Chemistry Textbook

Second-Order Reactions

The equations that relate the concentrations of reactants and the rate constant of second-order reactions can be fairly complicated. To illustrate the point with minimal complexity, only the simplest second-order reactions will be described here, namely, those whose rates depend on the concentration of just one reactant. For these types of reactions, the differential rate law is written as:

The integrated rate law for these second-order reactions is given by:

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

The integrated rate law for second-order reactions has the form of the equation of a straight line. A plot of $\frac{1}{[A{]}_t}$ versus t for a second-order reaction is a straight line with a slope of k and a y-intercept of $\frac{1}{{\left[A\right]}_0}.$ If the plot is not a straight line, then the reaction is not second order.

EXAMPLE 10.10

Graphical Determination of Reaction Order and Rate Constant The data below are for the same reaction described in Example 10.9. Prepare and compare two appropriate data plots to identify the reaction as being either first or second order. After identifying the reaction order, estimate a value for the rate constant.

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

Table 10.7

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

Table 10.8

The plots are shown in Figure 10.7, which clearly shows the plot of ln[C₄H₆]_t versus t is not linear, therefore the reaction is not first order. The plot of $\frac{1}{[{\text{C}}_4{\text{H}}_6{]}_t}$ versus t is linear, indicating that the reaction is second order.

According to the second-order integrated rate law, the rate constant is equal to the slope of the $\frac{1}{[A{]}_t}$ versus t plot. Using the data for t = 0 s and t = 6200 s, the rate constant is estimated as follows:

$$k=\text{slope}=\frac{(481M^{\mathrm{-1}}-100M^{\mathrm{-1}})}{(6200\text{s}-0\text{s})}=0.0614{\text{M}}^{\mathrm{-1}}{\text{s}}^{\mathrm{-1}}$$

Check Your Learning Do 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

Table 10.9

Answer:

Yes. The plot of $\frac{1}{[A{]}_t}$ 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 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 is a linear function:

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 10.8 shows a plot of [NH₃] 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 (SiO₂) 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.

The Half-Life of a Reaction

The half-life of a reaction (t_1/2) is the time required for the concentration of a reactant to decrease to one-half of it's original value; in other words, after a half-life, [A]_t = [A]_o/2. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide (Figure 10.4) 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

An equation relating the half-life of a first-order reaction to its rate constant may be derived from the integrated rate law as follows:

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

Invoking the definition of half-life, symbolized $t_{1\text{/}2},$ requires that the concentration of A at this point is one-half its initial concentration: $t=t_{1\text{/}2},$ $[A{]}_t=\frac{1}{2}{[A]}_0.$

Substituting these terms into the rearranged integrated rate law and simplifying yields the equation for half-life:

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

This equation describes an expected inverse relation between the half-life of the reaction and its rate constant, k. Faster reactions exhibit larger rate constants and correspondingly shorter half-lives. Slower reactions exhibit smaller rate constants and longer half-lives.

EXAMPLE 10.12

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 10.9.

Solution Inspecting the concentration/time data in Figure 10.9 shows the half-life for the decomposition of H₂O₂ is 2.16 $\times$ 10⁴ s:

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

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.