Rate Laws and Reaction Order

A pinch of powdered zinc reacts with acid in a flash. A zinc nail fizzes for minutes. Same reactants, same temperature, wildly different speeds. The missing variable is concentration. The precise way rate depends on concentration is called the rate law. Rate laws are the grammar of kinetics. Once you know how rate scales with each reactant's concentration, you can predict how a reaction responds to dilution. You can design a process that avoids a dangerous rate spike. You can also diagnose a reaction mechanism from data alone. This article defines the differential rate law, the rate constant $k$ , and reaction order. You will learn how order in each species combines into an overall order. You will see how the units of $k$ change with order. Later articles build on this to integrate these laws over time. They also show how to extract orders from real data.

Key takeaways

The rate law expresses reaction rate as a function of concentrations, typically $v = k[\mathrm{A}]^m[\mathrm{B}]^n$ .
Orders $m$ , $n$ are experimentally determined exponents, not necessarily equal to stoichiometric coefficients.
Overall order is the sum $m+n+\cdots$ ; it sets the units of $k$ .
The rate constant $k$ is temperature-dependent but concentration-independent.
Orders can be zero, positive integers, negative, or fractional, depending on mechanism.
Key idea: the rate law is an empirical fingerprint of concentration dependence, discovered by experiment, not read off the balanced equation.

The core insight

A rate law is an experimentally measured equation, $v = k[\mathrm{A}]^m[\mathrm{B}]^n\cdots$ , that states how the instantaneous rate of a reaction depends on the concentrations of species present; the exponents (orders) and the rate constant $k$ must be determined from data, since the stoichiometric coefficients in the balanced chemical equation do not, in general, dictate them.

The physical picture

Imagine reactant molecules as a crowd of dancers on a floor, colliding and occasionally reacting when they meet with the right energy and orientation. The more dancers per unit floor area — the higher the concentration — the more collisions per second, and the faster the reaction proceeds. This is the intuitive root of a rate law: rate tracks how "crowded" the reacting mixture is. But crowding is not always simple. If a reaction requires two molecules of A to collide with one molecule of B, you might guess the rate depends on $[\mathrm{A}]^2[\mathrm{B}]$ . Sometimes it does. Often it does not, because the balanced equation only describes overall mass balance, not the sequence of elementary steps that actually control the speed. A slow, rate-determining step buried in the mechanism might use only one A molecule, making the reaction first order in A even though the balanced equation shows a coefficient of 2. The differential rate law captures this dependence at a single instant: it relates the instantaneous rate, defined via the extent of reaction as introduced with rate definitions, to the concentrations present at that moment. Because concentrations change as reaction proceeds, the rate itself evolves — that evolution is described later by integrated rate laws, but the differential rate law is the starting relationship from which everything else follows.

Figure

Rate versus concentration for orders 0, 1, and 2 in a single reactant A

Figure

Overall order as the sum of individual orders, worked for v = k times A squared times B to the first power

The core relations

v = k\,[\mathrm{A}]^m[\mathrm{B}]^n\cdots

where $v$ is the instantaneous rate, $k$ the rate constant, and $m,n$ the experimentally determined orders in A and B.

n_{\text{overall}} = m + n + \cdots

where $n_{\text{overall}}$ is the overall reaction order, the sum of all individual orders.

[k] = \left(\text{mol}\,\text{dm}^{-3}\right)^{1-n_{\text{overall}}}\,\text{s}^{-1}

where $[k]$ denotes the units of the rate constant, fixed once the overall order and the concentration/time units are chosen.

v = -\frac{1}{\nu_i}\frac{d[i]}{dt} \quad (\text{reactant } i), \qquad v = \frac{1}{\nu_j}\frac{d[j]}{dt} \quad (\text{product } j)

where $\nu_i$ , $\nu_j$ are stoichiometric numbers, connecting the rate law back to the general rate definition.

Derivation AWhy the units of k depend on overall order

Goal. Show that the units of the rate constant $k$ are forced by dimensional consistency once the overall order of the rate law is fixed, and derive the general formula for those units.

Step 1. Start from the general rate law for a reaction with overall order $n_{\text{overall}} = m+n+\cdots$ ,

v = k\,[\mathrm{A}]^m[\mathrm{B}]^n\cdots

The instantaneous rate $v$ always carries units of concentration per time, typically mol dm⁻³ s⁻¹, regardless of the reaction's order, because it is defined as a derivative of concentration with respect to time (with a stoichiometric number as a dimensionless prefactor).

Step 2. Write the units explicitly, using square brackets for "units of."

[v] = [k]\cdot[\mathrm{A}]^{[m]}\cdot[\mathrm{B}]^{[n]}

Substituting concentration units (mol dm⁻³) for each bracketed species and mol dm⁻³ s⁻¹ for $[v]$ gives

\text{mol dm}^{-3}\,\text{s}^{-1} = [k]\cdot(\text{mol dm}^{-3})^{m}\cdot(\text{mol dm}^{-3})^{n}

Step 3. Combine the exponents on the right using the sum rule for powers, since both A and B carry the same base unit mol dm⁻³.

\text{mol dm}^{-3}\,\text{s}^{-1} = [k]\cdot(\text{mol dm}^{-3})^{m+n}= [k]\cdot(\text{mol dm}^{-3})^{n_{\text{overall}}}

Step 4. Solve for $[k]$ by dividing both sides by $(\text{mol dm}^{-3})^{n_{\text{overall}}}$ .

[k] = \frac{\text{mol dm}^{-3}\,\text{s}^{-1}}{(\text{mol dm}^{-3})^{n_{\text{overall}}}} = (\text{mol dm}^{-3})^{1-n_{\text{overall}}}\,\text{s}^{-1}

Step 5. Check the two most common cases. For $n_{\text{overall}}=1$ (first order), the exponent $1-n_{\text{overall}}=0$ , so $[k]=\text{s}^{-1}$ — a pure frequency, consistent with a rate constant describing a per-molecule decay probability per second. For $n_{\text{overall}}=2$ (second order), $1-n_{\text{overall}}=-1$ , giving $[k]=\text{mol}^{-1}\,\text{dm}^{3}\,\text{s}^{-1}$ , matching the familiar bimolecular rate constant units.

\boxed{[k] = (\text{mol dm}^{-3})^{1-n_{\text{overall}}}\text{s}^{-1}}

Worked examples

Example 1 — Foundational: reading orders and units directly from a rate law. Experiments on the gas-phase reaction

2\,\mathrm{NO}(g) + \mathrm{O_2}(g) \rightarrow 2\,\mathrm{NO_2}(g)

give the empirical rate law

v = k\,[\mathrm{NO}]^2[\mathrm{O_2}]

Determine (a) the order in NO, (b) the order in O₂, (c) the overall order, (d) the units of $k$ if concentrations are in mol dm⁻³ and time in seconds, and (e) the new rate if $[\mathrm{NO}]$ is tripled at constant $[\mathrm{O_2}]$ . (a) The exponent on $[\mathrm{NO}]$ is 2, so the reaction is second order in NO. Notably, this matches the stoichiometric coefficient here — a coincidence that happens for some elementary-like steps, not a general rule. (b) The exponent on $[\mathrm{O_2}]$ is 1, so the reaction is first order in O₂. (c) Overall order:

n_{\text{overall}} = 2 + 1 = 3

This is a third-order reaction overall. (d) Using the boxed result from the derivation with $n_{\text{overall}}=3$ :

[k] = (\text{mol dm}^{-3})^{1-3}\,\text{s}^{-1} = \text{mol}^{-2}\,\text{dm}^{6}\,\text{s}^{-1}

(e) Tripling $[\mathrm{NO}]$ while holding $[\mathrm{O_2}]$ fixed scales the rate by $3^2=9$ , since NO enters the rate law squared.

v_{\text{new}} = k(3[\mathrm{NO}])^2[\mathrm{O_2}] = 9\,k[\mathrm{NO}]^2[\mathrm{O_2}] = 9\,v_{\text{old}}

Sanity check: because O₂'s concentration did not change, only the NO-dependent factor changes, and squaring a threefold increase correctly gives a ninefold rate increase — consistent with second-order behavior in NO.

Example 2 — Applied: extracting orders from paired experiments (method of initial rates). The reaction

\mathrm{BrO_3^-} + 5\,\mathrm{Br^-} + 6\,\mathrm{H^+} \rightarrow 3\,\mathrm{Br_2} + 3\,\mathrm{H_2O}

was studied at 25 °C. Three experiments gave initial rates:

Run	[BrO₃⁻] / mol dm⁻³	[Br⁻] / mol dm⁻³	[H⁺] / mol dm⁻³	$v_0$ / mol dm⁻³ s⁻¹
1	0.10	0.10	0.10	$8.0\times10^{-4}$
2	0.20	0.10	0.10	$1.6\times10^{-3}$
3	0.10	0.30	0.10	$2.4\times10^{-3}$
4	0.10	0.10	0.20	$3.2\times10^{-3}$

Find the rate law and $k$ .

Order in BrO₃⁻. Compare runs 1 and 2: $[\mathrm{BrO_3^-}]$ doubles, other concentrations fixed.

\frac{v_2}{v_1} = \frac{1.6\times10^{-3}}{8.0\times10^{-4}} = 2.0 = \left(\frac{0.20}{0.10}\right)^{m} = 2^{m}

So $m=1$ : first order in BrO₃⁻.

Order in Br⁻. Compare runs 1 and 3: $[\mathrm{Br^-}]$ triples.

\frac{v_3}{v_1} = \frac{2.4\times10^{-3}}{8.0\times10^{-4}} = 3.0 = 3^{n}

So $n=1$ : first order in Br⁻.

Order in H⁺. Compare runs 1 and 4: $[\mathrm{H^+}]$ doubles.

\frac{v_4}{v_1} = \frac{3.2\times10^{-3}}{8.0\times10^{-4}} = 4.0 = 2^{p}

So $p=2$ : second order in H⁺.

Assemble the rate law and overall order.

v = k[\mathrm{BrO_3^-}][\mathrm{Br^-}][\mathrm{H^+}]^2, \qquad n_{\text{overall}} = 1+1+2 = 4

Solve for $k$ using run 1.

k = \frac{v_1}{[\mathrm{BrO_3^-}][\mathrm{Br^-}][\mathrm{H^+}]^2} = \frac{8.0\times10^{-4}}{(0.10)(0.10)(0.10)^2}

k = \frac{8.0\times10^{-4}}{1.0\times10^{-4}} = 8.0 \ \text{mol}^{-3}\,\text{dm}^{9}\,\text{s}^{-1}

Sanity check: with $n_{\text{overall}}=4$ , the boxed unit formula predicts $[k]=(\text{mol dm}^{-3})^{1-4}\text{s}^{-1}=\text{mol}^{-3}\text{dm}^9\text{s}^{-1}$ , matching exactly. As a further check, plug $k=8.0$ into run 4's concentrations: $8.0\times(0.10)(0.10)(0.20)^2 = 8.0\times4.0\times10^{-4}=3.2\times10^{-3}$ mol dm⁻³ s⁻¹, matching the tabulated value.

Limiting cases & common traps

Situation	What to remember
Order equals stoichiometric coefficient	Can happen, especially for elementary steps, but must still be verified experimentally
Zero order in a species	Rate is unaffected by that species' concentration over the range studied; often signals a catalyst or surface saturation
Negative order	Increasing that species' concentration slows the reaction, common when it appears in a pre-equilibrium as an inhibitor
Fractional order	Signals a multistep mechanism, often with a chain or radical process
Confusing order with molecularity	Molecularity describes the number of species colliding in a single elementary step; order is empirical and applies to overall (often multistep) reactions

The chemist behind it

The systematic idea that reaction rate depends on concentration through measurable, sometimes non-integer exponents grew out of late nineteenth-century physical chemistry, when chemists began treating kinetics as a quantitative science rather than a qualitative observation. Wilhelmy's 1850 work on sucrose inversion gave one of the first rate expressions, but it was the generation of van 't Hoff and Ostwald that formalized "order" as a concept distinct from stoichiometry, and applied it broadly across gas-phase and solution reactions.

Origins · 1 minJacobus Henricus van 't Hoff and how this was discovered

Jacobus Henricus van 't Hoff

1852-1911 · pioneered the mathematical treatment of reaction rates and orders, work recognized by the first Nobel Prize in Chemistry in 1901

Van 't Hoff's 1884 treatise Études de Dynamique Chimique classified reactions by their kinetic order and showed how to extract that order from concentration-time data, laying groundwork that every kinetics course still follows today.

:::quiz[A reaction has rate law v = k[A][B]^2. What is the overall order?]

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:::quiz[Doubling [A] in a reaction leaves the rate unchanged. What is the order in A?]

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Negative two

Ask the tutor: Want to see how this differential rate law gets integrated to predict concentration versus time for a first-order reaction?

Is the rate law the same as the rate of reaction?

No. The rate of reaction, $v$ , is the instantaneous speed at which reactants disappear or products appear, defined through the extent of reaction and stoichiometric numbers. The rate law is the mathematical expression, $v = k[\mathrm{A}]^m[\mathrm{B}]^n$ , that tells you how that rate depends on concentrations. The rate is a single measurable number at any instant; the rate law is the underlying functional relationship that predicts it across all concentration conditions.

Can reaction order be negative or a fraction?

Yes, both occur. A negative order means increasing that species' concentration slows the reaction, which often happens when the species is a product that inhibits a reverse or competing step. A fractional order, such as three-halves, usually signals a multistep mechanism, frequently involving a pre-equilibrium or a radical chain process where an intermediate concentration itself depends on the square root of a reactant.

Why doesn't the balanced chemical equation tell you the rate law?

The balanced equation only enforces conservation of mass and charge; it says nothing about the sequence of molecular collisions that actually occurs. Most reactions proceed through several elementary steps, and the observed rate law reflects primarily the slowest, rate-determining step among them. Only for genuinely elementary reactions, where the written equation is itself one molecular event, do stoichiometric coefficients and rate-law exponents coincide.

How do units of k help me check my rate law?

Once you propose an overall order, the units of $k$ are fixed by dimensional analysis, since rate always carries units of concentration per time. If your calculated $k$ from experimental data doesn't have the units predicted by the formula for your assumed order, you've likely misidentified the order, made an arithmetic slip, or mismatched concentration and time units somewhere in the calculation.

What comes after determining the rate law?

Once you know the differential rate law, you can integrate it over time to predict how concentration evolves — giving the integrated rate laws for zero, first, and second order reactions. You can also use systematic techniques, such as the method of initial rates or integrated-law fitting, formalized under methods for determining reaction order, to extract $m$ , $n$ , and $k$ from real experimental datasets.