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For vaporization of water at 1 atmospher...

For vaporization of water at 1 atmospheric pressure, the values of `DeltaH and DeltaS` are `40.63 "kJ mol"^(-1) and 108.8 "JK"^(-1) mol^(-1)` respectively. The temperature when Gibbs energy change `(DeltaG)` for this transformation will be zero , is :

Text Solution

Verified by Experts

The correct Answer is:
373.4

`H_(2)O(l) overset(1 atm) `DeltaH = 406.30 J mol^(-1)`
`DeltaS = 108.8 JK^(-1) mol^(-1)`
`DeltaG = DeltaH - T DeltaS`
When `DeltaG =0`,
`DeltaH -TDeltaS=0`
`T = (DeltaH)/(DeltaS) = (40630 J "mol"^(-1))/(108.8 J "mol"^(-1)) = 373.4` K
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