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If the enthalpy change for the transition of liquid water to steam is 30 kJ $mol^{-1}$ at $27^\circ C$, the entropy change for the process would be
A
100 J $mol^{-1}K^{-1}$
B
10 J $mol^{-1}K^{-1}$
C
1.0 J $mol^{-1}K^{-1}$
D
0.1 J $mol^{-1}K^{-1}$
Detailed Solution
For a phase transition at equilibrium, $\Delta S = \frac{\Delta H}{T}$
$\Delta H_{vap} = 30$ kJ $mol^{-1}$ = 30000 J $mol^{-1}$
T = 27 + 273 = 300 K
$\Delta S_{vap} = \frac{30000}{300}$
$\Delta S_{vap} = 100$ J $mol^{-1}K^{-1}$
$\Delta H_{vap} = 30$ kJ $mol^{-1}$ = 30000 J $mol^{-1}$
T = 27 + 273 = 300 K
$\Delta S_{vap} = \frac{30000}{300}$
$\Delta S_{vap} = 100$ J $mol^{-1}K^{-1}$