Standard entropies of X₂, Y₂ and XY₃ are 60, 40 and 50 J K⁻¹ mol⁻¹ respectively. For the reaction (1/2)X₂ + (3/2)Y₂ ⇌ XY₃, ΔH = −30 kJ to be at equilibrium, the temperature should be

Question

Dr. Abubacker · Accepted Answer

[Solution from source] ΔS° = 50 − (½ × 60 + 3/2 × 40) = 50 − 90 = −40 J K⁻¹ mol⁻¹. At equilibrium ΔG = 0, so T = ΔH/ΔS = (−30000)/(−40) = 750 K. Forgetting the stoichiometric halves gives a different ΔS and a wrong T. Answer (2).

Q. Standard entropies of X₂, Y₂ and XY₃ are 60, 40 and 50 J K⁻¹ mol⁻¹ respectively. For the reaction (1/2)X₂ + (3/2)Y₂ ⇌ XY₃, ΔH = −30 kJ to be at equilibrium, the temperature should be

Explanation

Reviewed by Dr. Abubacker

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