12.4 Integrated Rate Laws

First-Order Reactions

An equation relating the rate constant k to the initial concentration [A]₀ and the concentration [A]_t present after any given time t can be derived for a first-order reaction and shown to be:

or

or

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:

A plot of ln[A] versus t for a first-order reaction is a straight line with a slope of −k and an intercept of ln[A]₀. 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 A.

Example 12.7

Determination of Reaction Order by Graphing

Show that the data in Figure 12.2 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 Figure 12.2 with the addition of values of ln[H₂O₂] are given in Figure 12.10.

Figure 12.10 The linear relationship between the ln[H₂O₂] and time shows that the decomposition of hydrogen peroxide is a first-order reaction.

Trial	Time (h)	[H2O2] (M)	ln[H2O2]
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[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:

$$\text{slope}=\frac{\text{change in}y}{\text{change in}x}=\frac{\text{Δ}y}{\text{Δ}x}=\frac{\text{Δln}\left[{\text{H}}_2{\text{O}}_2\right]}{\text{Δ}t}$$

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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:

$$\begin{array}{ccc}\hfill \text{slope}& =& \frac{\mathrm{-1.386}-\left(\mathrm{-0.693}\right)}{\text{12.00 h}-\text{6.00 h}}\hfill \\ & =& \frac{\mathrm{-0.693}}{\text{6.00 h}}\hfill \\ & =& \mathrm{-1.155}\times {10}^{\mathrm{-1}}{\text{h}}^{\mathrm{-1}}\hfill \\ \hfill k& =& -\text{slope}=-\left(\mathrm{-1.155}\times {10}^{\mathrm{-1}}{\text{h}}^{\mathrm{-1}}\right)=1.155\times {10}^{\mathrm{-1}}{\text{h}}^{\mathrm{-1}}\hfill \end{array}$$

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Check Your Learning

Graph the following data to determine whether the reaction $A⟶B+C$ is first order.

Trial	Time (s)	[A]
1	4.0	0.220
2	8.0	0.144
3	12.0	0.110
4	16.0	0.088
5	20.0	0.074

Answer:

The plot of ln[A] vs. t is not a straight line. The equation is not first order:

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:

For these second-order reactions, the integrated rate law is:

where the terms in the equation have their usual meanings as defined earlier.

The integrated rate law for our second-order reactions has the form of the equation of a straight line:

A plot of $\frac{1}{\left[A\right]}$ versus t for a second-order reaction is a straight line with a slope of k and an intercept of $\frac{1}{{\left[A\right]}_0}.$ If the plot is not a straight line, then the reaction is not second order.

Example 12.9

Determination of Reaction Order by Graphing

The data below are for the same reaction described in Example 12.8. Test these data to confirm that this dimerization reaction is second-order.

Solution

Trial	Time (s)	[C4H6] (M)
1	0	1.00 $\times$ 10−2
2	1600	5.04 $\times$ 10−3
3	3200	3.37 $\times$ 10−3
4	4800	2.53 $\times$ 10−3
5	6200	2.08 $\times$ 10−3

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]
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 12.11. 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.

Figure 12.11 These two graphs show first- and second-order plots for the dimerization of C₄H₆. Since the first-order plot (left) is not linear, we know that the reaction is not first order. The linear trend in the second-order plot (right) indicates that the reaction follows second-order kinetics.

Check Your Learning

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

Trial	Time (s)	[A] (M)
1	5	0.952
2	10	0.625
3	15	0.465
4	20	0.370
5	25	0.308
6	35	0.230

Answer:

Yes. The plot of $\frac{1}{\left[A\right]}$ vs. t is linear:

Zero-Order Reactions

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 reactants.

The integrated rate law for a zero-order reaction also has the form of the equation of a straight line:

A plot of [A] versus t for a zero-order reaction is a straight line with a slope of −k and an intercept of [A]₀. Figure 12.12 shows a plot of [NH₃] versus t for the decomposition of ammonia on a hot tungsten wire and for the decomposition of ammonia on hot quartz (SiO₂). The decomposition of NH₃ on hot tungsten is zero order; the plot is a straight line. The decomposition of NH₃ on hot quartz is not zero order (it is first order). From the slope of the line for the zero-order decomposition, we can determine the rate constant:

Figure 12.12 The decomposition of NH₃ on a tungsten (W) surface is a zero-order reaction, whereas on a quartz (SiO₂) surface, the reaction is first order.

The Half-Life of a Reaction

The half-life of a reaction (t_1/2) 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 (Figure 12.2) as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of H₂O₂ 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 H₂O₂ 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:

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

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If we set the time t equal to the half-life, $t_{1\text{/}2},$ the corresponding concentration of A at this time is equal to one-half of its initial concentration. Hence, when $t=t_{1\text{/}2},$ $[A]=\frac{1}{2}{[A]}_0.$

Therefore:

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

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Thus:

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

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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 12.10

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 12.13.

Figure 12.13 The decomposition of H₂O₂ $({\text{2H}}_2{\text{O}}_2⟶{\text{2H}}_2\text{O}+{\text{O}}_2)$ at 40 °C is illustrated. The intensity of the color symbolizes the concentration of H₂O₂ at the indicated times; H₂O₂ is actually colorless.

Solution

The half-life for the decomposition of H₂O₂ is 2.16 $\times$ 10⁴ s:

$$\begin{array}{ccc}\hfill t_{1\text{/}2}& =& \frac{0.693}{k}\hfill \\ \hfill k& =& \frac{0.693}{t_{1\text{/}2}}=\frac{0.693}{2.16\times {10}^4\text{s}}=3.21\times {10}^{\mathrm{-5}}{\text{s}}^{\mathrm{-1}}\hfill \end{array}$$

12.76

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?

Answer:

5.02 d.