15 moles of $\mathrm{H}_2$ and 5.2 moles of $\mathrm{I}_2$ are mixed and allowed to attain equilibrium at…
15 moles of $\mathrm{H}_2$ and 5.2 moles of $\mathrm{I}_2$ are mixed and allowed to attain equilibrium at 773 K . At equilibrium, the number of moles of HI is found to be 10 . The equilibrium constant for the dissociation of HI is
$2 \times 10^{-2}$
50
$2 \times 10^{-1}$
5.0
Solution
\begin{array}{llll} & \mathrm{H}_2+\mathrm{I}_2 \rightleftharpoons & 2 HI \\Initial & 15 & 5.2 & 0 \\At equilibrium & 15-\mathrm{x} & 5.2-\mathrm{x} & 2 x\end{array}
Let ' V ' is volume in liter
Given, $2 \mathrm{x}=10$
$x=5$
Concentration of $\mathrm{H}_2$ at equilibrium $=15-5$
$\begin{aligned}
& {\left[\mathrm{H}_2\right]=10 / \mathrm{V}} \\
& {\left[\mathrm{I}_2\right]=\frac{0.2}{\mathrm{~V}}} \\
& {[\mathrm{HI}]=\left[\frac{10}{\mathrm{v}}\right]}
\end{aligned}$ $\begin{aligned}
& K_c=\frac{\left[\frac{10}{V}\right]^2}{\left[\frac{10}{V}\right] \times\left[\frac{0.2}{V}\right]} \\
& K_c=\frac{10 \times 10}{10 \times 0.2}=50
\end{aligned}$ $\mathrm{K}_{\mathrm{c}}^{\prime}$ for the dissociation of HI is $=\frac{\mathrm{I}}{\mathrm{K}_{\mathrm{c}}}$
$\begin{aligned}
& =\frac{1}{50}=0.2 \times 10^{-1} \\
& =2 \times 10^{-2}
\end{aligned}$