Question

$H_{2}O\left(ice\right)\rightleftharpoons H_{2}O\left(water\right)$

Enthalpy change for the above process is 6.01 kJ $mo{l}^{-1}$. The entropy change of melting of 1 mole of ice at its melting point will be :

• A. 12 J $K^{-1}mo{l}^{-1}$
• B. 22 J $K^{-1}mo{l}^{-1}$
• C. 100 J $K^{-1}mo{l}^{-1}$
• D. 30 J $K^{-1}mo{l}^{-1}$

Answer 1

The correct option is B 22 J $K^{-1}mo{l}^{-1}$

At melting point (Equilibrium Condition) ,$\Delta S_{fusion}=\frac{\Delta H_{fusion}}{T_{fusion}}$

$=\frac{6.01\times 1000}{273}$ = $22J{K}^{-1}mo{l}^{-1}$

Hence, the correct option is $\text{B}$