Chapter 18: Kinetics
18.4 Integrated Rate Laws
Learning Outcomes
- Determine the concentration of a reactant at a given time
- Use integrated rate laws to identify the orders of reactions and determine their rate constants
- Analyze plots of reaction data to identify reaction order and rate constants
- Calculate the half-life of a reactant
- Use the half-life of a reactant to calculate its concentration after a given amount of time (or calculate the amount of time it will take for a reactant to decay)
The rate laws we have seen thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called integrated rate laws. We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.
Using calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. This process can either be very straightforward or very complex, depending on the complexity of the differential rate law. For purposes of discussion, we will focus on the resulting integrated rate laws for first-, second-, and zero-order reactions.
First-Order Reactions
Integration of the rate law for a simple first-order reaction (rate = [latex]k[A][/latex]) results in an equation describing how the reactant concentration varies with time:
[latex][A]_{t}=[A]_{0}e^{-kt}[/latex]
where [latex][A]_{t}[/latex] is the concentration of [latex]A[/latex] at any time [latex]t[/latex], [latex][A]_{0}[/latex] is the initial concentration of [latex]A[/latex], and [latex]k[/latex] is the first-order rate constant.
For mathematical convenience, this equation may be rearranged to other formats, including direct and indirect proportionalities:
[latex]\text{ln}\left(\dfrac{{\left[A\right]}_{t}}{{\left[A\right]}_{0}}\right)=-kt[/latex]
or
[latex]\text{ln}\left(\dfrac{{\left[A\right]}_{0}}{{\left[A\right]}_{t}}\right)=kt[/latex]
and a format showing a linear dependence of concentration in time:
[latex]\text{ln}[A]_{t}=\text{ln}[A]_{0}-kt[/latex]
Example 18.4.1: The Integrated Rate Law for a First-Order Reaction
The rate constant for the first-order decomposition of cyclobutane, [latex]\ce{C4H8}[/latex] at 500 °C is 9.2 × 10−3 s−1:
[latex]\ce{C4H8}\rightarrow\ce{2C2H4}[/latex]
How long will it take for 80.0% of a sample of [latex]\ce{C4H8}[/latex] to decompose?
Show Solution
We use the integrated form of the rate law to answer questions regarding time:
[latex]\text{ln}\left(\dfrac{{\left[A\right]}_{0}}{\left[A\right]}_{t}\right)=kt[/latex]
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 [latex]\ce{C4H8}[/latex], [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:
[latex]\begin{array}{cc}\hfill t& =\mathrm{ln}\dfrac{\left[x\right]}{\left[0.200x\right]}\times \dfrac{1}{k}\hfill \\ & =\text{ln}5\times \dfrac{1}{9.2\times {10}^{-3}{\text{s}}^{-1}}\hfill \\ & =1.609\times \dfrac{1}{9.2\times {10}^{-3}{\text{s}}^{-1}}\hfill \\ & =1.7\times {10}^{2}\text{s}\hfill \end{array}[/latex]
Check Your Learning
We can use integrated rate laws with experimental data that consist of time and concentration information to determine the order and rate constant of a reaction. The integrated rate law can be rearranged to a standard linear equation format:
[latex]\begin{array}{ccc}\hfill \text{ln}\left[A\right]_{t}& =& \left(-k\right)\left(t\right)+\text{ln}{\left[A\right]}_{0}\hfill \\ \hfill y& =& mx+b\hfill \end{array}[/latex]
As shown in Figure 18.4.1, a plot of [latex]\text{ln}[A]_{t}[/latex] versus [latex]t[/latex] for a first-order reaction is a straight line with a slope of –k and a y-intercept of [latex]\text{ln}[A]_{0}[/latex]. If a set of rate data are plotted in this fashion but do not result in a straight line, the reaction is not first order in [latex]A[/latex].
![A graph of natural log of the concentration of A versus time. A diagonal line slopes down from a point on the y-axis labeled b = ln[A] sub 0 to a point on the x-axis. The line is labeled m = -k.](https://pressbooks.online.ucf.edu/app/uploads/sites/414/2021/06/Figure-18.4.1_First-Order.svg)
Example 18.4.2: Determination of Reaction Order by Graphing
Show that the data in Figure 18.4.2 can be represented by a first-order rate law by graphing ln[latex]\ce{H2O2}[/latex] versus time. Determine the rate constant for the rate of decomposition of [latex]\ce{H2O2}[/latex] from this data.
![A table with five columns is shown. The first column is labeled, “Time, h.” Beneath it the numbers 0.00, 6.00, 12.00, 18.00, and 24.00 are listed. The second column is labeled, “[ H subscript 2 O subscript 2 ], mol / L.” Below, the numbers 1.000, 0.500, 0.250, 0.125, and 0.0625 are double spaced. To the right, a third column is labeled, “capital delta [ H subscript 2 O subscript 2 ], mol / L.” Below, the numbers negative 0.500, negative 0.250, negative 0.125, and negative 0.062 are listed such that they are double spaced and offset, beginning one line below the first number listed in the column labeled, “[ H subscript 2 O subscript 2 ], mol / L.” The first two numbers in the second column have line segments extending from their right side to the left side of the first number in the third row. The second and third numbers in the second column have line segments extending from their right side to the left side of the second number in the third row. The third and fourth numbers in the second column have line segments extending from their right side to the left side of the third number in the third row. The fourth and fifth numbers in the second column have line segments extending from their right side to the left side of the fourth number in the third row. The fourth column in labeled, “capital delta t, h.” Below the title, the value 6.00 is listed four times, each single-spaced. The fifth and final column is labeled “Rate of Decomposition, mol / L / h.” Below, the following values are listed single-spaced: negative 0.0833, negative 0.0417, negative 0.0208, and negative 0.0103.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/219/2016/08/09033022/Table2.png)
Show Solution
The data from Figure 18.4.2 with the addition of values of ln[latex]\ce{H2O2}[/latex] are given in Figure 18.4.3.
![A graph is shown with the label “Time ( h )” on the x-axis and “l n [ H subscript 2 O subscript 2 ]” on the y-axis. The x-axis shows markings at 6, 12, 18, and 24 hours. The vertical axis shows markings at negative 3, negative 2, negative 1, and 0. A decreasing linear trend line is drawn through five points represented at the coordinates (0, 0), (6, negative 0.693), (12, negative 1.386), (18, negative 2.079), and (24, negative 2.772).](https://s3-us-west-2.amazonaws.com/courses-images-archive-read-only/wp-content/uploads/sites/887/2015/08/23214429/FrstOKin.jpg)
| Trial | Time (h) | [latex]\ce{[H2O2]}[/latex] (M) | ln[latex]\ce{[H2O2]}[/latex] |
|---|---|---|---|
| 1 | 0 | 1.000 | 0.0 |
| 2 | 6.00 | 0.500 | –0.693 |
| 3 | 12.00 | 0.250 | –1.386 |
| 4 | 18.00 | 0.125 | –2.079 |
| 5 | 24.00 | 0.0625 | –2.772 |
The plot of ln[latex]\ce{[H2O2]}[/latex] 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[latex]\ce{[H2O2]}[/latex] versus time where:
[latex]\text{slope}=\dfrac{\text{change in }y}{\text{change in }x}=\dfrac{\Delta y}{\Delta x}=\dfrac{\Delta\text{ ln}\left[{\ce{H}}_{2}{\ce{O}}_{2}\right]}{\Delta t}[/latex]
In order to determine the slope of the line, we need two values of ln[latex]\ce{[H2O2]}[/latex] at different values of t (one near each end of the line is preferable). For example, the value of ln[latex]\ce{[H2O2]}[/latex] when t is 6.00 h is -0.693; the value when t = 12.00 h is −1.386:
[latex]\begin{array}{ccc}\hfill \text{slope}& =& \dfrac{-1.386-\left(-0.693\right)}{\text{12.00 h}-\text{6.00 h}}\hfill \\ & =& \dfrac{-0.693}{\text{6.00 h}}\hfill \\ & =& -1.155\times {10}^{-2}{\text{h}}^{-1}\hfill \\ \hfill k& =& -\text{slope}=-\left(-1.155\times {10}^{-2}{\text{h}}^{-1}\right)=1.155\times {10}^{-2}{\text{h}}^{-1}\hfill \end{array}[/latex]
Check Your Learning
Second-Order Reactions
The equations that relate the concentrations of reactants and the rate constant of second-order reactions are fairly complicated. We will limit ourselves to the simplest second-order reactions, namely, those with rates that are dependent upon just one reactant’s concentration and described by the differential rate law:
[latex]\text{rate}=k{\left[A\right]}^{2}[/latex]
For these second-order reactions, the integrated rate law is:
[latex]\dfrac{1}{\left[A\right]_{t}}=kt+\dfrac{1}{{\left[A\right]}_{0}}[/latex]
where the terms in the equation have their usual meanings as defined above.
Example 18.4.3: The Integrated Rate Law for a Second-Order Reaction
The reaction of butadiene gas [latex]\ce{(C4H6)}[/latex] with itself produces [latex]\ce{C8H_{12}}[/latex] gas as follows:
[latex]\ce{2C4H6}(g)\rightarrow\ce{C8H_{12}}(g)[/latex]
The reaction is second order with a rate constant equal to 5.76 × 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?
Show Solution
We use the integrated form of the rate law to answer questions regarding time. For a second-order reaction, we have:
[latex]\dfrac{1}{\left[A\right]}=kt+\dfrac{1}{{\left[A\right]}_{0}}[/latex]
We know three variables in this equation: [A]0 = 0.200 mol/L, k = 5.76 × 10−2 L/mol/min, and t = 10.0 min. Therefore, we can solve for [A], the fourth variable:
[latex]\begin{array}{ccc}\hfill \dfrac{1}{\left[A\right]}& =& \left(5.76\times {10}^{-2}{\text{L mol}}^{-1}{\mathrm{min}}^{-1}\right)\left(10\text{min}\right)+\dfrac{1}{0.200{\text{mol}}^{-1}}\hfill \\ \hfill \dfrac{1}{\left[A\right]}& =& \left(5.76\times {10}^{-1}{\text{L mol}}^{-1}\right)+5.00{\text{L mol}}^{-1}\hfill \\ \hfill \dfrac{1}{\left[A\right]}& =& 5.58{\text{L mol}}^{-1}\hfill \\ \hfill \left[A\right]& =& 1.79\times {10}^{-1}{\text{mol L}}^{-1}\hfill \end{array}[/latex]
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
The integrated rate law for our second-order reactions has the form of the equation of a straight line:
[latex]\begin{array}{ccc}\hfill \dfrac{1}{\left[A\right]_{t}}& =& kt+\dfrac{1}{{\left[A\right]}_{0}}\hfill \\ \hfill y& =& mx+b\hfill \end{array}[/latex]
As shown in Figure 18.4.4, a plot of [latex]\dfrac{1}{\left[A\right]_{t}}[/latex] versus t for a second-order reaction is a straight line with a slope of k and an intercept of [latex]\dfrac{1}{{\left[A\right]}_{0}}[/latex]. If the plot is not a straight line, then the reaction is not second order.
![A graph of one over the concentration of A versus time. A diagonal line slopes up from a point on the y-axis labeled b = 1/[A] sub 0. The line is labeled m = k.](https://pressbooks.online.ucf.edu/app/uploads/sites/414/2021/06/Figure-18.4.1_Second-Order.svg)
Example 18.4.4: Determination of Reaction Order by Graphing
Test the data given to show whether the dimerization of [latex]\ce{C4H6}[/latex] is a first- or a second-order reaction.
Show Solution
| Trial | Time (s) | [latex]\ce{[C4H6]}[/latex] (M) |
|---|---|---|
| 1 | 0 | 1.00 × 10–2 |
| 2 | 1600 | 5.04 × 10–3 |
| 3 | 3200 | 3.37 × 10–3 |
| 4 | 4800 | 2.53 × 10–3 |
| 5 | 6200 | 2.08 × 10–3 |
In order to distinguish a first-order reaction from a second-order reaction, we plot [latex]\text{ln}[\ce{4H6}][/latex] versus t and compare it with a plot of [latex]\dfrac{\text{1}}{[{\ce{C}}_{4}{\ce{H}}_{6}]}[/latex] versus t. The values needed for these plots follow.
| Time (s) | [latex]\dfrac{1}{\left[{\text{C}}_{4}{\text{H}}_{6}\right]}\left({M}^{-1}\right)[/latex] | ln[latex]\ce{[C4H6]}[/latex] |
|---|---|---|
| 0 | 100 | –4.605 |
| 1600 | 198 | –5.289 |
| 3200 | 296 | –5.692 |
| 4800 | 395 | –5.978 |
| 6200 | 481 | –6.175 |
The plots are shown in Figure 18.4.5. As you can see, the plot of ln[latex]\ce{[C4H6]}[/latex] versus t is not linear, therefore the reaction is not first order. The plot of [latex]\dfrac{1}{\ce{[C4H6]_t}}[/latex] versus t is linear, indicating that the reaction is second order.
![Two graphs are shown, each with the label “Time ( s )” on the x-axis. The graph on the left is labeled, “l n [ C subscript 4 H subscript 6 ],” on the y-axis. The graph on the right is labeled “1 divided by [ C subscript 4 H subscript 6 ],” on the y-axis. The x-axes for both graphs show markings at 3000 and 6000. The y-axis for the graph on the left shows markings at negative 6, negative 5, and negative 4. A decreasing slightly concave up curve is drawn through five points at coordinates that are (0, negative 4.605), (1600, negative 5.289), (3200, negative 5.692), (4800, negative 5.978), and (6200, negative 6.175). The y-axis for the graph on the right shows markings at 100, 300, and 500. An approximately linear increasing curve is drawn through five points at coordinates that are (0, 100), (1600, 198), (3200, 296), and (4800, 395), and (6200, 481).](https://s3-us-west-2.amazonaws.com/courses-images-archive-read-only/wp-content/uploads/sites/887/2015/04/23212252/CNX_Chem_12_04_2OrdKin1.jpg)
According to the second-order integrated rate law, the rate constant is equal to the slope of the [latex]\dfrac{1}{\left[A\right]_{t}}[/latex] versus t plot. Using the data for t = 0s and t = 6200 s, the rate constant is estimated as follows:
[latex]k=\text{slope}=\dfrac{(481\text{ M}^{-1}-100\text{ M}^{-1})}{(6200\text{s}-0\text{s})}=0.0614\text{ M}^{-1}\text{s}^{-1}[/latex]
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Zero-Order Reactions
For zero-order reactions, the differential rate law is:
[latex]\text{rate}=k{\left[A\right]}^{0}=k[/latex]
A zero-order reaction thus exhibits a constant reaction rate, regardless of the concentration of its reactant(s). This may seem counterintuitive, since the reaction rate certainly can’t be finite when the reactant concentration is zero. For purposes of this introductory text, it will suffice to note that zero-order kinetics are observed for some reactions only under certain specific conditions. These same reactions exhibit different kinetic behaviors when the specific conditions aren’t met, and for this reason the more prudent term pseudo-zero-order is sometimes used.
The integrated rate law for a zero-order reaction also has the form of the equation of a straight line:
[latex]\begin{array}{ccc}\hfill \left[A\right]_{t}& =& {-}kt+{\left[A\right]}_{0}\hfill \\ \hfill y& =& mx+b\hfill \end{array}[/latex]
As shown in Figure 18.4.6, a plot of [A] versus t for a zero-order reaction is a straight line with a slope of −k and a y-intercept of [A]0.
![A graph of the concentration of A versus time. A diagonal line slopes down from a point on the y-axis labeled b = [A] sub 0 to a point on the x-axis. The line is labeled m = -k.](https://pressbooks.online.ucf.edu/app/uploads/sites/414/2021/06/Figure-18.4.1_Zero-Order.svg)
Figure 18.4.7 shows a plot of [latex]\ce{[NH3]}[/latex] versus t for the thermal decomposition of ammonia at the surface of two different heated solids. The decomposition reaction exhibits first-order behavior at a quartz [latex]\ce{(SiO2)}[/latex] surface, as suggested by the exponentially decaying plot of concentration versus time. On a tungsten surface, however, the plot is linear, indicating zero-order kinetics.
![A graph is shown with the label, “Time ( s ),” on the x-axis and, “[ N H subscript 3 ] M,” on the y-axis. The x-axis shows a single value of 1000 marked near the right end of the axis. The vertical axis shows markings at 1.0 times 10 superscript negative 3, 2.0 times 10 superscript negative 3, and 3.0 times 10 superscript negative 3. A decreasing linear trend line is drawn through six points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (200, 2.6 times 10 superscript negative 3), (400, 2.3 times 10 superscript negative 3), (600, 2.0 times 10 superscript negative 3), (800, 1.8 times 10 superscript negative 3), and (1000, 1.6 times 10 superscript negative 3). This line is labeled “Decomposition on W.” A decreasing slightly concave up curve is similarly drawn through eight points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (100, 2.5 times 10 superscript negative 3), (200, 2.1 times 10 superscript negative 3), (300, 1.9 times 10 superscript negative 3), (400, 1.6 times 10 superscript negative 3), (500, 1.4 times 10 superscript negative 3), and (750, 1.1 times 10 superscript negative 3), ending at about (1000, 0.7 times 10 superscript negative 3). This curve is labeled “Decomposition on S i O subscript 2.”](https://s3-us-west-2.amazonaws.com/courses-images-archive-read-only/wp-content/uploads/sites/887/2015/04/23212255/CNX_Chem_12_04_AmDecomK1.jpg)
If we use the data from the plot in Figure 18.4.7, we can graphically estimate the zero-order rate constant for ammonia decomposition at a tungsten surface. The integrated rate law for zero-order kinetics describes a linear plot of reactant concentration, [A]t, versus time, t, with a slope equal to the negative of the rate constant, −k. Following the mathematical approach of previous examples, the slope of the linear data plot (for decomposition on W) is estimated from the graph. Using the ammonia concentrations at t = 0 and t = 1000 s:
[latex]k=-\text{slope} = \dfrac{(0.0015\text{ mol L}^{-1}-0.0028\text{ mol L}^{-1})}{(1000\text{s}-0\text{s})}= 1.3\times{10}^{-6}\text{ mol L}^{-1}\text{s}^{-1}[/latex]
The zero-order plot in Figure 18.4.7 shows an initial ammonia concentration of 0.0028 mol L−1 decreasing linearly with time for 1000 s. Assuming no change in this zero-order behavior, you should be able to calculate the time (min) when the concentration will reach 0.0001 mol L−1. The correct time is 35 minutes!
The Half-Life of a Reaction
The half-life of a reaction is the time required for one-half of a given amount of reactant to be consumed. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide in Figure 18.4.1 as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of [latex]\ce{H2O2}[/latex] decreases from 1.000 M to 0.500 M. During the second half-life (from 6.00 hours to 12.00 hours), it decreases from 0.500 M to 0.250 M; during the third half-life, it decreases from 0.250 M to 0.125 M. The concentration of [latex]\ce{H2O2}[/latex] decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a first-order reaction, and, as can be shown, the half-life of a first-order reaction is independent of the concentration of the reactant. However, half-lives of reactions with other orders depend on the concentrations of the reactants.
First-Order Reactions
We can derive an equation for determining the half-life of a first-order reaction from the alternate form of the integrated rate law as follows:
[latex]\begin{array}{lll}\hfill \mathrm{ln}\dfrac{{\left[A\right]}_{0}}{\left[A\right]}& =& kt\hfill \\ \hfill t& =& \mathrm{ln}\dfrac{{\left[A\right]}_{0}}{\left[A\right]}\times \dfrac{1}{k}\hfill \end{array}[/latex]
If we set the time t equal to the half-life, [latex]{t}_{1\text{/}2},[/latex] the corresponding concentration of A at this time is equal to one-half of its initial concentration. Hence, when [latex]t={t}_{1\text{/}2},[/latex] [latex]\left[A\right]_{t}=\dfrac{1}{2}{\left[A\right]}_{0}[/latex].
Therefore:
[latex]\begin{array}{cc}\hfill {t}_{1\text{/}2}& =\mathrm{ln}\dfrac{{\left[A\right]}_{0}}{\frac{1}{2}{\left[A\right]}_{0}}\times \dfrac{1}{k}\hfill \\ & =\mathrm{ln}2\times \dfrac{1}{k}=0.693\times \dfrac{1}{k}\hfill \end{array}[/latex]
Thus:
[latex]{t}_{1\text{/}2}=\dfrac{0.693}{k}[/latex]
We can see that the half-life of a first-order reaction is inversely proportional to the rate constant k. A fast reaction (shorter half-life) will have a larger k; a slow reaction (longer half-life) will have a smaller k.
Example 18.4.5: 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 Figure 18.4.8.
Show Solution
The half-life for the decomposition of [latex]\ce{H2O2}[/latex] is 2.16 × 104 s:
[latex]\begin{array}{ccc}\hfill {t}_{1\text{/}2}& =& \dfrac{0.693}{k}\hfill \\ \hfill k& =& \dfrac{0.693}{{t}_{1\text{/}2}}=\dfrac{0.693}{2.16\times {10}^{4}\text{s}}=3.21\times {10}^{-5}{\text{s}}^{-1}\hfill \end{array}[/latex]
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Second-Order Reactions
We can derive the equation for calculating the half-life of a second order as follows:
[latex]\dfrac{1}{\left[A\right]_{t}}=kt+\dfrac{1}{{\left[A\right]}_{0}}[/latex]
or
[latex]\dfrac{1}{\left[A\right]}-\dfrac{1}{{\left[A\right]}_{0}}=kt[/latex]
Restrict t
[latex]t={t}_{1\text{/}2}[/latex]
then we can define
[latex]\left[A\right]_{t}=\dfrac{1}{2}{\left[A\right]}_{0}[/latex]
and substitute into the integrated rate law and simplify:
[latex]\begin{array}{ccc}\hfill \dfrac{1}{\frac{1}{2}{\left[A\right]}_{0}}-\dfrac{1}{{\left[A\right]}_{0}}& =& k{t}_{1\text{/}2}\hfill \\ \hfill 2{\left[A\right]}_{0}-\dfrac{1}{{\left[A\right]}_{0}}& =& k{t}_{1\text{/}2}\hfill \\ \hfill \dfrac{1}{{\left[A\right]}_{0}}& =& k{t}_{1\text{/}2}\hfill \end{array}[/latex]
Thus:
[latex]{t}_{1\text{/}2}=\dfrac{1}{k{\left[A\right]}_{0}}[/latex]
For a second-order reaction, [latex]{t}_{1\text{/}2}[/latex] is inversely proportional to the concentration of the reactant, and the half-life increases as the reaction proceeds because the concentration of reactant decreases. Consequently, we find the use of the half-life concept to be more complex for second-order reactions than for first-order reactions. Unlike with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration is known.
Zero-Order Reactions
We can derive an equation for calculating the half-life of a zero order reaction as follows:
[latex]\left[A\right]={-}kt+{\left[A\right]}_{0}[/latex]
When half of the initial amount of reactant has been consumed [latex]t={t}_{1\text{/}2}[/latex] and [latex]\left[A\right]=\dfrac{{\left[A\right]}_{0}}{2}.[/latex] Thus:
[latex]\begin{array}{ccc}\hfill \dfrac{{\left[\text{A}\right]}_{0}}{2}& =& {-}k{t}_{1\text{/}2}+{\left[\text{A}\right]}_{0}\hfill \\ \hfill k{t}_{1\text{/}2}& =& \dfrac{{\left[\text{A}\right]}_{0}}{2}\hfill \end{array}[/latex]
and
[latex]{t}_{1\text{/}2}=\text{}\dfrac{{\left[A\right]}_{0}}{2k}[/latex]
The half-life of a zero-order reaction increases as the initial concentration increases.
Equations for both differential and integrated rate laws and the corresponding half-lives for zero-, first-, and second-order reactions are summarized in Table 18.4.1.
Example 18.4.6: Half-Life for Zero-Oreder and Second-Order Reactions
The reaction of butadiene gas [latex]\ce{(C4H6)}[/latex] with itself produces [latex]\ce{C8H_{12}}[/latex] gas as follows:
[latex]\ce{2C4H6}(g)\rightarrow\ce{C8H_{12}}(g)[/latex]
The reaction is second order with an initial butadiene concentration of 0.200 M, and a rate constant equal to 5.76 × 10−2 L/mol/min under certain conditions.
What is the half-life (ms) for the butadiene dimerization reaction described above?
Show Solution
The reaction in question is second order, is initiated with a 0.200 mol L−1 reactant solution, and exhibits a rate constant of 0.0576 L mol−1 min−1. Substituting these quantities into the second-order half-life equation:
[latex]t_{1/2}=\dfrac{1}{[(0.0576\text{ L mol}^{-1}\text{min}^{-1})(0.200\text{ mol L}^{-1})]}=86.8\text{ min}[/latex]
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Key Concepts and Summary
Differential rate laws can be determined by the method of initial rates or other methods. We measure values for the initial rates of a reaction at different concentrations of the reactants. From these measurements, we determine the order of the reaction in each reactant. Integrated rate laws are determined by integration of the corresponding differential rate laws. Rate constants for those rate laws are determined from measurements of concentration at various times during a reaction.
The half-life of a reaction is the time required to decrease the amount of a given reactant by one-half. The half-life of a zero-order reaction decreases as the initial concentration of the reactant in the reaction decreases. The half-life of a first-order reaction is independent of concentration, and the half-life of a second-order reaction decreases as the concentration increases.
Key Equations
- integrated rate law for zero-order reactions: [latex]\left[A\right]={-}kt+{\left[A\right]}_{0},[/latex] [latex]{t}_{1\text{/}2}=\dfrac{{\left[A\right]}_{0}}{2k}[/latex]
- integrated rate law for first-order reactions: [latex]\mathrm{ln}\left[A\right]={-}kt+{\left[A\right]}_{0},\text{}{t}_{1\text{/}2}=\dfrac{0.693}{k}[/latex]
- integrated rate law for second-order reactions: [latex]\dfrac{1}{\left[A\right]}=kt+\dfrac{1}{{\left[A\right]}_{0}},[/latex] [latex]{t}_{1\text{/}2}=\dfrac{1}{{\left[A\right]}_{0}k}[/latex]
Try It
- The half-life of a reaction of compound A to give compounds D and E is 8.50 min when the initial concentration of A is 0.150 mol/L. How long will it take for the concentration to drop to 0.0300 mol/L if the reaction is (a) first order with respect to A or (b) second order with respect to A?
- What is the half-life for the decomposition of [latex]\ce{NOCl}[/latex] when the concentration of NOCl is 0.15 M? The rate constant for this second-order reaction is 8.0 × 10-8 L/mol/s.
Show Selected Solutions
- 2. In a second-order reaction, the rate is concentration-dependent, [latex]{t}_{1\text{/}2}=\frac{1}{k{\left[A\right]}_{0}}.[/latex]
- [latex]{t}_{1\text{/}2}=\frac{1}{k{\left[A\right]}_{0}}=\frac{1}{8.0\times {10}^{-8}{\text{L mol}}^{-1}{\text{s}}^{-1}\left[0.15M\right]}=8.3\times {10}^{7}\text{s}[/latex]
Glossary
half-life of a reaction (tl/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
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- First Order Reaction Chemistry Problems – Half Life, Rate Constant K, Integrated Rate Law Derivation. Authored by: The Organic Chemistry Tutor. Located at: https://youtu.be/eIliRRpJ-I0. License: Other. License Terms: Standard YouTube License
- Second Order Reaction Chemistry Problems – Half Life, Units of K, Integrated Rate Law Derivation. Authored by: The Organic Chemistry Tutor. Located at: https://youtu.be/qBqyzi5_1YU. License: Other. License Terms: Standard YouTube License
- Zero Order Reaction Chemistry Problems – Half Life, Graph, Slope, Units of K, & Integrated Rate Law. Authored by: The Organic Chemistry Tutor. Located at: https://youtu.be/8oQSIy20jno. License: Other. License Terms: Standard YouTube License
equation that relates the concentration of a reactant to elapsed time of reaction
time required for half of a given amount of reactant to be consumed
