The equilibrium constant for the reaction SO_3(~g) leftharpoons SO_2(~g)+\frac12 O_2(~g) is K_c=4.9 \times 10^-2 . The value of K_c for the reaction given below is 2 SO_2(~g)+O_2(~g) leftharpoons 2 SO_3(~g) is :

Chemical Equilibrium PYQ — JEE Main 2024 · 4 Apr · Shift 2 · Q17 with Solution

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Correct answer: B

The reaction is

\mathrm{SO}_3(g) \rightleftharpoons \mathrm{SO}_2(g)+\frac{1}{2}\mathrm{O}_2(g)

with

K_c = 4.9 \times 10^{-2} = 0.049

We need the equilibrium constant for

2\mathrm{SO}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2\mathrm{SO}_3(g)

First, reverse the given reaction:

\mathrm{SO}_2(g)+\frac{1}{2}\mathrm{O}_2(g) \rightleftharpoons \mathrm{SO}_3(g)

For reversing a reaction, the equilibrium constant becomes reciprocal:

K_c' = \frac{1}{0.049}

Now multiply the entire reversed reaction by $2$ :

2\mathrm{SO}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2\mathrm{SO}_3(g)

When a reaction is multiplied by $2$ , the equilibrium constant is squared:

K_c'' = \left(\frac{1}{0.049}\right)^2

\frac{1}{0.049} = \frac{1000}{49} \approx 20.408

Therefore,

K_c'' = (20.408)^2 \approx 416.5

So,

K_c \approx 416

Hence, the correct answer is B.