AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/5/5-3)
Unit 5 · Topic 5.3
5.3 Concentration Changes Over Time
Graphs of concentration against time reveal the order: [A] versus t is a straight line for zero order, ln[A] versus t for first order and 1/[A] versus t for second order. The slope of the straight line gives k, and first-order reactions have a constant half-life, t½ = 0.693/k.
Key terms
- integrated rate law
- zero order
- first order
- second order
- half-life
Integrated rate laws
A rate law tells you how fast a reaction goes at a given concentration. An integrated rate law tells you the concentration at any time. Each order has its own, and each is the equation of a straight line when you graph the right thing. All three are on the equation sheet.
| Order | Integrated rate law | Straight-line plot | Slope |
|---|---|---|---|
| Zero | [A]t − [A]₀ = −kt | [A] vs t | −k |
| First | ln[A]t − ln[A]₀ = −kt | ln[A] vs t | −k |
| Second | 1/[A]t − 1/[A]₀ = kt | 1/[A] vs t | +k |
Finding the order from data
Given concentrations at several times, calculate ln[A] and 1/[A] for each point, then see which of the three plots is a straight line. That plot's order is the order with respect to A.
The sign of the slope is a quick check. Zero-order and first-order plots slope downward. The second-order plot of 1/[A] slopes upward, because 1/[A] grows as [A] shrinks.
These plots tell you the order with respect to the one reactant you're monitoring. If other reactants are present in large excess, their concentrations barely change, so the plot still works for the reactant you're watching.
Half-life
The half-life, t½, is the time it takes for a reactant's concentration to fall to half its value. For a first-order reaction, t½ = 0.693/k, and it doesn't depend on concentration. Every half-life cuts the amount in half again: after 1 half-life, 50% remains; after 2, 25%; after 3, 12.5%.
That constant half-life is a signature of first order. If successive half-lives in a data set get longer, the reaction isn't first order. (For second order, each half-life is twice as long as the one before.)
Radioactive decay is a classic first-order process. Carbon-14 has a half-life of about 5730 years, so after about 11,460 years a sample has a quarter of its original carbon-14.
What a zero-order reaction looks like
In a zero-order reaction, the concentration falls in a straight line: the same amount is used up in each time interval, until the reactant runs out. This happens, for example, when a reaction takes place on a catalyst surface that is completely covered with reactant, so adding more reactant can't make it go faster.
A zero-order reaction's half-life isn't constant: it gets shorter as the concentration drops, because a fixed amount is used per second. Comparing half-lives is a fast way to tell the three orders apart: constant for first order, getting longer for second order and getting shorter for zero order.
Using natural logs
In first-order calculations, ln is the natural log. To get [A] back from ln[A], use eˣ. You can also write the first-order law as ln([A]₀/[A]t) = kt, which is handy for finding the time to reach a given fraction.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Concentration after a time
A first-order reaction has k = 2.5 × 10⁻³ s⁻¹ and [A]₀ = 0.800 M. Find [A] after 300. s and the half-life.
Show the solutionHide the solution
- Step 1: ln[A]t = ln[A]₀ − kt = ln(0.800) − (2.5 × 10⁻³)(300.) = −0.2231 − 0.750 = −0.9731.
- Step 2: [A]t = e^(−0.9731) = 0.378 M.
- Step 3: t½ = 0.693/k = 0.693 ÷ 2.5 × 10⁻³ s⁻¹ = 277 s.
Answer: [A] = 0.378 M after 300. s; t½ = 277 s
- Example 2Calculator allowed
Identifying the order from data
For a reaction A → products: t = 0 s, [A] = 0.100 M; t = 50 s, 0.0500 M; t = 100 s, 0.0333 M; t = 150 s, 0.0250 M. Determine the order and k.
Show the solutionHide the solution
- Step 1: Test first order: ln[A] = −2.303, −2.996, −3.401, −3.689. Slopes between points: −0.0139, −0.0081, −0.0058 s⁻¹. Not constant, so not a straight line. Not first order.
- Step 2: Another clue: the first half-life is 50 s (0.100 → 0.0500), but going from 0.0500 to 0.0250 takes 100 s. Half-life isn't constant.
- Step 3: Test second order: 1/[A] = 10.0, 20.0, 30.0, 40.0 M⁻¹. These rise by exactly 10.0 every 50 s, a straight line.
- Step 4: Slope = k = 10.0 M⁻¹ ÷ 50 s = 0.200 M⁻¹s⁻¹.
Answer: Second order in A; k = 0.200 M⁻¹s⁻¹
- Example 3
Half-lives without a calculator (classic trap)
A first-order reaction has a half-life of 30.0 min. How long until 12.5% of the reactant remains? A student answers 3.75 min, reasoning that 12.5% is one eighth of 30.
Show the solutionHide the solution
- Step 1: Each half-life halves the amount: 100% → 50% → 25% → 12.5%.
- Step 2: That's 3 half-lives.
- Step 3: 3 × 30.0 min = 90.0 min. The student divided the half-life instead of counting how many half-lives have passed.
Answer: 90.0 min
Common mistakes
- Expecting a positive slope for first order. ln[A] vs t slopes down with slope −k; only the second-order plot slopes up.
- Using t½ = 0.693/k for zero- or second-order reactions. It applies only to first order.
- Using log (base 10) instead of ln.
- Deciding order from one pair of points. Check that the whole plot is linear.
On the exam
- Free-response questions often give a data table or three graphs ([A], ln[A] and 1/[A] versus time). Identify the straight one, state the order and use its slope to find k with units.
- Expect quick half-life reasoning questions, like what fraction remains after a given time, that don't need a calculator.
Connected topics
Videos
Check yourself
4 questions on 5.3 Concentration Changes Over Time. Pick an answer to see if you got it, and why.
| Time (s) | [A] (M) |
|---|---|
| 0 | 0.800 |
| 20 | 0.400 |
| 40 | 0.200 |
| 60 | 0.100 |
Hypothetical data. Concentration of reactant A during the reaction A → products at constant temperature
Which of the following graphs of these data would give a straight line?
What is the rate constant for the reaction?
If the reaction continues in the same way, what will [A] be at 100 s?
For the decomposition of a gas X, a plot of 1/[X] versus time is a straight line with a slope of 0.45 M⁻¹ s⁻¹. Which of the following is true?
0 of 4 answered