For the following reactions, equilibrium constants are given: S(s) + O2(g) SO2(g);\ K1 = 1052 2S(s) + 3O2(g) 2SO3(g);\ K2 = 10129 The equilibrium constant for the reaction 2SO2(g) + O2(g) 2SO3(g) is:

1025. Step 1 — Target reaction: 2SO2 + O2 2SO3 Step 2 — Construct from given reactions: - Reverse reaction 1 twice: 2SO2 2S + 2O2, K = 1K12 = 10-104 - Use reaction 2: 2S + 3O2 2SO3, K = K2 = 10129 Adding: 2SO2 + O2 2SO3 Step 3 — Combined K: K = 10-104 10129 = 1025 Answer: Option (2) — 1025

JEE Main · 2019 · Shift-IImediumCEQ-048

For the following reactions, equilibrium constants are given: S(s) + O2(g) SO2(g);\ K1 = 1052 2S(s) + 3O2(g) 2SO3(g);\…

Chemical Equilibrium · Class 11 · JEE Main Previous Year Question

Question

For the following reactions, equilibrium constants are given: $\mathrm{S(s) + O_2(g) \rightleftharpoons SO_2(g)};\ K_1 = 10^{52}$ $\mathrm{2S(s) + 3O_2(g) \rightleftharpoons 2SO_3(g)};\ K_2 = 10^{129}$ The equilibrium constant for the reaction $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$ is:

Options

a
$10^{154}$
b
$10^{25}$
✓
c
$10^{181}$
d
$10^{77}$

Correct Answerb

$10^{25}$

Detailed Solution

Step 1 — Target reaction: $\mathrm{2SO_2 + O_2 \rightleftharpoons 2SO_3}$

Step 2 — Construct from given reactions:

Reverse reaction 1 twice: $\mathrm{2SO_2 \rightleftharpoons 2S + 2O_2}$ , $K = \frac{1}{K_1^2} = 10^{-104}$
Use reaction 2: $\mathrm{2S + 3O_2 \rightleftharpoons 2SO_3}$ , $K = K_2 = 10^{129}$

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

Step 3 — Combined K: $K = 10^{-104} \times 10^{129} = 10^{25}$

Answer: Option (2) — $10^{25}$

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