Ice at -5 ° C is heated to become vapor with temperature of 110 ° C at atmospheric pressure. The entropy change associated with this process can be obtained from :

int 268 K 273 K frac C p , m T dT +frac Δ H m , fusion T f +frac Δ H m , vaporisation T b +int 273 K 373 K frac C p , m dT T +int 373 K 383 K frac C p , m dT T

int 268 K 383 K C p dT +frac Δ H melting 273 +frac Δ H boiling 373

int 268 K 383 K C p dT +frac q rev T

int 268 K 273 K C p , m dT +frac Δ H m , fusion T f +frac Δ H m , vaporisation T b +int 273 K 373 K C p , m dT +int 373 K 383 K C p , m dT

Chemistry · Thermodynamics & Thermochemistry

JEE Main 2025 — 23 January, Morning Shift — Question 34

Ice at $-5^{\circ} \mathrm{C}$ is heated to become vapor with temperature of $110^{\circ} \mathrm{C}$ at atmospheric pressure. The entropy change associated with this process can be obtained from :

Option A:
$\int_{268 \mathrm{~K}}^{383 \mathrm{~K}} \mathrm{C}_{\mathrm{p}} \mathrm{dT}+\frac{\Delta \mathrm{H}_{\text {melting }}}{273}+\frac{\Delta \mathrm{H}_{\text {boiling }}}{373}$
Option B:
$\int_{268 \mathrm{~K}}^{273 \mathrm{~K}} \frac{\mathrm{C}{\mathrm{p}, \mathrm{m}}}{\mathrm{T}} \mathrm{dT}+\frac{\Delta \mathrm{H}{\mathrm{m}} \text {, fusion }}{\mathrm{T}{\mathrm{f}}}+\frac{\Delta \mathrm{H}{\mathrm{m}, \text { vaporisation }}}{\mathrm{T}{\mathrm{b}}}$$$+\int{273 \mathrm{~K}}^{373 \mathrm{~K}} \frac{\mathrm{C}{\mathrm{p}, \mathrm{~m}} \mathrm{dT}}{\mathrm{~T}}+\int{373 \mathrm{~K}}^{383 \mathrm{~K}} \frac{\mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT}}{\mathrm{~T}}$$
Correct
Option C:
$\int_{268 \mathrm{~K}}^{383 \mathrm{~K}} \mathrm{C}_{\mathrm{p}} \mathrm{dT}+\frac{\mathrm{q}_{\mathrm{rev}}}{\mathrm{T}}$
Option D:
$\int_{268 \mathrm{~K}}^{273 \mathrm{~K}} \mathrm{C}{\mathrm{p}, \mathrm{m}} \mathrm{dT}+\frac{\Delta \mathrm{H}{\mathrm{m}} \text {, fusion }}{\mathrm{T}{\mathrm{f}}}+\frac{\Delta \mathrm{H}{\mathrm{m}, \text { vaporisation }}}{\mathrm{T}{\mathrm{b}}}$$$+\int{273 \mathrm{~K}}^{373 \mathrm{~K}} \mathrm{C}{\mathrm{p}, \mathrm{~m}} \mathrm{dT}+\int{373 \mathrm{~K}}^{383 \mathrm{~K}} \mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT}$$

Answer: B

Step-by-step solution

$\Delta \mathrm{S}_{\text {overall }}=\Delta \mathrm{S}_{1}+\Delta \mathrm{S}_{2}+\Delta \mathrm{S}_{3}+\Delta \mathrm{S}_{4}+\Delta \mathrm{S}_{5}$

$\Delta \mathrm{S}_{2}=\frac{\Delta \mathrm{H}_{\mathrm{m} \text { fusion }}}{273} \quad \mathrm{~T}_{\mathrm{f}}=273 \mathrm{~K}^{\prime} \mathrm{K}^{\prime}$

$\Delta \mathrm{S}_{3}=\int_{273}^{373} \frac{\mathrm{C}_{\mathrm{p}, \mathrm{m}} \mathrm{dT}}{\mathrm{T}}$

$\Delta \mathrm{S}_{4}=\frac{\Delta \mathrm{H}_{\mathrm{m} \text { vaporisation }}}{373} \quad \mathrm{~T}_{\mathrm{b}}=373{ }^{\prime} \mathrm{K}^{\prime}$

$\Delta \mathrm{S}_{5}=\int_{373}^{383} \frac{\mathrm{C}_{\mathrm{p}, \mathrm{m}} \mathrm{dT}}{\mathrm{T}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 23 January, Morning Shift
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Entropy Change in Different Processes

The standard enthalpy and standard entropy of decomposition of N 2 O 4 to NO 2 are 55.0 kJ mol^-1 and 175.0 J / K / mol respectively. The…

Answer: B

Step-by-step solution

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