For the gaseous reaction: CO + H₂O ⇌ CO₂ + H₂, the following thermodynamical data are given
ΔH°₃₀₀ ₖ = -41.0 kJ mol⁻¹; ΔS°₃₀₀ ₖ = -0.04 kJ K⁻¹ mol⁻¹
ΔH°₁₂₀₀ ₖ = -33.0 kJ mol⁻¹; ΔS°₁₂₀₀ ₖ = -0.03

Question

For the gaseous reaction: CO + H₂O ⇌ CO₂ + H₂, the following thermodynamical data are given
ΔH°₃₀₀ ₖ = -41.0 kJ mol⁻¹; ΔS°₃₀₀ ₖ = -0.04 kJ K⁻¹ mol⁻¹
ΔH°₁₂₀₀ ₖ = -33.0 kJ mol⁻¹; ΔS°₁₂₀₀ ₖ = -0.03 kJ K⁻¹ mol⁻¹
Assuming partial pressure of each component is 1 bar, the direction of spontaneous reaction is/are

Prashant Kotian · Accepted Answer

Correct answer: A,D. Since all partial pressures are 1 bar, Q = 1, and ΔG = ΔG°. At 300 K, ΔG° = -41.0 - 300(-0.04) = -41.0 + 12.0 = -29.0 kJ/mol. Since ΔG < 0, it is spontaneous in the forward direction. At 1200 K, ΔG° = -33.0 - 1200(-0.03) = -33.0 + 36.0 = +3.0 kJ/mol. Since ΔG > 0, it is spontaneous in the backward direction. Therefore, correct answer is A,D.

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