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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 :

A

273.4 K

B

393.4 K

C

373.4 K

D

293.4 K

Text Solution

Verified by Experts

The correct Answer is:
C
Gibbs free energy for transformation=0
`DeltaG= DeltaH- TDeltaS=0`
`T= (DeltaH)/(DeltaS)`
`DeltaT= (40.63 xx 10^(3))/(108.8) = 373.4K`
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