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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 isA. `273.4K`B. `393.4K`C. `373.4K`D. `293.4K` |
Answer» Correct Answer - D `DeltaG=DeltaH-TDeltaS` `Delta=0` , `:. DeltaH=TDeltaS` `T=(DeltaH)/(DeltaS)=(40.63xx10^(3))/(108.8)=373.4K` |
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