Question

A first-order reaction is \(75.0 \%\) complete in \(320 . \mathrm{s}\). a. What are the first and second half-lives for this reaction? b. How long does it take for \(90.0 \%\) completion?

Short answer

a. The first and second half-lives for this reaction are both 255 seconds. b. It takes 448 seconds for the reaction to reach 90.0% completion.

Step-by-step solution

(Determine k from the given information)

As the reaction is 75% complete, it means that 25% is left. So, \(\frac{[A]}{[A]_0} = 0.25\). We can use this information to find the rate constant from the given time. \(k = \frac{1}{320 s} \ln{\frac{1}{0.25}} = 0.00272 s^{-1}\)

(Find the First Half-life)

Now that we have the rate constant, we can use the half-life equation to find the first half-life: \(t_{1/2} = \frac{0.693}{0.00272 s^{-1}} = 255 s\) The first half-life is 255 seconds.

(Find the Second Half-life)

Since this is a first-order reaction, the half-life remains constant throughout the reaction. So, the second half-life is equal to the first half-life. The second half-life is 255 seconds. #a. Answer:# The first and second half-lives for this reaction are both 255 seconds. #b. Time for 90.0% completion#

(Calculate the remaining concentration)

For 90% completion, it means that 10% is left. We can set up the equation to find the time it takes for the reaction to reach that point: \(k = \frac{1}{t} \ln{\frac{[A]_0}{[A]}}\) \(\frac{1}{t} = \frac{0.00272 s^{-1}}{\ln{\frac{1}{0.1}}}\)

(Solve for time)

Next, we solve for time: \(t = \frac{1}{0.00272 s^{-1}} \times \ln{\frac{1}{0.1}} = 448 s\) #b. Answer:# It takes 448 seconds for the reaction to reach 90.0% completion.