CBSE Class 11 Chemistry Thermodynamics Worksheet Set 06

Thermodynamics Worksheet: Short and Long Answer Questions

Question 1. What is a state function? Neither q nor w is a state function, but q + w is. Explain.
Answer:
A state function is a property whose value depends only on the present state of the system and not on the path by which that state was reached. Examples are p, V, T, U, H, S and G.
The heat absorbed (q) and the work done (w) change when the same initial state is joined to the same final state by different paths, so each is a path function.
By the first law, ΔU = q + w. Internal energy is a state function, so ΔU is fixed by the initial and final states. When q increases along one path, w changes by the opposite amount, so the sum stays the same.
In simple words: Heat and work depend on the route, but their sum depends only on where you start and finish.

Teacher's Note:
1. Write the first law ΔU = q + w in the answer.
2. Always name one path function and one state function as examples.

Question 2. 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:
Heat is absorbed by the system, so q = +701 J. Work is done by the system, so w = −394 J (IUPAC convention).
ΔU = q + w = 701 + (−394) = +307 J.
In simple words: The system gained 701 J as heat and lost 394 J as work, so it keeps 307 J.

Teacher's Note:
1. Decide the sign of each quantity from the direction of energy flow.
2. Work done by the system is negative in the IUPAC convention.

Question 3. A gas in a cylinder with a frictionless piston expands against a constant external pressure of 1 atm from 2 L to 6 L. During the expansion the gas absorbs 800 J of heat from the surroundings. Find the increase in internal energy. (1 L atm = 101.3 J)
Answer:
ΔV = 6 − 2 = 4 L and p_ex = 1 atm.
w = −p_exΔV = −1 × 4 = −4 L atm = −4 × 101.3 J ≈ −405 J.
q = +800 J, so ΔU = q + w = 800 − 405 = +395 J (about 394.7 J using 101.325 J per L atm).
In simple words: Of the 800 J absorbed, about 405 J is spent pushing the piston and the rest raises the internal energy.

Teacher's Note:
1. Convert L atm to joules before adding to q.
2. Expansion gives a negative w.

Question 4. Two moles of a perfect gas undergo the following reversible processes. (a) Isobaric expansion from (1.0 atm, 20.0 L) to (1.0 atm, 40.0 L). (b) Isochoric change from (1.0 atm, 40.0 L) to (0.5 atm, 40.0 L). (c) Isothermal compression from (0.5 atm, 40.0 L) to (1.0 atm, 20.0 L). The three processes are shown on the p–V diagram below as A→B, B→C and C→A. Calculate (i) the total work w and total heat q, and (ii) ΔU, ΔH and ΔS for the whole cycle.
Find (i) total work and heat, (ii) ΔU, ΔH and ΔS for the cycle.
Answer:
(i) A→B: w = −pΔV = −1.0 × (40 − 20) = −20 L atm = −2026.5 J.
B→C: ΔV = 0, so w = 0.
C→A: isothermal, so nRT = pV = 0.5 × 40 = 20 L atm. w = −2.303 nRT log (V_f/V_i) = −2.303 × 20 × log (20/40) = +13.86 L atm = +1404.5 J (work done on the gas).
Total w = −2026.5 + 0 + 1404.5 ≈ −622 J. The cycle returns to the start, so ΔU = 0 and q = −w = +622 J.
(ii) U, H and S are state functions and the gas returns to its initial state, so ΔU = ΔH = ΔS = 0.
In simple words: Over one complete loop the gas does about 622 J of net work, taking in the same amount as heat.

Teacher's Note:
1. Use nRT = pV to avoid finding the temperature.
2. The sum of the three works equals minus the enclosed area on the p–V diagram.

Question 5. A heated copper block at 130 °C loses 340 J of heat to the surroundings at room temperature 32 °C. Calculate the entropy change of (i) the block, (ii) the surroundings and (iii) the universe. Assume both temperatures stay constant.
Answer:
T_block = 130 + 273 = 403 K and T_surr = 32 + 273 = 305 K. The block loses heat, so q_sys = −340 J; the surroundings gain it, so q_surr = +340 J.
(i) ΔS_system = q/T = −340/403 = −0.84 J K⁻¹.
(ii) ΔS_surr = +340/305 = +1.11 J K⁻¹.
(iii) ΔS_total = −0.84 + 1.11 = +0.27 J K⁻¹. It is positive, so the heat flow is spontaneous, as the second law requires.
In simple words: The cooler surroundings gain more entropy than the hot block loses, so the total entropy rises.

Teacher's Note:
1. Use the temperature at which each body exchanges heat.
2. ΔS_total > 0 means a spontaneous process.

Question 6. 5.6 dm³ of an unknown gas at STP needs 52.25 J of heat to raise its temperature by 10 °C at constant volume. Calculate C_V and C_p in J K⁻¹ mol⁻¹. (R = 8.314 J K⁻¹ mol⁻¹)
Answer:
Moles of gas = 5.6/22.4 = 0.25 mol.
q_V = n C_V ΔT, so C_V = 52.25 / (0.25 × 10) = 20.9 J K⁻¹ mol⁻¹.
C_p = C_V + R = 20.9 + 8.314 = 29.2 J K⁻¹ mol⁻¹.
In simple words: The heat needed per mole per kelvin at constant volume is 20.9 J, and at constant pressure it is larger by R.

Teacher's Note:
1. At STP one mole of gas occupies 22.4 L.
2. C_p − C_V = R holds for an ideal gas.

Question 7. The combustion of 1 mol of benzene, C₆H₆(l), at 298 K and 1 atm gives CO₂(g) and H₂O(l), and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation of benzene. Δ_fH°(CO₂) = −393.5 kJ mol⁻¹ and Δ_fH°(H₂O, l) = −285.83 kJ mol⁻¹.
Answer:
Combustion: C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3H₂O(l); Δ_cH° = −3267.0 kJ mol⁻¹.
Δ_cH° = [6Δ_fH°(CO₂) + 3Δ_fH°(H₂O)] − Δ_fH°(C₆H₆)
−3267.0 = 6(−393.5) + 3(−285.83) − Δ_fH°(C₆H₆) = −2361.0 − 857.49 − Δ_fH°(C₆H₆)
Δ_fH°(C₆H₆) = −3218.49 + 3267.0 = +48.51 kJ mol⁻¹.
The positive sign shows that benzene is formed endothermically from graphite and hydrogen.
In simple words: Benzene stores about 48.5 kJ more enthalpy per mole than carbon and hydrogen do.

Teacher's Note:
1. Δ_fH° of O₂(g) is zero.
2. Check the sign: the combustion enthalpy of the compound is less negative than the sum for its elements, so Δ_fH° is positive.

Question 8. Calculate the standard enthalpy of formation of CS₂(l) from the standard enthalpies of combustion of C, S and CS₂, which are −393.3, −293.72 and −1108.76 kJ mol⁻¹ respectively (CS₂ burns to CO₂ and SO₂).
Answer:
(1) C + O₂ → CO₂; ΔH = −393.3 kJ
(2) S + O₂ → SO₂; ΔH = −293.72 kJ
(3) CS₂ + 3O₂ → CO₂ + 2SO₂; ΔH = −1108.76 kJ
Target: C + 2S → CS₂. Add (1) and 2 × (2): C + 2S + 3O₂ → CO₂ + 2SO₂; ΔH = −393.3 − 587.44 = −980.74 kJ.
Subtract (3): C + 2S → CS₂; ΔH = −980.74 − (−1108.76) = +128.02 kJ mol⁻¹.
In simple words: CS₂ is an endothermic compound, formed with absorption of 128 kJ per mole.

Teacher's Note:
1. Multiply the sulphur equation by 2 because two S atoms are needed.
2. Subtracting an equation means reversing it and adding.

Question 9. Calculate the entropy change for the rusting of iron, 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s), ΔH° = −1648 kJ. The standard entropies are Fe 27.3, O₂ 205.0 and Fe₂O₃ 87.4 J K⁻¹ mol⁻¹. Is the reaction spontaneous at 25 °C?
Answer:
Δ_rS° = 2(87.4) − [4(27.3) + 3(205.0)] = 174.8 − 724.2 = −549.4 J K⁻¹.
Δ_rG° = ΔH° − TΔS° = −1 648 000 J − 298 K × (−549.4 J K⁻¹) = −1 648 000 + 163 721 = −1 484 279 J ≈ −1484 kJ.
ΔG° is negative, so the reaction is spontaneous even though ΔS is negative. The large negative ΔH outweighs the unfavourable entropy term.
In simple words: Rusting loses entropy but releases so much heat that the Gibbs energy still falls.

Teacher's Note:
1. Convert kJ to J before combining with J K⁻¹.
2. Check the sign of ΔG, not of ΔS, to judge spontaneity.

Question 10. Calculate the enthalpy of formation of methane from CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), Δ_rH = −890.5 kJ. Given Δ_fH°(CO₂) = −393.5 kJ mol⁻¹ and Δ_fH°(H₂O, l) = −286.0 kJ mol⁻¹.
Answer:
Δ_rH = [Δ_fH°(CO₂) + 2Δ_fH°(H₂O)] − [Δ_fH°(CH₄) + 2Δ_fH°(O₂)]
−890.5 = −393.5 − 572.0 − Δ_fH°(CH₄) − 0
Δ_fH°(CH₄) = −965.5 + 890.5 = −75.0 kJ mol⁻¹.
In simple words: Forming methane from graphite and hydrogen releases 75 kJ per mole.

Teacher's Note:
1. O₂ is an element, so its formation enthalpy is zero.
2. Substitute carefully with signs.

Question 11. For the reaction NH₄Cl(s) → NH₃(g) + HCl(g) at 25 °C, ΔH = 177 kJ mol⁻¹ and ΔS = 285 J K⁻¹ mol⁻¹. Calculate ΔG and predict whether the reaction is spontaneous.
Answer:
T = 25 + 273 = 298 K. TΔS = 298 × 285 J = 84 930 J = 84.93 kJ.
ΔG = ΔH − TΔS = 177 − 84.93 = +92.07 kJ mol⁻¹.
ΔG is positive, so the decomposition is non-spontaneous at 25 °C.
In simple words: At 25 °C the entropy gain is too small to pay for the large enthalpy cost.

Teacher's Note:
1. Keep ΔH and TΔS in the same unit.
2. A positive ΔG means non-spontaneous in the forward direction.

Question 12. Calculate the temperature above which the reduction PbO(s) + C(s) → Pb(s) + CO(g) becomes spontaneous. ΔH = 108.4 kJ mol⁻¹ and ΔS = 190 J K⁻¹ mol⁻¹.
Answer:
At the changeover ΔG = 0, so ΔH = TΔS.
T = ΔH/ΔS = 108 400 J mol⁻¹ / 190 J K⁻¹ mol⁻¹ = 570.5 K.
ΔH and ΔS are both positive, so ΔG = ΔH − TΔS becomes negative above this temperature. The reaction is spontaneous above 570.5 K.
In simple words: A high temperature makes the TΔS term large enough to beat the positive ΔH.

Teacher's Note:
1. For ΔH > 0 and ΔS > 0, the reaction is spontaneous only at high temperature.
2. Use ΔH in J when ΔS is in J K⁻¹ mol⁻¹.

Question 13. What is entropy? Why is the entropy of a perfect crystal taken as zero at 0 K? Calculate Δ_rG° at 298 K for N₂(g) + 3H₂(g) → 2NH₃(g) if K = 6.6 × 10⁵. (R = 8.314 J K⁻¹ mol⁻¹)
Answer:
Entropy is a state function that measures the randomness (disorder) of a system.
At 0 K a perfect crystal has all particles in fixed, ordered positions with no thermal motion, so the disorder is zero. This is the third law of thermodynamics.
Δ_rG° = −2.303 RT log K = −2.303 × 8.314 × 298 × log (6.6 × 10⁵)
log (6.6 × 10⁵) = 0.8195 + 5 = 5.8195, so Δ_rG° = −5705.8 × 5.8195 = −33 205 J = −33.2 kJ mol⁻¹.
In simple words: A large K corresponds to a large negative standard Gibbs energy change.

Teacher's Note:
1. Δ_rG° = −RT ln K links thermodynamics with equilibrium.
2. Check that the sign is negative when K > 1.

Question 14. Calculate the equilibrium constant at 400 K for 2NOCl(g) → 2NO(g) + Cl₂(g), given ΔH° = 77.2 kJ mol⁻¹ and ΔS° = 122 J K⁻¹ mol⁻¹. (R = 8.314 J K⁻¹ mol⁻¹)
Answer:
ΔG° = ΔH° − TΔS° = 77.2 − (400 × 122)/1000 = 77.2 − 48.8 = 28.4 kJ mol⁻¹ = 28 400 J mol⁻¹.
ΔG° = −2.303 RT log K, so log K = −28 400 / (2.303 × 8.314 × 400) = −3.708.
K = antilog (−3.708) = 1.96 × 10⁻⁴.
K is much less than 1, so the products are not favoured at 400 K.
In simple words: A positive ΔG° gives a K smaller than 1.

Teacher's Note:
1. Take care with the exponent: a negative logarithm gives a number below 1.
2. Convert ΔG° to J before using R in J.

Question 15. Justify the statements. (i) An endothermic reaction is always thermodynamically non-spontaneous. (ii) The entropy always increases when a liquid changes to vapour at any temperature.
Answer:
(i) The statement is false. Spontaneity depends on ΔG = ΔH − TΔS. If ΔS is positive and TΔS is greater than ΔH, then ΔG is negative even when ΔH is positive. Example: NH₄Cl dissolving in water, or any decomposition at high temperature.
(ii) The statement is true. The molecules of a vapour move freely and are far more disordered than those of the liquid. So S(vapour) > S(liquid) and Δ_vapS = Δ_vapH/T_b is positive.
In simple words: Endothermic does not mean impossible, because a gain in disorder can drive the change.

Teacher's Note:
1. Quote ΔG = ΔH − TΔS in the answer.
2. Give one example for each justification.