Question

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

• A. 4.9
• B. 41.6
• C. 49
• D. 416

Answer 1

The Correct Option is D

Given 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 reaction, we calculate \( K_c' \) as:
\[K_c' = \left( \frac{1}{K_c} \right)^2 = \left( \frac{1}{4.9 \times 10^{-2}} \right)^2\]
\[K_c' = 416.49\]