JEE Advanced 2021 · Paper 1 · Chemistry Q7

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ChemistryNumerical+2Question 26 of 57

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Question Stem for Question Nos. 7 and 8

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

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

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 standard enthalpy, $\Delta H^\ominus$ (in kJ mol $^{-1}$ ) for the given reaction is ____.

Your answer

Answer and worked solution

Official answer: 166.2

Slope of the plot. The line goes from $-3$ to $-7$ as $\frac{10^4}T$ goes from 10 to 12, so
$\frac{d\ln K}{d(1/T)}=\frac{-4}{2\times10^{-4}}=-2\times10^4\text{ K}.$

Enthalpy.
$\Delta H^\ominus=2\times10^4\times8.314=1.663\times10^5\text{ J mol}^{-1}=166.28\text{ kJ mol}^{-1}.$

Topics:Equilibrium and Gibbs energy Gibbs energy and spontaneity

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JEE Advanced 2021 · Paper 1, Chemistry question 7