Question

The reaction:so2cl2(g)---so2(g)+cl2(g) is the first order reaction .if the initial concentration of so2cl2 was 0.0248 mol/L and the rate constant is 2.2×10-⁵/s.what is the concentration of so2cl2 after two hours?

Answer 1

To find the concentration of SO2Cl2 after two hours, use the integrated first-order rate law. The concentration of SO2Cl2 after two hours is approximately 0.0212 mol/L.

Step-by-step explanation:

The reaction SO2Cl2(g) → SO2(g) + Cl2(g) is a first-order reaction. Given an initial concentration of SO2Cl2 of 0.0248 mol/L and a rate constant k of 2.2×10−5/s, we can use the integrated first-order rate law to find the concentration after two hours. The integrated rate law for a first-order reaction is ln([A]t/[A]0) = -kt, where [A]0 is the initial concentration, [A]t is the concentration at time t, and k is the rate constant.

First, convert two hours into seconds because the rate constant is given in per second: 2 hours × 3600 seconds/hour = 7200 seconds. Then, substitute the known values into the equation to solve for [A]t.

ln([A]t/0.0248) = -2.2×10−5/s × 7200 s

ln([A]t/0.0248) = -0.1584

[A]t = 0.0248 × e−0.1584

[A]t ≈ 0.0248 × 0.8535 = 0.0212 mol/L

The concentration of SO2Cl2 after two hours is roughly 0.0212 mol/L.