + O_{2} (g) \to C O_{2} (g), Δ H = - 393.5 kJ mo l^{- 1}$
4
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g), Δ H = - 285.8 kJ mo l^{- 1}$
5The target formation equation is:
6
$C (s) + 2 H_{2} (g) \to C H_{4} (g)$
7Reverse methane combustion, so its sign changes to +890.3 kJ mol⁻¹, and add the carbon-combustion equation plus twice the dihydrogen-combustion equation.
8
$Δ_{f} H^{\circ} [C H_{4} (g)] = + 890.3 - 393.5 + 2 (- 285.8)$
9
$Δ_{f} H^{\circ} [C H_{4} (g)] = - 74.8 kJ mo l^{- 1}$

Final answer

The enthalpy of formation of CH₄(g) is −74.8 kJ mol⁻¹, so option (i) is correct.

NCERT Exercise 5.6 — Temperature Dependence of Spontaneity

1Exercise question

Step-by-step solution

1Heat on the product side means that the reaction is exothermic, so ΔH is negative; the given entropy change is also positive.
2Apply the Gibbs-energy criterion:
3
$Δ G = Δ H - T Δ S$
4Both terms are unfavourable to a positive result: ΔH is negative and −TΔS is negative for T > 0.
5
$Δ G = negative - T (positive) < 0$
6Hence the reaction is spontaneous at every positive temperature.

Final answer

The reaction is possible at any temperature, so the corrected fourth option (iv) is correct.

NCERT Exercise 5.7 — Internal Energy Change from the First Law

1Exercise question

Step-by-step solution

1Apply the first law in the chemistry sign convention:
2
$Δ U = q + w$
3Heat absorbed by the system is positive, but work done by the system is work on the surroundings and therefore w is negative for the system.
4
$q = + 701 J, w = - 394 J$
5
$Δ U = 701 + (- 394) = + 307 J$
6The positive result means that the internal energy of the system increases.

Final answer

The change in internal energy is ΔU = +307 J.

NCERT Exercise 5.8 — Enthalpy Change from a Bomb-Calorimeter Value

1Exercise question

Step-by-step solution

1Apply the relation between enthalpy and internal energy:
2
$Δ H = Δ U + Δ n_{g} R T$
3Only gaseous species count toward the change in gaseous amount:
4
$Δ n_{g} = 2 - (0 + \frac{3}{2}) = + 0.5 mol$
5Express R in kJ units and substitute:
6
$R = 8.314 J mo l^{- 1} K^{- 1} = 8.314 \times 1 0^{- 3} kJ mo l^{- 1} K^{- 1}$
7
$Δ H = - 742.7 + (0.5) (8.314 \times 1 0^{- 3}) (298)$
8
$Δ H = - 742.7 + 1.2388 = - 741.4612 kJ mo l^{- 1}$

Final answer

The standard reaction enthalpy at 298 K is ΔH° = −741.5 kJ mol⁻¹.

NCERT Exercise 5.9 — Heat Required to Warm Aluminium

1Exercise question

Step-by-step solution

1Use the molar form of the heat equation:
2
$q = n C_{p} Δ T$
3Convert mass to moles using the molar mass of aluminium, 27.0 g mol⁻¹:
4
$n = \frac{60.0 g}{27.0 g mo l ^{- 1}} = 2.222 mol$
5A temperature interval in kelvins has the same magnitude as one in degrees Celsius:
6
$Δ T = 55 - 35 = 20 K$
7
$q = (2.222) (24) (20) = 1066.7 J$
8Convert joules to kilojoules explicitly:
9
$q = \frac{1066.7 J}{1000} = 1.0667 kJ$

Final answer

The required heat is 1.07 kJ for the 60.0 g sample.

NCERT Exercise 5.10 — Enthalpy Change from Water to Ice

1Exercise question

Step-by-step solution

1Choose a three-step path: cool liquid water from 10 °C to 0 °C, freeze it at 0 °C, and cool the ice from 0 °C to −10 °C.
2At constant pressure, the sensible-heat law is q = nCₚΔT. For 1.0 mol, the liquid-cooling contribution is:
3
$q_{1} = (75.3) (0 - 10) = - 753 J mo l^{- 1}$
4Freezing is the reverse of fusion, so its enthalpy change is negative:
5
$q_{2} = - 6.03 kJ mo l^{- 1} = - 6030 J mo l^{- 1}$
6The ice-cooling contribution is:
7
$q_{3} = (36.8) (- 10 - 0) = - 368 J mo l^{- 1}$
8Add the three contributions and convert to kilojoules:
9
$Δ H = - 753 - 6030 - 368 = - 7151 J mo l^{- 1} = - 7.151 kJ mo l^{- 1}$

Final answer

The enthalpy change from liquid water at 10 °C to ice at −10 °C is −7.151 kJ mol⁻¹.

NCERT Exercise 5.11 — Heat Released in Forming Carbon Dioxide

1Exercise question

Step-by-step solution

1Write the formation reaction and use the molar mass of CO₂:
2
$C (s) + O_{2} (g) \to C O_{2} (g), M_{C O_{2}} = 44.0 g mo l^{- 1}$
3Convert the required mass to moles:
4
$n = \frac{35.2 g}{44.0 g mo l ^{- 1}} = 0.800 mol$
5Multiply the amount formed by the molar enthalpy of formation:
6
$Δ H = (0.800 mol) (- 393.5 kJ mo l^{- 1}) = - 314.8 kJ$
7The negative enthalpy change means that 314.8 kJ is released; it is not an amount per mole of the 35.2 g sample.

Final answer

Forming 35.2 g of CO₂ releases 314.8 kJ of heat, corresponding to ΔH = −314.8 kJ for the sample.

NCERT Exercise 5.12 — Reaction Enthalpy from Formation Enthalpies

1Exercise question

Step-by-step solution

1Apply Hess’s formation-enthalpy relation:
2
$Δ_{r} H^{\circ} = \sum ν Δ_{f} H^{\circ} (products) - \sum ν Δ_{f} H^{\circ} (reactants)$
3Insert the stoichiometric coefficients, giving extra weight to each repeated compound:
4
$Δ_{r} H^{\circ} = [81 + 3 (- 393)] - [9.7 + 3 (- 110)]$
5
$Δ_{r} H^{\circ} = (- 1098) - (- 320.3) = - 777.7 kJ mo l^{- 1}$

Final answer

The standard reaction enthalpy is ΔᵣH° = −777.7 kJ mol⁻¹.

NCERT Exercise 5.13 — Standard Enthalpy of Formation of Ammonia

1Exercise question

Step-by-step solution

1The formation enthalpy is defined for formation of one mole from elements in their standard states.
2Divide the given reaction and its enthalpy by two:
3
$\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \to N H_{3} (g)$
4
$Δ_{f} H^{\circ} [N H_{3} (g)] = \frac{1}{2} (- 92.4) = - 46.2 kJ mo l^{- 1}$

Final answer

The standard enthalpy of formation of NH₃(g) is −46.2 kJ mol⁻¹.

NCERT Exercise 5.14 — Standard Enthalpy of Formation of Methanol

1Exercise question

Step-by-step solution

1Write the target formation equation:
2
$C (s) + 2 H_{2} (g) + \frac{1}{2} O_{2} (g) \to C H_{3} OH (l)$
3By Hess’s law, obtain the target as equation (ii) + 2 × equation (iii) − equation (i). Reversing equation (i) changes its sign to +726 kJ mol⁻¹.
4
$Δ_{f} H^{\circ} [C H_{3} OH (l)] = (- 393) + 2 (- 286) - (- 726)$
5
$Δ_{f} H^{\circ} [C H_{3} OH (l)] = - 393 - 572 + 726 = - 239 kJ mo l^{- 1}$

Final answer

The standard enthalpy of formation of CH₃OH(l) is −239 kJ mol⁻¹.

NCERT Exercise 5.15 — Atomisation Enthalpy and C–Cl Bond Enthalpy

1Exercise question

Step-by-step solution

1List the required enthalpy changes with their signs:
2
$CC l_{4} (l) \to CC l_{4} (g), Δ H = + 30.5 kJ mo l^{- 1}$
3
$C (s) \to C (g), Δ H = + 715.0 kJ mo l^{- 1}$
4
$2C l_{2} (g) \to 4Cl (g), Δ H = 2 (242) = + 484 kJ mo l^{- 1}$
5
$C (g) + 4Cl (g) \to CC l_{4} (g), Δ H = - 135.5 kJ mo l^{- 1}$
6Atomise carbon and chlorine, condense CCl₄ vapour back to liquid, and reverse the last formation equation so that gaseous CCl₄ remains on the left. The enthalpy is:
7
$Δ H = 715.0 + 484 - 30.5 + 135.5 = 1304.0 kJ mo l^{- 1}$
8One CCl₄ molecule contains four C–Cl bonds, so divide the atomisation enthalpy by four:
9
$D (C - Cl) = \frac{1304.0}{4} = 326.0 kJ mo l^{- 1}$

Final answer

The reaction enthalpy is +1304 kJ mol⁻¹, and the C–Cl bond enthalpy is 326 kJ mol⁻¹.

NCERT Exercise 5.16 — Entropy Change of an Isolated System

1Exercise question

Step-by-step solution

1An isolated system exchanges neither matter nor energy with its surroundings, so it is also the whole universe for this process.
2The second law requires the entropy of an isolated system not to decrease:
3
$Δ S_{isolated} \geq 0$
4The inequality is strict for a spontaneous irreversible change; the zero value applies only to an ideal reversible change.

Final answer

For the expected spontaneous change, ΔS is positive; ΔS = 0 only for an ideal reversible change.

NCERT Exercise 5.17 — Temperature for a Spontaneous Reaction

1Exercise question

Step-by-step solution

1A reaction becomes spontaneous when ΔG becomes negative:
2
$Δ G = Δ H - T Δ S < 0$
3First find the equilibrium boundary by setting ΔG equal to zero:
4
$T_{eq} = \frac{Δ H}{Δ S} = \frac{400 kJ mo l ^{- 1}}{0.2 kJ K ^{- 1} mo l ^{- 1}} = 2000 K$
5Because ΔS is positive, increasing temperature makes −TΔS more negative, so the reaction is spontaneous above this boundary.

Final answer

The reaction is spontaneous for T > 2000 K; at 2000 K it is at equilibrium.

NCERT Exercise 5.18 — Signs of ΔH and ΔS during Bond Formation

1Exercise question

Step-by-step solution

1The reaction forms a covalent bond, and bond formation releases energy.
2
$Δ H < 0$
3Two moles of gaseous atoms combine to form one mole of gaseous molecules, reducing the number of independently moving particles and the disorder.
4
$2 mol Cl (g) \to 1 mol C l_{2} (g) \Rightarrow Δ S < 0$

Final answer

Both enthalpy and entropy decrease, so ΔH < 0 and ΔS < 0.

NCERT Exercise 5.19 — Gibbs Energy from ΔU and ΔS

1Exercise question

Step-by-step solution

1First convert ΔU° to ΔH° by counting the change in gaseous amount:
2
$Δ n_{g} = 2 - (2 + 1) = - 1 mol$
3
$Δ H^{\circ} = Δ U^{\circ} + Δ n_{g} R T$
4
$Δ H^{\circ} = - 10.5 + (- 1) (8.314 \times 1 0^{- 3}) (298) = - 12.9776 kJ$
5Convert the entropy change to the same energy unit:
6
$Δ S^{\circ} = - 44.1 J K^{- 1} = - 0.0441 kJ K^{- 1}$
7Apply the Gibbs-energy equation:
8
$Δ G^{\circ} = - 12.9776 - 298 (- 0.0441)$
9
$Δ G^{\circ} = - 12.9776 + 13.1418 = + 0.1642 kJ$

Final answer

ΔG° ≈ +0.16 kJ for the reaction as written, so it is not spontaneous under the stated standard conditions.

NCERT Exercise 5.20 — Gibbs Energy from the Equilibrium Constant

1Exercise question

Step-by-step solution

1Apply the equilibrium relation:
2
$Δ G^{\circ} = - R T ln K$
3Substitute K = 10, for which ln 10 ≈ 2.303:
4
$Δ G^{\circ} = - (8.314) (300) (2.303)$
5
$Δ G^{\circ} = - 5744.14 J mo l^{- 1}$
6Convert from joules to kilojoules:
7
$Δ G^{\circ} = - \frac{5744.14}{1000} = - 5.74414 kJ mo l^{- 1}$

Final answer

The standard Gibbs energy change is ΔG° = −5.744 kJ mol⁻¹.

NCERT Exercise 5.21 — Thermodynamic Stability of Nitric Oxide

1Exercise question

Step-by-step solution

1A positive standard formation enthalpy means that NO(g) lies 90 kJ mol⁻¹ higher in enthalpy than its constituent elements in their standard states.
2Thus NO is thermodynamically unstable relative to N₂(g) and O₂(g) on the enthalpy criterion.
3Oxidation of NO to NO₂ is exothermic, so NO₂ is lower in enthalpy than the NO plus O₂ reactants and NO is further stabilised by conversion to NO₂.
4The signs therefore explain the tendency of NO to undergo further oxidation, although spontaneity in every possible mixture would also require entropy and concentration data.

Final answer

NO(g) is unstable relative to its elements, while its conversion to the lower-enthalpy NO₂(g) is energetically favoured.

NCERT Exercise 5.22 — Entropy Change of the Surroundings

1Exercise question

Step-by-step solution

1The negative formation enthalpy means that the chemical system releases 286 kJ per mole.
2The surroundings absorb that heat, so q(surroundings) is positive:
3
$q_{surr} = + 286 kJ mo l^{- 1} = + 286000 J mo l^{- 1}$
4For a reservoir at constant temperature, apply the reversible heat-transfer relation:
5
$Δ S_{surr} = \frac{q _{surr, rev}}{T}$
6
$Δ S_{surr} = \frac{286000 J mo l ^{- 1}}{298 K} = 959.73 J mo l^{- 1} K^{- 1}$

Final answer

The entropy change of the surroundings is ΔS(surr) = +959.73 J mol⁻¹ K⁻¹.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Gibbs free energy

Δ G = Δ H - T Δ S

Gibbs energy and equilibrium

Δ G^{\circ} = - R T ln K

Heat capacity relation

C_{p} - C_{v} = R

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

A reaction is spontaneous when ΔG is negative, and that can be because ΔH is very negative or because TΔS is very positive — check both, not just enthalpy.
ΔH and ΔS are state functions, so Hess's law lets you build any reaction from known steps; the sign of ΔG flips with temperature, which is why some reactions run differently in summer.

FAQ

Frequently asked questions

How many questions are in NCERT Class 11 Chemistry Chapter 5 (Chemical Thermodynamics)?

There are 22 exercise questions in this chapter, numbered 5.1 to 5.22. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Chemical Thermodynamics Class 11 Chemistry?

The formulas this chapter's questions actually turn on are: Gibbs free energy, Gibbs energy and equilibrium, Heat capacity relation. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Chemical Thermodynamics important for JEE Main and NEET?

Very important — enthalpy, entropy and Gibbs energy form the thermodynamic core of both JEE and NEET, and ΔG = ΔH − TΔS sums are near-universal.

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