From rate and concentration to concentration and time
U5.2’s rate laws relate rate to concentration. But experiments usually measure concentrations at different times. Integrating a rate law over time (which takes calculus; AP doesn’t ask you to derive it) gives the relationship between concentration and time, called an integrated rate law积分速率定律.
Integrated rate laws for the three orders
For a reaction with a single reactant A:
| Order | Rate law | Integrated rate law | Linear plot | Slope |
|---|---|---|---|---|
| 0 | rate = k | [A]ₜ = −kt + [A]₀ | [A] vs. t | −k |
| 1 | rate = k[A] | ln[A]ₜ = −kt + ln[A]₀ | ln[A] vs. t | −k |
| 2 | rate = k[A]² | 1/[A]ₜ = kt + 1/[A]₀ | 1/[A] vs. t | k |
Each integrated rate law has the form y = mx + b, so the matching graph is a straight line. The AP exam’s equation sheet gives these three equations, but you need to know how to use them: which plot is linear, and what its slope means.
Finding the order by graphing
Measure [A] at several times, then plot [A], ln[A], and 1/[A] against t. Whichever plot is linear gives the order, and k comes from the slope. Note that the second-order slope is positive.
The three data sets below come from three reactions with different orders. Find the linear graph for each, then mark the successive half-lives and compare how they change:
| t (s) | 0 | 10 | 20 | 30 | 40 | 50 | 60 |
|---|---|---|---|---|---|---|---|
| [NH₃] (M) | 2.00 × 10⁻³ | 1.75 × 10⁻³ | 1.50 × 10⁻³ | 1.25 × 10⁻³ | 1.00 × 10⁻³ | 7.50 × 10⁻⁴ | 5.00 × 10⁻⁴ |
| t (s) | 0 | 200 | 400 | 600 | 800 | 1000 | 1200 |
|---|---|---|---|---|---|---|---|
| [H₂O₂] (M) | 1.00 | 0.793 | 0.629 | 0.499 | 0.395 | 0.313 | 0.249 |
| t (s) | 0 | 100 | 200 | 300 | 400 | 500 | 600 |
|---|---|---|---|---|---|---|---|
| [C₄H₆] (M) | 0.0100 | 0.00667 | 0.00500 | 0.00400 | 0.00333 | 0.00286 | 0.00250 |
Data were calculated from the integrated rate laws and rounded to 3 significant figures (hypothetical data); each reaction’s order matches experimental results in the textbook. The dashed line is the best-fit line.
🔧 Math tool card · Natural LogarithmsSkip it if you already know this
A natural log is “e to what power?” e ≈ 2.718 is a constant, like π. If eˣ = N, then x is the natural log of N, written x = ln N. So ln e = 1 and ln 1 = 0.
The rules are the same as for common logs (2.12.3):
So ln[A]ₜ − ln[A]₀ = ln([A]ₜ/[A]₀). To go back from ln[A] to [A], use the eˣ key on your calculator: if ln[A] = −2.30, then [A] = e⁻²·³⁰ ≈ 0.100 M.
The natural log of a number less than 1 is negative. Concentrations are usually below 1 M, so ln[A] is usually negative, and it becomes more negative as [A] falls.
Remember ln 2 ≈ 0.693. When half of a first-order reactant remains, ln([A]ₜ/[A]₀) = ln(1/2) = −0.693, which is where the 0.693 in the half-life formula comes from.
First-order calculations
The first-order integrated rate law can be written as:
Given three of the quantities, you can find the fourth: the concentration at some time, the time to reach some concentration, or k. Any quantity proportional to concentration can replace the concentration ratio, such as pressure, absorbance, or the number of radioactive nuclei.
Half-life
The time for a reactant’s concentration to fall to half is the half-life半衰期, t½. For a first-order reaction, substituting [A]ₜ = [A]₀/2 gives:
The half-life of a first-order reaction doesn’t depend on concentration: going from 1.00 to 0.50 takes as long as going from 0.50 to 0.25. After n half-lives, (1/2)ⁿ remains: 75% reacted (1/4 left) takes 2 half-lives, and 87.5% reacted (1/8 left) takes 3. Working backward, a known half-life gives k = 0.693/t½. Radioactive decay, from 2.14.2, is a first-order process, which is why each radioactive nuclide has a fixed half-life.
Half-lives of zero- and second-order reactions do depend on concentration (zero order: t½ = [A]₀/2k, getting shorter; second order: t½ = 1/(k[A]₀), getting longer). They’re for reference; the AP exam focuses on first-order half-lives.
Example 1: Finding the order from data
Hypothetical data for X → products are below. Determine the order and k.
| t (s) | 0 | 50 | 100 | 150 |
|---|---|---|---|---|
| [X] (M) | 0.500 | 0.333 | 0.250 | 0.200 |
-
Method: find which plot is linear. The time steps are all 50 s, so just check whether the change per 50 s is constant:
- Zero order: [X] vs. t is linear
- First order: ln[X] vs. t is linear
- Second order: 1/[X] vs. t is linear
-
Try zero order: changes in [X]. Every 50 s, [X] falls by 0.167, 0.083, then 0.050 M. Not equal, so [X] vs. t isn’t linear.
-
Try first order: changes in ln[X]. ln[X] is −0.693, −1.100, −1.386, −1.609, so the changes every 50 s are:
Also unequal, so not first order.
-
Try second order: changes in 1/[X]. 1/[X] is 2.00, 3.00, 4.00, 5.00 M⁻¹, rising by 1.00 every 50 s. That’s linear, so the reaction is second order.
-
k from the slope. For second order, 1/[X] = kt + 1/[X]₀, so the slope is k:
-
Check. At t = 100 s: 1/[X] = 0.0200 × 100 + 2.00 = 4.00, so [X] = 0.250 M, matching the data.
Example 2: First-order calculations
A first-order reaction A → products has k = 2.31 × 10⁻² min⁻¹ and [A]₀ = 0.400 M. Find (1) the half-life, (2) [A] after 75.0 min, and (3) the time for [A] to fall to 0.0500 M.
-
Half-life. For a first-order reaction it doesn’t depend on the starting concentration:
-
After 75.0 min: use the integrated rate law.
-
Solve for [A]. Raise e to both sides:
Check: 75.0 min is 2.5 half-lives, and (1/2)²·⁵ = 0.177. It agrees.
-
Down to 0.0500 M: count half-lives.
Exactly 3 half-lives: 3 × 30.0 = .
-
Check with the formula.
The same result.
- ln[A] vs. t is linear
ln[A]ₜ = −kt + ln[A]₀.
- 1/[A] vs. t is linear
1/[A]ₜ = kt + 1/[A]₀.
- [A] vs. t is linear
[A]ₜ = −kt + [A]₀.
- The half-life stays the same
t½ = 0.693/k, independent of concentration.
- Each half-life is twice as long as the one before
t½ = 1/(k[A]₀): the lower the concentration, the longer the half-life.
- Each half-life is half as long as the one before
t½ = [A]₀/2k: the lower the concentration, the shorter the half-life.
- k has units of s⁻¹
A first-order k has units of 1/time.
- k has units of M/s
A zero-order k has the same units as the rate.
- k has units of M⁻¹ s⁻¹
Second order.
- Decay of a radioactive nuclide
Radioactive decay is first order, with a fixed half-life.
- The rate stays the same as [A] decreases
A zero-order rate doesn't depend on concentration.
Common mistakes
Mistake 1: “Every reaction has a constant half-life.” Only first-order half-lives are independent of concentration. Zero-order half-lives get shorter and second-order half-lives get longer.
Mistake 2: “The slope of the line is k.” For zero and first order, the slope is −k, so k is the negative of the slope; only the second-order plot (1/[A] vs. t) has a slope of +k.
Mistake 3: “After two half-lives, a first-order reactant is all gone.” Each half-life leaves half of what was there: 1/4 remains after two half-lives and 1/8 after three. In principle it never reaches 0.
Quick check
Which plot is linear for a first-order reaction?
ln[A]ₜ = −kt + ln[A]₀ has the form y = mx + b.
A plot of ln[A] vs. t (min) is linear with a slope of −0.0346 min⁻¹. What is the half-life?
It's first order with k = 0.0346 min⁻¹, so t½ = 0.693 ÷ 0.0346 min⁻¹ = 20.0 min.
A first-order reaction has a half-life of 12 min. What fraction of the reactant remains after 36 min?
36 min is 3 half-lives, and (1/2)³ = 1/8.
A plot of 1/[A] vs. t (s) is linear with a slope of 0.50 M⁻¹ s⁻¹. Which statement is correct?
A linear 1/[A] vs. t plot means second order, and the slope is +k.
For a first-order reaction, how does the half-life change if [A]₀ is doubled?
The initial rate does double, but so does the amount to be used up, so the time is the same.
For a zero-order reaction, [A] vs. t is linear with a slope of −0.0020 M/s. How long does it take for [A] to fall by 0.10 M?
A zero-order rate is constant: 0.10 M ÷ 0.0020 M/s = 50 s.
A reaction's successive half-lives are 40 s, 20 s, and 10 s. What is its order?
The half-life shrinks as the concentration falls, matching t½ = [A]₀/2k: zero order.
A first-order reaction has k = 0.010 s⁻¹ and [A]₀ = 0.50 M. About what is [A] after 100 s?
ln([A]/0.50 M) = −0.010 s⁻¹ × 100 s = −1.0, so [A] = 0.50 M × e⁻¹ = 0.18 M. Option A is the value after one half-life (69.3 s).
After 2.0 min, only 1/16 of a first-order reactant remains. What is the half-life?
1/16 = (1/2)⁴, so 2.0 min = 120 s is 4 half-lives and t½ = 30 s.
After 40 h, 93.75% of a radioactive nuclide has decayed. What is its half-life?
6.25% = 1/16 remains, so 4 half-lives have passed: 40 h ÷ 4 = 10 h. B assumes 3 half-lives.
A medical radionuclide has a half-life of 6.0 h, and shipping it to a hospital takes 18 h. To have 10 mg on arrival, at least how much must be shipped?
18 h is 3 half-lives, leaving 1/8, so 10 mg × 8 = 80 mg must be shipped.
In the reaction 2A → B, [A] takes 25 s to fall from 0.80 M to 0.40 M, another 25 s to fall to 0.20 M, and another 25 s to fall to 0.10 M. Which conclusion and reason are both correct?
The time for the concentration to halve stays the same, so the half-life doesn’t depend on concentration: first order. The reasons in A and C don’t hold, because orders come from data, not from the number of reactants or the coefficients.
A first-order reaction has a half-life of 23.1 s. What is the rate constant k?
k = 0.693/t½ = 0.693 ÷ 23.1 s = 0.0300 s⁻¹. B uses 1/t½ and leaves out 0.693.
The partial pressure of a gaseous reactant in a rigid container is measured over time. A plot of ln P vs. t is linear with a slope of −2.0 × 10⁻³ s⁻¹. Which statement is correct?
At constant volume and temperature, partial pressure is proportional to concentration, so a linear ln P vs. t plot means first order. k is the magnitude of the slope, and t½ = 0.693 ÷ 0.0020 s⁻¹ ≈ 350 s.
AP-style free response
Compound X decomposes: X → Y + Z. These hypothetical data were collected:
| t (min) | 0 | 10 | 20 | 30 | 40 |
|---|---|---|---|---|---|
| [X] (M) | 0.250 | 0.168 | 0.112 | 0.0753 | 0.0505 |
(a) Calculate ln[X] at each time, and show that the reaction is first order.
(b) Determine k, with units.
(c) Calculate the half-life.
(d) Calculate [X] at 60 min.
(e) If [X]₀ were 0.500 M instead (at the same temperature), how would the half-life and the initial rate change? Justify your answers.
Model answer and scoring
(a) ln[X] is −1.386, −1.784, −2.189, −2.586, −2.986. It falls by about 0.40 every 10 min, so ln[X] vs. t is linear and the reaction is first order.
(b) Slope = (−2.986 − (−1.386)) ÷ 40 min = −0.0400 min⁻¹, so k = 0.0400 min⁻¹.
(c) t½ = 0.693 ÷ 0.0400 min⁻¹ = 17.3 min.
(d) ln([X]/0.250 M) = −0.0400 min⁻¹ × 60 min = −2.40, so [X] = 0.250 M × e⁻²·⁴⁰ = 0.0227 M.
(e) The half-life stays 17.3 min: for a first-order reaction t½ = 0.693/k, and k depends only on temperature. The initial rate doubles: rate = k[X], and [X] doubles.
Scoring: (a) uses the even change in ln[X], or a linear plot, to show first order; (b) value, positive sign, and units of k; (c) 0.693/k; (d) uses the integrated rate law; (e) both conclusions with reasons.
Summary
- Integrated rate laws: zero order [A]ₜ = −kt + [A]₀; first order ln[A]ₜ = −kt + ln[A]₀; second order 1/[A]ₜ = kt + 1/[A]₀.
- Whichever of [A], ln[A], or 1/[A] vs. t is linear tells you the reaction is zero, first, or second order; zero- and first-order slopes are −k, and the second-order slope is k.
- First order: ln([A]ₜ/[A]₀) = −kt; the half-life t½ = 0.693/k doesn’t depend on concentration, and (1/2)ⁿ remains after n half-lives.
- Radioactive decay is a first-order process.
AP tip: When you justify an order, say which plot is linear rather than just stating the order; when you find k from a graph, watch the sign of the slope. For half-life questions, first check how many half-lives have passed; for a whole number, (1/2)ⁿ is quickest.