QJEEAdv21Chb830: JEE Advanced 2021 Chemistry - Class 11 JEEAdv21Chb830 Solution: The value of standard enthalpy, H o (in kJ mol 1 ) for the…

For the given reaction,

$\mathrm{X}(\mathrm{s}) \rightleftharpoons \mathrm{X}(\mathrm{s})+\mathrm{Z}(\mathrm{g})$

We have

$\frac{d(\ln K)}{d(1 / T)}=\frac{-\Delta H^{\circ}}{R}$

And $K=\frac{p_{z}}{p^{\circ}}$

Substituting the value of $K$ in Eq. (1), we get

$\frac{d\left(\ln \frac{p_{z}}{p^{\circ}}\right)}{d\left(\frac{1}{T}\right)}=\frac{-\Delta H^{\circ}}{R}$

Or

$d\left(\ln \frac{p_{z}}{p^{\circ}}\right)=\frac{-\Delta H^{\circ}}{R} d\left(\frac{1}{T}\right)$

Integrating both sides of Eq. (2), we get

$\ln \frac{p_{z}}{p^{\circ}}=\frac{-\Delta H^{\circ}}{R T}+\ln A$

Where $\ln A=$ Integration constant

From the given plot of $\ln \frac{p_{z}}{p^{\circ}} \mathrm{vs} \frac{10^{4}}{T}$


The slope of the plot is

$\text { Slope }=\frac{-\Delta H^{\circ}}{R}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-7-(-3)}{\left(\frac{12}{10^{4}}-\frac{10}{10^{4}}\right)}$

$\frac{-\Delta H^{\circ}}{R}=\frac{-4}{2} \times 10^{4}$

$\Delta H^{\circ}=-2 \times 10^{4} \times R$

$\Delta H^{\circ}=2 \times 8.314 \times 10^{4}$

$\Delta H^{\circ}=16.628 \times 10^{4} \mathrm{~J} \mathrm{~mol}^{-1}$

$\Delta H^{\circ}=166.28 \mathrm{~kJ} \mathrm{~mol}^{-1}$