QJEEAdv21Chd78d: JEE Advanced 2021 Chemistry - Class 11 JEEAdv21Chd78d Solution: The value of S (in J K 1 mol 1 ) for the given reaction, at…

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

We can also calculate the intercept $(\ln A)$ from the plot of $\ln \frac{p_{z}}{p^{\circ}} \mathrm{vs} \frac{1}{T}$ .

Now, from the equation of straight line, we have

or

$\begin{aligned}&y=m x+C \\&C=y-m x\end{aligned}$

From the plot, we have

$m=-2 \times 10^{4} \mathrm{~K}, y=-3, x=\frac{10}{10^{4}} \mathrm{~K}$

Now, substituting the above values in Eq. (4), we get

$\begin{aligned}&C=-3-\left(-2 \times 10^{4} \times \frac{10}{10^{4}}\right) \\&C=-3+20 \\&C=17\end{aligned}$

$\Rightarrow C=17$

This intercept is equal to $\ln A$ in Eq. (3), so we have

$\ln A=C=17$

From the thermodynamic equations, we have

$\Delta G^{\circ}=\Delta H^{\circ}-7 \Delta S^{\circ}$

For any equilibrium reaction, the Gibb's energy is,

$\Delta G^{\circ}=-R T \ln K$

On comparing both the equations, we get

$-R T \ln K=\Delta H^{\circ}-T \Delta S^{\circ}$

Or

$\begin{aligned}\ln K &=\frac{-\Delta H^{\circ}}{R T}+\frac{\Delta S^{\circ}}{R} \\\left(\frac{p_{z}}{p^{\circ}}\right) &=\frac{-\Delta H^{\circ}}{R T}+\frac{\Delta S^{\circ}}{R}\end{aligned}$

On comparing Eq. (3) and (6), we have

$\frac{\Delta S^{\circ}}{R}=\ln A=17$

Or

$\begin{aligned}\Delta S^{\circ} &=17 R \\&=17 \times 8.314 \\&=141.34 \,\mathrm{JK}^{-1}\end{aligned}$