Half-life (t₁/₂) is the time it takes for a reactant's concentration to fall to half its initial value. In AP Chem, the big idea is that only first-order reactions have a constant half-life, given by t₁/₂ = 0.693/k, which makes it independent of how much reactant you start with.
Half-life is exactly what it sounds like. Start a stopwatch, and t₁/₂ is the time it takes for [A] to drop from its starting value to half of that value. After one half-life you have 50% left, after two half-lives 25%, after three 12.5%, and so on.
Here's the part the AP exam actually cares about. For a first-order reaction, half-life is constant, no matter the concentration. Whether you start with 2.0 M or 0.001 M, it takes the same amount of time to cut the amount in half. That's because the rate is proportional to [A], so as the concentration shrinks, the rate slows down by exactly the right amount to keep the halving time fixed. The relationship is t₁/₂ = 0.693/k (that 0.693 is ln 2), and it's on your AP equation sheet. Zeroth- and second-order reactions also have half-lives, but theirs change as the reaction proceeds, so a constant half-life is fingerprint evidence for first-order kinetics.
Half-life lives in Topic 5.3 (Concentration Changes Over Time) in Unit 5: Kinetics, supporting learning objective 5.3.A, which asks you to identify a reaction's rate law from concentration-versus-time data. Half-life is one of the fastest tools for that job. If a data table shows the concentration halving in equal time intervals, the reaction is first order and you can write rate = k[A] without ever making a graph. It also connects directly to the integrated rate laws, since t₁/₂ = 0.693/k comes straight from setting [A] = ½[A]₀ in the first-order integrated rate law. And because radioactive decay is first order, half-life is the bridge between kinetics math and nuclear applications like carbon dating.
Keep studying AP® Chemistry Unit 5
First-Order Reaction (Unit 5)
This is the pairing that matters most. Constant half-life and first-order kinetics imply each other on the AP exam. See equal halving times in a data table, conclude first order. Told a reaction is first order, you immediately get t₁/₂ = 0.693/k.
Integrated Rate Law (Unit 5)
The half-life equation isn't a separate fact to memorize blind. It falls out of the first-order integrated rate law, ln[A] = ln[A]₀ − kt, when you plug in [A] = ½[A]₀. The ln(½) is where the 0.693 comes from.
Rate Constant (Unit 5)
Half-life and k are two sides of the same coin for first-order reactions. A big k means a fast reaction and a short half-life. Exam questions love giving you one and asking for the other through t₁/₂ = 0.693/k.
Radioactive Decay (Unit 5)
All radioactive decay follows first-order kinetics, which is exactly why every isotope has one fixed, characteristic half-life. This is the real-world payoff of the 'first order means constant half-life' rule, and it's how carbon-14 dating works.
Half-life shows up in a few predictable ways. Multiple-choice questions give you a concentration-versus-time table and expect you to spot equal halving times as proof of first-order kinetics, or they ask 'what fraction remains after n half-lives?' (answer: (½)ⁿ, so 25% after two, 12.5% after three). Calculation questions go back and forth between t₁/₂ and k using t₁/₂ = 0.693/k, which is provided on the equation sheet. In free-response kinetics problems, half-life often appears as one part of a larger Unit 5 question, where you might determine the order from data, calculate k, and then find the half-life or the time needed to reach a given concentration. The classic trap is applying the constant-half-life shortcut to a reaction that isn't first order, so always confirm the order first.
Every reaction has a half-life in the literal sense (a time for [A] to halve), but only first-order reactions have a constant one. For second-order reactions, the half-life gets longer as concentration drops, and for zeroth-order it gets shorter. The formula t₁/₂ = 0.693/k applies to first order only. Using it on a second-order problem is one of the most common Unit 5 errors.
Half-life (t₁/₂) is the time required for a reactant's concentration to drop to half of its initial value.
For first-order reactions, half-life is constant and independent of the starting concentration, with t₁/₂ = 0.693/k.
A data table showing the concentration halving over equal time intervals is direct evidence that the reaction is first order, which supports LO 5.3.A.
After n half-lives, the fraction of reactant remaining is (½)ⁿ, so 25% remains after two half-lives and 12.5% after three.
The half-life equation comes from plugging [A] = ½[A]₀ into the first-order integrated rate law, and ln 2 is where 0.693 comes from.
Radioactive decay is first order, which is why isotopes have fixed half-lives and why dating techniques like carbon-14 dating work.
Half-life (t₁/₂) is the time it takes for a reactant's concentration to fall to half its initial value. It's covered in Topic 5.3 of Unit 5 (Kinetics), and the headline result is that first-order reactions have a constant half-life equal to 0.693/k.
No. Only first-order reactions have a constant half-life. For second-order reactions the half-life increases as the reaction proceeds, and for zeroth-order it decreases, so t₁/₂ = 0.693/k is first-order only.
Yes. The first-order half-life equation t₁/₂ = 0.693/k is on the reference sheet, along with the integrated rate laws. You don't need to memorize it, but you do need to know it only works for first-order reactions.
The rate constant k tells you how fast the reaction goes in the rate law, while half-life tells you how long it takes the concentration to halve. For first-order reactions they're inversely linked through t₁/₂ = 0.693/k, so a larger k means a shorter half-life.
One eighth, or 12.5%, of the original amount. Each half-life cuts the remaining concentration in half, so after n half-lives the fraction left is (½)ⁿ. This is a common quick-math MCQ setup.
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