Rate law

If the Law of Mass Action is the conceptual seed, the rate law is the fully grown, experimentally verified equation connecting reaction rate to reactant concentrations — the equation every kinetics numerical in this chapter ultimately relies on. Rate law appears in Unit 3 (Chemical Kinetics) of ISC Class 12 Physical Chemistry, right after the Law of Mass Action, under "effect of concentration on rate — qualitative treatment based on the Law of Mass Action; general rate law Rate equals k times concentration raised to n." You must be able to write the general rate law, identify the rate constant $k$ and the order $n$ , and grasp the exam-defining distinction: unlike the Law of Mass Action applied to a single elementary step, the rate law of an overall multi-step reaction is never assumed from the balanced equation — it is always determined by experiment.

The core insight

The rate law, $\text{Rate}=k[A]^{x}[B]^{y}$ , gives the experimentally measured dependence of rate on concentration, and its exponents equal the balanced-equation coefficients only when the reaction is a single elementary step.

The core idea

The rate law (rate equation) expresses the rate of a reaction as a function of the molar concentrations of the reactants, each raised to a power found by experiment:

\text{Rate} = k[A]^{x}[B]^{y}\ldots

where: $k$ is the rate constant, a proportionality constant fixed for a given reaction at a given temperature (its numerical value and units depend on the overall order); $x, y, \ldots$ are the individual orders, the powers to which each reactant's concentration is raised; and the sum $x+y+\ldots$ is the overall order of the reaction.

Why exponents are not stoichiometric coefficients. For an elementary (single-step) reaction, the Law of Mass Action guarantees that the rate-law exponents equal the balanced-equation coefficients. But most reactions proceed through a sequence of elementary steps — a mechanism — and the observed rate law reflects only the slowest, rate-determining step. So the measured $x, y$ need not match the coefficients $a, b$ of the overall equation; they must come from data, never from inspection of the balanced equation.

How chemists find it: the initial rate method. Run the reaction several times, changing the starting concentration of one reactant while holding the others fixed, and measure the initial rate each time. Comparing how the rate scales when one concentration is varied in isolation reveals that reactant's order.

Units of $k$ . Since $\text{Rate}$ always carries units $\text{mol L}^{-1}\text{s}^{-1}$ and $[\text{concentration}]^n$ carries units $(\text{mol L}^{-1})^n$ , rearranging $k = \text{Rate}/[\text{concentration}]^n$ shows the units of $k$ must be $(\text{mol L}^{-1})^{1-n}\text{s}^{-1}$ — they change with overall order $n$ , so they must be re-derived each time, never memorised as one fixed unit.

Figure

A schematic showing three experiments where doubling concentration of A doubles the rate while doubling concentration of B quadruples the rate, revealing first order in A and second order in B

Derivation AExtracting an order from initial-rate data

The goal is a general algebraic rule for pulling an exponent out of two experiments in which only one reactant's concentration has changed, so that "rate doubles when concentration doubles" or "rate quadruples when concentration doubles" can be converted into a definite number for $x$ rather than eyeballed, and so the same rule works even when the scaling factor is not a clean integer.

Step 1. Write the rate law for two experiments that differ only in $[A]$ , with $[B]$ held fixed: $\text{Rate}_1 = k[A]_1^{x}[B]^{y}$ and $\text{Rate}_2 = k[A]_2^{x}[B]^{y}$ .

Step 2. Divide the second equation by the first. The constants $k$ and $[B]^y$ are identical in both experiments, so they cancel exactly, leaving $\dfrac{\text{Rate}_2}{\text{Rate}_1} = \left(\dfrac{[A]_2}{[A]_1}\right)^{x}$ .

Step 3. Take logarithms of both sides to bring the exponent down: $x = \dfrac{\log(\text{Rate}_2/\text{Rate}_1)}{\log([A]_2/[A]_1)}$ . This is the general formula; when the concentration ratio is a clean $2$ and the rate ratio is a power of $2$ , $x$ can be read off by inspection (rate doubles for a doubled concentration means $x=1$ , quadruples means $x=2$ ), but the logarithmic form handles any ratio.

Step 4. Repeat the same division-and-log procedure holding $[A]$ fixed and varying $[B]$ across a different pair of experiments to get $y$ , then add the exponents to get the overall order $n = x+y$ . Result: $x = \log(\text{Rate}_2/\text{Rate}_1)\big/\log([A]_2/[A]_1)$ , the general order-extraction formula behind every initial-rate table.

Worked example

Question. For $A + B \rightarrow \text{Products}$ , the following initial-rate data were obtained at constant temperature:

Expt	$[A]_0$ (mol L $^{-1}$ )	$[B]_0$ (mol L $^{-1}$ )	Initial rate (mol L $^{-1}$ s $^{-1}$ )
1	0.10	0.10	$2.0\times10^{-3}$
2	0.20	0.10	$4.0\times10^{-3}$
3	0.20	0.20	$16.0\times10^{-3}$

Determine the rate law and evaluate $k$ .

Step 1. Compare Expt 1 and 2: only $[A]$ doubles, $[B]$ is constant. The rate goes from $2.0\times10^{-3}$ to $4.0\times10^{-3}$ — it doubles. Since $2^{x}=2$ , $x=1$ : first order in $A$ .

Step 2. Compare Expt 2 and 3: only $[B]$ doubles, $[A]$ is constant. The rate goes from $4.0\times10^{-3}$ to $16.0\times10^{-3}$ — it becomes $4$ times larger. Since $2^{y}=4$ , $y=2$ : second order in $B$ .

Step 3. The rate law is $\text{Rate}=k[A]^{1}[B]^{2}$ , with overall order $x+y=1+2=3$ (third order).

Step 4. Substitute Expt 1's values to find $k$ : $2.0\times10^{-3}=k(0.10)(0.10)^{2}=k\times1.0\times10^{-3}$ , giving $k=2.0\ \text{mol}^{-2}\text{L}^{2}\text{s}^{-1}$ .

Competitive stretch

For $2A \rightarrow B+C$ , doubling $[A]$ is found to double the rate, even though the balanced equation carries a coefficient of $2$ for $A$ . The experimental scaling gives $\text{Rate}\propto[A]^{1}$ , so the correct rate law is $\text{Rate}=k[A]$ — first order overall, not second order as the coefficient might naively suggest. The mismatch arises because the balanced equation only records overall mass balance, not the mechanism; the reaction very likely proceeds through more than one elementary step, and the observed first-order law reflects only the slowest step, which may consume just one molecule of $A$ even though two are used up by the time the whole mechanism completes. The rule this reinforces: order is always read from data, never from stoichiometric coefficients.

Common traps

Assuming rate-law exponents automatically equal the stoichiometric coefficients of the balanced equation — true only for elementary (single-step) reactions.
Forgetting that the rate law is found by experiment, typically the initial rate method, not derived theoretically from the equation alone.
Confusing "order with respect to a reactant" (an individual exponent like $x$ ) with "overall order" (the sum $x+y+\ldots$ ).
Quoting a fixed unit for $k$ instead of re-deriving it from $k=\text{Rate}/[\text{conc}]^{n}$ , since the units change with the overall order $n$ .

Ask the tutor: if a fresh initial-rate table has you stuck on which pair of experiments to compare, ask your AI tutor to generate one and walk through isolating each order step by step.

Why can't I just use the coefficients from the balanced equation as the exponents in the rate law?

Because the balanced equation only guarantees mass balance — the same number of each atom on both sides — not the actual step-by-step mechanism. That shortcut only works for an elementary reaction, one that genuinely happens in a single step, where the Law of Mass Action applies directly. Most reactions you meet in this chapter proceed through several elementary steps, and the observed rate law reflects only the slowest of those steps. So you must always get $x$ and $y$ from experimental data, such as an initial-rate table, never by inspection of the overall equation.

How is "order with respect to a reactant" different from "overall order"?

"Order with respect to a reactant" is the individual exponent on that reactant's concentration in the rate law — for example, $x$ for $A$ in $\text{Rate}=k[A]^{x}[B]^{y}$ . "Overall order" is the sum of all these individual exponents, $x+y+\ldots$ , for the whole rate law. In the worked example, the reaction is first order in $A$ , second order in $B$ , and third order overall. Exam questions often ask for both, so always report which one you are giving.

Why do the units of the rate constant k keep changing between problems?

Because $k$ absorbs whatever units are needed to make $\text{Rate}=k[\text{concentration}]^{n}$ balance dimensionally, and that depends on the overall order $n$ , which differs from reaction to reaction. Rate is always in $\text{mol L}^{-1}\text{s}^{-1}$ , while $[\text{concentration}]^{n}$ is in $(\text{mol L}^{-1})^{n}$ ; dividing gives $k$ the units $(\text{mol L}^{-1})^{1-n}\text{s}^{-1}$ . Rather than memorising a table of unit patterns for each order, just re-derive them from this ratio every time — it takes seconds and avoids a common slip.

Rate law

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads