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. `10 J mo1^(-1)K^(-1)`
B. ` 1.0 J mo1^(-1) K^(-1)`
C. `0.1 J mol^(-1) K^(-1)`
D. `100 J mo1^(-1) K^(-1)`

Answer

D

Explanation

We know that `Delta G = Delta H- T Delta S`
`0 = Delta H-T Delta S[ therefore Delta G=0]`
`Delta S = (Delta H^@)/(T)= (30 xx 10^3)/(300)= 100 J " mol "^(-1) K^(-1)`