At $T(\mathrm{~K})$, the equilibrium constant of $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g})…
At $T(\mathrm{~K})$, the equilibrium constant of $\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ is 49 . If $\left[\mathrm{H}_2\right],\left[\mathrm{I}_2\right]$ at equilibrium at the same temperature are $2.0 \times 10^{-2} \mathrm{M}$ and $8.0 \times 10^{-2} \mathrm{M}$ respectively, the [HI] at equilibrium in $\mathrm{mol} \mathrm{L}^{-1}$ is
2.8
0.28
0.14
1.4
Solution
Given,
$\begin{aligned}
& K_C=49 \\
& {\left[\mathrm{H}_2\right]=2.0 \times 10^{-2} \mathrm{M}} \\
& {\left[\mathrm{I}_2\right]=8.0 \times 10^{-2} \mathrm{M}}
\end{aligned}$
For, the reaction,
$\begin{aligned}
\mathrm{H}_2(g)+\mathrm{I}_2(g) & \rightleftharpoons 2 \mathrm{HI}(g) \\
& K_C=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]} \text { or }[\mathrm{HI}]^2=K_C\left[\mathrm{H}_2\left[\mathrm{I}_2\right]\right.
\end{aligned}$
On putting the values in the above equation, we have
$\begin{aligned}
{[\mathrm{HI}]^2 } & =49 \times 16 \times 10^{-4} \\
\mathrm{HI} & =\sqrt{49 \times 16+10^{-4}}=0.28 \mathrm{~mol}
\end{aligned}$