Q4 / 160 · AP EAMCET 2010 · Chemistry

If the equilibrium constant for the reaction, $\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \text { is } K$ what is the equilibrium constant of $\mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g) ?$

  • A
    $\frac{1}{K}$
  • B
    $\sqrt{K}$
  • C
    $K$
  • Correct answer
    D
    $\frac{1}{\sqrt{K}}$

Solution

$\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) ; K$ $\therefore \quad K=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}$ ...(i) For the reaction, $\mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g)$ Equilibrium constant, $K^{\prime}=\frac{\left[\mathrm{H}_2\right]^{1 / 2}\left[\mathrm{I}_2\right]^{1 / 2}}{[\mathrm{HI}]}$ On squaring both sides, we get $\left(K^{\prime}\right)^2=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2}$ ...(ii) On multiplying Eqs. (i) and (ii), we get $K \cdot\left(K^{\prime}\right)^2=1$ $K^{\prime}=\frac{1}{\sqrt{K}}$

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

Relevant Questions