![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)
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.
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:
where [latex][A]_[/latex] is the concentration of [latex]A[/latex] at any time [latex]t[/latex], [latex][A]_[/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:
and a format showing a linear dependence of concentration in time:
The rate constant for the first-order decomposition of cyclobutane, [latex]\ce[/latex] at 500 °C is 9.2 × 10 −3 s −1 :
How long will it take for 80.0% of a sample of [latex]\ce[/latex] to decompose?
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 [latex]\ce[/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:
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\hfill \text\left[A\right]_& =& \left(-k\right)\left(t\right)+\text<\left[A\right]>_\hfill \\ \hfill y& =& mx+b\hfill \end[/latex]
As shown in Figure 18.4.1, a plot of [latex]\text[A]_[/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[A]_[/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].
Show that the data in Figure 18.4.2 can be represented by a first-order rate law by graphing ln[latex]\ce[/latex] versus time. Determine the rate constant for the rate of decomposition of [latex]\ce[/latex] from this data.
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The data from Figure 18.4.2 with the addition of values of ln[latex]\ce[/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[/latex] (M) | ln[latex]\ce[/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[/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[/latex] versus time where:
In order to determine the slope of the line, we need two values of ln[latex]\ce[/latex] at different values of t (one near each end of the line is preferable). For example, the value of ln[latex]\ce[/latex] when t is 6.00 h is -0.693; the value when t = 12.00 h is −1.386:
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:
For these second-order reactions, the integrated rate law is:
where the terms in the equation have their usual meanings as defined above.
The reaction of butadiene gas [latex]\ce[/latex] with itself produces [latex]\ce>[/latex] gas as follows:
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?
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 × 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.
The integrated rate law for our second-order reactions has the form of the equation of a straight line:
As shown in Figure 18.4.4, a plot of [latex]\dfrac<\left[A\right]_
Test the data given to show whether the dimerization of [latex]\ce[/latex] is a first- or a second-order reaction.
| Trial | Time (s) | [latex]\ce[/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[\ce][/latex] versus t and compare it with a plot of [latex]\dfrac><[<\ce
| Time (s) | [latex]\dfrac<\left[<\text | ln[latex]\ce[/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[/latex] versus t is not linear, therefore the reaction is not first order. The plot of [latex]\dfrac>[/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<\left[A\right]_
For zero-order reactions, the differential rate law is:
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:
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.
Figure 18.4.7 shows a plot of [latex]\ce[/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[/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:
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 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[/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[/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.
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:
If we set the time t equal to the half-life, [latex]_2>,[/latex] the corresponding concentration of A at this time is equal to one-half of its initial concentration. Hence, when [latex]t=_2>,[/latex] [latex]\left[A\right]_=\dfrac<\left[A\right]>_[/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.
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.
5 beakers is shown, each approximately half-filled with colored substances. Beneath each beaker are three rows of text. The first beaker contains a bright green substance and is labeled below as, “1.000 M, 0 s, and ( 0 h ).” The second beaker contains a slightly lighter green substance and is labeled below as, “0.500 M, 2.16 times 10 superscript 4 s, and ( 6 h ).” The third beaker contains an even lighter green substance and is labeled below as, “0.250 M, 4.32 times 10 superscript 4 s, and ( 12 h ).” The fourth beaker contains a green tinted substance and is labeled below as, “0.125 M, 6.48 times 10 superscript 4 s, and ( 18 h ).” The fifth beaker contains a colorless substance and is labeled below as, “0.0625 M, 8.64 times 10 superscript 4 s, and ( 24 h ).”" width="650" />
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