Integrated Rate Laws

By the end of this section, you will be able to:

• Explain the form and function of an integrated rate law
• Perform integrated rate law calculations for zero-, first-, and second-order reactions
• Define half-life and carry out related calculations
• Identify the order of a reaction from concentration/time data

$\text{ln}\left(\frac{{\left[A\right]}_t}{{\left[A\right]}_0}\right)=\text{−}kt$

$\text{ln}\left(\frac{{\left[A\right]}_0}{{\left[A\right]}_t}\right)=kt$

$\left[A\right]={\left[A\right]}_0{e}^{-kt}$

The Integrated Rate Law for a First-Order ReactionThe rate constant for the first-order decomposition of cyclobutane, C₄H₈ at 500 °C is 9.2 $\times$

10⁻³ s⁻¹:

How long will it take for 80.0% of a sample of C₄H₈ to decompose?

Solution We use the integrated form of the rate law to answer questions regarding time:

There are four variables in the rate law, so if we know three of them, we can determine the fourth. In this case we know [A]₀, [A], and k, and need to find t.

The initial concentration of C₄H₈, [A]₀, is not provided, but the provision that 80.0% of the sample has decomposed is enough information to solve this problem. Let x be the initial concentration, in which case the concentration after 80.0% decomposition is 20.0% of x or 0.200x. Rearranging the rate law to isolate t and substituting the provided quantities yields:

Check Your Learning Iodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:

The decay is first-order with a rate constant of 0.138 d⁻¹. All radioactive decay is first order. How many days will it take for 90% of the iodine−131 in a 0.500 M solution of this substance to decay to Xe-131?

$\begin{array}{ccc}\hfill \text{ln}\left[A\right]& =& (\text{−}k)(t)+\text{ln}{\left[A\right]}_0\hfill \\ \hfill y& =& mx+b\hfill \end{array}$

Determination of Reaction Order by Graphing Show that the data in [link] can be represented by a first-order rate law by graphing ln[H₂O₂] versus time. Determine the rate constant for the rate of decomposition of H₂O₂ from this data.

Solution The data from [link] with the addition of values of ln[H₂O₂] are given in [link].

| Trial | Time (h) | [H₂O₂] (M) | ln[H₂O₂] | {: valign=”middle”}|———- | 1 | 0 | 1.000 | 0.0 | {: valign=”middle”}| 2 | 6.00 | 0.500 | −0.693 | {: valign=”middle”}| 3 | 12.00 | 0.250 | −1.386 | {: valign=”middle”}| 4 | 18.00 | 0.125 | −2.079 | {: valign=”middle”}| 5 | 24.00 | 0.0625 | −2.772 | {: valign=”middle”}{: summary=”This table contains four columns and six rows. The first row is a header row, and it labels each column, “Trial,” “Time ( h ),” “[ H subscript 2 O subscript 2 ] ( M ),” and “l n [ H subscript 2 O subscript 2 ].” Under the “Trial” column are the numbers: 1, 2, 3, 4, and 5. Under the column, “Time ( h )” are the numbers 0, 6.00, 12.00, 18.00, and 24.00. Under the column “[ H subscript 2 O subscript 2 ] ( M ),” are the numbers 1.000, 0.500, 0.250, 0.125, and 0.0625. Under the column, “l n [ H subscript 2 O subscript 2 ],” are the numbers: 0.0, negative 0.693, negative 1.386, negative 2.079, and negative 2.772.” .medium .unnumbered data-label=””}

The plot of ln[H₂O₂] versus time is linear, thus we have verified that the reaction may be described by a first-order rate law.

The rate constant for a first-order reaction is equal to the negative of the slope of the plot of ln[H₂O₂] versus time where:

In order to determine the slope of the line, we need two values of ln[H₂O₂] at different values of t (one near each end of the line is preferable). For example, the value of ln[H₂O₂] when t is 6.00 h is −0.693; the value when t = 12.00 h is −1.386:

Check Your Learning Graph the following data to determine whether the reaction $A⟶B+C$

is first order.

| Trial | Time (s) | [A] | {: valign=”middle”}|———- | 1 | 4.0 | 0.220 | {: valign=”middle”}| 2 | 8.0 | 0.144 | {: valign=”middle”}| 3 | 12.0 | 0.110 | {: valign=”middle”}| 4 | 16.0 | 0.088 | {: valign=”middle”}| 5 | 20.0 | 0.074 | {: valign=”middle”}{: summary=”This table has three columns and six rows. The first row is a header row, and it labels each column, “Trial,” “Time ( s ),” and, “[ A ].” Under the “Trial” column are the numbers: 1, 2, 3, 4, and 5. Under the “Time ( s )” column are the numbers: 4.0, 8.0, 12.0, 16.0, and 20.0. Under the “ [ A ]” column are the numbers: 0.220, 0.144, 0.110, 0.088, and 0.074.” .medium .unnumbered data-label=””}

$\text{Rate}=k{\left[A\right]}^2$

$\frac{1}{\left[A\right]}=kt+\frac{1}{{\left[A\right]}_0}$

The Integrated Rate Law for a Second-Order Reaction The reaction of butadiene gas (C₄H₆) with itself produces C₈H₁₂ gas as follows:

The reaction is second order with a rate constant equal to 5.76 $\times$

10⁻² L/mol/min under certain conditions. If the initial concentration of butadiene is 0.200 M, what is the concentration remaining after 10.0 min?

Solution We use the integrated form of the rate law to answer questions regarding time. For a second-order reaction, we have:

We know three variables in this equation: [A]₀ = 0.200 mol/L, k = 5.76 $\times$

10⁻² L/mol/min, and t = 10.0 min. Therefore, we can solve for [A], the fourth variable:

Therefore 0.179 mol/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol/L that was originally present.

Check Your Learning If the initial concentration of butadiene is 0.0200 M, what is the concentration remaining after 20.0 min?

$\begin{array}{ccc}\hfill \frac{1}{\left[A\right]}& =& kt+\frac{1}{{\left[A\right]}_0}\hfill \\ \hfill y& =& mx+b\hfill \end{array}$

The data below are for the same reaction described in [link]. Test these data to confirm that this dimerization reaction is second-order.

Solution

Trial	Time (s)	[C4H6] (M)
{: valign=”middle”}	———-
1	0	1.00 $\times$

10−2
{: valign=”middle”}	2	1600	5.04 $\times$

10−3
{: valign=”middle”}	3	3200	3.37 $\times$

10−3
{: valign=”middle”}	4	4800	2.53 $\times$

10−3
{: valign=”middle”}	5	6200	2.08 $\times$

10⁻³ | {: valign=”middle”}{: summary=”This table contains three columns and six rows. The first row is a header row and it labels each column, “Trial,” “Time ( s ),” and “[ C subscript 4 H subscript 6 ] ( M ).” Under the “Trial” column are the numbers: 1, 2, 3, 4, and 5. Under the “Time ( s )” column are the numbers: 0, 1600, 3200, 4800, and 6200. Under the column “[ C subscript 4 H subscript 6 ] ( M )” are the numbers: 1.00 times ten to the negative 2; 5.04 times ten to the negative 3; 3.37 times ten to the negative 3; 2.53 times ten to the negative 3; and 2.08 times ten to the negative 3.” .medium .unnumbered data-label=””}

In order to distinguish a first-order reaction from a second-order reaction, we plot ln[C₄H₆] versus t and compare it with a plot of $\frac{\text{1}}{\left[{\text{C}}_4{\text{H}}_6\right]}$

versus t. The values needed for these plots follow.

Time (s)

$\frac{1}{\left[{\text{C}}_4{\text{H}}_6\right]}(M^{\mathrm{-1}})$

| ln[C₄H₆] | {: valign=”middle”}|———- | 0 | 100 | −4.605 | {: valign=”middle”}| 1600 | 198 | −5.289 | {: valign=”middle”}| 3200 | 296 | −5.692 | {: valign=”middle”}| 4800 | 395 | −5.978 | {: valign=”middle”}| 6200 | 481 | −6.175 | {: valign=”middle”}{: summary=”This table contains three columns and six rows. The first row is a header row and it labels each column, “Time ( s ),” “1 over [ C subscript 4 H subscript 6 ] ( M superscript negative 1 ),” and “l n [ C subscript 4 H subscript 6 ].” Under the column “Time ( s )” are the numbers: 0, 1600, 3200, 4800, and 6200. Under the “1 over [ C subscript 4 H subscript 6 ] ( M superscript negative 1 )” column are the numbers: 100, 198, 296, 395, and 481. Under the “l n [ C subscript 4 H subscript 6 ]” column are the numbers: negative 4.605, negative 5.289, negative 5.692, negative 5.978, and negative 6.175.” .medium .unnumbered data-label=””}

The plots are shown in [link]. As you can see, the plot of ln[C₄H₆] versus t is not linear, therefore the reaction is not first order. The plot of $\frac{1}{\left[{\text{C}}_4{\text{H}}_6\right]}$

versus t is linear, indicating that the reaction is second order.

Check Your Learning Does the following data fit a second-order rate law?

| Trial | Time (s) | [A] (M) | {: valign=”middle”}|———- | 1 | 5 | 0.952 | {: valign=”middle”}| 2 | 10 | 0.625 | {: valign=”middle”}| 3 | 15 | 0.465 | {: valign=”middle”}| 4 | 20 | 0.370 | {: valign=”middle”}| 5 | 25 | 0.308 | {: valign=”middle”}| 6 | 35 | 0.230 | {: valign=”middle”}{: summary=”This table contains three columns and seven rows. The first row is a header row, and it labels each column, “Trial,” “Time ( s ),” and, “[ A ] ( M ).” Under the “Trial” column are the numbers: 1, 2, 3, 4, 5, and 6. Under the “Time ( s )” column are the numbers: 5, 10, 15, 20, 25, and 35. Under the “[ A ] ( M )” column are the numbers 0.952, 0.625, 0.465, 0.370, 0.308, and 0.230.” .medium .unnumbered data-label=””}

$\text{Rate}=k{\left[A\right]}^0=k$

$\begin{array}{ccc}\hfill \left[A\right]& =& \text{−}kt+{\left[A\right]}_0\hfill \\ \hfill y& =& mx+b\hfill \end{array}$

$\text{slope}=\text{−}k={1.3110}^{\mathrm{-6}}\text{mol/L/s}$

$\begin{array}{}\\ \hfill \text{ln}\frac{{\left[A\right]}_0}{\left[A\right]}& =& kt\hfill \\ \hfill t& =& \text{ln}\frac{{\left[A\right]}_0}{\left[A\right]}\times \frac{1}{k}\hfill \end{array}$

$\begin{array}{cc}\hfill t_{1\text{/}2}& =\text{ln}\frac{{\left[A\right]}_0}{\frac{1}{2}{\left[A\right]}_0}\times \frac{1}{k}\hfill \\ & =\text{ln}2\times \frac{1}{k}=0.693\times \frac{1}{k}\hfill \end{array}$

$t_{1\text{/}2}=\frac{0.693}{k}$

Calculation of a First-order Rate Constant using Half-Life Calculate the rate constant for the first-order decomposition of hydrogen peroxide in water at 40 °C, using the data given in [link].

Solution The half-life for the decomposition of H₂O₂ is 2.16 $\times$

10⁴ s:

Check Your Learning The first-order radioactive decay of iodine-131 exhibits a rate constant of 0.138 d⁻¹. What is the half-life for this decay?

$\frac{1}{\left[A\right]}=kt+\frac{1}{{\left[A\right]}_0}$

$\frac{1}{\left[A\right]}-\frac{1}{{\left[A\right]}_0}=kt$

$t=t_{1\text{/}2}$

$\left[A\right]=\frac{1}{2}{\left[A\right]}_0$

$\begin{array}{ccc}\hfill \frac{1}{\frac{1}{2}{\left[A\right]}_0}-\frac{1}{{\left[A\right]}_0}& =& k{t}_{1\text{/}2}\hfill \\ \hfill 2{\left[A\right]}_0-\frac{1}{{\left[A\right]}_0}& =& k{t}_{1\text{/}2}\hfill \\ \hfill \frac{1}{{\left[A\right]}_0}& =& k{t}_{1\text{/}2}\hfill \end{array}$

$t_{1\text{/}2}=\frac{1}{k{\left[A\right]}_0}$

$\left[A\right]=\text{−}kt+{\left[A\right]}_0$

$\begin{array}{ccc}\hfill \frac{{\left[\text{A}\right]}_0}{2}& =& \text{−}k{t}_{1\text{/}2}+{\left[\text{A}\right]}_0\hfill \\ \hfill k{t}_{1\text{/}2}& =& \frac{{\left[\text{A}\right]}_0}{2}\hfill \end{array}$

$t_{1\text{/}2}=\frac{{\left[A\right]}_0}{2k}$

Glossary

half-life of a reaction (t_l/2)

time required for half of a given amount of reactant to be consumed

integrated rate law

equation that relates the concentration of a reactant to elapsed time of reaction