4.5: Integrated Rate Laws

First-Order Reactions

An equation relating the rate constant [latex]k[/latex] to the initial concentration [latex][A]_0[/latex] and the concentration [latex][A]_t[/latex] present after any given time [latex]t[/latex] can be derived for a first-order reaction and shown to be:

[latex]ln\left(\dfrac{[A]_t}{[A]_0}\right)=−kt[/latex]

or alternatively

[latex]\ln\left(\dfrac{[A]_0}{[A]_t}\right)=kt \nonumber[/latex]

or

[latex][A]=[A]_0e^{−kt} \nonumber[/latex]

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]ln[A]=(−k)(t)+\ln[A]_0[/latex]
[latex]y=mx+b \notag[/latex]

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

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]\ce{Rate}=k[A]^2 \nonumber[/latex]

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

[latex]\dfrac{1}{[A]}=kt+\dfrac{1}{[A]_0} \label{int2nd}[/latex]

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:

[latex]\begin{align*}
\dfrac{1}{[A]}&=kt+\dfrac{1}{[A]_0}\[latex]4pt]
y&=mx+b
\end{align*} \nonumber[/latex]

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

Example 4.5.4: Determination of Reaction Order by Graphing

Test the data given to show whether the dimerization of C₄H₆ is a first- or a second-order reaction.

Solution

Solutions to Example 12.4.4

Trial	Time (s)	[C4H6] (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 ln[C₄H₆] versus t and compare it with a plot of [latex]\mathrm{\dfrac{1}{[C_4H_6]}}[/latex] versus t. The values needed for these plots follow.

Time (s)	[latex]\dfrac{1}{[\ce{C4H6}]}\:(M^{−1})[/latex]	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 4.5.3. As you can see, the plot of ln[C₄H₆] versus t is not linear, therefore the reaction is not first order. The plot of [latex]\dfrac{1}{[\ce{C4H6}]}[/latex] versus t is linear, indicating that the reaction is second order.

Zero-Order Reactions

For zero-order reactions, the differential rate law is:

[latex]\ce{Rate}=k[A]^0=k \nonumber[/latex]

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:

[latex][A]=−kt+[A]_0[/latex]
[latex]y =mx+b \notag[/latex]

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

Figure 4.5.5 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:

[latex]\ce{slope}=−k=\mathrm{1.3110^{−6}\:mol/L/s} \nonumber[/latex]

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 as an example (see data in above Example 4.5.2), 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:

[latex]\begin{align*}
\ln\dfrac{[A]_0}{[A]}&=kt\[latex]4pt]
t&=\ln\dfrac{[A]_0}{[A]}×\dfrac{1}{k}
\end{align*} \nonumber[/latex]

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

Therefore:

[latex]\begin{align*}
t_{1/2}&=\ln\dfrac{[A]_0}{\dfrac{1}{2}[A]_0}×\dfrac{1}{k}\[latex]4pt]
&=\ln 2×\dfrac{1}{k}=0.693×\dfrac{1}{k}
\end{align*} \nonumber[/latex]

Thus:

[latex]t_{1/2}=\dfrac{0.693}{k} \nonumber[/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 4.5.5: 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 4.5.4.

Solution The half-life for the decomposition of H₂O₂ is 2.16 × 10⁴ s:

[latex]\begin{align*}
t_{1/2}&=\dfrac{0.693}{k}\[latex]4pt]
k&=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{2.16×10^4\:\ce s}=3.21×10^{−5}\:\ce s^{−1}
\end{align*} \nonumber[/latex]

Exercise 4.5.1

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.

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.