Exercises
Exercise 1. A reaction, A + B → P, has a rate constant of 3.2 x 10-3 s-1. If the experiment begins with a concentration of A equal to 0.25 M, how long, in s, will it take for the concentration of A to decrease to 0.047 M?
From the units of the rate constant, we are first order with respect to [A]. We use the first-order integrated rate law.
\(\displaystyle ln\biggl(\frac{[A]_t}{[A]_0}\biggr)\;=\;-kt\)
k = 3.2 x 10-3 s-1, [A]0 = 0.25 M, [A]t = 0.047 M, t = ?
Solve the equation for t
\(\displaystyle t\;=\;-\frac{ln\biggl(\frac{[A]_t}{[A]_0}\biggr)}{k}\;=\;-\frac{ln\biggl(\frac{[0.047\;M]}{[0.25\;M]}\biggr)}{3.2\times 10^{-3}s^{-1}}\;=\;\mathbf{522\;s}\)
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Exercise 2. The half-life for the radioactive decay of K-40 is 1.30 x 109 years. How long will it take for K-40 to decay to 20% of its initial concentration? All radioactive decay follows first order kinetics.
We need to find the rate constant, k, using the first order half-life equation.
\(\displaystyle k\;=\;\frac{0.693}{t_{1/2}}\;=\;\frac{0.693}{1.30\times 10^9\;yr}\;=\;5.3\times 10^{-10}\;yr^{-1}\)
Now, we can use the first order integrated rate law to find t.
k = 5.3 x 10-10 yr-1, [A]t = 20%, [A]0 = 100%
\(\displaystyle t\;=\;-\frac{ln\biggl(\frac{[A]_t}{[A]_0}\biggr)}{k}\;=\;-\frac{ln\biggl(\frac{[20\%]}{[100\%]}\biggr)}{5.3\times 10^{-10}yr^{-1}}\;=\;\mathbf{1.3\times 10^9\;yr}\)
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Exercise 3. The decomposition of cyclobutane (C4H8) is a first order process. The half-life is 7.96 x 10-3 seconds. If the initial concentration of cyclobutane is 8.50 M, what is the concentration after 0.045 seconds?
First, find the rate constant (k) from the half-life.
\(\displaystyle k\;=\;\frac{0.693}{t_{1/2}}\;=\;\frac{0.693}{7.96\times 10^{-3}\;s}\;=\;87.1\;s^{-1}\)
The first-order integrated rate law is
\(\displaystyle ln\biggl(\frac{[A]_t}{[A]_0}\biggr)\;=\;-kt\)
[A]t = ?, [A]0 = 8.50 M, k = 87.1 s-1, t = 0.045 s
Solve for ln[C_4H_8]t in the integrated rate law
\(\displaystyle ln[C_4H_8]_t\;=\;-kt\;+\;ln[C_4H_8]_0\)
\(\displaystyle ln[C_4H_8]\;=\;-87.1 s^{-1}\times 0.045\;s\;+\;ln[8.50\;M]\;=\;-1.78\)
Take the antilog ln[C4H8] = -1.78
e-1.78 = 0.169 M
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Exercise 4. Consider the following reaction at 250oC.
2 NO2 (g) → 2 NO (g) + O2 (g)
The rate constant was determined to be 0.483 M-1s-1. If the initial concentration is 8.65 M, what is the concentration after 2.3 minutes? What is the half-life of NO2?
The units of k tell us the reaction is second order. The second order integrated rate law is:
\(\displaystyle \frac{1}{[NO]_t}\;=\;kt\;+\;\frac{1}{[NO]_0}\)
[A]t = ?, [A]0 = 8.65 M, k = 0.483 M-1s-1, t = 2.3 min = 138 s
\(\displaystyle \frac{1}{[NO]_t}\;=\;0.483\;M^{-1}s^{-1}\times \;138\;s\;+\;\frac{1}{[8.65\;M]}\;=\;66.8\;\;\;\mathrm{and}\;\;\;\frac{1}{66.8}\;=\;0.0150\;M\)
The half-life is given by:
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Exercise 5. Consider the following reaction of butadiene (C4H6) gas.
2 C4H6 → C8H12 (g)
At a given temperature, the reaction is second order and the half-life of butadiene is 236 seconds. If the initial concentration of butadiene was 0.985 M, what is the concentration after 1.5 minutes?
Find k using the half-life equation for a second order reaction.
\(\displaystyle k\;=\;\frac{1}{t_{1/2}[A]_0}\)
Convert 1.5 minutes to seconds. 1.5 min = 90 s
\(\displaystyle k\;=\;\frac{1}{236\;s\times\;0.985\;M}\;=\;4.30\times 10^{-3}\;M^{-1}s^{-1}\)
The second order rate law is:
\(\displaystyle \frac{1}{[A]_t}\;=\;kt\;+\;\frac{1}{[A]_0}\)
[C4H6]t = ?, [C4H6]0 = 0.985 M, k = 4.30 x 10-3 M-1s-1, t = 90 s
\(\displaystyle \frac{1}{[C_4H_6]_t}\;=\;4.30\times 10^{-3}\;M^{-1}s^{-1}\times\;90\;s\;+\;\frac{1}{0.985\;M}\;=\;1.40\)
\(\displaystyle [C_4H_6]_t\;=\;\frac{1}{1.40}\;=\;\mathbf{0.713\;M}\)