Q21 / 90 · KVPY 2013 (SB/SX) · Chemistry
Consider the equilibria (1) and (2) with equilibrium constants $\mathrm{K}_{1}$ and $\mathrm{K}_{2}$, respectively
$$
\begin{array}{l}
\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \leftrightharpoons \mathrm{SO}_{3}(\mathrm{~g}) ...(1) \\
2 \mathrm{SO}_{3}(\mathrm{~g}) \leftrightharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) ...(2)
\end{array}
$$
$\mathrm{K}_{1}$ and $\mathrm{K}_{2}$ are related as
- A$2 \mathrm{~K}_{1}=\mathrm{K}_{2}{ }^{2}$
- Correct answerB$\mathrm{K}_{1}{ }^{2}=\frac{1}{\mathrm{~K}_{2}}$
- C$\mathrm{K}_{2}^{2}=\frac{1}{\mathrm{~K}_{1}}$
- D$\mathrm{K}_{2}=\frac{2}{\mathrm{~K}_{1}^{2}}$
Solution
$\left.\mathrm{K}_{1}=\frac{\left[\mathrm{sO}_{3}\right]}{\left[\mathrm{O}_{2}\right]^{\frac{1}{2}}\left[\mathrm{SO}_{2}\right]}\right]$
$\mathrm{K}_{2}=\frac{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}{\left[\mathrm{SO}_{3}\right]^{2}}$
Hence: $\mathrm{K}_{2}=\frac{1}{\mathrm{~K}_{1}^{2}}$
$\mathrm{~K}_{1}^{2}=\frac{1}{\mathrm{~K}_{2}}$
Hence, the correct answer is option: 'B'.