Integrated rate laws include time as a variable. Here I will derive integrated rate laws for zero, first, and second order rate laws.

Zero-Order Integrated Rate Law

For a zeroth order reaction, A → P

\(\displaystyle \mathrm{Rate}\;=\;-\frac{\Delta[A]}{\Delta t}\;=\;k[A]^0\;=\;k\)
 

Integrate both sides of the equation.

d[A] = -k dt

\(\displaystyle \int_{[A]_0}^{[A]_t}\;d[A]\;=\;-k\int_{t_0}^{t}\;dt\)
 
Both [A]₀ and t₀ are initial conditions. Here we will evaluate over the limits.

[A]_t – [A]₀ = -k (t – t₀)

t₀ is equal to 0. The integrated rate law is below.

[A]_t – [A]₀ = -kt

We rearrange the equation to the straight-line form.

[A]_t = -kt + [A]₀

where [A]_t is the concentration of [A] at some time, t, k is the rate constant, t is time, and [A]₀ is the initial concentration of [A]. The equation is linear, and corresponds to, y = mx + b. The value of y is equal [A]_t, the slope, m, is equal to the negative of the rate constant, [A]₀ is the y intercept, and x is equal to t. The slope of the line is negative and is equal to -k as shown in the figure below.

First-Order Integrated Rate Law

For a first-order reaction, A → P, the rate can be written as:

\(\displaystyle \mathrm{Rate}\;=\;-\frac{\Delta[A]}{\Delta t}\;=\;k[A]\)
 

Separate the variables and integrate.

\(\displaystyle\frac{d[A]}{dt}\;=\;-k[A]\)
 
\(\displaystyle\frac{d[A]}{[A]}\;=\;-k dt\)
 
\(\displaystyle \int_{[A]_0}^{[A]_t}\;\frac{d[A]}{[A]}\;=\;-k\int_{t_0}^{t}\;dt\)
 

Recall,

\(\displaystyle \int\;\frac{1}{x}\;dx\;=\;ln\;x\mathrm\;\;\;and\;\;\;\int dx\; =\; x\)
 

Evaluate over the limits and t₀ is equal to 0.

ln[A]_t – ln[A]₀ = -k (t – t₀)

Below is the integrated rate law for a first order reaction.

ln [A]_t – ln [A]₀ = -kt

We can rearrange the equation to the straight-line form.

ln [A]_t = -kt + ln [A]₀

The concentration of A at some time t is, [A]_t, the initial concentration of A is [A]₀, k is the rate constant and t is time. The data is plotted as ln [A] vs time for a first order reaction. The slope of the line is equal to -k. The plot for a first order reaction is below.

Using the properties of logarithms, the first order integrated rate law equation can also be written as:

\(\displaystyle ln\Biggl(\frac{[A]_t}{[A]_0}\Biggl)\;=\;-kt\)

Second-Order Integrated Rate Law

For the second order reaction, A → P

\(\displaystyle\mathrm{Rate}\;=\;-\frac{\Delta [A]}{\Delta t}\;=\;k[A]^2\)
 
Separate the variables in then integrate.

\(\displaystyle-\frac{d[A]}{dt}\;=\;-k[A]^2\)
 
and

\(\displaystyle\frac{d[A]}{[A]^2}\;=\;-k\;dt\)
 
\(\displaystyle \int_{[A]_0}^{[A]_t}\;\frac{d[A]}{[A]^2}\;=\;-k\int_{t_0}^{t}\;dt\)
 

Recall from calculus

\(\displaystyle\;X^n\;dx\;=\;\frac{1}{n\;+\;1}X^{n+1}\)   in this case n = -2
 
\(\displaystyle\int X^{-2}\;dx\;=\;-X^{-1}\)
 
and

\(\displaystyle -\frac{1}{[A]}\bigg|_{[A]_0}^{[A]_t}\;=\;-kt\bigg|_{t_0}^{t}\)
 
Evaluate over the limits

\(\displaystyle \frac{1}{[A]_t}\;-\;\frac{1}{[A]_0}\;=\;k\;(t\;-\;t_0)\)
 
t₀ is equal to 0, and our second order integrated rate law is

\(\displaystyle\frac{1}{[A]_t}\;-\;\frac{1}{[A]_0}\;=\;kt\)
 
The equation can be rearranged into its straight line form.

\(\displaystyle\frac{1}{[A]_t}\;=\;kt\;+\;\frac{1}{[A]_0}\)
 
Below, is a plot of 1/[A]_t vs time for a second order reaction. The slope is positive and is equal to k. The y intercept is 1/[A]₀

The integrated rate laws can be used to determine how long it takes for a reactant to from its initial concentration to some other concentration, or to determine the concentration after some time t. When using the integrated rate laws, you will need to know the order with respect to the reactant you are working with. This will either be stated in the problem or you can determine the order from the units of k. You can then use the appropriate integrated rate law.

For example, The decomposition of hydrogen peroxide is first order in H₂O₂ with a rate constant of 3.7 x 10^-3 s^-1 and an initial concentration of 1.00 M.

2 H₂O₂ → 2 H₂O (l) + O₂ (g)

What is the concentration after two hours? First, we have to make sure we are using the correct integrated rate law. This is first order because the units of the rate constant are s^-1. We use the following first order integrated rate law:

\(\displaystyle ln\Biggl(\frac{[A]_t}{[A]_0}\Biggl)\;=\;-kt\)
 
First we can rearrange the equation:

ln[H₂O₂]_t = -kt + ln[H₂O₂]₀

Substitute in the known quantities. The quantity 2 hours is 120 seconds.

ln[H₂O₂]_t = -3.7 x 10^-3 s^-1 x 120 s + ln[1.00 M]

ln[H₂O₂]_t = -0.444

Take the antilog of both sides

e^{ln[H₂O₂]_t} = e^-0.444

[H₂O₂]_{120 s} = 0.641 M. The concentration of hydrogen peroxide after 2 hours is 0.641 M.

Half-Life

The half-life of a reaction is the amount of time it will take for the concentration of a reactant to decrease to one-half of its initial amount. Half-life is represented by t_1/2. Consider the following first order reaction:

A → P

The integrated rate law is:

\(\displaystyle ln\Biggl(\frac{[A]_t}{[A]_0}\Biggl)\;=\;-kt\)
 
At t_1/2, [A]_t will be one-half of the initial concentration [A]₀. We put this into the integrated rate law.

\(\displaystyle ln\Biggl(\frac{1/2[A]_0}{[A]_0}\Biggl)\;=\;-kt_{1/2}\)
 
Because [A]₀ cancels out and the properties of logarithms we are left with

ln (2) = kt_1/2

ln(2) = 0.693. Solve the equation for t_1/2. The equation for half-life for a first order reaction is:

t_1/2 = \(\displaystyle\frac{0.693}{k}\)

The half-life for a first order reaction is only dependent on k. It does not depend on the initial concentration of the reactant.

For a zero order reaction the integrated rate law is [A]_t = -kt + [A]₀

Again, we substitute 1/2[A]₀ for [A]_t.

1/2[A]₀ – [A]₀ = -kt_1/2

-1/2[A]₀ = -kt_1/2

t_1/2 = \(\displaystyle\frac{[A]_0}{2k}\)

The half-life for a zero order reaction is directly proportional to the initial concentration of reactant, and is inversely proportional to k.

The half life for a second order reaction is derived from the integrated rate law.

\(\displaystyle\frac{1}{[A]_t}\;=\;kt\;+\;\frac{1}{[A]_0}\)
 
Replace [A]_t with 1/2[A]₀.

\(\displaystyle\frac{1}{1/2[A]_0}\;=\;kt_{1/2}\;+\;\frac{1}{[A]_0}\)

t_1/2 = \(\displaystyle\frac{1}{[A]_0k}\)
 
For a second order reaction, the half-life is inversely proportional to the initial concentration of the reactant and to k.

Below is a summary of integrated rate laws and half-lives for zero, first, and second order reactions.

Worksheet: Integrated Rate Laws and Half-Life

Please watch the following videos:

Second Order Decomposition of NO₂ Gas
Integrated Rate Law and Half-Life
Use Half-Life to determine the Age of a Wooden Relic

Exercises

Exercise 1. A reaction, A + B → P, has a rate constant of 3.2 x 10^-3 s^-1. If the experiment begins with a concentration of A equal to 0.25 M, how long, in s, will it take for the concentration of A to decrease to 0.047 M?

Check Solution for Exercise 1

Exercise 2. The half-life for the radioactive decay of K-40 is 1.30 x 10⁹ years. How long will it take for K-40 to decay to 20% of its initial concentration? All radioactive decay follows first order kinetics.

Check Solution for Exercise 2

Exercise 3. The decomposition of cyclobutane (C₄H₈) is a first order process. The half-life is 7.96 x 10^-3 seconds. If the initial concentration of cyclobutane is 8.50 M, what is the concentration after 0.045 seconds?

Check Solution for Exercise 3

Exercise 4. Consider the following reaction at 250^oC.

2 NO₂ (g) → 2 NO (g) + O₂ (g)

The rate constant was determined to be 0.483 M^-1s^-1. If the initial concentration is 8.65 M, what is the concentration after 2.3 minutes? What is the half-life of NO₂?

Check Solution for Exercise 4

Exercise 5. Consider the following reaction of butadiene (C₄H₆) gas.

2 C₄H₆ → C₈H₁₂ (g)

At a given temperature, the reaction is second order and the half-life of butadiene is 236 seconds. If the initial concentration of butadiene was 0.985 M, what is the concentration after 1.5 minutes?

Check Solution for Exercise 5

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