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 be

Options

A. `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

C

Explanation

`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)`