A zeroth order reaction is one whose rate stays constant no matter how much reactant is present, so the rate law is rate = k and a plot of reactant concentration versus time gives a straight line with slope equal to -k.
A zeroth order reaction proceeds at a constant rate regardless of reactant concentration. In rate-law terms, the reactant's concentration is raised to the power of zero, and anything to the zeroth power equals 1, so the rate law collapses to rate = k. Doubling the concentration changes nothing. The reaction just chugs along at the same speed until the reactant runs out.
The graphical signature is the part AP Chem actually tests. Because the rate never changes, concentration drops by the same amount in each time interval, so a plot of [A] versus time is a straight line with slope = -k (per EK 5.3.A.1 and 5.3.A.4). Compare that to first order, where ln[A] vs. time is linear, and second order, where 1/[A] vs. time is linear. Zeroth order reactions usually happen when something other than concentration limits the rate, like a saturated catalyst surface or an enzyme working at full capacity. Adding more reactant doesn't help because the bottleneck is already maxed out.
Zeroth order lives in Unit 5 (Kinetics), specifically Topics 5.2 and 5.3. It supports learning objective 5.2.A, representing experimental data with a consistent rate law, and 5.3.A, identifying the rate law from concentration-versus-time data. The whole point of these topics is that reaction order is determined experimentally, not by reading coefficients off the balanced equation, and zeroth order is the clearest proof of that. A coefficient of 1 in the equation can still mean an exponent of 0 in the rate law. Zeroth order also completes the three-plot pattern (linear [A], linear ln[A], linear 1/[A]) that the exam uses constantly to test whether you can match a graph to an order.
Keep studying AP Chemistry Unit 5
First Order Reaction (Unit 5)
First order is the contrast case the exam loves. A first order reaction's rate scales directly with concentration and its ln[A] vs. time plot is linear, while zeroth order's plain [A] vs. time plot is linear. If you can tell these two graphs apart, you've got the core skill of Topic 5.3.
Half-Life (Unit 5)
Half-life behaves completely differently across orders. First order half-life is constant, but zeroth order half-life shrinks as the reaction proceeds because the rate stays the same while less reactant remains. A 'constant half-life' clue in a problem points to first order, never zeroth.
Integrated Rate Law (Unit 5)
The zeroth order integrated rate law, [A]t = -kt + [A]0, is literally the equation of its straight-line graph. Slope is -k and the y-intercept is the starting concentration, which is why pulling k off a concentration-time plot is a one-step move.
Rate Law (Unit 5)
Zeroth order is the edge case that proves rate laws come from experiments, not stoichiometry. A reactant can appear in the balanced equation and still have zero effect on the rate, which is exactly the kind of trap MCQ writers build.
Zeroth order shows up almost entirely as graph and data interpretation. The classic multiple-choice stem describes a plot of concentration versus time that comes out as a straight line and asks for the order; the answer is zeroth, because first order needs the ln[A] plot to be linear and second order needs the 1/[A] plot. You should also expect data-table questions where changing a reactant's concentration produces no change in rate, which means the exponent on that reactant is zero. On the kinetics FRQ, you may need to justify an order claim by naming which transformed plot is linear, calculate k from a slope (for zeroth order, k = -slope of [A] vs. time), or use the integrated rate law to find a concentration at a given time. Watch for the half-life trap too. A problem stating that half-life is constant regardless of initial amount (like radioactive decay) is describing first order, not zeroth.
Zeroth order means the rate ignores concentration entirely (rate = k), while first order means the rate is directly proportional to concentration (rate = k[A]). The graphs flip accordingly. Zeroth order gives a linear [A] vs. time plot; first order gives a linear ln[A] vs. time plot. Half-life is the other giveaway: zeroth order half-life gets shorter as reactant is used up, while first order half-life never changes, which is why radioactive decay is the textbook first order example.
A zeroth order reaction has the rate law rate = k, meaning the rate stays constant no matter how the reactant concentration changes.
The graphical fingerprint of zeroth order is a straight line on a plot of concentration versus time, with slope equal to -k.
The integrated rate law for zeroth order is [A]t = -kt + [A]0, which is just the equation of that straight line.
Zeroth order half-life is not constant; it gets shorter as the reaction proceeds, unlike first order where half-life never changes.
Reaction order is determined from experimental data, so a reactant can appear in the balanced equation and still be zeroth order in the rate law.
Zeroth order behavior often appears when a catalyst or surface is saturated, so adding more reactant can't speed things up.
It's a reaction whose rate doesn't depend on reactant concentration at all, so the rate law simplifies to rate = k. Concentration falls by equal amounts in equal time intervals, producing a linear [A] vs. time graph.
No, radioactive decay is first order. Its constant half-life (the same 5.7 years whether you start with 10 grams or 10 kilograms, for example) is the signature of first order kinetics, since a zeroth order half-life would shrink as material is used up.
In a zeroth order reaction the rate ignores concentration (rate = k), so [A] vs. time is linear. In a first order reaction the rate is proportional to concentration (rate = k[A]), so ln[A] vs. time is the linear plot and half-life stays constant.
The plain concentration versus time graph. It's a straight line with slope = -k and y-intercept [A]0, which comes straight from the integrated rate law [A]t = -kt + [A]0.
No. The reaction is happening at a perfectly steady rate; it just doesn't speed up or slow down when concentration changes. This usually happens when something else, like a saturated catalyst surface, controls how fast the reaction can go.