Question

The reaction: SO2Cl2(g) →SO2(g) + Cl2(g) is a first order reaction. If the initial concentration of SO2Cl2 was 0.0248 mol/L and the rate constant is 2.2 X 10-5/s. What is the concentration of SO2Cl2 after 2 hours?

Answer 1

The concentration of SO2Cl2 after 2 hours is calculated using the first-order integrated rate law and is found to be approximately 0.021 mol/L.

Step-by-step explanation:

The reaction SO2Cl2(g) → SO2(g) + Cl2(g) is a first-order reaction with a given rate constant. To calculate the concentration of SO2Cl2 after 2 hours, we use the first-order integrated rate law:

ln([SO2Cl2] / [SO2Cl2]0) = -kt

With k being the rate constant (2.2 x 10-5/s) and t being the time in seconds (2 hours or 7200 seconds). Plugging in the values, we get:

ln([SO2Cl2] / 0.0248) = -(2.2 x 10-5/s)(7200 s)

Solving this equation gives us the concentration of SO2Cl2 after 2 hours. Let's perform the calculation:

ln([SO2Cl2]/0.0248) = -0.1584

[SO2Cl2]/0.0248 = e-0.1584

[SO2Cl2] = 0.0248 x e-0.1584

[SO2Cl2] ≈ 0.021 mol/L

Therefore, the concentration of SO2Cl2 after 2 hours is approximately 0.021 mol/L.