NCERT Solutions for Class 11 Chemistry Chapter 5: Chemical Thermodynamics – Free PDF Download

Chapter 5 of NCERT Class 11 Chemistry, titled “Thermodynamics” (Unit 5 in the current 2026-27 reprint of Chemistry Part-I), builds on the idea of energy changes in chemical and physical processes, introducing state functions, the first law, enthalpy, Hess’s law, entropy and Gibbs energy so that students can predict whether a reaction is spontaneous.

NCERT Exercise Solutions

Question 5.1

Question: 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: The correct answer is (ii) whose value is independent of path. A state function depends only on the initial and final states of the system and not on the path taken to reach that state — internal energy (U), enthalpy (H), entropy (S) and Gibbs energy (G) are all state functions, whereas heat (q) and work (w) are path functions.

Question 5.2

Question: 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: The correct answer is (iii) q = 0. An adiabatic process is one in which no heat (q) is exchanged between the system and the surroundings, i.e. q = 0. (Note: options (i)–(iv) as printed in the textbook list (iv) as “w = 0”; the defining condition of an adiabatic process is q = 0.)

Question 5.3

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

Solution: The correct answer is (ii) zero. By convention, the standard enthalpy of formation of an element in its most stable (reference) state at 298 K and 1 bar pressure is taken as zero, since it forms the baseline from which enthalpies of formation of compounds are measured.

Question 5.4

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

Solution: The correct answer is (iii) < ΔU°. For the combustion of methane, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), the change in moles of gas is Δn_g = 1 − (1 + 2) = −2 (water is formed as a liquid, so it is not counted as a gas). Using ΔH° = ΔU° + Δn_gRT, since Δn_g is negative, ΔH° = ΔU° − 2RT, which is more negative than ΔU°. Given ΔU° = −X, ΔH° is therefore less than ΔU°.

Question 5.5

Question: 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: The correct answer is (i) −74.8 kJ mol^-1. The formation reaction is C(graphite) + 2H₂(g) → CH₄(g). By Hess’s law, Δ_fH(CH₄) = Δ_cH(C) + 2Δ_cH(H₂) − Δ_cH(CH₄) = (−393.5) + 2(−285.8) − (−890.3) = −393.5 − 571.6 + 890.3 = −74.8 kJ mol^-1.

Question 5.6

Question: 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: The correct answer is (iv) possible at any temperature. Since heat (q) is released, the reaction is exothermic, so ΔH is negative; it is also given that ΔS is positive. From ΔG = ΔH − TΔS, with ΔH negative and −TΔS also negative (since ΔS is positive), ΔG is negative at every temperature, so the reaction is spontaneous at any temperature.

Question 5.7

Question: 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: By the first law of thermodynamics, ΔU = q + w. Heat absorbed by the system, q = +701 J. Since work is done by the system (on the surroundings), w = −394 J. Therefore ΔU = 701 J + (−394 J) = 307 J.

Question 5.8

Question: 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: In a bomb calorimeter, volume is constant so the heat measured equals ΔU. To convert to ΔH, use ΔH = ΔU + Δn_gRT. Moles of gas: products = N₂ + CO₂ = 1 + 1 = 2 (H₂O is liquid, not counted); reactants = O₂ = 3/2. So Δn_g = 2 − 1.5 = 0.5. Using R = 8.314 × 10^-3 kJ mol^-1 K^-1 and T = 298 K: Δn_gRT = 0.5 × 8.314 × 10^-3 × 298 = 1.239 kJ mol^-1. Therefore ΔH = −742.7 + 1.239 = −741.5 kJ mol^-1 (approximately).

Question 5.9

Question: 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: Molar mass of Al ≈ 27 g mol^-1, so moles of Al = 60.0/27 = 2.22 mol. Rise in temperature, ΔT = 55 − 35 = 20°C = 20 K. Heat required, q = n × C_m × ΔT = 2.22 mol × 24 J mol^-1 K^-1 × 20 K = 1066.7 J ≈ 1.07 kJ.

Question 5.10

Question: Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at −10.0°C. Δ_fusH = 6.03 kJ mol^-1 at 0°C. C_p[H₂O(l)] = 75.3 J mol^-1 K^-1, C_p[H₂O(s)] = 36.8 J mol^-1 K^-1.

Solution: This is done in three steps using Hess’s law:
Step 1 – cool liquid water from 10°C to 0°C: ΔH₁ = C_p(l) × ΔT = 75.3 J mol^-1 K^-1 × (0 − 10) K = −753 J = −0.753 kJ.
Step 2 – freeze water to ice at 0°C: this is the reverse of fusion, so ΔH₂ = −Δ_fusH = −6.03 kJ.
Step 3 – cool ice from 0°C to −10°C: ΔH₃ = C_p(s) × ΔT = 36.8 J mol^-1 K^-1 × (−10 − 0) K = −368 J = −0.368 kJ.
Total enthalpy change = ΔH₁ + ΔH₂ + ΔH₃ = −0.753 − 6.03 − 0.368 = −7.15 kJ mol^-1 (approximately).

Question 5.11

Question: 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: Molar mass of CO₂ = 44 g mol^-1, so moles of CO₂ formed = 35.2/44 = 0.8 mol. Since 1 mol of CO₂ formation releases 393.5 kJ, heat released for 0.8 mol = 0.8 × 393.5 = 314.8 kJ.

Question 5.12

Question: 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 Δ_rH for the reaction: N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)

Solution: Δ_rH = [Δ_fH(N₂O) + 3Δ_fH(CO₂)] − [Δ_fH(N₂O₄) + 3Δ_fH(CO)] = [81 + 3(−393)] − [9.7 + 3(−110)] = [81 − 1179] − [9.7 − 330] = (−1098) − (−320.3) = −777.7 kJ mol^-1.

Question 5.13

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

Solution: The given reaction produces 2 mol of NH₃, so the enthalpy of formation (per mole) is half of Δ_rH°: Δ_fH°(NH₃) = −92.4/2 = −46.2 kJ mol^-1.

Question 5.14

Question: Calculate the standard enthalpy of formation of CH₃OH(l) from the following data: CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l); Δ_rH° = −726 kJ mol^-1. C(graphite) + O₂(g) → CO₂(g); Δ_cH° = −393 kJ mol^-1. H₂(g) + 1/2 O₂(g) → H₂O(l); Δ_fH° = −286 kJ mol^-1.

Solution: The target reaction is C(graphite) + 2H₂(g) + 1/2 O₂(g) → CH₃OH(l). By Hess’s law: Δ_fH°(CH₃OH) = Δ_cH°(C) + 2Δ_fH°(H₂O) − Δ_rH°(combustion of methanol) = (−393) + 2(−286) − (−726) = −393 − 572 + 726 = −239 kJ mol^-1.

Question 5.15

Question: Calculate the enthalpy change for the process CCl₄(g) → C(g) + 4 Cl(g) and calculate bond enthalpy of C–Cl in CCl₄(g). Δ_vapH°(CCl₄) = 30.5 kJ mol^-1. Δ_fH°(CCl₄) = −135.5 kJ mol^-1. Δ_aH°(C) = 715.0 kJ mol^-1, where Δ_aH° is enthalpy of atomisation. Δ_aH°(Cl₂) = 242 kJ mol^-1.

Solution: Combine the given steps by Hess’s law: reverse of vaporisation, CCl₄(g) → CCl₄(l), ΔH = −30.5 kJ; reverse of formation, CCl₄(l) → C(graphite) + 2Cl₂(g), ΔH = +135.5 kJ; atomisation of carbon, C(graphite) → C(g), ΔH = +715.0 kJ; atomisation of chlorine (2 mol Cl₂), 2Cl₂(g) → 4Cl(g), ΔH = 2 × 242 = +484 kJ. Adding these: ΔH = −30.5 + 135.5 + 715.0 + 484 = 1304 kJ mol^-1. This enthalpy corresponds to breaking 4 C–Cl bonds, so the average bond enthalpy of C–Cl = 1304/4 = 326 kJ mol^-1.

Question 5.16

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

Solution: In an isolated system there is no exchange of matter or energy with the surroundings. For any spontaneous process to occur in such a system, the entropy must increase, so ΔS will be positive (ΔS > 0).

Question 5.17

Question: 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: A reaction becomes spontaneous when ΔG = ΔH − TΔS < 0, i.e. when T > ΔH/ΔS. The boundary (equilibrium) temperature is T = ΔH/ΔS = 400 kJ mol^-1 / 0.2 kJ K^-1 mol^-1 = 2000 K. So the reaction becomes spontaneous at temperatures above 2000 K.

Question 5.18

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

Solution: Formation of a Cl–Cl bond from two free chlorine atoms releases energy, so ΔH is negative. Also, two moles of gaseous atoms combine to give one mole of gaseous molecules, which decreases the randomness/disorder of the system, so ΔS is negative.

Question 5.19

Question: 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: First find ΔH° from ΔU°. Δn_g = 2 (product D) − (2 + 1) (reactants A and B) = −1. At T = 298 K, ΔH° = ΔU° + Δn_gRT = −10.5 kJ + (−1)(8.314 × 10^-3 kJ mol^-1 K^-1)(298 K) = −10.5 − 2.478 = −12.98 kJ. Now ΔG° = ΔH° − TΔS° = −12.98 kJ − (298 K)(−44.1 × 10^-3 kJ K^-1) = −12.98 + 13.14 = +0.16 kJ (approximately). Since ΔG° is positive (though close to zero), the reaction is not spontaneous under standard conditions at 298 K.

Question 5.20

Question: The equilibrium constant for a reaction is 10. What will be the value of ΔG°? R = 8.314 JK^-1 mol^-1, T = 300 K.

Solution: Using ΔG° = −RT ln K = −2.303 RT log K: ΔG° = −2.303 × 8.314 J K^-1 mol^-1 × 300 K × log(10) = −2.303 × 8.314 × 300 × 1 = −5744.6 J mol^-1 ≈ −5.74 kJ mol^-1.

Question 5.21

Question: Comment on the thermodynamic stability of NO(g), given 1/2 N₂(g) + 1/2 O₂(g) → NO(g); Δ_rH° = 90 kJ mol^-1. NO(g) + 1/2 O₂(g) → NO₂(g); Δ_rH° = −74 kJ mol^-1.

Solution: The formation of NO(g) from N₂ and O₂ is endothermic (Δ_rH° = +90 kJ mol^-1), which means NO(g) is at a higher energy level than the elements it is formed from and is therefore thermodynamically unstable with respect to decomposition back into N₂ and O₂. Additionally, the oxidation of NO to NO₂ is exothermic (Δ_rH° = −74 kJ mol^-1), showing that NO is also unstable with respect to further oxidation and readily combines with more oxygen to form the more stable NO₂.

Question 5.22

Question: Calculate the entropy change in surroundings when 1.00 mol of H₂O(l) is formed under standard conditions. Δ_fH° = −286 kJ mol^-1.

Solution: Since the reaction is exothermic, heat equal to 286 kJ is released to the surroundings, i.e. q_surroundings = +286 kJ = 286000 J (at standard temperature, T = 298 K). Entropy change of the surroundings, ΔS_surr = −ΔH_system/T = 286000 J / 298 K = 959.7 J K^-1 mol^-1 (approximately).

Notes and Extra Questions

This chapter lays the conceptual foundation of chemical thermodynamics for Class 11: the distinction between system, surroundings, state functions and path functions; the first law of thermodynamics (ΔU = q + w) and its application to enthalpy changes at constant pressure; the use of Hess’s law to calculate reaction enthalpies indirectly (formation, combustion, atomisation and bond enthalpies); and finally entropy and Gibbs energy as criteria for predicting the spontaneity of a process. Numerical questions in this chapter regularly combine two or more of these ideas (for example, converting ΔU to ΔH using Δn_gRT, or using both ΔH and ΔS to compute ΔG), so students should practise unit consistency (J vs kJ) and sign conventions carefully, since these are the most common sources of error.

Frequently Asked Questions

How many exercise questions are there in NCERT Class 11 Chemistry Chapter 5, Thermodynamics?
The current (2026-27 reprint) NCERT textbook has 22 exercise questions in this chapter, numbered 5.1 to 5.22, covering multiple-choice conceptual questions as well as numerical problems on the first law, enthalpy, Hess’s law, entropy and Gibbs energy.

Is this chapter called “Thermodynamics” or “Chemical Thermodynamics” in the current NCERT book?
In the current, live NCERT PDF (file kech105.pdf, Chemistry Part-I, Reprint 2026-27) the chapter is titled simply “Thermodynamics” (Unit 5). “Chemical Thermodynamics” is not the title used in this current edition, though the chapter is sometimes referred to informally by that name since it deals specifically with energy changes in chemical systems.

What is the difference between ΔU and ΔH, and why does it matter in this chapter’s numericals?
ΔU is the change in internal energy (heat exchanged at constant volume, as measured in a bomb calorimeter), while ΔH is the change in enthalpy (heat exchanged at constant pressure, the usual laboratory condition). They are related by ΔH = ΔU + Δn_gRT, where Δn_g is the change in moles of gaseous species; several exercise questions (5.4, 5.8, 5.19) specifically test this conversion.

Which formula is used to decide whether a reaction is spontaneous?
Spontaneity is determined using the Gibbs energy change, ΔG = ΔH − TΔS. A reaction is spontaneous when ΔG is negative, non-spontaneous when ΔG is positive, and at equilibrium when ΔG = 0 (which also gives the relation ΔG° = −RT ln K linking Gibbs energy to the equilibrium constant, used in Question 5.20).