First-order kinetics links how concentration changes over time to a simple exponential model. These ideas let you recognise first-order behaviour from graphs, extract the rate constant, and interpret a constant half-life.

What “first-order” means

In a first-order reaction, the instantaneous rate depends linearly on the reactant concentration.

This model is commonly written for a single reactant $A$ as “rate is proportional to $[A]$,” which implies that $[A]$ decreases exponentially rather than by a fixed amount each second.

Integrated rate law (first-order)

The integrated rate law connects concentration and time for a first-order process. It is the key tool for using concentration–time data to test first-order behaviour and to relate a measured concentration at time $t$ to the initial concentration.

A defining graphical feature from the AP syllabus is that $\ln[A]$ versus time is linear.

Plot of $\ln[\mathrm{H_2O_2}]$ versus time showing a straight-line trend for a first-order process. The linearity supports the integrated rate-law form $\ln[A]_t = -kt + \ln[A]_0$, where the slope is negative and equal to $-k$. Source

Linearity here means points fall on a straight line (within experimental uncertainty), supporting first-order kinetics under the same conditions.

Because the equation is in the form “$y = mx + b$,” a plot of $y=\ln[A]$ versus $x=t$ gives:

• Slope $m = -k$ (so $k$ is the positive magnitude of the slope)

• Intercept $b = \ln[A]_0$

What the straight-line test tells you

When $\ln[A]$ vs. $t$ is linear:

• The reaction is consistent with first-order dependence on $A$ under those conditions.

• The rate constant $k$ is constant across the dataset (same temperature, catalyst status, solvent, etc.).

• A larger $k$ corresponds to a steeper negative slope and faster decay of $[A]$.

Half-life for first-order reactions

The AP syllabus emphasises that the half-life is constant for first-order kinetics. “Constant” means it does not depend on the starting concentration; it depends only on $k$ (and thus the conditions that affect $k$).

How to interpret “constant half-life”

For a first-order process:

• The time to go from $[A]$ to $\tfrac{1}{2}[A]$ is always $t_{1/2}$.

• The next halving, from $\tfrac{1}{2}[A]$ to $\tfrac{1}{4}[A]$, takes the same $t_{1/2}$ again.

Three coordinate plots of exponential decay illustrating how halving occurs at equal time intervals (constant half-life behavior). The figure also highlights how using a semilog-style plot makes exponential decay appear linear, mirroring the “straight-line test” used for first-order kinetics. This helps connect the constant-$t_{1/2}$ concept to the graphical cue that $\ln[A]$ vs. $t$ is linear. Source

• After $n$ half-lives, $[A] = [A]_0\left(\tfrac{1}{2}\right)^n$ (useful for reasoning, even when you are not doing full calculations).

What changes half-life

Since $t_{1/2} = 0.693/k$:

• Increasing $k$ decreases $t_{1/2}$ (faster reaction, shorter half-life).

• Decreasing $k$ increases $t_{1/2}$ (slower reaction, longer half-life).

Common AP Chemistry graph-reading cues

When analysing experimental outputs for a single reactant $A$:

• If $[A]$ vs. $t$ curves downward (exponential-looking), that can be consistent with first-order behaviour.

• If $\ln[A]$ vs. $t$ is a straight line, that is the syllabus-aligned diagnostic.

• The line’s negative slope indicates depletion of reactant over time; a perfectly horizontal line would imply no reaction under those conditions.