The equilibrium constant for the reaction $$$SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2}O_2(g)$$$ is $$K_c = 4.9 \times 10^{-2}$$. The value of $$K_c$$ for the reaction $$$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$$$ will be

The given equilibrium is

$$SO_3(g) \rightleftharpoons SO_2(g) + \tfrac12\,O_2(g)$$

with $$K_c = 4.9 \times 10^{-2}$$.

Step 1 - Reverse the reaction.
Reversing an equation inverts its equilibrium constant:

$$SO_2(g) + \tfrac12\,O_2(g) \rightleftharpoons SO_3(g)$$

$$K_{c,\text{rev}} = \frac{1}{K_c} = \frac{1}{4.9 \times 10^{-2}} = 20.4$$ (to three significant figures).

Step 2 - Multiply coefficients by 2.
Multiplying every stoichiometric coefficient by $$n$$ raises the equilibrium constant to the $$n^{\text{th}}$$ power:

$$2\left(SO_2(g) + \tfrac12\,O_2(g) \right) \rightleftharpoons 2SO_3(g)$$

which simplifies to

$$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g).$$

Here $$n = 2$$, so

$$K_{c,\text{new}} = \left(K_{c,\text{rev}}\right)^2 = (20.4)^2 \approx 416.$$

Hence the required equilibrium constant is 416.

Option A which is: 416