Question

For the reaction, $H_2+I_2\rightleftharpoons 2HI,K=47.6$. If the initial number of moles of each reactant and product is 1 mole then at equilibrium

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• A. $\left[I_2\right]=\left[H_2\right],\left[I_2\right]>\left[HI\right]$
• B. $\left[I_2\right]<\left[H_2\right],\left[I_2\right]=\left[HI\right]$
• C. $\left[I_2\right]>\left[H_2\right],\left[I_2\right]=\left[HI\right]$
• D. $\left[I_2\right]=\left[H_2\right],\left[I_2\right]<\left[HI\right]$

Answer 1

The correct option is D $\left[I_2\right]=\left[H_2\right],\left[I_2\right]<\left[HI\right]$

For the given reaction,
$K=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}$

As 1 mole of $H_2$ reacts with 1 mole of $I_2$, even at equilibrium, $\left[H_2\right]=\left[I_2\right]$.

Hence, $K=\frac{[HI{]}^2}{[I_2{]}^2}$
(or)
$\sqrt{K}=\frac{\left[HI\right]}{\left[I_2\right]}$
i.e., $\left[HI\right]>\left[I_2\right]$
Hence, option (b) is correct.