For a reaction, A rightleftharpoons P , the plots of [ A ] and [ P ] with time at temperatures T 1 and T 2 are given below. If T 2 T 1 , the correct statement(s) is (are) (Assume Δ H 0 and Δ S 0 are independent of temperature and ratio of ln K at T 1 to ln K at T 2 is greater than T 2 / T 1 . Here H, S, G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

Chemistry · Thermodynamics & Thermochemistry

JEE Advanced 2018 — Paper 2 — Question 21

For a reaction, $\mathrm{A} \rightleftharpoons \mathrm{P}$ , the plots of $[\mathrm{A}]$ and $[\mathrm{P}]$ with time at temperatures $T_{1}$ and $T_{2}$ are given below.

If T $_2$ > T $_1$ , the correct statement(s) is (are) (Assume $\Delta$ H $^0$ and $\Delta$ S $^0$ are independent of temperature and ratio of $\ln K$ at $\mathrm{T}_{1}$ to $\ln K$ at $\mathrm{T}_{2}$ is greater than $\mathrm{T}_{2} / \mathrm{T}_{1}$ . Here $H, S, G$ and $K$ are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

Option A:
$\Delta \mathrm{H}^{\ominus}<0, \Delta \mathrm{~S}^{\mathrm{\theta}}<0$
Correct
Option B:
$\Delta \mathrm{G}^{\ominus}<0, \Delta \mathrm{H}^{\theta}>0$
Option C:
$\Delta G^{\theta}<0, \Delta S^{\ominus}<0$
Correct
Option D:
$\Delta \mathrm{G}^{\ominus}<0, \Delta \mathrm{~S}^{\ominus}>0$

Answer: A, C

Step-by-step solution

$A \rightleftharpoons P$ $(\mathrm{P})_{\text {eq }}>5,(\mathrm{~A})_{\text {eq }}<5$ $\mathrm{K}_{\mathrm{eq}}=\frac{[\mathrm{P}]}{[\mathrm{A}]}>1$ $\Delta \mathrm{G}^{0}=-\mathrm{RT} \ln \mathrm{K}_{\mathrm{eq}}, \Delta \mathrm{G}^{0}<0$ $\frac{\ln \mathrm{K}_{\mathrm{T}_{1}}}{\ln \mathrm{~K}_{\mathrm{T}_{2}}}>\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}>1$ $\Rightarrow \frac{\mathrm{K}_{\mathrm{T}_{1}}}{\mathrm{~K}_{\mathrm{T}_{2}}}>1$ $\Rightarrow \mathrm{K}_{\mathrm{T}_{2}}<\mathrm{K}_{\mathrm{T}_{1}}$ (exothermic) $\Delta \mathrm{H}^{0}<0$ , since $(\mathrm{P})$ at $\mathrm{T}_{2}<$ at $\mathrm{T}_{1}$ . $\Delta \mathrm{G}^{0}=\Delta \mathrm{H}^{0}-\mathrm{T} \Delta \mathrm{S}^{0}$ $\mathrm{T} \Delta \mathrm{S}^{0}=\Delta \mathrm{H}^{0}-\Delta \mathrm{G}^{0}$ $\Delta \mathrm{S}^{0}=\frac{\Delta \mathrm{H}^{0}-\Delta \mathrm{G}^{0}}{\mathrm{~T}} ;\left(\Delta \mathrm{H}^{0}\right)>\left(\Delta \mathrm{G}^{0}\right)$ $\Delta S^{0}<0$ Also, $-\mathrm{T}_{1} \ln \mathrm{~K}_{\mathrm{T}_{1}}<-\mathrm{T}_{2} \ln \mathrm{~K}_{\mathrm{T}_{2}} \mathrm{Il}$ $\Delta \mathrm{G}_{\mathrm{T}_{1}}^{0}<\Delta \mathrm{G}_{\mathrm{T}_{2}}^{0}$ $\Delta \mathrm{H}_{\mathrm{T}_{1}}^{0}-\mathrm{T} \Delta \mathrm{S}_{\mathrm{T}_{1}}^{0}<\Delta \mathrm{H}_{\mathrm{T}_{2}}^{0}-\mathrm{T} \Delta \mathrm{S}_{\mathrm{T}_{2}}^{0}$ It is possible only if $\Delta \mathrm{S}^{0}<0$ .

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Exam: JEE Advanced 2018
Paper: Paper 2
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Enthalpy and Calorimetry

Answer: A, C

Step-by-step solution

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