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KCET · Chemistry · Chemical Equilibrium

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

  1. A \(2 K_{1}=K_{2}^{2}\)
  2. B \(K_{1}^{2}=\frac{1}{K_{2}}\)
  3. C \(K_{2}^{2}=\frac{1}{K_{1}}\)
  4. D \(K_{2}=\frac{2}{K_{1}^{2}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(K_{1}^{2}=\frac{1}{K_{2}}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)\)
... (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) \(K_{2}\) \(\therefore \quad K_{2}=\frac{1}{K_{1}^{2}}\)
\[
K_{1}^{2}=\frac{1}{K_{2}}
\]
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