Question

For vaporization of water at 1 atmospheric pressure, the values of Δ H and Δ S are 40.63kJ mol⁻1 and 108.8J K⁻1mol⁻1, respectively. The temperature when Gibbs free energy change (Δ G) for this transformation will be zero, is:

• A. \(293.4\,\text{K}\)
• B. \(273.4\,\text{K}\)
• C. \(393.4\,\text{K}\)
• D. 373.4K

Hint:

At phase equilibrium, Δ G = 0.

Answer 1

The Correct Option is D

Step 1: At equilibrium, Δ G = 0 = Δ H - TΔ S
Step 2: T = (Δ H)/(Δ S) = (40.63×10³)/(108.8) ≈ 373.4K