The equilibrium constant for the reaction
\( \text{SO}_3 \, (g) \rightleftharpoons \text{SO}_2 \, (g) + \frac{1}{2} \text{O}_2 \, (g) \)
is \( K_C = 4.9 \times 10^{-2} \). The value of \( K_C \) for the reaction given below is
\( 2 \text{SO}_2 \, (g) + \text{O}_2 \, (g) \rightleftharpoons 2 \text{SO}_3 \, (g) \) is
\( \text{SO}_3 \, (g) \rightleftharpoons \text{SO}_2 \, (g) + \frac{1}{2} \text{O}_2 \, (g) \)
is \( K_C = 4.9 \times 10^{-2} \). The value of \( K_C \) for the reaction given below is
\( 2 \text{SO}_2 \, (g) + \text{O}_2 \, (g) \rightleftharpoons 2 \text{SO}_3 \, (g) \) is
- 4.9
- 41.6
- 49
- 416
The Correct Option is D
Solution and Explanation
The equilibrium constant for the reaction \( \text{SO}_3 \, (g) \rightleftharpoons \text{SO}_2 \, (g) + \frac{1}{2} \text{O}_2 \, (g) \) is \( K_c = 4.9 \times 10^{-2} \). For the reaction \( 2\text{SO}_2 \, (g) + \text{O}_2 \, (g) \rightleftharpoons 2\text{SO}_3 \, (g) \), we need the equilibrium constant \( K_c' \). Since the second reaction is the reverse and doubled version of the original, \( K_c' \) is calculated as: \(K_c' = \left( \frac{1}{K_c} \right)^2 = \left( \frac{1}{4.9 \times 10^{-2}} \right)^2\), resulting in \(K_c' = 416.49\).
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