Order of reaction

Look at a balanced chemical equation and you might guess you can read off how fast the reaction runs by just glancing at the coefficients. You cannot. The order of a reaction is not a piece of arithmetic done on a balanced equation — it is a fact about the world, discovered only by running the reaction and measuring how its rate actually responds when concentrations are changed. Two reactions can have identical-looking balanced equations and completely different orders, because order reflects the slow, rate-determining step of a mechanism that the overall equation does not show. This article builds the definition of order from the experimentally found rate law, shows how individual orders combine into an overall order, and works through the classic contrast between a reaction where order happens to match stoichiometry and one where it strikingly does not — the hydrogen–bromine reaction, whose fractional order is one of the most quoted results in Class 12 kinetics.

The core insight

Order of reaction is the sum of the powers to which reactant concentrations are raised in the experimentally determined rate law, and it can only be found by experiment, never assumed from the balanced equation.

The core idea

Definition. For a reaction whose rate law has been measured to be

\text{Rate} = k[A]^{x}[B]^{y}, \qquad (1)

where: $k$ is the rate constant, $[A]$ and $[B]$ are molar concentrations of the reactants, and $x$ , $y$ are the powers found by experiment — the reaction is said to be of order $x$ with respect to $A$ , order $y$ with respect to $B$ , and the overall order is $x + y$ .

Order comes only from experiment. Never from the stoichiometric coefficients. Most reactions occur through a multi-step mechanism, and the observed rate law reflects only the slowest step of that mechanism, which the balanced equation gives no clue about.
Order can be zero, a positive integer, a fraction, or even negative with respect to an individual species. This syllabus restricts detailed integrated-equation work to zero and first order reactions, but the definition of order itself covers all these cases.
Zero order: rate does not change when that reactant's concentration is changed — for example, catalytic decomposition of $\text{NH}_3$ on a hot platinum surface at high pressure, where the surface is already saturated with reactant molecules and adding more makes no difference to the rate.
First order: rate is directly proportional to one reactant's concentration — radioactive decay and the decomposition of $\text{N}_2\text{O}_5$ are the standard examples.
Fractional or higher order: signals a multi-step mechanism; the worked example below shows exactly this.

Order versus stoichiometric coefficient. The stoichiometric coefficient is a bookkeeping number from mass balance — it tells you the mole ratio in which reactants combine and products form, with no built-in connection to how fast the reaction goes. Order is an empirical rate quantity. The two coincide only when a reaction happens to be elementary (a genuine single-step collision), which is the exception rather than the rule for the reactions tested at this level.

Figure

Two boxes comparing stoichiometric coefficient read from the balanced equation with order read from the experimental rate law, joined by a question mark to show they need not match

Derivation ACombining partial orders into the overall order

Given an experimentally established rate law of the form Rate equals k times concentration terms raised to individual powers for each reactant, show how the powers on each species combine to give the single number called the overall order of the reaction, and confirm this against a real measured rate law rather than an assumed one.

Step 1. Start from the general experimental rate law for a reaction involving reactants $A$ and $B$ : $\text{Rate} = k[A]^{x}[B]^{y}$ . Here $x$ and $y$ are not chosen in advance — they are the numbers that make the equation fit the measured initial-rate data, typically obtained by holding one concentration fixed and varying the other, then reading off how the rate scales.

Step 2. By definition, the reaction is order $x$ in $A$ (because doubling $[A]$ while $[B]$ is held fixed multiplies the rate by $2^{x}$ ) and order $y$ in $B$ (because doubling $[B]$ while $[A]$ is held fixed multiplies the rate by $2^{y}$ ). Each exponent is tied to how sensitively the rate responds to that one species alone.

Step 3. The overall order is defined as the sum of these individual powers: overall order $= x + y$ . This addition rule extends to any number of species appearing in the rate law — a reaction with three species in its rate law simply sums three exponents.

Step 4. Apply this to the measured rate law for $\text{H}_2(g) + \text{Br}_2(g) \rightarrow 2\text{HBr}(g)$ , which experiment shows to be $\text{Rate} = k[\text{H}_2][\text{Br}_2]^{1/2}$ . Here $x = 1$ (order 1 in $\text{H}_2$ ) and $y = \tfrac{1}{2}$ (order $\tfrac{1}{2}$ in $\text{Br}_2$ ), so overall order $= 1 + \tfrac{1}{2} = \tfrac{3}{2}$ , a fractional overall order that could never have been guessed from the stoichiometric coefficients of 1 and 1 in the balanced equation.

Worked example

Question. The decomposition of $\text{N}_2\text{O}_5(g) \rightarrow 2\text{NO}_2(g) + \tfrac{1}{2}\text{O}_2(g)$ obeys the experimentally determined rate law $\text{Rate} = k[\text{N}_2\text{O}_5]$ . (a) State the order with respect to $\text{N}_2\text{O}_5$ and the overall order. (b) Compare with the stoichiometric coefficient of $\text{N}_2\text{O}_5$ and comment on whether this match can be assumed in general. (c) For comparison, using the rate law $\text{Rate} = k[\text{H}_2][\text{Br}_2]^{1/2}$ for $\text{H}_2 + \text{Br}_2 \rightarrow 2\text{HBr}$ , state the order with respect to each reactant and the overall order.

Step 1. In the $\text{N}_2\text{O}_5$ rate law, $[\text{N}_2\text{O}_5]$ carries the power 1, so the reaction is first order with respect to $\text{N}_2\text{O}_5$ .

Step 2. Since $\text{N}_2\text{O}_5$ is the only species in the rate law, the overall order is also 1.

Step 3. The stoichiometric coefficient of $\text{N}_2\text{O}_5$ in the balanced equation is 1, which numerically matches the order found in Step 1 — but this match is a coincidence of this particular mechanism's slow step, not a rule; it was confirmed only by experiment and must never be assumed for a different reaction.

Step 4. For $\text{H}_2 + \text{Br}_2 \rightarrow 2\text{HBr}$ with $\text{Rate} = k[\text{H}_2][\text{Br}_2]^{1/2}$ : order with respect to $\text{H}_2$ is 1, order with respect to $\text{Br}_2$ is $\tfrac{1}{2}$ , and overall order $= 1 + \tfrac{1}{2} = \tfrac{3}{2}$ .

Final answer: the $\text{N}_2\text{O}_5$ decomposition is first order overall (order coincidentally equals its coefficient of 1), while the H2 + Br2 reaction is order 1 in H2, order 1/2 in Br2, and 3/2 overall — a fractional order that proves the reaction is not a single elementary step.

Competitive stretch

If a rate law were instead found to be $\text{Rate} = k[A]^2[B]^{-1}$ , the order with respect to $B$ is $-1$ (negative order — increasing $[B]$ actually slows the reaction, as happens when $B$ acts to consume a reactive intermediate), and the overall order is $2 + (-1) = 1$ , even though two species appear in the rate law. Overall order counts the algebraic sum of powers, not the number of species present.

Common traps

Reading order directly off the balanced equation's coefficients — the single most heavily tested mistake; order must come from the measured rate law.
Assuming order must be a whole number — fractional orders like $\tfrac{3}{2}$ are real and signal a multi-step mechanism, not an error in the experiment.
Confusing "order with respect to one species" with "overall order" — always state clearly which is being asked for, since a question may want either.
Thinking zero order means the reaction does not happen — it only means the rate is insensitive to that reactant's concentration; the reaction still proceeds, often limited by something else, such as a fixed number of catalyst surface sites.

Ask the tutor: if you want to practise deriving overall order from a given rate law, or want the reasoning for why a particular reaction turns out to have a fractional order, ask your AI tutor for more worked rate laws to analyse.

Why can't I just use the balanced equation to find the order?

Because the balanced equation only tells you the overall mass balance — how many moles of each species react and form — not the mechanism by which the reaction actually proceeds. Most reactions happen through several elementary steps, and the rate law is controlled entirely by the slowest of these steps, which is invisible in the overall balanced equation. That is why order must always be measured experimentally, for example by the initial rate method.

Is it possible for order to be negative?

Yes. A negative order with respect to a species means that increasing its concentration actually decreases the rate. This happens in some mechanisms where that species reacts with (and removes) a reactive intermediate that the main reaction pathway needs, so more of it means less rate. It is less common at this syllabus level but is a valid outcome of the general definition, as shown in the competitive stretch example above.

If order and molecularity are different ideas, how do I tell them apart in an exam answer?

Molecularity is a theoretical count — the number of reactant molecules that come together in one elementary step — and it is always a positive whole number (1, 2, or rarely 3) for that single step; it cannot be zero or fractional. Order is an experimental quantity for the overall reaction's rate law, and can be zero, fractional, or in special cases negative. Molecularity applies to a single elementary step; order applies to the whole observed reaction.

Order of reaction

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads