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At $783 \mathrm{~K}$ in the reaction $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$, the molar concentrations $\left(\mathrm{mol}^{-1}\right.$ ) of $\mathrm{H}_{2}, \mathrm{I}_{2}$ and HI at some instant of time are $0.1,0.2$ and $0.4$, respectively. If the equilibrium constant is 46 at the same temperature, then as the reaction proceeds
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Verified Answer
The correct answer is:
the amount of HI will increase
Reaction quotient
$\begin{array}{l}
\mathrm{Q}=\frac{[\mathrm{HI}]^{2}}{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}=\frac{0.4 \times 0.4}{0.1 \times 0.2} \\
\mathrm{Q}=8 \\
\mathrm{Q} < \mathrm{K}
\end{array}$
So reaction will proceeds in forward direction.
Hence amount of HI increases.
$\begin{array}{l}
\mathrm{Q}=\frac{[\mathrm{HI}]^{2}}{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}=\frac{0.4 \times 0.4}{0.1 \times 0.2} \\
\mathrm{Q}=8 \\
\mathrm{Q} < \mathrm{K}
\end{array}$
So reaction will proceeds in forward direction.
Hence amount of HI increases.
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