Indian Forest Service (Mains), Chemistry optional · 2024 · Paper I
Q1(b). Entropy of water-vapour formation at 373 K
Question
For H₂(g) + ½O₂(g) → H₂O(g), calculate the standard entropy change at 373 K. Given S°₂₉₈ for H₂O(g), O₂(g) and H₂(g) are 188.825, 205.138 and 130.684 J K⁻¹ mol⁻¹. Their constant-pressure molar heat capacities are 33.58, 29.37 and 28.84 J K⁻¹ mol⁻¹, respectively.
Answer in brief
Assuming constant heat capacities, ΔᵣS°(373 K) = −46.66 J mol⁻¹ K⁻¹.
Key points
- At 298 K, ΔᵣS° = 188.825 − 130.684 − ½(205.138) = −44.428 J mol⁻¹ K⁻¹.
- The reaction heat-capacity change is ΔCₚ = 33.58 − 28.84 − ½(29.37) = −9.945 J mol⁻¹ K⁻¹.
- Since (∂S/∂T)ₚ = Cₚ/T, ΔᵣS°(T₂) = ΔᵣS°(T₁) + ∫ (ΔCₚ/T) dT from T₁ to T₂.
- Thus ΔᵣS°(373) = −44.428 − 9.945 ln(373/298) = −46.6605 J mol⁻¹ K⁻¹.
- The negative entropy change is consistent with converting 1.5 mol gaseous reactants into 1 mol gaseous product.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
Coversentropy change temperature dependenceheat capacitystandard molar entropyformation of water vapourKirchhoff-type relation for entropy
All parts of Question 1
- 1(a) Free-particle waves and de Broglie relation8 marks
- 1(b) Entropy of water-vapour formation at 373 K8 marks
- 1(c) Primary quantum yield8 marks
- 1(d) Components in ammonium chloride dissociation8 marks
- 1(e) Aluminium–magnesium galvanic cell8 marks