Question

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

[JEE 2021]

(Given, $\frac{d(lnk)}{d(\frac{1}{T})} = -\frac{\Delta H^{\theta}}{R}$, where the equiliberium constant, $K = \frac{p_z}{p^{\theta}}$ and the gas constant, $R = 8.314$ JK$^{-1}$mol$^{-1}$)

Answer 1

13. 166.28 kJ mol$^{-1}$14. 141.34 JK$^{-1}$ mol$^{-1}$

Answer 2

13. The slope of the plot of $\ln K$ vs $\frac{10^4}{T}$ is $m = -\frac{\Delta H^{\theta}}{10^4 R}$. From the graph, $m=-2$. Thus, $\Delta H^{\theta} = -m \cdot 10^4 \cdot R = -(-2) \cdot 10^4 \cdot 8.314 = 166280$ J/mol $= 166.28$ kJ/mol.

14. The y-intercept of the plot of $\ln K$ vs $\frac{10^4}{T}$ corresponds to $\frac{\Delta S^{\theta}}{R}$. The line equation is $y = -2x + 17$, so the intercept is $17$. Thus, $\frac{\Delta S^{\theta}}{R} = 17$, which gives $\Delta S^{\theta} = 17 \cdot R = 17 \cdot 8.314 = 141.338$ JK$^{-1}$mol$^{-1}$.