Rate Law & Order of Reaction

What the Rate Law Says

The rate of a chemical reaction at a given temperature depends on the concentration of one or more of its reactants. The mathematical statement of that dependence is the rate law — also called the rate equation or rate expression. NCERT defines it crisply: the rate law is the expression in which the reaction rate is given in terms of the molar concentration of reactants, with each term raised to some power which may or may not be the same as the stoichiometric coefficient of that species in the balanced equation.

Consider the general reaction with stoichiometric coefficients $a, b, c, d$:

$\ce{aA + bB -> cC + dD}$

Experiment shows that the rate is proportional to the concentrations of the reactants raised to certain powers:

$\text{Rate} \propto [\text{A}]^{x}\,[\text{B}]^{y} \qquad\Longrightarrow\qquad \text{Rate} = k\,[\text{A}]^{x}\,[\text{B}]^{y}$

Here $x$ and $y$ are the exponents that tell us how sensitive the rate is to a change in the concentration of A and B respectively, and $k$ is the proportionality constant called the rate constant. Written in differential form, the same statement reads:

$-\dfrac{d[\text{R}]}{dt} = k\,[\text{A}]^{x}\,[\text{B}]^{y}$

This is the differential rate equation. It is the most important sentence in chemical kinetics, because every other result — half-life, the shape of a concentration–time plot, the catalytic effect — flows from knowing the values of $x$, $y$ and $k$.

The Rate Constant k

The rate constant $k$ is the proportionality factor in the rate law. Set every reactant concentration equal to unity and the rate law collapses to $\text{Rate} = k$. In other words, $k$ is numerically the rate of the reaction when all concentrations are one molar. A large $k$ signals a fast reaction; a small $k$ signals a slow one.

Two properties of $k$ are repeatedly tested. First, $k$ is independent of concentration — it does not change as the reactants are consumed. Second, $k$ depends on temperature (and on the presence of a catalyst), rising sharply as temperature increases. The distinction between the rate of reaction and the rate constant is a favourite trap, summarised below.

Order of a Reaction, Defined

In the rate law $\text{Rate} = k\,[\text{A}]^{x}\,[\text{B}]^{y}$, the exponent $x$ is the order with respect to A and $y$ is the order with respect to B. The sum of these exponents,

$\text{overall order} = x + y$

is the overall order of the reaction. NCERT states it precisely: the sum of the powers of the concentrations of the reactants in the rate law expression is called the order of that chemical reaction. The order with respect to each individual species tells you how the rate scales with that one reactant; the overall order tells you the combined sensitivity.

A defining feature — and a frequent point of confusion with molecularity — is that order is an experimental quantity that can be zero, a whole number, or even a fraction. A zero order reaction is one whose rate is independent of the concentration of the reactant. The decomposition of ethanal is a textbook fractional-order case:

$\ce{CH3CHO(g) ->[723\,K] CH4(g) + CO(g)} \qquad \text{Rate} = k\,[\ce{CH3CHO}]^{3/2}$

an overall order of $1.5$. Worked through directly from a rate law, NCERT Example 3.3 gives two more: $\text{Rate} = k[\text{A}]^{1/2}[\text{B}]^{3/2}$ is second order ($\tfrac{1}{2}+\tfrac{3}{2}=2$), while $\text{Rate} = k[\text{A}]^{3/2}[\text{B}]^{-1}$ is half order ($\tfrac{3}{2}-1=\tfrac{1}{2}$). Negative exponents are permitted, and they mean the species actually retards the rate.

Exponents Are Not Coefficients

The single most important caution in this topic: the order of a reaction cannot be predicted by looking at the balanced equation. The exponents $x$ and $y$ in the rate law are determined experimentally, and they may or may not equal the stoichiometric coefficients $a$ and $b$. Sometimes they match; often they do not.

For the oxidation of nitric oxide, experiment gives a rate law whose exponents happen to coincide with the coefficients:

$\ce{2NO(g) + O2(g) -> 2NO2(g)} \qquad \text{Rate} = k\,[\ce{NO}]^{2}[\ce{O2}]$

Here the order is $2$ in NO and $1$ in $\ce{O2}$, overall third order, and the powers happen to equal the coefficients $2$ and $1$. But this is a coincidence, not a rule. Contrast two reactions where the exponents flatly contradict the stoichiometry:

The reason exponents diverge from coefficients is that a balanced equation rarely shows how a reaction actually proceeds. Most reactions occur through a sequence of elementary steps (a mechanism), and the overall rate is governed by the slowest of those steps — the rate determining step. The apparent tenth-order $\ce{KClO3}$ reaction above is in reality second order precisely because it unfolds in several steps. The molecularity of that slowest step equals the observed order of the overall reaction, which is why mechanism, not stoichiometry, dictates the rate law.

Units of k for Each Order

Because the rate always carries the units $\text{mol L}^{-1}\,\text{s}^{-1}$ but the concentration product is raised to the overall order $n$, the units of the rate constant change with order. Rearranging the rate law for a reaction of overall order $n$:

$k = \dfrac{\text{Rate}}{[\text{A}]^{x}[\text{B}]^{y}} = \dfrac{(\text{concentration})\,(\text{time})^{-1}}{(\text{concentration})^{n}} = (\text{mol L}^{-1})^{1-n}\,\text{s}^{-1}$

This single formula, $k = (\text{mol L}^{-1})^{1-n}\,\text{s}^{-1}$, generates every entry in the table below. It is worth committing to memory because NEET routinely supplies a value of $k$ with units and asks you to read off the order — exactly NCERT Example 3.4.

Two reverse-reading examples from NCERT make the point. A rate constant $k = 2.3\times10^{-5}\ \text{L mol}^{-1}\,\text{s}^{-1}$ carries the second-order units and so identifies a second order reaction. A rate constant $k = 3\times10^{-4}\ \text{s}^{-1}$ carries first-order units and so identifies a first order reaction. The units alone betray the order — no rate law needed.

Finding Order from Initial Rates

The cleanest experimental route to the rate law is the initial-rate method. The principle is simple: hold the concentration of every reactant constant except one, change that one by a known factor, and watch how the initial rate responds. The ratio of rates reveals the exponent for that species. Repeating the procedure for each reactant builds up the complete rate law.

NCERT applies this exact reasoning to the formation of $\ce{NO2}$ from the data in Table 3.2:

Comparing Experiments 1 and 2, $[\ce{O2}]$ is fixed while $[\ce{NO}]$ is doubled; the rate jumps four-fold ($0.096 \to 0.384$), so the rate depends on the square of $[\ce{NO}]$ — second order in NO. Comparing Experiments 1 and 3, $[\ce{NO}]$ is fixed while $[\ce{O2}]$ is doubled; the rate exactly doubles ($0.096 \to 0.192$) — first order in $\ce{O2}$. The complete rate law is therefore $\text{Rate} = k[\ce{NO}]^{2}[\ce{O2}]$, overall third order. Experiment 4 is a useful check: doubling both reactants multiplies the rate by $2^2 \times 2^1 = 8$, and indeed $0.096 \times 8 = 0.768$.

Worked Example

The procedure becomes mechanical once the rate ratios are set up as powers. The following is modelled on the NCERT exercise data for the reaction $\ce{2A + B -> C + D}$.

Notice the discipline of the method: never compare two experiments in which more than one concentration has changed unless you have already pinned down the other exponents. Pairing experiments so that exactly one reactant varies is what isolates each exponent cleanly.

Traps & Quick Recap

Two further confusions reliably cost marks. The first is mistaking the order in one reactant for the overall order; the second is forgetting that a reactant of order zero — such as water in the hydrolysis of ethyl acetate — drops out of the rate law entirely even though it appears in the equation.