12.4 Integrated Rate Laws - Chemistry 2e | OpenStax

First-Order Reactions

Integration of the rate law for a simple first-order reaction (rate = k[A]) results in an equation describing how the reactant concentration varies with time:

[A]t=[A]0e−kt[A]t=[A]0e−kt

where [A]t is the concentration of A at any time t, [A]0 is the initial concentration of A, and k is the first-order rate constant.

For mathematical convenience, this equation may be rearranged to other formats, including direct and indirect proportionalities:

ln([A]t[A]0)=−ktorln([A]0[A]t)=ktln([A]t[A]0)=−ktorln([A]0[A]t)=kt

and a format showing a linear dependence of concentration in time:

ln[A]t=ln[A]0−ktln[A]t=ln[A]0−kt

The Integrated Rate Law for a First-Order Reaction

The rate constant for the first-order decomposition of cyclobutane, C4H8 at 500 °C is 9.2 ×× 10−3 s−1: C4H8⟶2C2H4C4H8⟶2C2H4

How long will it take for 80.0% of a sample of C4H8 to decompose?

Solution

Since the relative change in reactant concentration is provided, a convenient format for the integrated rate law is: ln([A]0[A]t)=ktln([A]0[A]t)=kt

The initial concentration of C4H8, [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:

t=ln[x][0.200x]×1k=ln5×19.2×10−3s−1=1.609×19.2×10−3s−1=1.7×102st=ln[x][0.200x]×1k=ln5×19.2×10−3s−1=1.609×19.2×10−3s−1=1.7×102s

Check Your Learning

Iodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation: I-131⟶Xe-131+electronI-131⟶Xe-131+electron

The decay is first-order with a rate constant of 0.138 d−1. How many days will it take for 90% of the iodine−131 in a 0.500 M solution of this substance to decay to Xe-131?

16.7 days

In the next example exercise, a linear format for the integrated rate law will be convenient:

ln[A]t=(−k)(t)+ln[A]0y=mx+bln[A]t=(−k)(t)+ln[A]0y=mx+b

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

Graphical Determination of Reaction Order and Rate Constant

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 decomposition of H2O2 from these data.

Solution

The data from Figure 12.2 are tabulated below, and a plot of ln[H2O2] is shown in Figure 12.9.

Time (h)	[H2O2] (M)	ln[H2O2]
0.00	1.000	0.000
6.00	0.500	−0.693
12.00	0.250	−1.386
18.00	0.125	−2.079
24.00	0.0625	−2.772

Figure 12.9 A linear relationship between ln[H2O2] and time suggests the decomposition of hydrogen peroxide is a first-order reaction.

The plot of ln[H2O2] versus time is linear, indicating that the reaction may be described by a first-order rate law.

According to the linear format of the first-order integrated rate law, the rate constant is given by the negative of this plot’s slope.

slope=change inychange inx=ΔyΔx=Δln[H2O2]Δtslope=change inychange inx=ΔyΔx=Δln[H2O2]Δt

The slope of this line may be derived from 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 0.00 h is 0.000; the value when t = 24.00 h is −2.772

slope=−2.772−0.00024.00−0.00 h =−2.77224.00 h =−0.116h−1 k=−slope=−(−0.116h−1)=0.116h−1slope=−2.772−0.00024.00−0.00 h =−2.77224.00 h =−0.116h−1 k=−slope=−(−0.116h−1)=0.116h−1

Check Your Learning

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

Time (s)	[A]
4.0	0.220
8.0	0.144
12.0	0.110
16.0	0.088
20.0	0.074

The plot of ln[A]t vs. t is not linear, indicating the reaction is not first order: