Question

The first order reaction, SO2Cl2--> SO2 + Cl2, has a rate constant equal to 2.20 X 10^-5 s^-1 at 593 K. What percentage of the initial amount of So2Cl2 will remain after 6.00 hours? Please show your work thoroughly.

Answer 1

To find the percentage of the initial amount of SO2Cl2 that remains after 6.00 hours, use the first-order integrated rate law equation [A] = [A]0 * e^(-kt). Plugging in the given values, the remaining amount is approximately 99.9208% of the initial amount.

Step-by-step explanation:

To find the percentage of the initial amount of SO2Cl2 that remains after 6.00 hours, we need to use the first-order integrated rate law equation:

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

Where [A] is the concentration of SO2Cl2 at a given time, [A]0 is the initial concentration of SO2Cl2, k is the rate constant, t is the time, and e is the base of the natural logarithm. Plugging in the given values:

[A] = 100% * e^(-2.20x10^-5 s^-1 * 6.00 hours)

[A] = 100% * e^(-7.92x10^-4)

[A] ≈ 100% * 0.999208

[A] ≈ 99.9208%

Therefore, approximately 99.9208% of the initial amount of SO2Cl2 will remain after 6.00 hours.