A second order reaction is one whose rate law has reactant concentration powers summing to 2, either rate = k[A]² (doubling [A] quadruples the rate) or rate = k[A][B]. It is part of writing rate laws from experimental data in AP Chem Topic 5.2 (Unit 5: Kinetics).
A second order reaction is any reaction whose overall order is 2, meaning the exponents in the rate law add up to 2. That can happen two ways. The rate might depend on the square of one reactant, rate = k[A]², or on two different reactants each to the first power, rate = k[A][B]. Either way, the math behaves the same. If a reaction is second order in a single reactant, doubling that concentration multiplies the rate by 2² = 4, and tripling it multiplies the rate by 9.
The key idea from the CED (5.2.A.2 and 5.2.A.3) is that the order is an exponent in the rate law, and you find it from experimental data, never from the balanced equation's coefficients. You compare trials where one concentration changes while the others stay fixed and see what happens to the rate. A second order reaction also has a tell in its rate constant. Since rate (M/s) = k[A]², the units of k must be M⁻¹s⁻¹ for everything to cancel correctly. That units trick is a fast way to identify the overall order on a multiple-choice question.
This term lives in Topic 5.2 (Introduction to Rate Law) in Unit 5: Kinetics, supporting learning objective 5.2.A, which asks you to represent experimental data with a consistent rate law expression. "Second order" is one of the three orders (zeroth, first, second) the AP exam expects you to recognize and manipulate fluently. It also sets up everything that follows in Unit 5. In Topic 5.3, the second order integrated rate law (1/[A] vs. time gives a straight line) is one of the three linear plots you have to match to data. Later, in reaction mechanisms, a bimolecular elementary step is second order by definition, which is one of the few times you CAN read order straight from an equation. If you can identify second order behavior from a data table, a graph, or the units of k, you've covered three of the most common kinetics question formats.
Keep studying AP Chemistry Unit 5
First Order Reaction (Unit 5)
These are the two orders you'll compare constantly. First order means doubling the concentration doubles the rate; second order means doubling it quadruples the rate. Spotting that 2x-vs-4x difference in a data table is the classic Topic 5.2 move.
Integrated Rate Law (Unit 5)
Each order has its own linear plot. For second order, graphing 1/[A] versus time gives a straight line with slope k. If a question hands you concentration-vs-time data instead of initial rates, this is how you prove the reaction is second order.
Reaction Mechanism (Unit 5)
A bimolecular elementary step, where two particles collide, has a rate law of k[A][B] or k[A]², so it's automatically second order. Elementary steps are the one exception where coefficients DO give you the order.
Rate Constant (Unit 5)
The units of k change with order, and for second order they're M⁻¹s⁻¹. This works in reverse too. If a question tells you k has units of M⁻¹s⁻¹, you immediately know the overall order is 2 without seeing any data.
Second order reactions show up most often in method-of-initial-rates questions. You get a table of trials, and you have to notice that doubling a reactant's concentration quadrupled the rate, which means the reaction is second order in that reactant. Fiveable practice questions in this topic also test overall order, like a rate = k[X]¹[Y]¹ law where the exponents sum to 2, making the reaction second order overall even though it's first order in each reactant. Expect MCQs asking for the units of k (M⁻¹s⁻¹ for second order) and graph-matching questions where the linear 1/[A] vs. time plot identifies the order. On FRQs, kinetics questions typically ask you to determine the rate law from data, justify each order with a specific comparison of trials, and calculate k with correct units. Showing the 4x rate change explicitly is what earns the point.
Both describe how rate responds to concentration, but the response is different. In a first order reaction, rate scales linearly with concentration, so doubling [A] doubles the rate, and the linear plot is ln[A] vs. time. In a second order reaction, rate scales with the square, so doubling [A] quadruples the rate, and the linear plot is 1/[A] vs. time. Also watch the units of k. First order k is s⁻¹, while second order k is M⁻¹s⁻¹. Mixing up which plot goes with which order is one of the most common Unit 5 errors.
A reaction is second order overall when the exponents in its rate law sum to 2, which can mean rate = k[A]² or rate = k[A][B].
If doubling a reactant's concentration quadruples the rate, the reaction is second order in that reactant.
Reaction order comes from experimental data, not from the coefficients of the balanced equation (the one exception is an elementary step in a mechanism).
The rate constant for a second order reaction has units of M⁻¹s⁻¹, so units alone can tell you the order.
For a second order reaction, a plot of 1/[A] versus time is a straight line with slope equal to k.
A bimolecular elementary step is always second order because the rate depends on two particles colliding.
It's a reaction whose rate law exponents sum to 2, either rate = k[A]² for one reactant or rate = k[A][B] for two. It appears in Topic 5.2 (Introduction to Rate Law) in Unit 5: Kinetics.
No. Order comes from experimental data, not coefficients (CED 5.2.A.3). A reaction written 2A → B could be zeroth, first, or second order depending on its mechanism. The only exception is an elementary step, where coefficients do equal the order.
In first order, doubling concentration doubles the rate and ln[A] vs. time is linear with k in s⁻¹. In second order, doubling concentration quadruples the rate and 1/[A] vs. time is linear with k in M⁻¹s⁻¹.
M⁻¹s⁻¹ (or L·mol⁻¹·s⁻¹). Since rate in M/s equals k times concentration squared (M²), k must carry M⁻¹s⁻¹ for the units to work out. The exam uses k's units as a quick way to test whether you know the order.
Yes. A rate law like rate = k[X]¹[Y]¹ is first order in X, first order in Y, and second order overall because the exponents sum to 2. This exact setup shows up in practice questions, so add the exponents before answering.
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