The molar enthalpy of fusion and molar entropy of fusion for ice at `0^(@)C` and 1 atm are 6.01 `kJ*mol^(-1)` and `22.0J*K^(-1)*mol^(-1)`, respectively. Assuming `DeltaH and DeltaS` are independent of temperature, show that the melting of ice at 1 atm is not spontaneous, while the reverse process is spotantaneous.

Explanation

A process is spontaneous when the change in entropy of the universe `(DeltaS_("univ")=DeltaS_("sys")+DeltaS_("surr"))` is positive. The transformation of ice into water involves the process.
`H_(2)O(s) to H_(2)O(l)`
`DeltaH=6.01kJ*mol^(-1),DeltaS=22.0J*K^(-1)*mol^(-1)`
Now, `DeltaS_("sys")=22.0J*mol^(-1)`
`DeltaS_("surr")=(DeltaH)/(T)=(-6010)/(271)=-22.17J*mol^(-1)`
Thus, `DeltaS_("univ")=DeltaS_("sys")+DeltaS_("surr")=22.0+(-22.17)J*mol^(-1)`
`=-0.17J*mol^(-1)`
Hence, the total entropy change in the process is negative at 271 K. therefore, the transformation of ice into water at `-2^(@)C` is not spontaneous.
The reverse process i.e., the conversion of water to ice at `-2^(@)C` is spontaneous. this is because at `-2^(@)C` and 1 atm pressure the overall entropy `(DeltaS_("univ"))` is +0.17 `J*mol^(-1)`.