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JEE Advanced 2021 · Paper 1, Chemistry question 8

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Question Stem for Question Nos. 7 and 8

For the reaction, X(s)⇌Y(s)+Z(g)\mathbf X(s) \rightleftharpoons \mathbf Y(s) + \mathbf Z(g), the plot of ln⁡pZp⊖\ln\dfrac{p_{\mathbf Z}}{p^\ominus} versus 104T\dfrac{10^4}{T} is given below (in solid line), where pZp_{\mathbf Z} is the pressure (in bar) of the gas Z\mathbf Z at temperature TT and p⊖=1p^\ominus = 1 bar.

(Given, d(ln⁡K)d(1T)=−ΔH⊖R\dfrac{d(\ln K)}{d\left(\frac1T\right)} = -\dfrac{\Delta H^\ominus}{R}, where the equilibrium constant, K=pZp⊖K = \dfrac{p_Z}{p^\ominus} and the gas constant, R=8.314R = 8.314 J K−1^{-1} mol−1^{-1})

Straight line of ln(p_Z/p°) (vertical axis) against 10^4/T in K^-1 (horizontal axis, broken near the origin). Dashed guide lines: at 10^4/T = 10 the line is at ln(p_Z/p°) = −3, and at 10^4/T = 12 it is at −7, so the slope is −2 per unit of 10^4/T.

The value of ΔS⊖\Delta S^\ominus (in J K−1^{-1} mol−1^{-1}) for the given reaction, at 1000 K is ____.

Answer and worked solution

Official answer: 141

Gibbs energy at 1000 K. Here 104T=10\frac{10^4}T=10, so ln⁡K=−3\ln K=-3 and
ΔG⊖=−RTln⁡K=8.314×1000×3=24 942 J mol−1.\Delta G^\ominus=-RT\ln K=8.314\times1000\times3=24\,942\text{ J mol}^{-1}.

Entropy.
ΔS⊖=ΔH⊖−ΔG⊖T=166 280−24 9421000≈141.3 J K−1 mol−1.\Delta S^\ominus=\frac{\Delta H^\ominus-\Delta G^\ominus}T=\frac{166\,280-24\,942}{1000}\approx141.3\text{ J K}^{-1}\text{ mol}^{-1}.

Topics:Equilibrium and Gibbs energyGibbs energy and spontaneity
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