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At 0^@C, ice and water are in equilibriu...

At `0^@C`, ice and water are in equilibrium and `DeltaH = 6.0 kJ mol^(-1)` for the process
`H_2O (s) hArr H_2O (l)`
What will be `Delta S and DeltaG` for the conversion of ice to liquid water?

Text Solution

Verified by Experts

For the given process
`DeltaH = 6.0 kJ mol^(-1) and T = 0^@C = 273 K`
`:. Delta S = (DeltaH)/(T) = (6.0)/(273) = 0.022 kJ K^(-1) mol^(-1)`
`DeltaG` is given by
`DeltaG = Delta H - T Delta S`
` = 6.0 - (273 xx 6.0/273)`
= 0
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