The enthalpy of fusion of water is $1.435 \mathrm{kcal} / \mathrm{mol}$. The molar entropy change for the…

The enthalpy of fusion of water is $1.435 \mathrm{kcal} / \mathrm{mol}$. The molar entropy change for the melting of ice at $0^{\circ} \mathrm{C}$ is
  1. $10.52 \mathrm{cal} /(\mathrm{mol} \mathrm{K})$
  2. $21.04 \mathrm{cal} /(\mathrm{mol} \mathrm{K})$
  3. $5.260 \mathrm{cal} /(\mathrm{mol} \mathrm{K})$
  4. $0.526 \mathrm{cal} /(\mathrm{mol} \mathrm{K})$

Solution

Molar entropy change for the melting of ice, $\begin{aligned} \Delta S_{\text {melt }} & =\frac{\Delta H_{\text {fusion }}}{T} \\ & =\frac{1.435 \mathrm{kcal} / \mathrm{mol}}{(0+273) \mathrm{K}} \\ & =5.26 \times 10^{-3} \mathrm{kcal} / \mathrm{mol} \mathrm{K} \\ & =5.26 \mathrm{cal} / \mathrm{mol} \mathrm{K} \end{aligned}$

Asked in: NEET 2012 (Screening)

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