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 SO2Cl2 will remain after 6.00 hours?

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

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

Where:
[A]t is the concentration of the reactant at time t
[A]0 is the initial concentration of the reactant
k is the rate constant
t is the time

We can rearrange this equation to solve for [A]t:

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

Now, we have the equation to find the concentration of SO2Cl2 at a given time. We can substitute the given values:

[A]0 = initial concentration of SO2Cl2
k = rate constant = 2.20 x 10^-5 s^-1
t = 6.00 hours = 6.00 * 60 * 60 seconds

We need to convert the given rate constant from s^-1 to hours^-1.

k_hours^-1 = k_s^-1 * 1 hour/3600 seconds

Using the given rate constant:
k_hours^-1 = 2.20 x 10^-5 s^-1 * 1 hour/3600 seconds = 6.11 x 10^-9 hours^-1

Now, we can substitute the values into the equation to find [A]t:

[A]t = [A]0 * e^(-k_hours^-1 * t)

[A]t = [A]0 * e^(-6.11 x 10^-9 hours^-1 * 6.00 hours)

Solving this equation will give us the concentration of SO2Cl2 after 6.00 hours. To find the percentage remaining, we divide this concentration by the initial concentration and multiply by 100:

Percentage remaining = ([A]t / [A]0) * 100

Now, we have the equation and the method to calculate the percentage of the initial amount of SO2Cl2 that remains after 6.00 hours.

To solve this problem, we can use the first-order reaction equation:

ln([A]/[A0]) = -kt

Where [A] is the concentration of the reactant at a given time, [A0] is the initial concentration of the reactant, k is the rate constant, t is the time, and ln represents the natural logarithm.

We can rearrange the equation to solve for [A]:

[A]/[A0] = e^(-kt)

Now let's plug in the given values:

k = 2.20 x 10^-5 s^-1
t = 6.00 hours

First, we need to convert the time from hours to seconds:

6.00 hours = 6.00 x 60 x 60 seconds = 21600 seconds

Now we can calculate [A]/[A0]:

[A]/[A0] = e^(-2.20 x 10^-5 s^-1 * 21600 seconds)

[A]/[A0] = e^(-475.2)

Using a calculator:

[A]/[A0] ≈ 0.624

To find the percentage remaining, we can multiply by 100:

Percentage remaining = 0.624 * 100 ≈ 62.4%

Therefore, approximately 62.4% of the initial amount of SO2Cl2 will remain after 6.00 hours.

62%