Concentration changes over time

AP Chemistry· AP Chemistry CED — Kinetics· 14 min read

1. Integrated Rate Laws by Reaction Order★★☆☆☆⏱ 4 min

An integrated rate law expresses reactant concentration as an explicit function of time, derived by integrating the differential rate law for a given reaction order. For each common order, the integrated rate law can be rearranged to the linear form $y = m x + b$ to enable experimental identification of reaction order. Standard notation: $[A]_{0}$ = initial concentration of reactant A at $t = 0$ , $[A]_{t}$ = concentration at time $t$ , $k$ = rate constant, $t_{1/2}$ = half-life.

📘 Definition

Integrated Rate Law

A mathematical relationship that gives the concentration of a reactant or product as a direct function of time, derived from the reaction's differential rate law.

For a zero-order reaction with differential rate law $rate = - \frac{d [ A ]}{d t} = k$ , the integrated form is:

[A]_t = -kt + [A]_0

This is linear when plotting $[A]$ (y-axis) vs $t$ (x-axis), with slope equal to $- k$ and intercept equal to $[A]_{0}$ .

For a first-order reaction with differential rate law $rate = - \frac{d [ A ]}{d t} = k [A]$ , the integrated form is:

\ln[A]_t = -kt + \ln[A]_0

This is linear when plotting $ln [A]$ (y-axis) vs $t$ (x-axis), with slope equal to $- k$ and intercept equal to $ln [A]_{0}$ .

For a second-order reaction in a single reactant with differential rate law $rate = - \frac{d [ A ]}{d t} = k [A]^{2}$ , the integrated form is:

\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}

This is linear when plotting $1/ [A]$ (y-axis) vs $t$ (x-axis), with slope equal to $+ k$ and intercept equal to $1/ [A]_{0}$ .

📐 Worked Example

Experimental concentration-time data for the reaction $A \to products$ is given below. Identify the reaction order and calculate the rate constant $k$ .

| $t$ (s) | 0 | 10 | 20 | 30 |
|---|---|---|---|---|
| $[A]$ (M) | 0.80 | 0.40 | 0.20 | 0.10 |

1
First, test for patterns in concentration change: the concentration halves every 10 seconds. A constant half-life across the reaction is a hallmark of first-order reactions.
2
Confirm using the first-order integrated rate law by calculating $ln [A]$ for each data point:
3
$\ln(0.80) = -0.223,\ \ln(0.40) = -0.916,\ \ln(0.20) = -1.609$
4
Calculate the slope between the first two points, which equals $- k$ for first-order reactions:
5
$slope = \frac{\Delta \ln[A]}{\Delta t} = \frac{-0.916 - (-0.223)}{10 - 0} = -0.0693\ \text{s}^{-1} = -k$
6
Slopes between all other points are identical, confirming linearity. Therefore, $k = 0.069 s^{- 1}$ and the reaction is first-order.

Exam tip:

When asked to determine reaction order from concentration-time data, always confirm which transformed concentration gives a straight line; constant half-life is a useful shortcut only for first-order reactions.

2. Half-Life of Reactions★★☆☆☆⏱ 3 min

📘 Definition

Half-Life ($t_{1/2}$)

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 also increases as the reaction proceeds and $[A]$ decreases.

📐 Worked Example

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

1
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:
3
$t_{1/2} = \frac{0.693}{0.0231\ \text{day}^{-1}} = 30\ \text{days}$
4
Multiply the half-life by the number of half-lives to get total time:
5
$t = 3 \times 30\ \text{days} = 90\ \text{days}$
6
Confirm with the integrated first-order rate law:
7
$\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt \rightarrow \ln(0.125) = -0.0231t \rightarrow t = 90\ \text{days}$

Exam tip:

When calculating time to reach a given percentage of initial concentration for first-order reactions, convert the fraction to powers of 1/2 to avoid logarithm calculation errors.

3. Graphical Determination of Reaction Order★★★☆☆⏱ 3 min

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 matching reaction order.

AP exam questions may 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.

📐 Worked Example

A student collects concentration-time data for the reaction $B \to products$ and plots three transformed graphs: 1. $[B]$ vs $t$ : curved decreasing line, 2. $ln [B]$ vs $t$ : curved decreasing line, 3. $1/ [B]$ vs $t$ : a straight line 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, matching the second-order integrated rate law form.
2
Per the second-order integrated rate law $\frac{1}{[ B ] _{t}} = k t + \frac{1}{[ B ] _{0}}$ , the slope of the linear graph equals $k$ .
3
Calculate the slope from the given points to get $k$ :
4
$k = \frac{\Delta (1/[B])}{\Delta t} = \frac{12.5\ M^{-1} - 2.5\ M^{-1}}{50\ s - 0\ s} = 0.20\ M^{-1}s^{-1}$
5
The y-intercept of 2.5 M⁻¹ equals $1/ [B]_{0}$ , which aligns with the formula, confirming the calculation is correct.

Exam tip:

Remember that for second-order reactions, the slope of $1/ [A]$ vs $t$ is positive and equal to $k$ , while zero and first-order plots have negative slopes equal to $- k$ . Sign errors for $k$ are extremely common in graph questions.

4. AP-Style Practice Check★★★☆☆⏱ 4 min

✓ Quick check

Test your understanding of core concentration-time relationships:

Which of the following relationships is correct for a reaction that follows the rate law $rate = k [X]^{2}$ , where $[X]_{0} = 0.20 M$ ?
- 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 $k = 0.014 M^{- 1} s^{- 1}$ .
- 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 $0.20 M^{- 1}$ .
Reveal answer
B —
Correct. For second-order reactions, $t_{1/2} = \frac{1}{k [ A ] _{0}} = \frac{1}{( 0.014 ) ( 0.20 )} \approx 357$ s, which rounds to 350 s. All other options are incorrect: (A) is for first-order, (C) is only true for first-order, (D) intercept is $1/ [A]_{0} = 5 M^{- 1}$ .

📐 Worked Example

Radioactive carbon-14 decay follows first-order kinetics and has a half-life of 5730 years. A bone fragment has 12.5% of the carbon-14 activity of living bone (activity is proportional to concentration). Estimate the age of the fragment.

1
Carbon-14 decay is first-order, so half-life is constant throughout the process. 12.5% of initial activity equals $(1/2)^{3}$ , so 3 half-lives have passed.
2
Calculate the age by multiplying half-life by number of half-lives:
3
$t = 3 \times 5730\ \text{years} = 17190\ \text{years}$
4
Confirm with the integrated rate law:
5
$k = \frac{0.693}{5730} \approx 1.21 \times 10^{-4}\ \text{year}^{-1}, \quad t = -\frac{1}{k}\ln(0.125) \approx 17190\ \text{years}$
6
In context, this means the bone fragment from the dig is approximately 17,200 years old.

5. Common Pitfalls

Wrong move:

Using the first-order half-life formula $t_{1/2} = 0.693/ k$ for zero or second-order reactions.

Why:

Students often memorize the simple first-order half-life formula and forget that other orders have concentration-dependent half-life formulas.

Correct move:

Always confirm reaction order before applying a half-life formula; if you forget the specific formula, derive it by substituting $[A]_{t} = [A]_{0} /2$ into the integrated rate law for the reaction.

Wrong move:

Reporting the slope of a $ln [A]$ vs $t$ or $[A]$ vs $t$ graph directly as $k$ , leaving the negative sign intact.

Why:

The integrated rate law for zero and first order gives slope = $- k$ , but students often copy the slope value without adjusting for sign.

Correct move:

After calculating slope for these plots, drop the negative sign to get the positive value of $k$ , since rate constants are always positive.

Wrong move:

Assuming constant half-life for all reaction orders, e.g., that if concentration halves in 10 s it will halve again in another 10 s.

Why:

Only first-order reactions have constant half-life independent of initial concentration; other orders have changing half-life as concentration drops.

Correct move:

Only assume constant half-life if the reaction is confirmed to be first-order. For both zero and second order, half-life increases as concentration drops.

Wrong move:

Mixing up the correct y-axis transformation for linear plots, e.g., plotting $ln [A]$ vs $t$ for second-order reactions.

Why:

Confusion between the three linear forms of integrated rate laws, especially between first and second order.

Correct move:

Remember the y-axis is always the term on the left side of the integrated rate law: zero order = $[A]$ vs $t$ , first order = $ln [A]$ vs $t$ , second order = $1/ [A]$ vs $t$ .

Wrong move:

Applying the single-reactant second-order integrated rate law to second-order reactions with two different reactants at unequal initial concentrations.

Why:

The standard $1/ [A]_{t} = k t + 1/ [A]_{0}$ formula is only derived for second order in a single reactant or two reactants with equal initial concentrations.

Correct move:

Only use the standard single-reactant second-order integrated rate law for the cases it is derived for; other second-order systems require more complex calculations.

6. Quick Reference Cheatsheet

Category	Formula	Key Notes
Zero Order Integrated Rate Law	$[A]_{t} = - k t + [A]_{0}$	Linear plot: $[A]$ vs $t$ ; slope = $- k$
First Order Integrated Rate Law	$ln [A]_{t} = - k t + ln [A]_{0}$	Linear plot: $ln [A]$ vs $t$ ; slope = $- k$ ; applies to radioactive decay
Second Order Integrated Rate Law	$\frac{1}{[ A ] _{t}} = k t + \frac{1}{[ A ] _{0}}$	Linear plot: $1/ [A]$ vs $t$ ; slope = $+ k$
Zero Order Half-Life	$t_{1/2} = \frac{[ A ] _{0}}{2 k}$	Depends directly on $[A]_{0}$ ; increases as reaction proceeds
First Order Half-Life	$t_{1/2} = \frac{0.693}{k}$	Independent of $[A]_{0}$ ; constant for entire reaction
Second Order Half-Life	$t_{1/2} = \frac{1}{k [ A ] _{0}}$	Depends inversely on $[A]_{0}$ ; increases as reaction proceeds
Graphical Reaction Order ID	Straight line = matching order	Zero: $[A]$ vs $t$ ; First: $ln [A]$ vs $t$ ; Second: $1/ [A]$ vs $t$
First Order Concentration after n Half-Lives	$[A]_{t} = [A]_{0} (\frac{1}{2})^{n}$	Only valid for first-order reactions; n = number of half-lives

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Reaction order identification from graph
2022 · FRQ
Half-life and concentration calculation

What's Next

Mastering concentration changes over time and integrated rate laws lays the foundational experimental method for identifying reaction order and rate constants, which is the basis for all subsequent topics in AP Chemistry Unit 5 Kinetics. Without a solid understanding of the relationships covered here, you will struggle to interpret experimental kinetic data to justify proposed reaction mechanisms, a common high-weight free-response question on the AP exam. This topic also connects directly to first-order radioactive decay in nuclear chemistry, and helps build intuition for tracking concentration changes as systems approach equilibrium in later units.