Q94 / 200 · NEET 2008 (Screening) · Chemistry

The value of equilibrium constant of the reaction $\mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2$ is 8.0 The equilibrium constant of the reaction $\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g)$ will be

  • A
    $\frac{1}{16}$
  • Correct answer
    B
    $\frac{1}{64}$
  • C
    16
  • D
    $\frac{1}{8}$

Solution

$\begin{gathered} \mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g) \\ \mathrm{K}_1=\frac{\left[\mathrm{H}_2\right]^{1 / 2}\left[\mathrm{I}_2\right]^{1 / 2}}{[\mathrm{HI}]} \\ \mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \\ K_2=\frac{\left[\mathrm{HI}^2\right.}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]} \end{gathered}$ From Eqs (i) and (ii) $\begin{array}{l} & K_1^2=\frac{1}{K_2} \\ \because & K_1=8.0 \\ \therefore & K_2=\frac{1}{K_1^2}=\frac{1}{8^2}=\frac{1}{64} \end{array}$

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

Relevant Questions