For vapourisation of water at 1 atm,values of `DeltaH `&`DeltaS`are `40.6 Kjmol^(-1)` and `108JK^(-1) mol^(-1)` ,respectively will be. The temperature when `DeltaG for this transition zero is :-

Options

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

Answer

D

Explanation

`H_(2) O_((l)) overset(1 "atm")(hArrr) H_(2) O_((g))`
`Delta H = 40630 "J mol"^(-1)`
`Delta S = 108.8 "JK mol"^(-1)`
`Delta G = Delta H - T Delta S`.
(When `Delta G = 0, Delta H - T Delta S= 0`)
`T= (Delta H)/( Delta S)= (40630 "J mol"^(-1) )/( 108.8 "J mol"^(-1) )`
`= 373.4` K
`therefore` Correct answer is (D)