JEE Advanced 2021 — Paper 1 — Question 40

\begin{array}{*{35}{r}}{} & X\left( s \right)\rightleftharpoons Y\left( s \right)+Z\left( g \right) \\ {} & \begin{array}{*{35}{r}}{} & {{K}_{p}}=\frac{{{p}_{z}}}{{{p}^{0}}},\text{ }\!\!~\!\!\text{ also }\!\!~\!\!\text{ }\!\!\Delta\!\!\text{ }{{G}^{0}}=-RT\text{ln}{{k}_{p}} \\{} & ~=-RT\text{ln}\left(\frac{{{p}_{z}}}{{{p}^{0}}} \right) \\\end{array} \\\end{array}

Now, $\text{ }\!\!\Delta\!\!\text{ }{{\text{G}}^{0}}=\text{ }\!\!\Delta\!\!\text{ }{{\text{H}}^{0}}-\text{T }\!\!\Delta\!\!\text{ }{{\text{S}}^{0}}$ -RT kn $\left( \frac{{{p}_{z}}}{{{p}^{0}}} \right)=\text{ }\!\!\Delta\!\!\text{ }{{H}^{0}}-T\text{ }\!\!\Delta\!\!\text{ }{{S}^{0}}$

$\begin{matrix}ln\left( \frac{{{p}_{z}}}{{{p}^{0}}} \right)=-\left( \frac{\text{ }\!\!\Delta\!\!\text{ }{{H}^{0}}}{R} \right)\frac{1}{T}+\frac{\text{ }\!\!\Delta\!\!\text{ }{{S}^{0}}}{R}\#\left( 1 \right) \\\end{matrix}$

(1) $\Rightarrow \text{ln}\left( \frac{{{p}_{z}}}{{{\text{p}}^{0}}} \right)=-\left( \frac{\text{ }\!\!\Delta\!\!\text{ }{{H}^{0}}}{{{10}^{4}}\text{R}} \right)\times \frac{{{10}^{4}}}{\text{ }\!\!~\!\!\text{ T}}+\frac{\text{ }\!\!\Delta\!\!\text{ }{{\text{S}}^{0}}}{\text{ }\!\!~\!\!\text{ T}}\ldots$. (2)

Slope of the line $=-\frac{\text{ }\!\!\Delta\!\!\text{ }{{H}^{0}}}{{{10}^{4}}\text{R}}=\frac{\left[ -7-\left( -3 \right) \right]}{12-10}=-2$ $\therefore \text{ }\!\!\Delta\!\!\text{ }{{H}^{0}}=2\text{R}\times {{10}^{4}}$ $=2\times 8.314\times {{10}^{-3}}\times {{10}^{4}}=1.66.28\text{ }\!\!~\!\!\text{ kJ }\!\!~\!\!\text{ mo}{{\text{l}}^{-1}}\text{ }\!\!~\!\!\text{ }{{\text{K}}^{-1}}$