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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

1 Answer

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Best answer
Correct Answer - C
`DeltaG=DeltaH-TDeltaS`
`DeltaG=0`
`DeltaH=TDeltaS`,
`T=(40.63xx10^(3))/(108.8)=373.4 K`.

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