For vaporization of water at 1 atmospheric pressure the values of `DeltaH` and `DeltaS` are `40.63KJmol^(-1)` and `108JK^(-1)mol^(-1)` , respectively. The temperature when Gibbs energy change `(DeltaG)` for this transformation will be zero is

Options

A. 273.4 K
B. 393.4 K
C. 373.4 K
D. 293.4 K

Answer

C

Explanation

According to Gibbs equation `Delta G = Delta H - T Delta S`
Given `Delta H = 40.63 "KJ mol"^(-1) = 40.63 xx 10^(-3) " J mol "^(-1) "and " Delta S = 108.8 JK^(-1) "mol "^(-1)`
`therefore R (Delta H)/(Delta S) = (40 .63 xx 10^3)/(108.8) =373.43 K`