A first order reaction is a reaction whose rate is directly proportional to the concentration of one reactant (rate = k[A]), so a plot of ln[A] versus time is linear with slope -k, and the half-life stays constant no matter how much reactant you start with.
A first order reaction has a rate law of rate = k[A]¹. The exponent (the order) is 1, which means the rate scales directly with concentration. Double [A] and the rate doubles. Cut [A] in half and the rate drops by half. That exponent comes from experimental data, not from the balanced equation's coefficients, which is exactly what learning objective 5.2.A is testing.
The fingerprint of a first order reaction shows up when you graph concentration over time. A regular [A] vs. time plot curves downward, but a plot of ln[A] vs. time gives a straight line with slope equal to -k (EK 5.3.A.2). That linearity comes from the integrated rate law, ln[A]ₜ = -kt + ln[A]₀. The other signature is a constant half-life. Because t½ = 0.693/k, the time it takes for the concentration to drop by half never changes, whether you're going from 0.80 M to 0.40 M or from 0.40 M to 0.20 M. No other reaction order has that property.
First order reactions live in Unit 5 (Kinetics), specifically Topics 5.2 and 5.3. They support two learning objectives. 5.2.A asks you to build a rate law from experimental data, so you need to recognize when doubling a concentration doubles the rate (that's first order behavior). 5.3.A asks you to identify reaction order from concentration-vs-time graphs, where 'ln[A] vs. t is linear' is the first order tell (EK 5.3.A.2). First order is also the one order the AP exam connects to the real world most often, because radioactive decay is always first order. That's why half-lives of isotopes are constant, and it's why first order math shows up in everything from carbon dating contexts to drug metabolism examples.
Keep studying AP Chemistry Unit 5
Integrated Rate Law (Unit 5)
The integrated rate law is the first order rate law solved over time. Instead of telling you the rate right now, ln[A]ₜ = -kt + ln[A]₀ tells you the concentration at any future time t. It's also why ln[A] vs. time graphs straight.
Half-Life (Unit 5)
Only first order reactions have a constant half-life. Since t½ = 0.693/k, the half-life depends only on the rate constant, not on concentration. If a problem says 'the half-life stays the same regardless of starting amount,' the answer is first order, full stop.
Second Order Reaction (Unit 5)
Second order is first order's main rival on graph-identification questions. For second order, doubling [A] quadruples the rate, and the linear plot is 1/[A] vs. time instead of ln[A] vs. time. Knowing which transformed plot is straight is how you tell them apart.
Rate Constant (Unit 5)
For a first order reaction, k has units of s⁻¹ (or time⁻¹). Checking the units of k is a quick sanity check on order, since zeroth, first, and second order each force different units on k.
First order reactions are mostly multiple-choice and short-answer math. Expect three classic setups. First, rate-data tables where you compare trials. If doubling [X] doubles the rate, X is first order, and adding up exponents gives the overall order (a rate = k[X]¹[Y]¹ reaction is second order overall). Second, graph identification. You'll see (or be told about) plots of [A], ln[A], and 1/[A] versus time and have to pick which one is linear. A straight ln[A] plot means first order, and a straight plain [A] vs. time plot means zeroth order, not first. Third, half-life problems, like a reactant dropping from 0.80 M to 0.40 M in 45 minutes (that 45 minutes is one half-life). On FRQs, first order math typically appears as a calculation step, using ln[A]ₜ = -kt + ln[A]₀ or t½ = 0.693/k, with the equations provided on the AP equation sheet.
Both depend on reactant concentration, but the math diverges. First order means rate = k[A], doubling [A] doubles the rate, ln[A] vs. time is linear, and half-life is constant. Second order means rate = k[A]², doubling [A] quadruples the rate, 1/[A] vs. time is linear, and the half-life gets longer as the reaction proceeds. The fastest tiebreaker is the half-life behavior, since only first order keeps it constant.
A first order reaction has rate = k[A], so the rate is directly proportional to concentration. Doubling the concentration doubles the rate.
For a first order reaction, a plot of ln[A] versus time is a straight line with slope -k. That linear ln plot is the graphical fingerprint AP questions test.
First order reactions have a constant half-life, t½ = 0.693/k, that does not depend on the starting concentration.
Radioactive decay is always first order, which is why isotopes have fixed half-lives like 5.7 years no matter how much sample you start with.
Reaction order comes from experimental data (rate tables or linear plots), never from the coefficients of the balanced equation.
The rate constant k for a first order reaction has units of s⁻¹, which you can use to double-check the order.
It's a reaction whose rate depends linearly on one reactant's concentration, written rate = k[A]. Its giveaway signs are a linear ln[A] vs. time plot (slope = -k) and a half-life that stays constant throughout the reaction.
No, and this trips up a lot of people. A straight line on a plain [A] vs. time plot means zeroth order. First order gives a straight line only when you plot ln[A] vs. time (EK 5.3.A.2).
First order means doubling [A] doubles the rate and ln[A] vs. time is linear with a constant half-life. Second order means doubling [A] quadruples the rate, 1/[A] vs. time is linear, and the half-life grows as concentration drops.
Because t½ = 0.693/k, which contains no concentration term at all. That's why a reaction taking 45 minutes to go from 0.80 M to 0.40 M will take another 45 minutes to reach 0.20 M, and why radioactive isotopes have fixed half-lives.
No. Per EK 5.2.A, the order comes only from experimental data, like rate tables or which transformed concentration plot is linear. A coefficient of 1 in the equation does not mean the reaction is first order in that species.