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The values of Delta H and Delta S in vap...

The values of `Delta H and Delta S` in vaporisation of water at 1 atm pressure are `40.63 " kJ mol"^(-1) and 108.8 J. K^(-1). " mol"^(-1)` respectively. At what temperature the free energy change of the reaction will be zero ? What will be the sign of `Delta G` below that temperature ?

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`T= (Delta H)/(Delta S)= (40.63xx10^(3) J. " mol"^(-1))/(108.8 J. K^(-1) " mol"^(-1))= 373.43 K`
`Delta G= Delta H- T Delta S= (40.63xx10^(3)- Txx108.8) J`
At `373.43 K, Delta G= 0 , " So", 40.63xx10^(3)= Txx108.8`
At ` T lt 373.43 K, 40.63xx10^(3) gt T xx108.8. " So", Delta G gt 0`
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