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At 0^(@)C ice and water are in equilibri...

At `0^(@)C` ice and water are in equilibrium and `DeltaH=6kJ" "mol^(-1)` for this process:
`H_(2)OhArrH_(2)O(l)`
The values of `DeltaS and DeltaG` for conversion of ice into liquid water at `0^(@)C` are: (Answer in J/K/mol and multiply your answer with 10)

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The correct Answer is:
219
For equilibrium reaction,
`H_(2)O(s) iff H_(2)O(l)`
`DeltaG = 0`
`DeltaG = DeltaH -T DeltaS`
`0 = DeltaH - TDeltaS`
or `DeltaS = (DeltaH)/(T) = (6.00 kJ "mol"^(-1))/(273)`
` = 0.0219 kJ K^(-1) "mol"^(-1) = 21.9 J K^(-1) "mol"^(-1)`.
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