Rate laws: the mathematical form of speed
Learn what a rate law is, how exponents are found from experimental data, and why you can never guess a rate law from a balanced equation.
Quick answer
A rate law is an experimentally determined equation that expresses the reaction rate as the rate constant k multiplied by reactant concentrations raised to powers called orders.
Example: For data where doubling [A] doubles the rate and doubling [B] quadruples it, the rate law is rate = k[A][B]², first order in A and second order in B.
What you will learn
- Write the general form of a rate law and identify each symbol.
- Determine reaction orders from experimental rate data using the method of initial rates.
- Calculate the rate constant k from data once the orders are known.
- Explain why rate laws must be measured, not derived from balanced equations.
What a rate law says
A rate law connects the speed of a reaction to the concentrations you can actually control. Its general form for a reaction of A and B is:
- k is the rate constant, a number specific to the reaction at a given temperature.
- m and n are the orders with respect to A and B — usually small whole numbers (0, 1, or 2), sometimes fractions or negative numbers.
- m + n is the overall order of the reaction.
The rate law answers a practical question: if I double the amount of A, what happens to the speed? The answer depends entirely on m, and m must be measured.
The method of initial rates
Orders are found by changing one concentration at a time and watching the initial rate, before products build up and complicate things. Consider this data for A + B → products:
- Experiment 1: [A] = 0.10 M, [B] = 0.10 M, rate = 0.020 M/s
- Experiment 2: [A] = 0.20 M, [B] = 0.10 M, rate = 0.040 M/s
- Experiment 3: [A] = 0.10 M, [B] = 0.20 M, rate = 0.080 M/s
Comparing experiments 1 and 2, [A] doubles while [B] is held fixed, and the rate doubles (0.020 → 0.040). Doubling the concentration doubles the rate, so the reaction is first order in A (m = 1).
Comparing experiments 1 and 3, [B] doubles while [A] is fixed, and the rate quadruples (0.020 → 0.080). Doubling the concentration multiplies the rate by four (), so the reaction is second order in B (n = 2).
The rate law is therefore rate = k[A][B]², third order overall.
Finding the rate constant k
Once the orders are known, plug any single experiment into the rate law and solve for k. Using experiment 1 above:
Note the units of k: they change with the overall order so that the right side always comes out in M/s. Checking units is a good way to catch order mistakes — first order gives s⁻¹, second order M⁻¹s⁻¹, third order M⁻²s⁻¹, and zeroth order M/s.
Why you cannot read orders from the equation
It is tempting to look at 2A + B → products and write rate = k[A]²[B]. That is wrong in general. Reaction orders reflect the actual molecular mechanism, not the overall stoichiometry.
The balanced equation only describes the net change; the rate law describes the slowest step of the hidden sequence of collisions (the mechanism). Only if you independently know the mechanism can you predict the rate law — and even then, chemists verify it experimentally. The classic cautionary example: the reaction H₂ + I₂ → 2HI happens to be first order in each reactant, matching its equation, but the superficially similar H₂ + Br₂ → 2HBr has a complicated fractional-order rate law that no one could guess from the equation.
What rate laws reveal about mechanism
Because the rate law mirrors the slow step, it is one of the best clues to how a reaction actually happens at the molecular level. If a proposed mechanism predicts rate = k[A]² but experiments show rate = k[A][B], the mechanism is wrong and must be revised.
This detective work — proposing elementary steps, deriving the predicted rate law, and testing it against data — is how chemists uncover reaction mechanisms, which matter for designing catalysts and controlling industrial processes.
Variables and units
| Symbol | Name | Unit | Meaning |
|---|---|---|---|
| Rate constant | varies with order | Proportionality constant linking rate to concentrations at a given temperature | |
| Reaction orders | unitless | Exponents showing how the rate depends on each reactant's concentration |
Common mistakes
Coefficients come from atom balance; orders come from experiment. They coincide only for a single elementary step. For an overall reaction, assuming rate = k[A]²[B] just because the equation is 2A + B → products is one of the most common errors in kinetics.
k is constant at a fixed temperature. Changing concentrations changes the rate (through the concentration terms), not k. Only temperature — or a catalyst — changes k itself.
Zeroth order means the rate does not depend on that reactant's concentration, not that the reactant is absent. It often appears when a surface is saturated with reactant or when the reactant enters only after the slow step.
Real-world applications
Pharmaceuticals: rate laws predict how quickly a drug degrades on the shelf, setting expiration dates.
Environmental science: atmospheric chemists use rate laws to model ozone depletion step by step.
Industrial design: knowing the true orders lets engineers size reactors correctly instead of overbuilding.
Forensics: rate laws for decomposition reactions help estimate time since death or time since a fire.
Frequently asked questions
Can reaction orders be fractions or negative numbers?
Yes. Fractional orders arise in chain reactions and surface processes, and negative orders mean increasing that reactant actually slows the reaction — for instance, when a product inhibits the reaction or a reactant ties up the catalyst.
Why use initial rates instead of measuring the whole curve?
As a reaction proceeds, concentrations change and products accumulate, which can cause the reverse reaction or side effects that muddy the analysis. Initial rates are measured before any of that happens, so each rate corresponds to a known, controlled set of concentrations.
What does it mean if a reactant is missing from the rate law entirely?
It is zeroth order in that reactant: changing its concentration does not affect the rate. This usually means the reactant is involved only in fast steps after the rate-determining step, or that a catalyst surface is already saturated with it.
How is a rate law different from the equilibrium constant expression?
They look similar — both multiply concentrations with exponents — but the rate law's exponents are measured orders (kinetics: how fast), while K's exponents are stoichiometric coefficients (thermodynamics: how far). Never confuse the two.