A plot of [A] versus time is a straight line with slope -2.0 × 10^-3\ M s^-1. State the order and the value of k. [2 points]

What is sign of the slope?

Zero- and first-order plots have negative slopes; the second-order reciprocal plot has a positive slope.

College Board AP 2023 (style)-style practice: Section II (long FRQ, part). The decomposition of A is first order with k = 0.0250\ s^-1, starting at [A]_0 = 0.800 M. (a) Calculate [A] after 60.0 s. (b) Calculate the half-life of the reaction. (c) A plot of ln[A] versus time is found to be linear. Determine what the slope of that line equals. (d) Justify why the half-life of a first-order reaction is independent of starting concentration.

A 4-point quantitative FRQ on first-order kinetics. (a) Concentration (1 point): ln[A] = ln[A]_0 - kt = ln(0.800) - (0.0250)(60.0) = -0.223 - 1.50 = -1.723, so [A] = e^-1.723 = 0.179 M. (b) Half-life (1 point): t_1/2 = 0.693k = 0.6930.0250 = 27.7 s. (c) Slope (1 point): the integrated law ln[A] = -kt + ln[A]_0 is linear with slope -k = -0.0250\ s^-1. (d) Justify (1 point): t_1/2 = 0.693k contains only k, not [A]_0; the fraction of A remaining after one half-life is always one half regardless of where you start. Markers reward the concentration from the integrated law, the half-life, identifying the slope as -k, and the constant-half-life reasoning.

United StatesChemistrySyllabus dot point

How does the concentration of a reactant change with time, and how can the shape of that change reveal the reaction order?

Topic 5.3 Concentration Changes Over Time: use the integrated rate laws for zero-, first- and second-order reactions, identify order from a linear plot, and use the half-life of a first-order reaction.

A focused answer to AP Chemistry Topic 5.3, covering the integrated rate laws for zero-, first- and second-order reactions, identifying order from linear plots, and the first-order half-life, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Each reaction order has an integrated rate law that gives concentration as a function of time, and each becomes a straight line when the right quantity is plotted against time. Zero order: $[\text{A}] = [\text{A}]_0 - kt$ , so a plot of $[\text{A}]$ versus $t$ is linear with slope $-k$ . First order: $\ln[\text{A}] = \ln[\text{A}]_0 - kt$ , so $\ln[\text{A}]$ versus $t$ is linear with slope $-k$ . Second order: $\frac{1}{[\text{A}]} = \frac{1}{[\text{A}]_0} + kt$ , so $\frac{1}{[\text{A}]}$ versus $t$ is linear with slope $+k$ . You identify the order by finding which plot is straight. The half-life of a first-order reaction is $t_{1/2} = \frac{0.693}{k}$ , independent of starting concentration.

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What this topic is asking
The three integrated rate laws
Identifying order from a linear plot
Half-life
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What this topic is asking

The College Board (Topic 5.3) wants you to use the integrated rate laws that give concentration as a function of time for zero-, first- and second-order reactions, recognize the linear plot that identifies each order, and use the half-life of a first-order reaction. This is the time-domain partner of the rate law: where Topic 5.2 measured how rate depends on concentration, this topic tracks how concentration falls as the clock runs.

The three integrated rate laws

These come from integrating the differential rate laws, but on the AP exam they are given on the equations sheet, so you apply them rather than derive them. The skill is choosing the right one for the order and rearranging for the unknown.

Identifying order from a linear plot

This graphical test is a favorite AP question. Given a table of concentration and time, you decide the order by seeing which transformation straightens the data, then read $k$ from the slope. The sign of the slope is negative for zero and first order (concentration falls) and positive for second order (reciprocal rises).

Half-life

The half-life $t_{1/2}$ is the time for the concentration to fall to half its value. For a first-order reaction it has a uniquely simple form:

t_{1/2} = \frac{0.693}{k}

This is constant: each successive half-life takes the same time, so after $n$ half-lives the fraction remaining is $\left(\tfrac{1}{2}\right)^n$ . This concentration-independence is the signature of first-order kinetics and underlies radioactive decay. Zero- and second-order half-lives do depend on starting concentration and are not required to be memorized.

A first-order decay

$\text{N}_2\text{O}_5$ decomposes by first-order kinetics with $k = 6.93 \times 10^{-3}\ \text{s}^{-1}$ , starting at $0.100$ M. Find the half-life and the concentration after $200.$ s.

step 1 Half-life

$t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{6.93 \times 10^{-3}} = 100.$ s.

step 2 Number of half-lives in 200 s

$\dfrac{200.}{100.} = 2$ half-lives.

step 3 Concentration by halving

After 2 half-lives, $[\text{A}] = 0.100 \times \left(\tfrac{1}{2}\right)^2 = 0.0250$ M.

step 4 Check with the integrated law

$\ln[\text{A}] = \ln(0.100) - (6.93 \times 10^{-3})(200.) = -2.303 - 1.386 = -3.689$ , so $[\text{A}] = e^{-3.689} = 0.0250$ M, which matches.

Try this

Q1. A first-order reaction has $t_{1/2} = 40.$ s. Calculate the rate constant. [2 points]

Cue. $k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{40.} = 0.017\ \text{s}^{-1}$ .

Q2. A plot of $[\text{A}]$ versus time is a straight line with slope $-2.0 \times 10^{-3}\ \text{M s}^{-1}$ . State the order and the value of $k$ . [2 points]

Cue. Linear in $[\text{A}]$ means zero order; $k = 2.0 \times 10^{-3}\ \text{M s}^{-1}$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). The decomposition of

\text{A}

is first order with

k = 0.0250\ \text{s}^{-1}

, starting at

[\text{A}]_0 = 0.800

M. (a) Calculate

[\text{A}]

after

60.0

s. (b) Calculate the half-life of the reaction. (c) A plot of

\ln[\text{A}]

versus time is found to be linear. Determine what the slope of that line equals. (d) Justify why the half-life of a first-order reaction is independent of starting concentration.

Show worked answer →

A 4-point quantitative FRQ on first-order kinetics.

(a) Concentration (1 point): $\ln[\text{A}] = \ln[\text{A}]_0 - kt = \ln(0.800) - (0.0250)(60.0) = -0.223 - 1.50 = -1.723$ , so $[\text{A}] = e^{-1.723} = 0.179$ M.
(b) Half-life (1 point): $t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{0.0250} = 27.7$ s.
(c) Slope (1 point): the integrated law $\ln[\text{A}] = -kt + \ln[\text{A}]_0$ is linear with slope $-k = -0.0250\ \text{s}^{-1}$ .
(d) Justify (1 point): $t_{1/2} = \dfrac{0.693}{k}$ contains only $k$ , not $[\text{A}]_0$ ; the fraction of A remaining after one half-life is always one half regardless of where you start.

Markers reward the concentration from the integrated law, the half-life, identifying the slope as $-k$ , and the constant-half-life reasoning.

AP 2021 (style)1 marksSection I (multiple choice). A plot of

\dfrac{1}{[\text{A}]}

versus time is a straight line. The reaction is (A) zero order in A (B) first order in A (C) second order in A (D) third order in A. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (C).

The integrated rate law for a second-order reaction is $\dfrac{1}{[\text{A}]} = kt + \dfrac{1}{[\text{A}]_0}$ , which is linear when $\dfrac{1}{[\text{A}]}$ is plotted against time. Zero order gives a straight line for $[\text{A}]$ itself, and first order for $\ln[\text{A}]$ . The trap is mixing up which function of concentration is linear for each order.

Related dot points

Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)