In a closed container maintained at 1 atm pressure and 25°C, 2 moles of SO₂(g) and 1 mole of O₂(g) were allowed to react to form SO₃(g) under the influence of a catalyst.
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

Question

In a closed container maintained at 1 atm pressure and 25°C, 2 moles of SO₂(g) and 1 mole of O₂(g) were allowed to react to form SO₃(g) under the influence of a catalyst.
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
At equilibrium, it was found that 50% of SO₂(g) was converted to SO₃(g). The partial pressure of O₂(g) at equilibrium will be

Prashant Kotian · Accepted Answer

Correct answer: D. Initial SO₂ = 2, O₂ = 1. 50% of SO₂ reacts = 1 mole. O₂ reacted = 0.5 mole. Eq moles: SO₂ = 1, O₂ = 0.5, SO₃ = 1. Total eq moles = 2.5. Partial pressure of O₂ = (Moles of O₂ / Total moles) × Total P = (0.5 / 2.5) × 1 atm = 0.20 atm. Therefore, correct answer is D.

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