SO₂Cl₂ decomposes 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^-6 s^-1at 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

To solve this problem, we need to use the integrated rate law for a first-order reaction: ln([SO₂Cl₂]/[SO₂Cl₂]0) = -kt where [SO₂Cl₂]0 is the initial concentration, k is the rate constant, and t is time. Rearranging this equation, we get: [SO₂Cl₂]/[SO₂Cl₂]0 = e^(-kt) At t = 3.0 × 10^-5 s, we can plug in the given values: [SO₂Cl₂]/[SO₂Cl₂]0 = e^(-(3.0 × 10^-6 s^-1)(3.0 × 10^-5 s)) = e^-0.09 Now, we can use the ideal gas law to find the partial pressure of SO₂Cl₂ at 600 K: PV = nRT n = moles of SO₂Cl₂ = P(total) x V / RT n = (2.4 atm) x (1 L) / ((0.08206 L atm K^-1 mol^-1)(600 K)) = 0.048 mol The initial partial pressure of SO₂Cl₂ is therefore: P(SO₂Cl₂)0 = (0.048 mol)/(1 L) = 0.048 atm Multiplying this by the ratio we found above gives: P(SO₂Cl₂) = (0.048 atm) x e^-0.09 = 0.98 atm So the answer is choice C, 0.98 atm.

SO2Cl2 Decomposes in the Gas Phase by the Reaction SO2Cl2(g)→

SO₂Cl₂ Decomposes in the Gas Phase by the Reaction SO₂Cl₂(g)→