Q12 / 160 · TS EAMCET 2020 (09 Sep Shift 2) · Chemistry

If the equilibrium constant for the reaction, $2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3$ is 64 at $500 \mathrm{~K}$, then the equilibrium constant for the reaction $\mathrm{SO}_3 \rightleftharpoons \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2$ at the same temperature is

  • A
    $8$
  • Correct answer
    B
    $\frac{1}{8}$
  • C
    $32$
  • D
    $\frac{1}{64}$

Solution

Given, $2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3, K=64$ $\mathrm{SO}_3 \rightleftharpoons \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2, K^{\prime}=?$ $\because$ New $(K)=[K]^{1 / n}$ i.e. $K^{\prime}$ where, $n=$ factor to the new equilibrium constant, which is $n$th root of the previous value and $K^{\prime}=\left[\frac{1}{K}\right]^n,\left(n=\frac{1}{2}\right)$ $\therefore$ Hence, new $K$ i.e. $\left(K^{\prime}\right)=\left[\frac{1}{64}\right]^{1 / 2}=\frac{1}{8} \quad\left(\because n=\frac{1}{2}\right)$ $\therefore$ Option (b) is the correct answer.

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

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