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 is ________ atm.

Question

Quizplus · Accepted Answer

We can use the first-order rate law equation to solve for the partial pressure of SO₂Cl₂ after a certain amount of time has passed: ln(P0/P) = kt where P0 is the initial partial pressure of SO₂Cl₂, P is the partial pressure at a time t, k is the rate constant, and ln denotes the natural logarithm. We can rearrange this equation to solve for P: P = P0 * e^(-kt) Plugging in the given values, we get: P = 2.4 atm * e^(-(3.0 × 10^15 s^-1) * (5.0 min * 60 s/min)) P = 0.978 atm Rounding to two significant figures, the answer is 0.98 atm (choice C).

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

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