Q64 / 180 · AIIMS 2017 · Chemistry

For the reaction, $\mathrm{H}_2+\mathrm{I}_2 \rightleftharpoons 2 \mathrm{HI}, \mathrm{K}=47.6$. If the initial number of moles of each reactant and product is 1 mole then at equilibrium

  • A
    $\left[\mathrm{I}_2\right]=\left[\mathrm{H}_2\right],\left[\mathrm{I}_2\right]>[\mathrm{HI}]$
  • Correct answer
    B
    $\left[\mathrm{I}_2\right]=\left[\mathrm{H}_2\right],\left[\mathrm{I}_2\right] < [\mathrm{HI}]$
  • C
    $\left[\mathrm{I}_2\right] < \left[\mathrm{H}_2\right],\left[\mathrm{I}_2\right]=[\mathrm{HI}]$
  • D
    $\left[\mathrm{I}_2\right]>\left[\mathrm{H}_2\right],\left[\mathrm{I}_2\right]=[\mathrm{HI}]$

Solution

For the given reaction, $K=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}$ As 1 mole of $\mathrm{H}_2$ reacts with 1 mole of $\mathrm{I}_2$, even at equilibrium, $\left[\mathrm{H}_2\right]=\left[\mathrm{I}_2\right]$ Hence, $K=\frac{[\mathrm{HI}]^2}{\left[\mathrm{I}_2\right]^2} \quad$ or $\quad \sqrt{K}=\frac{[\mathrm{HI}]}{\left[\mathrm{I}_2\right]}=\sqrt{47.6}$ i.e., $[\mathrm{HI}]>\left[\mathrm{I}_2\right]$

Hence, the correct answer is option: 'B'.

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