For the following reactions, equilibrium constants are given: $\mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g}) ; \mathrm{K}_{1}=10^{52}$ $2 \mathrm{~S}(\mathrm{~s})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) ; \mathrm{K}_{2}=10^{129}$ The equilibrium constant for the reaction $2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})$ is:

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Equilibrium

For the following reactions, equilibrium constants are given:

S (s) + O_{2} (g) ⇌ SO_{2} (g); K_{1} = 1 0^{52}

2 S (s) + 3 O_{2} (g) ⇌ 2 SO_{3} (g); K_{2} = 1 0^{129}

The equilibrium constant for the reaction

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

is:

(A) $1 0^{77}$
(B) $1 0^{25}$ ✓ Correct
(C) $1 0^{181}$
(D) $1 0^{154}$

Answer: (B)

Solution

We want

K

for

2 SO X_{2} (g) + O X_{2} (g) 2 SO X_{3} (g)

. Doubling the first given reaction gives

2 S (s) + 2 O X_{2} (g) 2 SO X_{2} (g)

with constant

K_{1}^{2} = 1 0^{104}

. Reversing this doubled reaction and adding it to the second given reaction

2 S (s) + 3 O X_{2} (g) 2 SO X_{3} (g)

cancels the solid sulfur and leaves exactly our target equation. Reversing a reaction inverts its constant and adding reactions multiplies the constants, so

K = \frac{K _{2}}{K _{1}^{2}} = \frac{1 0 ^{129}}{1 0 ^{104}} = 1 0^{25}

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Equilibrium

Solution

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