Concentration changes over time — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Integrated rate laws for zero, first, and second order reactions, half-life calculations for each order, graphical methods for reaction order identification, and half-life order testing aligned to AP Chemistry CED Unit 5 Kinetics.

You should already know: Differential rate law notation and definitions, reaction order and rate constant conventions, plotting linear relationships to find slope and intercept.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Concentration changes over time?

Concentration changes over time describes how reactant (and product) concentrations evolve from their initial values as a reaction proceeds, connecting rate law relationships to measurable experimental data. In the AP Chemistry CED, this topic is a core part of Unit 5 Kinetics, which makes up 7-9% of the total AP exam score, with concentration-time content representing approximately 1-2% of total exam points. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often as part of a multi-part FRQ that asks for order identification and rate constant calculation. Standard notation used across this topic: $[A{]}_0$ is the initial concentration of reactant A at time $t=0$, $[A{]}_t$ is the concentration of A at time $t$, $k$ is the temperature-dependent rate constant, and $t_{1/2}$ is the half-life of the reaction. Unlike differential rate laws that give reaction rate as a function of concentration, this topic gives concentration as an explicit function of time, which matches what researchers measure when tracking reactions in lab, making it a critical tool for experimental kinetics.

2. Integrated Rate Laws by Reaction Order

An integrated rate law is a mathematical relationship that expresses reactant concentration as a direct function of time, derived by integrating the differential rate law for a given reaction order. For each common reaction order (zero, first, second in a single reactant), the integrated rate law can be rearranged into the linear form $y=mx+b$, which allows for easy experimental identification of reaction order.

For a zero-order reaction with differential rate law $\text{rate}=-\frac{d[A]}{dt}=k$, integrating from $t=0$ to $t$ gives: $$[A{]}_t=-kt+[A{]}_0$$ This is linear when plotting $[A]$ (y-axis) vs $t$ (x-axis), with slope equal to $-k$ and y-intercept equal to $[A{]}_0$.

For a first-order reaction with differential rate law $\text{rate}=-\frac{d[A]}{dt}=k[A]$, integrating gives: $$\ln [A{]}_t=-kt+\ln [A{]}_0$$ This is linear when plotting $\ln [A]$ vs $t$, with slope $-k$ and intercept $\ln [A{]}_0$.

For a second-order reaction in one reactant with differential rate law $\text{rate}=-\frac{d[A]}{dt}=k[A{]}^2$, integrating gives: $$\frac{1}{[A{]}_t}=kt+\frac{1}{[A{]}_0}$$ This is linear when plotting $1/[A]$ vs $t$, with slope $+k$ and intercept $1/[A{]}_0$.

Worked Example

Experimental concentration-time data for the reaction $A\to \text{products}$ is given below:

Identify the reaction order and calculate the rate constant $k$.

1. First, test for pattern in concentration changes: The concentration halves every 10 seconds, from 0.80→0.40 in 10 s, 0.40→0.20 in another 10 s, 0.20→0.10 in a third 10 s. Constant half-life is a hallmark of first-order reactions.
2. Confirm with the first-order integrated rate law: Calculate $\ln [A]$ for each point: $\ln (0.80)=-0.223$, $\ln (0.40)=-0.916$, $\ln (0.20)=-1.609$.
3. Calculate the slope between the first two points: $slope=\frac{\Delta \ln [A]}{\Delta t}=\frac{-0.916-(-0.223)}{10-0}=-0.0693{\text{s}}^{-1}=-k$.
4. Confirm the slope between the second and third points is identical, confirming linearity. This gives $k=0.069{\text{s}}^{-1}$, so the reaction is first order.

Exam tip: When asked to determine reaction order from concentration-time data, always test which transformed variable ($\ln [A]$ or $1/[A]$) gives a straight line; constant half-life is a quick shortcut only for first order, always confirm before answering.

3. Half-Life of Reactions

Half-life ($t_{1/2}$) is defined as the time required for the initial concentration of a reactant to decrease to half its original value. Each reaction order has a unique half-life relationship derived directly from its integrated rate law, which can be used to quickly identify order and calculate time to reach a given concentration.

For first-order reactions, substituting $[A{]}_t=[A{]}_0/2$ into the integrated rate law gives: $$t_{1/2}=\frac{\ln 2}{k}\approx \frac{0.693}{k}$$ The key unique property of first-order half-life is that it is independent of initial concentration, so it remains constant throughout the entire reaction. For zero-order reactions, the half-life relationship is $t_{1/2}=\frac{[A{]}_0}{2k}$: half-life depends directly on initial concentration, so it increases as the reaction proceeds and $[A]$ decreases. For second-order reactions, the half-life relationship is $t_{1/2}=\frac{1}{k[A{]}_0}$: half-life depends inversely on initial concentration, so it decreases as the reaction proceeds.

Worked Example

The first-order decomposition of a toxic industrial pollutant in a river has a rate constant $k=0.0231{\text{day}}^{-1}$. How many days will it take for the pollutant concentration to drop to 12.5% of its initial concentration?

1. First, convert the final concentration to a fraction of the initial value: 12.5% = 1/8 = (1/2)³, meaning the concentration has halved 3 times.
2. Calculate the half-life for this first-order reaction: $t_{1/2}=\frac{0.693}{0.0231{\text{day}}^{-1}}=30\text{days}$.
3. Multiply the half-life by the number of half-lives to get total time: $t=3\times 30\text{days}=90\text{days}$.
4. Confirm with the integrated first-order rate law: $\ln \left(\frac{[A{]}_t}{[A{]}_0}\right)=-kt\to \ln (0.125)=-0.0231t\to -2.079=-0.0231t\to t=90\text{days}$, which matches.

Exam tip: When calculating time to reach a given percentage of initial concentration for first order, always convert the fraction to powers of 1/2 to calculate with half-lives directly; this avoids logarithm calculation errors.

4. Graphical Determination of Reaction Order

A common AP Chemistry exam task is to determine reaction order from experimental data using graphical methods, based on the linear form of each integrated rate law. The core rule is: whichever transformation of concentration gives a straight line when plotted against time confirms the reaction order matching that transformation. AP exam questions often ask you to identify order from provided graphs, draw the correct transformed graph from raw data, or calculate the rate constant from the slope of the correct linear graph.

Graphical identification avoids the algebraic testing of multiple rate laws and directly leverages the integrated rate law structure, making it the most common method for determining reaction order from experimental concentration-time data in both lab and on the AP exam.

Worked Example

A student collects concentration-time data for the reaction $B\to \text{products}$ and plots three transformed graphs:

1. $[B]$ vs $t$: a curved decreasing line with decreasing slope
2. $\ln [B]$ vs $t$: a curved decreasing line with increasing negative slope
3. $1/[B]$ vs $t$: a straight line increasing from (0 s, 2.5 M⁻¹) to (50 s, 12.5 M⁻¹)

Identify the reaction order and calculate the rate constant $k$.

1. A straight line for $1/[B]$ vs $t$ confirms the reaction is second order in B, per the second-order integrated rate law form.
2. The second-order integrated rate law is $\frac{1}{[B{]}_t}=kt+\frac{1}{[B{]}_0}$, so the slope of the linear graph equals $k$.
3. Calculate the slope from the two given points: $slope=k=\frac{\Delta (1/[B])}{\Delta t}=\frac{12.5M^{-1}-2.5M^{-1}}{50s-0s}=0.20M^{-1}{s}^{-1}$.
4. Confirm the y-intercept matches the expected value: the intercept is $2.5M^{-1}=1/[B{]}_0$, which aligns with the formula, so the calculation is correct.

Exam tip: Always remember that for second order, the slope of the $1/[A]$ vs $t$ graph is positive and equal to $k$, whereas zero and first order have negative slopes equal to $-k$; it is common to mix up sign and magnitude for $k$ in graph questions.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using the first order half-life formula $t_{1/2}=0.693/k$ for zero or second order reactions. Why: Students memorize the first order half-life formula because it is independent of concentration, and forget that the other orders have different half-life formulas that depend on $[A{]}_0$. Correct move: Always confirm reaction order before applying a half-life formula; write the general integrated rate law and substitute $[A{]}_t=[A{]}_0/2$ to derive the half-life if you forget the specific formula.
• Wrong move: Reporting the slope of the $\ln [A]$ vs t graph as $k$ directly, leaving the negative sign intact. Why: The integrated first order law rearranges to $\ln [A{]}_t=-kt+\ln [A{]}_0$, so slope = $-k$, but students often just copy the slope value from the graph without adjusting the sign. Correct move: After calculating the slope of a $\ln [A]$ vs t or $[A]$ vs t graph, drop the negative sign to get the positive value of $k$ (rate constants are always positive).
• Wrong move: Assuming that a reaction that halves its concentration in 10 seconds will halve it again in another 10 seconds for all reaction orders. Why: Students associate constant half-life with any reaction, but only first order has constant half-life independent of concentration. Correct move: Check if the order is first order before assuming half-life is constant; for zero order, half-life increases as concentration drops, for second order, half-life decreases.
• Wrong move: Mixing up the y-axis for linear plots: plotting $\ln [A]$ vs t for second order to get a straight line. Why: Confusion between the three integrated rate law linear forms, especially mixing up first and second order. Correct move: Remember the correct transformed y-axis is always the term on the left side of the integrated rate law: zero = $[A]$ vs t, first = $\ln [A]$ vs t, second = $1/[A]$ vs t.
• Wrong move: Applying the standard second order integrated rate law to a second order reaction with two different reactants at unequal initial concentrations. Why: Textbooks most commonly present the second order integrated rate law for a single reactant, but students apply it incorrectly to other second order systems. Correct move: Only use $\frac{1}{[A{]}_t}=kt+\frac{1}{[A{]}_0}$ for reactions that are second order in a single reactant, or second order overall with equal initial concentrations of two reactants.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following relationships is correct for a reaction that follows the rate law $\text{rate}=k[X{]}^2$, where $[X{]}_0=0.20M$? A) A plot of $\ln [X]$ vs $t$ is linear with slope $-k$. B) The half-life of the reaction is approximately 350 s when $k=0.014M^{-1}{s}^{-1}$. C) After two half-lives, the total time elapsed is twice the length of the first half-life. D) A plot of $1/[X]$ vs $t$ is linear with intercept $0.20M^{-1}$.

Worked Solution: The reaction is second order in X, so we evaluate each option against the rules for second order reactions. Option A is incorrect because a linear plot of $\ln [X]$ vs $t$ is characteristic of first order, not second order. Option B: The half-life for second order is $t_{1/2}=\frac{1}{k[A{]}_0}=\frac{1}{(0.014M^{-1}{s}^{-1})(0.20M)}\approx 357s$, which rounds to 350 s, so this is correct. Option C is incorrect because for second order, half-life increases as concentration decreases, so the second half-life is longer than the first, making total time more than twice the first half-life. Option D is incorrect because the intercept of $1/[X]$ vs $t$ is $1/[X{]}_0=5M^{-1}$, not $0.20M^{-1}$. Correct answer: B.

Question 2 (Free Response)

The decomposition of a prescription drug in the human body follows first order kinetics. The initial concentration of the drug immediately after injection is $1.2\text{mg/dL}$. After 4 hours, the concentration drops to $0.6\text{mg/dL}$. (a) Calculate the rate constant $k$ for the decomposition reaction. (b) What is the concentration of the drug after 6 hours from injection? (c) A patient needs a top-up injection when the drug concentration drops below $0.15\text{mg/dL}$. How many hours after the first injection should the top-up be given?

Worked Solution: (a) The concentration drops from $1.2\text{mg/dL}$ to $0.6\text{mg/dL}$ in 4 hours, so the half-life $t_{1/2}=4$ hours. For first order, $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{4h}=0.173h^{-1}$. This can also be confirmed via the integrated rate law, which gives the same result. (b) Use the integrated first order rate law: $\ln [A{]}_t=-kt+\ln [A{]}_0=-(0.173h^{-1})(6h)+\ln (1.2mg/dL)=-1.038+0.182=-0.856$. $[A{]}_t=e^{-0.856}\approx 0.42\text{mg/dL}$. (c) Solve for $t$ when $[A{]}_t=0.15\text{mg/dL}$: $t=-\frac{1}{k}\ln \left(\frac{[A{]}_t}{[A{]}_0}\right)=-\frac{1}{0.173h^{-1}}\ln \left(\frac{0.15}{1.2}\right)=-\frac{(-2.079)}{0.173h^{-1}}=12\text{hours}$. This makes sense because $0.15\text{mg/dL}$ is 1/8 of the initial concentration, which is 3 half-lives, $3\times 4=12$ hours.

Question 3 (Application / Real-World Style)

Radioactive carbon-14 decay, used to date ancient organic samples, follows first order kinetics. The half-life of carbon-14 is 5730 years. A bone fragment found at an archaeological dig has 12.5% of the carbon-14 activity measured in living bone (activity is proportional to carbon-14 concentration). Estimate the age of the bone fragment, and interpret the result in context.

Worked Solution: Carbon-14 decay is first order, so half-life is constant throughout the decay process. 12.5% of the initial activity is equal to $(1/2{)}^3$, so 3 half-lives have passed. Calculate age: $t=3\times t_{1/2}=3\times 5730\text{years}=17190\text{years}$. Confirm with the integrated rate law: $k=0.693/5730\approx 1.21\times 10^{-4}{\text{year}}^{-1}$, $t=-\frac{1}{k}\ln (0.125)\approx 17190$ years, which matches. In context, this means the bone fragment from the dig is approximately 17,200 years old, based on the amount of radioactive carbon-14 remaining.

7. Quick Reference Cheatsheet

8. What's Next

This topic lays the foundational experimental method for identifying reaction order and rate constants, which is the basis for all subsequent topics in unit 5 Kinetics. Without mastering integrated rate laws and half-life relationships from this chapter, you will not be able to interpret experimental kinetic data to justify a proposed reaction mechanism, a common high-weight FRQ task on the AP exam. This topic also connects to radioactive decay in nuclear chemistry, where first-order kinetics are universally applied, and to equilibrium problems where you track concentration changes as a system approaches equilibrium. Mastery of this topic also helps you distinguish between reaction rate (how fast) and reaction extent (how far), a core distinction across AP Chemistry. Next topics to study after this chapter are: Collision theory and reaction rates, Reaction mechanisms and rate laws, Nuclear decay kinetics