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At $500 \mathrm{~K}$, equilibrium constant, $K_c$, for the following reaction is $5$ .
$\frac{1}{2} \mathrm{H}_2(\mathrm{g})+\frac{1}{2} \mathrm{I}_2(\mathrm{g}) \rightleftharpoons \mathrm{HI}(\mathrm{g})$
What would be the equilibrium constant $\mathrm{K}_{\mathrm{C}}$ for the reaction?
$2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g})$
Options:
$\frac{1}{2} \mathrm{H}_2(\mathrm{g})+\frac{1}{2} \mathrm{I}_2(\mathrm{g}) \rightleftharpoons \mathrm{HI}(\mathrm{g})$
What would be the equilibrium constant $\mathrm{K}_{\mathrm{C}}$ for the reaction?
$2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g})$
Solution:
1246 Upvotes
Verified Answer
The correct answer is:
$0.04$
$0.04$
For the reaction, $\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{I}_2(\mathrm{~g}) \rightleftharpoons \mathrm{HI}(\mathrm{g})$
$K_c=\frac{[\mathrm{HI}]}{\left[\mathrm{H}_2\right]^{1 / 2}\left[\mathrm{I}_2\right]^{1 / 2}}=5$
Thus, for the reaction,
$\begin{aligned}
&2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \\
&K_c^{\prime}=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{\left[\mathrm{HI}]^2\right.} \\
&K_c^{\prime}=\left(\frac{1}{K_c}\right)^2=\left(\frac{1}{5}\right)^2=\frac{1}{25}=0.04
\end{aligned}$
$K_c=\frac{[\mathrm{HI}]}{\left[\mathrm{H}_2\right]^{1 / 2}\left[\mathrm{I}_2\right]^{1 / 2}}=5$
Thus, for the reaction,
$\begin{aligned}
&2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \\
&K_c^{\prime}=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{\left[\mathrm{HI}]^2\right.} \\
&K_c^{\prime}=\left(\frac{1}{K_c}\right)^2=\left(\frac{1}{5}\right)^2=\frac{1}{25}=0.04
\end{aligned}$
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