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Chapter 11: Problem 55

For the reaction \(\mathrm{A} \rightarrow\) products, successive half-lives are observed to be \(10.0,20.0,\) and 40.0 min for an experiment in which \([\mathrm{A}]_{0}=0.10 \mathrm{M} .\) Calculate the concentration of \(\mathrm{A}\) at the following times. a. \(80.0 \mathrm{min}\) b. \(30.0 \mathrm{min}\)

Short Answer

Expert verified
a. At 80.0 min, the concentration of A is \(0.0123 M\). b. At 30.0 min, the concentration of A is \(0.0323 M\).

Step by step solution

01

Determine the reaction order

Observe the relationship between the successive half-lives given for the reaction. Compare each half-life with the previous one to see if they increase, decrease, or remain constant. In this case: For 1st half-life: 10.0 min For 2nd half-life: 20.0 min (which is 2 x 10.0 min) For 3rd half-life: 40.0 min (which is 2 x 20.0 min) The half-lives are increasing in a doubling pattern. Hence, the reaction is most likely a second-order reaction.
02

Calculate the rate constant

For a second-order reaction, we know that the half-life (t1/2) is equal to \(\frac{1}{k[A]_0}\). We will use the first half-life and initial concentration value to find the rate constant k. Given t1/2 = 10.0 min and \([A]_0\) = 0.10 M, we set up the equation: 10.0 = \(\frac{1}{k(0.10)}\) Solve for k: k = \(\frac{1}{10.0(0.10)}\) k = 1.00 min\(^{-1}\)M\(^{-1}\)
03

Calculate the concentration of A at 80.0 min

For a second-order reaction, the integrated rate law is given by: \(\frac{1}{[A]} = kt + \frac{1}{[A]_0}\) We want to find the concentration of A at 80.0 min. Plug in the values for k, t, and \([A]_0\): \(\frac{1}{[A]} = (1.00)(80.0) + \frac{1}{0.10}\) Solve for [A]: \(\frac{1}{[A]} = 81\) \([A] = \frac{1}{81}\) \([A] = 0.0123 M\) The concentration of A at 80.0 min is 0.0123 M.
04

Calculate the concentration of A at 30.0 min

Using the second-order integrated rate law again, plug in the values for k, t, and \([A]_0\) to find the concentration of A at 30.0 min: \(\frac{1}{[A]} = (1.00)(30.0) + \frac{1}{0.10}\) Solve for [A]: \(\frac{1}{[A]} = 31\) \([A] = \frac{1}{31}\) \([A] = 0.0323 M\) The concentration of A at 30.0 min is 0.0323 M. To summarize: a. At 80.0 min, the concentration of A is 0.0123 M. b. At 30.0 min, the concentration of A is 0.0323 M.

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Most popular questions from this chapter

A reaction of the form $$aA \longrightarrow Products$$gives a plot of \(\ln [\mathrm{A}]\) versus time (in seconds), which is a straight line with a slope of \(-7.35 \times 10^{-3} .\) Assuming \([\mathrm{A}]_{0}=\) \(0.0100 M,\) calculate the time (in seconds) required for the reaction to reach \(22.9 \%\) completion.

How does temperature affect \(k,\) the rate constant? Explain.

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