SO₂Cl₂decomposes in the gas phase by the reaction SO₂Cl₂ (g) → SO₂ (g) + Cl₂ (g)
The reaction is first order in SO₂Cl₂ and the rate constant is 3.0 × 10^-6s^-1 at 600 K. A vessel is charged with 2.4 atm of SO₂Cl₂ at 600 K. The partial pressure of SO₂Cl₂ at 3.0 × 10⁵ s is __________ atm.

Question

Quizplus · Accepted Answer

The given reaction is first order in SO₂Cl₂. Therefore, we can use the following equation to calculate the partial pressure of SO₂Cl₂ at 3.0 × 10⁵ s. ln(P0/Pt) = kt Where, P0 = Initial pressure of SO₂Cl₂ = 2.4 atm Pt = Pressure of SO₂Cl₂ at 3.0 × 10⁵ s = ? k = Rate constant = 3.0 × 10^-6s^-1 t = Time = 3.0 × 10⁵ s Substituting the given values in the above equation, we get: ln(2.4/Pt) = (3.0 × 10^-6s^-1) × (3.0 × 10⁵) ln(2.4/Pt) = -0.08514 Pt = 0.98 atm (approx) Therefore, the partial pressure of SO₂Cl₂ at 3.0 × 10⁵ s is 0.98 atm. Hence, the correct answer is (C).

SO2Cl2 decomposes in the Gas Phase by the Reaction SO2Cl2

SO₂Cl₂decomposes in the Gas Phase by the Reaction SO₂Cl₂