Rate Law (Reaction Order)
The rate law links the reaction rate to the reactant concentrations; the exponents m and n are the reaction orders and are determined experimentally.
Free · no credit card · in your study plan in 2 minutes
Formula
v = k \cdot [A]^m \cdot [B]^nVariables & units – Rate Law (Reaction Order)
| Symbol | Meaning | Unit |
|---|---|---|
| v | Reaction rate | mol/(L·s) |
| k | Rate constant (temperature-dependent) | depending on the overall reaction order |
| [A], [B] | Concentrations of the reactants | mol/L |
| m, n | Reaction orders with respect to A and B (experimental) | dimensionless |
Derivation & background – Rate Law (Reaction Order)
The exponents do not follow from the reaction equation but from the mechanism; only for elementary reactions do they equal the coefficients. m + n is the overall order. For a first-order reaction the integrated form [A] = [A]₀·e^(−kt) holds, with the half-life t½ = ln 2/k. The temperature dependence of k is described by the Arrhenius equation.
Exam blueprint
Validity range
Applies at fixed temperature; m and n are experimental quantities and equal the stoichiometric coefficients only for elementary reactions.
Derivation steps
The rate depends on the collision frequency and thus on the particle concentrations.
- 1Measurement series show how v responds to concentration changes: v ∝ [A]^m·[B]^n.
- 2The proportionality constant k bundles temperature and collision factors.
Rearrangements
Rate constant
The unit of k depends on the overall order m + n.
Order from two measurements
All other concentrations must stay constant.
Task variant
Doubling [A] quadruples v, [B] changes nothing: what is the rate law?
v = k·[A]²: from 2^m = 4 follows m = 2, from the unchanged rate follows n = 0. The overall order is 2.
A reaction is second order in A. At [A] = 0.2 mol/L, v = 4.0×10⁻³ mol/(L·s). Calculate k.
k = v/[A]² = 4.0×10⁻³/(0.2)² = 4.0×10⁻³/0.04 = 0.10 L/(mol·s). The unit follows from the overall order 2.
Common mistakes
Reading the exponents off the reaction equation.
m and n are determined experimentally; only elementary reactions mirror the stoichiometry.
Giving k a fixed unit.
The unit of k changes with the overall order: s⁻¹ for first, L/(mol·s) for second order.
Confusing reaction order with molecularity.
The order is a measured property of the overall reaction; molecularity counts particles of one elementary step.
Treating k as independent of temperature.
k grows strongly with temperature; the Arrhenius equation describes this.
Exam context
- Determining the order from data tables, calculating k, concentration-time diagrams and the first-order half-life.
These mistakes cost points in real exams. The set drills them until they stick.
Formula cluster
Reaction kinetics
Connects concentration, time and temperature into the rate.
Worked example
Reaction first order in A and in B (k = 0.5 L·mol⁻¹·s⁻¹): at [A] = 0.2 mol/L and [B] = 0.1 mol/L, v = 0.5·0.2·0.1 = 0.01 mol/(L·s). Doubling [A] doubles v to 0.02 mol/(L·s).
Applications
Elucidating reaction mechanisms, sizing reactors, drug shelf life (first-order kinetics), radioactive decay, enzyme kinetics
Quanta exam set
Curated exam set for "Rate Law (Reaction Order)":
Question (front)
Which formula describes Rate Law (Reaction Order)?
Answer in your set
Question (front)
How do you rearrange v = k·[A]^m·[B]^n for Rate constant?
Answer in your set
Question (front)
Which common mistake happens with Rate Law (Reaction Order)?
Answer in your set
+ 7 more cards: units, variables, derivation, example, exam task
These 10 cards are ready. One click and they sit in your deck, FSRS schedules the reviews until exam day.
Scientific sources
Common notations & search queries
Related formulas
More Chemistry formulas
Frequently asked questions about Rate Law (Reaction Order)
How do you determine the reaction order experimentally?+
With the method of initial rates: you measure v for several runs in which only one concentration is changed while all others stay constant. Comparing two measurements yields the exponent: if v doubles when [A] doubles, m = 1; if it quadruples, m = 2; if v stays the same, m = 0. Formally m = lg(v₁/v₂)/lg([A]₁/[A]₂). Alternatively you test integrated rate laws: if ln[A] versus t is linear, the reaction is first order; if 1/[A] versus t is linear, second order. The order can never be read off the reaction equation, it is purely a measured result.
Why can you not read the exponents off the reaction equation?+
Because the reaction equation only describes the overall balance, not the pathway. Most reactions proceed through several elementary steps, and the rate is set by the slowest one. Its rate law contains only the particles involved in that step. A classic example is the reaction of hydrogen with bromine: the equation H₂ + Br₂ → 2 HBr suggests order 1 in each, but the measured rate law is fractional and complicated because a radical chain mechanism operates. Only for elementary reactions that really occur in a single collision do order and stoichiometric coefficients coincide. That is why, conversely, the measured rate law gives valuable clues about the mechanism.
What is the unit of the rate constant k?+
The unit of k depends on the overall order of the reaction, because the product k·[A]^m·[B]^n must always come out in mol/(L·s). For a zero-order reaction, k itself is in mol/(L·s). For first order one concentration cancels: k has the unit s⁻¹, which is convenient because k then relates directly to the half-life via t½ = ln 2/k. Second order requires L/(mol·s), third order L²/(mol²·s). Conversely, the unit of a given k immediately reveals the overall order; this is a popular exam trick and a good self-check when calculating.
What do zero, first and second order mean intuitively?+
For zero order the rate is constant and independent of concentration, for example when a catalyst surface or an enzyme is saturated: material is processed as fast as possible. For first order, v is proportional to the concentration of a single substance; radioactive decay and many decompositions follow this pattern with a constant half-life. For second order two particles must collide, either two identical ones (v = k·[A]²) or two different ones (v = k·[A]·[B]); here the half-life grows as the reaction proceeds. The order thus describes how strongly the reaction responds to concentration changes and reveals something about the underlying mechanism.
How are the rate law and the Arrhenius equation connected?+
The two equations split the description of the reaction rate between them. The rate law v = k·[A]^m·[B]^n describes, at fixed temperature, how v depends on the concentrations. The entire temperature dependence sits in the rate constant k, and that is exactly what the Arrhenius equation k = A·e^(−E_A/(R·T)) describes. If you raise the temperature, the orders m and n stay the same, but k grows exponentially, roughly doubling per 10 K as a rule of thumb. For complete problems you combine both: first determine k at the desired temperature via Arrhenius, then insert the concentrations into the rate law. This is how chemists cleanly separate concentration and temperature effects.
Retain Rate Law (Reaction Order) for exams
Create a curated FSRS exam set for v = k·[A]^m·[B]^n: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.
Free · curated formula set · LaTeX · FSRS spaced repetition
How do you calculate with Rate Law (Reaction Order)?
Here is how to work through a typical Rate Law (Reaction Order) (v = k·[A]^m·[B]^n) task step by step:
- 1
Task
Doubling [A] quadruples v, [B] changes nothing: what is the rate law?
Solution path
v = k·[A]²: from 2^m = 4 follows m = 2, from the unchanged rate follows n = 0. The overall order is 2.
- 2
Task
A reaction is second order in A. At [A] = 0.2 mol/L, v = 4.0×10⁻³ mol/(L·s). Calculate k.
Solution path
k = v/[A]² = 4.0×10⁻³/(0.2)² = 4.0×10⁻³/0.04 = 0.10 L/(mol·s). The unit follows from the overall order 2.