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

To find the concentration of SO2Cl2 after 2 hours, the integrated first-order rate law is used with the provided initial concentration and rate constant. After converting 2 hours to seconds and substituting the values, the final SO2Cl2 concentration is calculated to be approximately 0.0212 mol/L.

Step-by-step explanation:

To calculate the concentration of SO2Cl2 after 2 hours for a first order reaction, we use the integrated rate law for first order reactions:

ln([A]_t/[A]_0) = -kt
Where:

• [A]_0 is the initial concentration of SO2Cl2
• [A]_t is the concentration of SO2Cl2 at time t
• k is the first-order rate constant
• t is time

Given:

• [A]_0 = 0.0248 mol/L
• k = 2.2 x 10-5 s-1
• t = 2 hours (which is 7200 seconds)

We convert the time from hours to seconds because the rate constant is in seconds. Then we can substitute these values into the integrated rate law and solve for [A]_t.

ln([A]_t/0.0248 mol/L) = - (2.2 x 10-5 s-1) (7200 s)
ln([A]_t/0.0248 mol/L) = -0.1584
[A]_t/0.0248 mol/L = e-0.158

[A]_t = 0.0248 mol/L * e-0.1584

Calculating the final concentration yields:

[A]_t ≈ 0.0248 mol/L * 0.8536
[A]_t ≈ 0.0212 mol/L

Answer 2

The concentration of SO2Cl2 after 2 hours is approximately 0.0245 mol/L.

Step-by-step explanation:

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

ln([A]t/[A]0) = -kt

where [A]t is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.

Substituting the given values, [A]0 = 0.0248 mol/L, k = 2.2 x 10-5/s, and t = 2 hours = 7200 seconds, we can find the concentration of SO2Cl2 after 2 hours.

[A]t = [A]0 * e^(-kt)

[A]t = 0.0248 * e^(-2.2 x 10-5/s * 7200s)

[A]t ≈ 0.0245 mol/L