Q38 / 120 · NEET 2012 (Mains) · Chemistry
Given that the equilibrium constant for the reaction,
$$
2 \mathrm{SO}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{SO}_3(g)
$$
has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature?
$$
\mathrm{SO}_3(g) \rightleftharpoons \mathrm{SO}_2(g)+\frac{1}{2} \mathrm{O}_2(g)
$$
- A$1.8 \times 10^{-3}$
- B$3.6 \times 10^{-3}$
- Correct answerC$6.0 \times 10^{-2}$
- D$1.3 \times 10^{-5}$
Solution
$$
2 \mathrm{SO}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{SO}_3(g)
$$
Equilibrium constant for this reaction,
$$
\begin{gathered}
K=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]} \\
\mathrm{SO}_3(g) \rightleftharpoons \mathrm{SO}_2(g)+\frac{1}{2} \mathrm{O}_2(g)
\end{gathered}
$$
Equilibrium constant for this reaction,
$$
K^{\prime}=\frac{\left[\mathrm{SO}_2\right]\left[\mathrm{O}_2\right]^{\frac{1}{2}}}{\left[\mathrm{SO}_3\right]}
$$
On squaring Eq. (ii) both sides, we have
$$
\left(K^{\prime}\right)^2=\frac{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}{\left[\mathrm{SO}_3\right]^2}
$$
$\begin{aligned} & =\frac{1}{K} \\ & =\frac{1}{278} \\ \therefore \quad K^{\prime} & =\sqrt{\frac{1}{278}} \\ & =\sqrt{0.003597} \\ & =5.99 \times 10^{-2} \\ & \approx 6 \times 10^{-2}\end{aligned}$
Hence, the correct answer is option: 'C'.