Total enthalpy change for freezing of 1 mol of water at 10 ° C to ice at -10 ° C is \ \ \ \ \ (Given : Δ fus H =x kJ / mol C p ( H 2 O (ell) ]= y J mol -1 K -1 C p ( H 2 O ( s ) ]= zJ mol -1 K -1

Chemistry · Thermodynamics & Thermochemistry

JEE Main 2025 — 7 April, Morning Shift — Question 11

Total enthalpy change for freezing of 1 mol of water at $10^{\circ} \mathrm{C}$ to ice at $-10^{\circ} \mathrm{C}$ is $\_\_\_\_\_$

(Given : $\Delta_{\text {fus }} \mathrm{H}=x \mathrm{~kJ} / \mathrm{mol}$ $\mathrm{C}_{\mathrm{p}}\left(\mathrm{H}_{2} \mathrm{O}(\ell)\right]=\mathrm{y} \mathrm{J} \mathrm{mol}{ }^{-1} \mathrm{~K}^{-1}$ $\mathrm{C}_{\mathrm{p}}\left(\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\right]=\mathrm{zJ} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$

Option A:
$x-10 y-10 z$
Option B:
$10(100 x+y+z)$
Option C:
$-x-10 y-10 z$
Option D:
$-10(100 x+y+z)$
Correct

Answer: D

Step-by-step solution

$\underset{10^{\circ} \mathrm{C}}{\mathrm{H}_{2} \mathrm{O}(\ell)} \longrightarrow \underset{(1)}{ } \underset{0^{\circ} \mathrm{C}}{\mathrm{H}_{2} \mathrm{O}}(\ell) \underset{(2)}{\rightleftharpoons} \underset{0^{\circ} \mathrm{C}}{\mathrm{H}_{2} \mathrm{O}(\mathrm{s})} \longrightarrow(3) \underset{-10^{\circ} \mathrm{C}}{\mathrm{H}_{2} \mathrm{O}(\mathrm{s})}$

(1) $\rightarrow-y \times 10$

(2) $\rightarrow-x \times 1000$

(3) $\rightarrow-z \times 10$

$\Delta H_{\text {net }}=-10(y+100 x+z)$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Morning Shift
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Thermochemistry and Enthalpy Changes