For vaporization of water at 1 atmospher...

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

`273.4K`

`393.4K`

`373.4K`

`293.4K`

Text Solution

Verified by Experts

The correct Answer is:

`DeltaG=DeltaH-TDeltaS`
`Delta=0` , `:. DeltaH=TDeltaS`
`T=(DeltaH)/(DeltaS)=(40.63xx10^(3))/(108.8)=373.4K`

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