Kinetics is the chemistry of how fast — not whether a reaction can happen, but how quickly it does. The rate law is the equation that ties the reaction rate to the concentrations of the reactants, and the most common exam task is to work out the rate law from a table of initial-rates data. This guide on reaction kinetics and rate laws covers what a rate law is, what "order" means, and the ratio method that turns a four-row data table into a rate law in about a minute.
What Is a Reaction Rate?
The rate of a reaction is how fast a reactant is consumed or a product is formed, measured in concentration per time — usually mol/(L·s) or M/s.
For a generic reaction aA + bB → cC + dD, the rate is defined consistently across all species using the coefficients:
rate = −(1/a) d[A]/dt = −(1/b) d[B]/dt = +(1/c) d[C]/dt = +(1/d) d[D]/dt
The negative signs on the reactants are there because their concentrations decrease over time, so d[A]/dt is negative, and the rate itself must be positive. Dividing each derivative by its coefficient gives one consistent rate regardless of which species you watch.
In an intro course you almost always work with the initial rate — the rate measured immediately at t = 0 when only reactants are present. That avoids complications from the reverse reaction or product buildup.
The Rate Law
The rate law is an empirical equation that relates the reaction rate to the concentrations of the reactants:
rate = k [A]^m [B]^n
- k is the rate constant, with units that depend on the overall order. It varies with temperature (Arrhenius equation) but not with concentration.
- m is the order with respect to A. n is the order with respect to B.
- m + n is the overall order of the reaction.
Two things to know. First, the exponents are not the same as the stoichiometric coefficients in the balanced equation. m and n must be determined experimentally — you cannot read them off the equation. Second, m and n are usually 0, 1, or 2, occasionally a fraction, almost never larger than 2.
What the orders mean physically
- Zero order (rate ∝ [A]⁰ = 1) — concentration of A does not affect the rate. Doubling [A] does nothing. Common when a surface or enzyme is saturated.
- First order (rate ∝ [A]¹) — rate is directly proportional to [A]. Doubling [A] doubles the rate. Radioactive decay and many decomposition reactions are first order.
- Second order (rate ∝ [A]²) — rate is proportional to [A] squared. Doubling [A] quadruples the rate. Many bimolecular reactions are second order overall.
The total order m + n determines how fast the rate climbs as you scale concentrations and dictates the units of k.
The Initial-Rates Method
The standard way to determine m and n is the initial-rates method. Run the reaction multiple times, varying one reactant's concentration at a time, and measure the initial rate each time. Then use ratios.
The key move: pick two experiments where only one reactant's concentration changes. The ratio of their rates depends only on that one reactant.
(rate₂ / rate₁) = ([A]₂ / [A]₁)^m when [B] is constant
Take the log, or just inspect: if doubling [A] doubles the rate, m = 1. If doubling [A] quadruples the rate, m = 2. If the rate does not change, m = 0.
Worked Example: Determining Order from Initial-Rates Data
Suppose for the reaction 2 NO + O₂ → 2 NO₂, you measure these initial rates:
| Trial | [NO] (M) | [O₂] (M) | Initial rate (M/s) |
|---|---|---|---|
| 1 | 0.020 | 0.010 | 0.0080 |
| 2 | 0.020 | 0.020 | 0.0160 |
| 3 | 0.040 | 0.020 | 0.0640 |
Step 1: Find m (order in NO) using Trials 2 and 3
In Trials 2 and 3, [O₂] is constant at 0.020 M; [NO] doubles from 0.020 to 0.040. The rate goes from 0.0160 to 0.0640 — a factor of 4. Doubling [NO] gives 4× the rate, so m = 2. NO is second order.
Step 2: Find n (order in O₂) using Trials 1 and 2
In Trials 1 and 2, [NO] is constant at 0.020 M; [O₂] doubles from 0.010 to 0.020. The rate goes from 0.0080 to 0.0160 — a factor of 2. Doubling [O₂] doubles the rate, so n = 1. O₂ is first order.
Step 3: Write the rate law
rate = k [NO]² [O₂]
Overall order: m + n = 3.
Step 4: Compute k
Plug in any trial. Use Trial 1: 0.0080 = k × (0.020)² × (0.010) = k × 0.0004 × 0.010 = k × 4.0 × 10⁻⁶.
k = 0.0080 / (4.0 × 10⁻⁶) = 2.0 × 10³ M⁻² s⁻¹.
The units on k come from the equation: rate is M/s, [NO]² is M², [O₂] is M, so k must carry M⁻² s⁻¹ to make the units balance. For any rate law, k's units = M^(1 − total order) × s⁻¹. Always check the units.
Reading the Rate Law Back
With rate = (2.0 × 10³) [NO]² [O₂] in hand, you can answer any "what if" question on this reaction.
- Triple [NO] and the rate goes up by 3² = 9×.
- Halve [O₂] and the rate goes down by 2× (since it is first order).
- Triple [NO] and halve [O₂] simultaneously and the rate changes by 9 × (1/2) = 4.5×.
That predictive power is what the rate law buys you, and it is the answer to almost every kinetics-table problem on an exam.
Getting Help
Rate laws describe how fast a reaction approaches its destination. The destination itself is set by chemical equilibrium — K tells you where the reaction lands, the rate law tells you how quickly it gets there. The two are independent: a catalyst changes the rate without changing K.
Conclusion
A rate law is the equation that ties the speed of a reaction to the concentrations of its reactants, with exponents that have to be measured rather than read off the balanced equation. Determine each exponent with the initial-rates method: hold every reactant constant except one, double its concentration, and see what the rate does. Doubling produces 2× rate means first order, 4× means second order, no change means zero order. Once you have the exponents, plug any trial in to solve for k, and the rate law is set.
