Unit 5 Kinetics

AP Chemistry Integrated Rate Laws

How concentration changes with time for zero-, first-, and second-order reactions, how to read the order and k from a graph, and how half-life fits in, in four worked examples.

A rate law tells you how fast a reaction goes at a given concentration. An integrated rate law tells you how much reactant is left after a given time, which is what most free-response kinetics questions actually ask. This is Topic 5.3, Concentration Changes Over Time, in the AP course. Finding orders from initial-rate tables, the earlier step, is covered in AP Chemistry rate law.

The Three Integrated Rate Laws on the Equations Sheet

Order	Integrated rate law	Straight-line plot	Slope	Half-life	Units of k
Zero	[A]_t − [A]₀ = −kt	[A] vs. t	−k	[A]₀ ÷ 2k	M s⁻¹
First	ln[A]_t − ln[A]₀ = −kt	ln[A] vs. t	−k	0.693 ÷ k	s⁻¹
Second	1/[A]_t − 1/[A]₀ = kt	1/[A] vs. t	+k	1 ÷ (k[A]₀)	M⁻¹ s⁻¹

All three integrated rate laws, and t_1/2 = 0.693/k, are printed on the AP Chemistry equations sheet. The zero- and second-order half-life expressions are not, but they come straight from setting [A]_t = ½[A]₀.
Each law has the form of a straight line, y = mx + b, with time on the x-axis. That is why one plot of the data is linear for each order.
For a first-order reaction you can also write [A]_t = [A]₀e^−kt, which is often faster for finding a concentration.

How to Find the Order From a Concentration-Time Graph

Make three columns from the data: [A], ln[A], and 1/[A].
Plot each against time, or check whether each column changes by equal amounts over equal time steps.
The straight line gives the order: [A] for zero, ln[A] for first, 1/[A] for second.
Read k from the slope, changing the sign for zero and first order.

Fig. 1 · One data set plotted three ways (Worked Example 1)

The dashed lines join the first and last points. Where the data sag below that line the plot is curved, so only the 1/[A] plot, where every point lies on the line, passes the test. On the exam you may be given the three graphs already drawn and asked to justify the order: name the plot that is linear and say that the order follows from it.

How Half-Life Depends on the Reaction Order

The half-life, t_1/2, is the time for half of the reactant to be used up. Only first-order reactions have a constant half-life, which is why the AP course treats it as the key quantity for first-order kinetics and for radioactive decay:

Zero order: each half-life is half as long as the one before, because the reactant is used at a constant rate.
First order: every half-life is the same length, so 1/2, 1/4, 1/8 of the reactant remains after 1, 2, and 3 half-lives.
Second order: each half-life is twice as long as the one before.

That gives a quick test without any graph: if the time for the concentration to halve stays the same, the reaction is first order. To try different values, the half-life calculator solves for the amount, the time, or the half-life, and works any of the three integrated rate laws with every step shown.

Worked Example 1: Finding the Order and k From a Data Table

The concentration of reactant A was measured over time. Determine the order and the rate constant.

Time (s)	[A] (M)	ln[A]	1/[A] (M⁻¹)
0	0.500	−0.693	2.00
10	0.250	−1.386	4.00
20	0.167	−1.792	6.00
30	0.125	−2.079	8.00
40	0.100	−2.303	10.0

[A]: drops by 0.250 M in the first 10 s but only 0.025 M in the last 10 s. Not constant, so not zero order.
ln[A]: drops by 0.693 in the first 10 s but only 0.224 in the last 10 s. Not constant, so not first order.
1/[A]: rises by exactly 2.00 M⁻¹ every 10 s, a straight line, so the reaction is second order.
Rate constant: slope = (10.0 − 2.00) M⁻¹ ÷ (40 − 0) s = k = 0.200 M⁻¹ s⁻¹.

The half-life confirms it: [A] falls from 0.500 to 0.250 M in 10 s, but from 0.250 to 0.125 M takes another 20 s. A half-life that doubles is the mark of a second-order reaction, and 1 ÷ (k[A]₀) = 1 ÷ (0.200 × 0.500) = 10.0 s matches the first one.

Worked Example 2: Concentration and Time for a First-Order Reaction

A first-order reaction has k = 6.00 × 10⁻⁴ s⁻¹ and [A]₀ = 0.0500 M. Find (a) [A] after 20.0 minutes, (b) the time for 90.0% of A to react, and (c) the half-life.

(a) Convert the time to match k: 20.0 min = 1200. s. Then kt = (6.00 × 10⁻⁴)(1200.) = 0.720, and [A] = 0.0500 × e^−0.720 = 0.0500 × 0.4868 = 0.0243 M.
(b) 10.0% remains, so ln([A]₀/[A]) = ln(10.0) = 2.303, and t = 2.303 ÷ (6.00 × 10⁻⁴) = 3.84 × 10³ s, about 64.0 minutes.
(c) t_1/2 = 0.693 ÷ (6.00 × 10⁻⁴) = 1.16 × 10³ s, about 19.3 minutes.

Part (b) does not need [A]₀ at all: for a first-order reaction, the time to use up a given fraction is the same at any starting concentration. Check the units of k before substituting; a k in s⁻¹ needs time in seconds.

Worked Example 3: When a Zero-Order Reactant Runs Out

Some reactions on a surface are zero order, such as ammonia decomposing on a hot metal surface, because the rate is limited by the surface, not by how much reactant is left. A zero-order reaction has k = 2.50 × 10⁻⁴ M s⁻¹ and [A]₀ = 0.0500 M. Find (a) [A] after 120. s, (b) when A runs out, and (c) the first two half-lives.

(a) [A] = 0.0500 − (2.50 × 10⁻⁴)(120.) = 0.0500 − 0.0300 = 0.0200 M.
(b) [A] = 0 when t = [A]₀ ÷ k = 0.0500 ÷ (2.50 × 10⁻⁴) = 200. s.
(c) First half-life: 0.0500 ÷ (2 × 2.50 × 10⁻⁴) = 100. s. The second, starting from 0.0250 M, is 0.0250 ÷ (5.00 × 10⁻⁴) = 50.0 s.

The zero-order law stops applying when the reactant is gone, so it never gives a negative concentration; a negative answer means the reactant ran out earlier.

Worked Example 4: Radioactive Decay of Phosphorus-32

Radioactive decay is always first order, and the AP course uses it as the main example of a constant half-life. Phosphorus-32 has a half-life of 14.3 days. How much of an 8.00 mg sample is left after 60.0 days, and when will 1.00 mg remain?

Half-lives elapsed: n = 60.0 ÷ 14.3 = 4.196.
Amount left: 8.00 × (1/2)^4.196 = 0.437 mg. Keep the unrounded n; rounding it to 4.20 first gives 0.435 mg.
Time to 1.00 mg: 8.00 to 1.00 mg is a factor of 8 = 2³, so 3 half-lives: 3 × 14.3 = 42.9 days.

The same answer comes from the first-order law with k = 0.693 ÷ 14.3 days = 0.0485 day⁻¹. Counting half-lives is faster whenever the fraction left is a power of 1/2, such as 1/2, 1/4, or 1/8.

Five Integrated Rate Law Practice Questions With Answers

A first-order reaction has a half-life of 25.0 min. What is k?
Show answer
k = 0.693 ÷ 25.0 min = 0.0277 min⁻¹.
A second-order reaction has k = 0.0500 M⁻¹ s⁻¹ and [A]₀ = 0.200 M. Find the first half-life and [A] after 300. s.
Show answer
t_1/2 = 1 ÷ (0.0500 × 0.200) = 100. s. 1/[A] = 1/0.200 + (0.0500)(300.) = 5.00 + 15.0 = 20.0 M⁻¹, so [A] = 0.0500 M.
75.0% of a first-order reactant is used up in 40.0 min. What is the half-life?
Show answer
25.0% remains, which is (1/2)², so 40.0 min is two half-lives: t_1/2 = 20.0 min.
A plot of [A] against time is a straight line with slope −3.0 × 10⁻³ M s⁻¹. Give the order, k, and the half-life when [A]₀ = 0.60 M.
Show answer
Zero order, k = 3.0 × 10⁻³ M s⁻¹, and t_1/2 = 0.60 ÷ (2 × 3.0 × 10⁻³) = 1.0 × 10² s.
Carbon-14 has a half-life of 5730 years. How old is a sample with 30.0% of its original carbon-14?
Show answer
n = ln(100 ÷ 30.0) ÷ 0.693 = 1.737 half-lives, so t = 1.737 × 5730 = 9.95 × 10³ years.

Common Integrated Rate Law Mistakes

Using log instead of ln in the first-order law.
Getting the sign of the slope wrong. k is always positive; the slope is −k for zero and first order and +k for second order.
Mixing time units, such as a k in s⁻¹ with a time in minutes.
Using t_1/2 = 0.693/k for a zero- or second-order reaction.
Rounding the number of half-lives before raising 1/2 to that power.
Mixing up the two kinds of data. Initial-rate tables give orders through the rate law; concentration-time data give them through the integrated rate laws.

The rate constant itself changes with temperature, which is the subject of AP Chemistry activation energy. To see how stronger kinetics answers could move your overall estimate, try the AP Chem Score Calculator.

Frequently Asked Questions

What is the difference between a rate law and an integrated rate law?

A rate law, rate = k[A]^n, links the rate to the concentration at one moment and comes from initial-rate experiments. An integrated rate law links the concentration to time, so it tells you how much reactant is left after a given time, and it comes from concentration-versus-time data.

How do you tell the order of a reaction from a graph?

Plot the same data three ways: [A], ln[A], and 1/[A] against time. Whichever plot is a straight line gives the order: [A] for zero order, ln[A] for first order, and 1/[A] for second order. The slope of that line is -k for zero and first order and +k for second order.

Are the integrated rate laws on the AP Chemistry equations sheet?

Yes. The kinetics section prints all three, [A]t - [A]0 = -kt, ln[A]t - ln[A]0 = -kt, and 1/[A]t - 1/[A]0 = kt, along with t1/2 = 0.693/k. You still need to know which plot is linear for each order and what the slope means.

Why is the half-life constant only for first-order reactions?

In the first-order law, setting [A]t to half of [A]0 cancels the concentration and leaves t1/2 = 0.693/k. In the zero- and second-order laws the concentration does not cancel, so the half-life is [A]0/2k or 1/(k[A]0) and changes as the reaction proceeds.

Do you use ln or log in the first-order integrated rate law?

The natural log, ln. Using log base 10 gives a value 2.303 times too small and the wrong rate constant. Exam answers can be checked quickly: the ln of 2 is 0.693, which is where the half-life equation comes from.

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