The equilibrium constant for the reaction $$\mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})$$ is $\mathrm{K}_{\mathrm{c}}=4.9 \times 10^{-2}$. The value of $\mathrm{K}_{\mathrm{c}}$ for the reaction $$2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})$$ will be

JEE Main · 2006 · 30 Apr 2006 · Offline · Equilibrium

The equilibrium constant for the reaction

SO_{3} (g) ⇌ SO_{2} (g) + \frac{1}{2} O_{2} (g)

K_{c} = 4.9 \times 1 0^{- 2}

. The value of

K_{c}

for the reaction

2 SO_{2} (g) + O_{2} (g) ⇌ 2 SO_{3} (g)

will be

(A)416✓ Correct
(B) $2.40 \times 1 0^{- 3}$
(C) $9.8 \times 1 0^{- 2}$
(D) $4.9 \times 1 0^{- 2}$

Answer: (A)

Solution

The target reaction is the first one reversed and then doubled. Reversing a reaction inverts its equilibrium constant, and multiplying the coefficients by two squares it, so the required constant is

(\frac{1}{4.9 \times 1 0 ^{- 2}})^{2}

. Since

\frac{1}{0.049} \approx 20.4

, its square is about

416

, the equilibrium constant for the formation of

S O_{3}

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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