99% of a first order reaction, was completed in 32 minute. When will 99.9% of the reaction complete?

To solve this problem, we need to use the formula for a first order reaction, which is:

t = (1/k) * ln([A]0/[A])

where:

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

From the problem, we know that 99% of the reaction is completed in 32 minutes. This means that [A] is 1% of [A]0. So we can write:

32 = (1/k) * ln([A]0/(0.01*[A]0))

Solving for k gives:

k = ln(100)/32

Now, we want to find out when 99.9% of the reaction will be complete. This means that [A] is 0.1% of [A]0. So we can write:

t = (1/k) * ln([A]0/(0.001*[A]0))

Substituting the value of k we found earlier gives:

t = 32 * ln(1000)/ln(100)

Calculating this gives:

t ≈ 48 minutes

So, 99.9% of the reaction will be complete in approximately 48 minutes.