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

Question

Dr. Abubacker · Accepted Answer

[Solution from source] ΔG = ΔH − TΔS = 0 gives T = ΔH/ΔS = 40630/108.8 = 373.4 K. Forgetting to convert kJ to J (or mixing units) gives nonsensical values; 373 K is also the normal boiling point of water, as expected. Answer (3).

Q. For vaporization of water at 1 atmospheric pressure, the values of ΔH and ΔS are 40.63 kJ mol⁻¹ and 108.8 J K⁻¹ mol⁻¹, respectively. The temperature when Gibbs energy change (ΔG) for this transformation will be zero, is

Explanation

Reviewed by Dr. Abubacker

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