Rate Laws and Reaction Kinetics: How to Determine Reaction Order and Calculate Rate Constants

A rate law is a mathematical equation that relates the rate of a chemical reaction to the concentrations of the reactants. For a reaction aA + bB → products, the rate law takes the form: Rate = k[A]^m[B]^n, where k is the rate constant, [A] and [B] are molar concentrations, and m and n are the reaction orders with respect to each reactant.

The crucial point that trips up students on every kinetics exam: the reaction orders m and n are NOT the stoichiometric coefficients a and b from the balanced equation. They must be determined experimentally. A reaction with the balanced equation 2A + B → C might be first order in A and first order in B, or second order in A and zero order in B, or any other combination. The balanced equation tells you the stoichiometric ratios for mole calculations — it tells you nothing about the rate law. The rate law reflects the reaction mechanism (how molecules actually collide and react at the molecular level), not the overall stoichiometry.

The overall reaction order is the sum of the individual orders: m + n. A reaction that is first order in A and first order in B is second order overall. A reaction that is second order in A and zero order in B is second order overall. The overall order determines the units of the rate constant k, which is an important check on your work — if your calculated k has the wrong units for the reaction order you determined, something is wrong.

The method of initial rates is the most common technique for determining reaction order, and it appears on virtually every general chemistry and AP Chemistry exam. You are given a table of experimental data showing different initial concentrations of reactants and the resulting initial rate, and you need to determine the order with respect to each reactant.

The approach is systematic comparison. Find two experiments where the concentration of one reactant changes while all other concentrations stay constant. The ratio of rates equals the ratio of concentrations raised to the power of the reaction order for that reactant.

Example: for the reaction A + B → C, given these data:

Experiment 1: [A] = 0.10 M, [B] = 0.10 M, Rate = 2.0 × 10⁻³ M/s Experiment 2: [A] = 0.20 M, [B] = 0.10 M, Rate = 8.0 × 10⁻³ M/s Experiment 3: [A] = 0.10 M, [B] = 0.20 M, Rate = 4.0 × 10⁻³ M/s

Compare Experiments 1 and 2: [B] is constant, [A] doubles (0.10 → 0.20), and the rate quadruples (2.0 → 8.0). Since (Rate₂/Rate₁) = ([A]₂/[A]₁)^m, we get 4 = 2^m, so m = 2. The reaction is second order in A.

Compare Experiments 1 and 3: [A] is constant, [B] doubles (0.10 → 0.20), and the rate doubles (2.0 → 4.0). Since 2 = 2^n, n = 1. The reaction is first order in B.

The rate law is: Rate = k[A]²[B]. Overall order is 2 + 1 = 3 (third order).

To find k, plug any experiment's data into the rate law: 2.0 × 10⁻³ = k(0.10)²(0.10), so k = 2.0 × 10⁻³ / (0.010 × 0.10) = 2.0 M⁻²s⁻¹. The units M⁻²s⁻¹ are correct for a third-order reaction, which confirms our answer.

The ChemistryIQ app generates unlimited initial rate problems with randomized data so you can drill this technique until it becomes automatic.

While the differential rate law (Rate = k[A]^n) tells you the instantaneous rate at any moment, the integrated rate law tells you the concentration of a reactant at any time t. Each reaction order has its own integrated rate law, and identifying which one applies is a common exam question.

For a zero-order reaction (Rate = k): the integrated rate law is [A]t = [A]₀ − kt. Concentration decreases linearly with time. A plot of [A] vs. time gives a straight line with slope = −k. The half-life is t₁/₂ = [A]₀/(2k), which depends on initial concentration — this is unique to zero-order reactions.

For a first-order reaction (Rate = k[A]): the integrated rate law is ln[A]t = ln[A]₀ − kt, or equivalently [A]t = [A]₀e^(−kt). Concentration decreases exponentially. A plot of ln[A] vs. time gives a straight line with slope = −k. The half-life is t₁/₂ = 0.693/k, which is constant and does not depend on initial concentration. Radioactive decay follows first-order kinetics, which is why radioactive half-lives are constant regardless of the amount of material.

For a second-order reaction (Rate = k[A]²): the integrated rate law is 1/[A]t = 1/[A]₀ + kt. A plot of 1/[A] vs. time gives a straight line with slope = k. The half-life is t₁/₂ = 1/(k[A]₀), which increases as concentration decreases — meaning each successive half-life is twice as long as the previous one.

To determine reaction order from concentration-vs-time data, make all three plots (concentration, ln(concentration), and 1/concentration vs. time) and see which one gives a straight line. This is the graphical method of determining order, and it complements the method of initial rates.

Reaction rates almost always increase with temperature, and the Arrhenius equation quantifies this relationship: k = Ae^(−Ea/RT), where k is the rate constant, A is the pre-exponential factor (also called the frequency factor), Ea is the activation energy, R is the gas constant (8.314 J/mol·K), and T is the absolute temperature in Kelvin.

The intuition behind the Arrhenius equation: Ea/RT is the ratio of the activation energy barrier to the average thermal energy available. When temperature increases, RT increases, making Ea/RT smaller, making e^(−Ea/RT) larger, making k larger. The rate constant increases exponentially with temperature, not linearly — this is why a 10°C increase can double or triple a reaction rate.

The linear form is more useful for calculations: ln(k) = ln(A) − Ea/(RT). A plot of ln(k) vs. 1/T gives a straight line with slope = −Ea/R and y-intercept = ln(A). This is the Arrhenius plot, and it is used to determine Ea from experimental data.

The two-point form eliminates the need for A: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂). This is the equation used most often in exam problems where you are given rate constants at two temperatures and asked to find the activation energy, or given Ea and one rate constant and asked to find k at a different temperature.

Common mistake: using Celsius instead of Kelvin. The Arrhenius equation requires absolute temperature (Kelvin). Using Celsius gives nonsensical results. Always convert: T(K) = T(°C) + 273.15.

Another common mistake: confusing activation energy with the overall energy change of the reaction (ΔH or ΔG). Activation energy is the energy barrier that must be overcome for the reaction to proceed — it determines how fast the reaction goes. ΔH determines whether the reaction is exothermic or endothermic — it determines the net energy change. An exothermic reaction can have a high activation energy (slow despite being thermodynamically favorable), and an endothermic reaction can have a low activation energy (fast despite being thermodynamically unfavorable).

Most reactions occur in multiple elementary steps, and the rate law for the overall reaction is determined by the rate-determining step — the slowest step in the mechanism, which acts as a bottleneck for the entire process.

For elementary reactions (reactions that occur in a single molecular step), the rate law CAN be written from the stoichiometry. A unimolecular elementary step (A → products) is first order. A bimolecular elementary step (A + B → products) is first order in A and first order in B. A termolecular elementary step (A + B + C → products) is first order in each. This is the one case where stoichiometry equals reaction order — but only for elementary steps, not for the overall reaction.

When a mechanism has a slow first step, writing the rate law is straightforward: use the slow step's molecularity. If step 1 (slow) is A + B → intermediate, the rate law is Rate = k[A][B].

When the slow step involves an intermediate (a species produced in an earlier step), the problem is harder because intermediates cannot appear in the final rate law — their concentrations are not measurable or controllable. You must use the equilibrium from a preceding fast step to express the intermediate's concentration in terms of reactant concentrations.

Example: Step 1 (fast, equilibrium): A ⇌ B, with equilibrium constant K = [B]/[A]. Step 2 (slow): B + C → D. The rate law from step 2 is Rate = k₂[B][C]. But B is an intermediate, so substitute from the equilibrium: [B] = K[A]. The final rate law is Rate = k₂K[A][C] = k_obs[A][C], where k_obs = k₂K.

This is a high-yield exam topic because it tests your understanding of mechanisms, equilibrium, and rate laws simultaneously. The key skill is recognizing when an intermediate needs to be eliminated and knowing how to use the pre-equilibrium assumption to do it.