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KCET · Chemistry · Chemical Equilibrium

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

  1. A \(\mathrm{K}_{1}^{2}=\frac{1}{\mathrm{~K}_{2}}\)
  2. B \(2 \mathrm{~K}_{1}=\mathrm{K}_{2}^{2}\)
  3. C \(\mathrm{K}_{2}=\frac{2}{\mathrm{~K}_{1}^{2}}\)
  4. D \(\mathrm{K}_{2}^{2}=\frac{1}{\mathrm{~K}_{1}}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\mathrm{K}_{1}^{2}=\frac{1}{\mathrm{~K}_{2}}\)

Step-by-step Solution

Detailed explanation

For the reaction,
\[
\begin{aligned}
&\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g}) \\
&\text { Equilibrium constant, } \mathrm{K}_{1}=\frac{\left[\mathrm{SO}_{3}\right]}{\left[\mathrm{SO}_{2}\right]\left[\mathrm{O}_{2}\right]^{1 / 2}} \quad \ldots \text { (i) }
\end{aligned}
\]
For the reaction,
\[
\begin{aligned}
&2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \\
&\text { equilibrium constant, } \mathrm{K}_{2}=\frac{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}{\left[\mathrm{SO}_{3}\right]^{2}} \quad \ldots \text { (ii) }
\end{aligned}
\]
On squaring both sides in Eq (i), we get
\[
\mathrm{K}_{1}^{2}=\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]} \quad \ldots \text { (iii) }
\]
Eqs. (ii) \(\times\) Eq (iii), we get
\[
\mathrm{K}_{1}^{2} \times \mathrm{K}_{2}=1
\]
or
\[
\begin{aligned}
\mathrm{K}_{2} &=\frac{1}{\mathrm{~K}_{1}^{2}} \\
\mathrm{~K}_{1}^{2} &=\frac{1}{\mathrm{~K}_{2}}
\end{aligned}
\]