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If the enthaply change for the transition of liquid water to steam is 30 KJ `"mol"^(-1)` at `27^(@)` C . The entropy change for the process would beA. `1.0 J "mol"^(-1) K^(-1)`B. `0.1 J "mol"^(-1) K^(-1)`C. `100 J "mol"^(-1) K^(-1)`D. `10 J "mol"^(-1) K^(-1)` |
Answer» Correct Answer - C `DeltaG^(@) = DeltaH^(@)-TDeltaS^(@)` Given, `DeltaH_("vap")=30 KJ "mol"^(-1)` T=27 + 273 =300 K `DeltaG^(@)=0` at equilibrium `DeltaS_("vap")=(DeltaH_("vap"))/(T)=(30xx10^(3) J "mol"^(-1))/(300 K)` `=100 J "mol"^(-1) K^(-1)` |
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