Question

At $373\, K$, steam and water are in equilibrium and $\Delta H = 40.98 \,kJ\, mol^{-1}$. What will be $\Delta S$ for conversion of water into steam?
$H_2O_{(l)} \to H_2O_{(g)}$

• A. $109.8 \,J\,K^{-1}\, mol^{-1}$
• B. $31 \,J\,K^{-1}\, mol^{-1}$
• C. $21.98 \,J\,K^{-1}\, mol^{-1}$
• D. $326 \,J\,K^{-1}\, mol^{-1}$

Answer 1

Answer: (A)

$\Delta S_{vap} = \frac{\Delta H_{vap}}{T_{b}} = \frac{40.98 \times 1000}{373}$
$= 109.8\,J\,K^{-1}\,mol^{-1}$