Question

The standard enthalpy and entropy changes for the reaction in equilibrium for the forward reaction are given below.

CO (g) + H₂O (g) CO₂ (g) + H₂ (g)

DHŗ_300K = – 41.16 kJ mol^–1

DSŗ_300K = – 4.24 × 10^–2 kJ mol^–1

DHŗ_1200K = – 32.93 kJ mol^–1

Then, the incorrect statement is –

• A. The reaction proceeds in the forward direction at 300 K
• B. At 1200 K, reaction proceeds in the reverse direction
• C. At 1200 K, Kp> 1
• D. At 300 K, the products will be favoured more than reactants at equilibrium

Answer 1

Answer: (C)

At 1200 K, Kp> 1

Answer 2

At 300 K, DG⁰ = – 41.16

– (300 × – 4.24 × 10^–2) < 0 Ž spontaneous

At 1200K, DG⁰ = – 32.93

– (1200 × (–4.24 × 10^–2)) > 0

Ž Non-spontaneous