The enthalpy of fusion of water is 1.435 kcal / mol . The molar entropy change for the melting of ice at 0^ C is

Correct answer: C. 5.260 cal /(mol K) Molar entropy change for the melting of ice, aligned Δ S_ melt & = Δ H_ fusion T & = 1.435 kcal / mol (0+273) K & =5.26 × 10^ -3 kcal / mol K & =5.26 cal / mol K aligned

NEET 2012

Exam: NEET
Year: 2012
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2012 · Chemistry · Thermodynamics (C)

The enthalpy of fusion of water is 1.435 kcal / mol . The molar...

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}

Options

A $10.52 \mathrm{cal} /(\mathrm{mol} \mathrm{K})$
B $21.04 \mathrm{cal} /(\mathrm{mol} \mathrm{K})$
C $5.260 \mathrm{cal} /(\mathrm{mol} \mathrm{K})$
D $0.526 \mathrm{cal} /(\mathrm{mol} \mathrm{K})$

Correct answer

C. $5.260 \mathrm{cal} /(\mathrm{mol} \mathrm{K})$

Step-by-step 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}

Practice Thermodynamics (C) on Quantrex Academy →

More from Thermodynamics (C)