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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:

A

`-21.9JK^(-1)mol^(-1)` and 0

B

`0.219JK^(-1) mol^(-1) and 0`

C

`21.9JK^(-1)mol^(-1) and 0`

D

`0.0219JK^(-1)mol^(-1) and 0`

Text Solution

Verified by Experts

The correct Answer is:
C

`DeltaG=0`
`thereforeH-TDeltaS=0`
`DeltaS=(DeltaH)/(T)=(6000)/(273)=21.9JK^(-1)mol^(-1)`
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