The equilibrium constant for the reaction SO3(g) SO2(g) + 12O2(g) is Kc = 4.9 10-2. The value of Kc for the reaction…

Step 1 — Relate the two reactions:

Given: $\mathrm{SO_3 \rightleftharpoons SO_2 + \frac{1}{2}O_2}$, $K_1 = 4.9 \times 10^{-2}$

Target: $\mathrm{2SO_2 + O_2 \rightleftharpoons 2SO_3}$

Step 2 — Manipulate:

• Reverse the given reaction: $\mathrm{SO_2 + \frac{1}{2}O_2 \rightleftharpoons SO_3}$, $K = \frac{1}{K_1} = \frac{1}{4.9 \times 10^{-2}}$
• Multiply by 2: $\mathrm{2SO_2 + O_2 \rightleftharpoons 2SO_3}$, $K_2 = \left(\frac{1}{K_1}\right)^2$

Step 3 — Calculate: $K_2 = \left(\frac{1}{4.9 \times 10^{-2}}\right)^2 = \left(20.41\right)^2 \approx 416$

Answer: Option (4) — 416