AP Chemistry · Lesson 2

Integrated rate laws and half-life

Know Know it

A rate law tells you the rate at one moment. An integrated rate law tells you the concentration at any time. Each order has its own, and each turns a different graph into a straight line.

First order: ln⁡[A]t=−kt+ln⁡[A]0\ln[\text{A}]_t = -kt + \ln[\text{A}]_0. A plot of ln[A] against t is a straight line with slope −k. The half-life is t1/2=0.693/kt_{1/2} = 0.693/k, and it stays the same however much A is left. Radioactive decay is first order.

Second order (in one reactant): 1[A]t=kt+1[A]0\frac{1}{[\text{A}]_t} = kt + \frac{1}{[\text{A}]_0}. A plot of 1/[A] against t is a straight line with slope k. Each half-life is longer than the one before.

Zero order: [A]t=−kt+[A]0[\text{A}]_t = -kt + [\text{A}]_0. A plot of [A] against t is a straight line with slope −k.

Which plot is straight for each order
OrderStraight-line plotSlopeHalf-life
0[A] against t−kshrinks as [A] falls
1ln[A] against t−k0.693/k, constant
21/[A] against t+kgrows as [A] falls

Calculator On the exam

You may use a scientific or graphing calculator on both sections of the exam; a four-function calculator is allowed but not recommended. The ln key does the work in first-order problems: ln(2) is about 0.693. A reference booklet of equations and constants is provided, but know which plot is straight for each order without looking it up.

Worked example Easy

A first-order reaction has a half-life of 15 min. If the starting concentration of the reactant is 0.800 M, what is its concentration after 45 min?

  1. 0.267 M
  2. 0.200 M
  3. 0.0500 M
  4. 0.100 MAnswer

How to solve it

  1. Count half-lives: 45 min ÷ 15 min = 3.
  2. Halve three times: 0.800 × (½)³ = 0.100 M.

Why each choice is right or wrong

  • A. Incorrect. 0.267 M divides 0.800 M by 3, but 45 min is three half-lives, so the concentration is halved three times, not divided by 3.
  • B. Incorrect. 0.200 M is the concentration after two half-lives (30 min), not after 45 min.
  • C. Incorrect. 0.0500 M is the concentration after four half-lives (60 min), one half-life too many.
  • D. Correct. 45 min is 45 ÷ 15 = 3 half-lives, so 0.800 → 0.400 → 0.200 → 0.100 M.

Trap Constant half-life means first order

Only a first-order reaction has a half-life that stays the same as the reactant is used up. Do not use t1/2=0.693/kt_{1/2} = 0.693/k for a zero- or second-order reaction, whose half-lives change as [A] changes.

Explain Explain it

Task verbs: analyze, justify

To find the order from concentration-time data, make all three plots (or check the three columns): [A], ln[A] and 1/[A] against time. Whichever is a straight line tells you the order, and its slope gives k.

A quick test for first order: in equal time intervals, [A] falls by the same fraction each time. For zero order, it falls by the same amount each time.

Worked example Medium

A first-order reaction has k = 0.0462 min⁻¹. How long does it take for the reactant concentration to fall from 0.600 M to 0.150 M?

  1. 15.0 min
  2. 9.74 min
  3. 30.0 minAnswer
  4. 45.0 min

How to solve it

  1. Half-life: 0.693 ÷ 0.0462 min⁻¹ = 15.0 min.
  2. 0.600 → 0.300 → 0.150 M is two half-lives.
  3. Time = 2 × 15.0 = 30.0 min, which matches ln(4) ÷ 0.0462 = 30.0 min.

Why each choice is right or wrong

  • A. Incorrect. 15.0 min is one half-life (0.693 ÷ 0.0462); falling from 0.600 M to 0.150 M takes two half-lives.
  • B. Incorrect. 9.74 min divides the drop of 0.450 M by k as if the reaction were zero order, but k here has first-order units of min⁻¹.
  • C. Correct. The concentration falls to one quarter, which is two half-lives: t = ln(0.600/0.150) ÷ 0.0462 = 30.0 min.
  • D. Incorrect. 45.0 min is three half-lives, which would take the concentration down to 0.0750 M, not 0.150 M.

Justify Your turn

A student follows the concentration of reactant B over time, taking the rate as the rate at which B is used up. A plot of ln[B] against time curves, but a plot of 1/[B] against time is a straight line with a slope of 0.25 M⁻¹ min⁻¹. Give the order of the reaction in B, justify it from the plots, and give the rate constant with its units.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • States that the reaction is second order in B
  • Justifies the order with the straight-line plot of 1/[B] against time
  • Gives k = 0.25 M⁻¹ min⁻¹, with units

Model answer

The reaction is second order in B. For a second-order reaction, 1/[B] against time is a straight line, and that is the plot that is straight here, while ln[B] against time, the first-order test, curves. The slope of the 1/[B] plot equals k, so k = 0.25 M⁻¹ min⁻¹.

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