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NCERT Solutions · Class 11 Chemistry · NCERT Chemistry Part I and II Part I · Chapter 5

Chapter 5: Thermodynamics (Thermodynamics)

Step-by-step solutions to all 22 exercises of Unit 5, Thermodynamics (NCERT Class 11 Chemistry Part I, 2026-27): state functions, first law, ΔH and ΔU, heat capacity, Hess's law and enthalpies of formation and combustion, bond enthalpy, entropy and spontaneity, Gibbs energy and equilibrium constant. All 22 questions are answered, with the key answer highlighted.

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First law: $Δ U = q + w$ (w is work done on the system). $Δ H = Δ U + Δ n_{g} R T$ . Hess's law: $Δ_{r} H = \sum Δ_{f} H (products) - \sum Δ_{f} H (reactants)$ . $Δ G = Δ H - T Δ S$ ; spontaneous if ΔG < 0. $Δ G^{\circ} = - 2.303 R T lo g K$ . R = 8.314 J K⁻¹ mol⁻¹.

Exercises

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 path (iii) used to determine pressure volume work (iv) whose value depends on temperature only.

Solution

A state function depends only on the initial and final states of the system, not on how the change is made.

(ii)

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

Solution

Adiabatic means no heat exchange between system and surroundings.

(iii) q = 0

5.3

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

Solution

By convention, the standard enthalpy of formation of an element in its reference state is zero.

(ii) zero

5.4

ΔU° of combustion of methane is – X kJ mol⁻¹. The value of ΔH° is (i) = ΔU° (ii) > ΔU° (iii) < ΔU° (iv) = 0

Solution

$CH X_{4} (g) + 2 O X_{2} (g) CO X_{2} (g) + 2 H X_{2} O (l)$ : $Δ n_{g} = 1 - 3 = - 2$ .

$Δ H^{\circ} = Δ U^{\circ} + Δ n_{g} R T = - X - 2 R T$ , which is more negative than $Δ U^{\circ}$ .

(iii) ΔH° < ΔU°

5.5

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.

Solution

$C + 2 H X_{2} CH X_{4}$ : $Δ_{f} H = Δ_{c} H (C) + 2 Δ_{c} H (H X_{2}) - Δ_{c} H (CH X_{4})$

$= - 393.5 + 2 (- 285.8) + 890.3 = - 393.5 - 571.6 + 890.3 = - 74.8$ kJ mol⁻¹.

(i) −74.8 kJ mol⁻¹

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 (iv) possible at any temperature

Solution

"+ q" on the product side means heat is released: ΔH < 0. With ΔS > 0, $Δ G = Δ H - T Δ S$ is negative at every temperature.

(iv) possible at any temperature

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?

Solution

$q = + 701$ J (absorbed), $w = - 394$ J (work done by the system).

$Δ U = q + w = 701 - 394$

ΔU = +307 J

5.8

The reaction of cyanamide, NH₂CN (s), with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be –742.7 kJ mol⁻¹ at 298 K. Calculate enthalpy change for the reaction at 298 K. NH₂CN(g) + 3/2 O₂(g) → N₂(g) + CO₂(g) + H₂O(l)

Solution

Using the equation as written: $Δ n_{g} = (1 + 1) - (1 + 1.5) = - 0.5$ .

$Δ H = Δ U + Δ n_{g} R T = - 742.7 + (- 0.5) (8.314 \times 1 0^{- 3}) (298) = - 742.7 - 1.24$

(If cyanamide is taken as a solid, $Δ n_{g} = + 0.5$ and ΔH = −741.5 kJ mol⁻¹.)

ΔH = −743.9 kJ mol⁻¹

5.9

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.

Solution

$n = \frac{60.0}{27} = 2.22$ mol, $Δ T = 20$ K. $q = n C_{m} Δ T = 2.22 \times 24 \times 20 = 1066.7$ J.

1.067 kJ

5.10

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0 °C to ice at –10.0 °C. ΔfusH = 6.03 kJ mol⁻¹ at 0 °C. Cp [H₂O(l)] = 75.3 J mol⁻¹ K⁻¹, Cp [H₂O(s)] = 36.8 J mol⁻¹ K⁻¹

Solution

H is a state function, so use a path through 0 °C:

(1) Water 10 °C → 0 °C: $75.3 \times (- 10) = - 753$ J.

(2) Freezing at 0 °C: $- 6030$ J.

(3) Ice 0 °C → −10 °C: $36.8 \times (- 10) = - 368$ J.

$Δ H = - 753 - 6030 - 368 = - 7151$ J.

ΔH = −7.151 kJ mol⁻¹

5.11

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.

Solution

$n (CO X_{2}) = \frac{35.2}{44} = 0.8$ mol. Heat released $= 0.8 \times 393.5$ .

314.8 kJ

5.12

Enthalpies of formation of CO(g), CO₂(g), N₂O(g) and N₂O₄(g) are –110, – 393, 81 and 9.7 kJ mol⁻¹ respectively. Find the value of ΔrH for the reaction: N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)

Solution

$Δ_{r} H = [81 + 3 (- 393)] - [9.7 + 3 (- 110)] = (81 - 1179) - (9.7 - 330) = - 1098 + 320.3$

−777.7 kJ mol⁻¹

5.13

Given N₂(g) + 3H₂(g) → 2NH₃(g); ΔrH° = –92.4 kJ mol⁻¹. What is the standard enthalpy of formation of NH₃ gas?

Solution

The enthalpy of formation refers to 1 mol of NH₃: $\frac{- 92.4}{2}$ .

−46.2 kJ mol⁻¹

5.14

Calculate the standard enthalpy of formation of CH₃OH(l) from the following data: CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l); ΔrH° = –726 kJ mol⁻¹. C(graphite) + O₂(g) → CO₂(g); ΔcH° = –393 kJ mol⁻¹. H₂(g) + ½O₂(g) → H₂O(l); ΔfH° = –286 kJ mol⁻¹.

Solution

Target: $C + 2 H X_{2} + \frac{1}{2} O X_{2} CH X_{3} OH (l)$ = (eq 2) + 2 × (eq 3) − (eq 1).

$Δ_{f} H^{\circ} = - 393 + 2 (- 286) - (- 726) = - 393 - 572 + 726$

−239 kJ mol⁻¹

5.15

Calculate the enthalpy change for the process CCl₄(g) → C(g) + 4 Cl(g) and calculate bond enthalpy of C – Cl in CCl₄(g). ΔvapH°(CCl₄) = 30.5 kJ mol⁻¹. ΔfH°(CCl₄) = –135.5 kJ mol⁻¹. ΔaH°(C) = 715.0 kJ mol⁻¹, where ΔaH° is enthalpy of atomisation. ΔaH°(Cl₂) = 242 kJ mol⁻¹

Solution

ΔfH° of CCl₄(g) $= - 135.5 + 30.5 = - 105.0$ kJ mol⁻¹ (liquid → gas).

Cycle: CCl₄(g) → C(graphite) + 2Cl₂(g): +105.0; C(graphite) → C(g): +715.0; 2Cl₂ → 4Cl: $2 \times 242 = + 484$ .

$Δ H = 105.0 + 715.0 + 484 = 1304$ kJ mol⁻¹. Four C–Cl bonds are broken: bond enthalpy $= \frac{1304}{4} = 326$ kJ mol⁻¹.

ΔH = 1304 kJ mol⁻¹; C–Cl bond enthalpy = 326 kJ mol⁻¹

5.16

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

Solution

An isolated system exchanges neither energy nor matter (ΔU = 0). Any change in it can only be spontaneous, so its entropy increases: ΔS > 0 (ΔS = 0 at equilibrium). E.g. two gases mixing in an isolated container.

ΔS > 0 (positive)

5.17

For the reaction at 298 K, 2A + B → C, ΔH = 400 kJ mol⁻¹ and ΔS = 0.2 kJ K⁻¹ mol⁻¹. At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range.

Solution

ΔG = 0 at $T = \frac{Δ H}{Δ S} = \frac{400}{0.2} = 2000$ K. Both ΔH and ΔS are positive, so ΔG < 0 above this temperature.

Above 2000 K

5.18

For the reaction, 2 Cl(g) → Cl₂(g), what are the signs of ΔH and ΔS?

Solution

ΔH is negative: a bond forms, releasing energy. ΔS is negative: two free atoms become one molecule, so randomness decreases.

ΔH < 0 and ΔS < 0

5.19

For the reaction 2 A(g) + B(g) → 2D(g), ΔU° = –10.5 kJ and ΔS° = –44.1 J K⁻¹. Calculate ΔG° for the reaction, and predict whether the reaction may occur spontaneously.

Solution

$Δ n_{g} = 2 - 3 = - 1$ . $Δ H^{\circ} = Δ U^{\circ} + Δ n_{g} R T = - 10.5 + (- 1) (8.314 \times 1 0^{- 3}) (298) = - 12.98$ kJ.

$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} = - 12.98 - 298 \times (- 0.0441) = - 12.98 + 13.14 = + 0.164$ kJ.

ΔG° = +0.164 kJ: not spontaneous

5.20

The equilibrium constant for a reaction is 10. What will be the value of ΔG°? R = 8.314 J K⁻¹ mol⁻¹, T = 300 K.

Solution

$Δ G^{\circ} = - 2.303 R T lo g K = - 2.303 \times 8.314 \times 300 \times lo g 10 = - 5744$ J mol⁻¹.

−5.744 kJ mol⁻¹

5.21

Comment on the thermodynamic stability of NO(g), given ½N₂(g) + ½O₂(g) → NO(g); ΔrH° = 90 kJ mol⁻¹. NO(g) + ½O₂(g) → NO₂(g): ΔrH° = –74 kJ mol⁻¹

Solution

The formation of NO is endothermic (+90 kJ): NO lies at a higher energy than its elements and is thermodynamically unstable. Its oxidation to NO₂ is exothermic (−74 kJ), so NO readily changes into the more stable NO₂ (ΔfH° of NO₂ = 90 − 74 = +16 kJ mol⁻¹).

NO is unstable (endothermic formation); it tends to change into NO₂.

5.22

Calculate the entropy change in surroundings when 1.00 mol of H₂O(l) is formed under standard conditions. ΔfH° = –286 kJ mol⁻¹.

Solution

The heat released goes to the surroundings: $q_{surr} = + 286$ kJ mol⁻¹ at 298 K.

$Δ S_{surr} = \frac{q _{surr}}{T} = \frac{286000}{298}$

+959.7 J K⁻¹ mol⁻¹

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