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 :

273.4K

393.4K

373.4K

293.4K

Text Solution

Verified by Experts

The correct Answer is:

`Delta G = Delta H- T Delta S`
`Delta G= 0`
`Delta H= T Delta S`
`T= (40.63 xx 10^(3))/(108.8) = 373.4K`

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