Q18 / 180 · KCET 2011 · Chemistry

Consider the following gaseous equilibria with equilibrium constants $K_{1}$ and $K_{2}$ respectively, $$ \begin{aligned} &\mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) \\ &2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ The equilibrium constants are related as

  • A
    $2 K_{1}=K_{2}^{2}$
  • Correct answer
    B
    $K_{1}^{2}=\frac{1}{K_{2}}$
  • C
    $K_{2}^{2}=\frac{1}{K_{1}}$
  • D
    $K_{2}=\frac{2}{K_{1}^{2}}$

Solution

$\mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)$ ... (i) $K_{1}$ Equation (i) is reversed and multiplied by 2 $2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \quad$ (i) $\frac{1}{K_{1}^{2}}$ $2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$ (ii) $K_{2}$ $\therefore \quad K_{2}=\frac{1}{K_{1}^{2}}$ $$ K_{1}^{2}=\frac{1}{K_{2}} $$

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

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