If a rate law matches the stoichiometry, is the reaction elementary?

Not necessarily, and assuming so is a standard trap. $\mathrm{2NO + O_2 \rightarrow 2NO_2}$ is third order overall, rate $= k[\mathrm{NO}]^2[\mathrm{O_2}]$, which looks exactly like a termolecular step. Termolecular steps are rare enough to be almost never the answer, and the accepted mechanism is two bimolecular steps: a fast equilibrium $\mathrm{NO + NO \rightleftharpoons N_2O_2}$ followed by a slow $\mathrm{N_2O_2 + O_2}$. Substituting $[\mathrm{N_2O_2}] = K[\mathrm{NO}]^2$ into the slow step reproduces the observed law exactly. A rate law is consistent with a mechanism and never proof of one, which is why questions ask what a rate law reveals rather than what it establishes.

Chapter 8 of 18

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JEE Main Chemistry★ ~1 question (4 marks) in JEE Main, almost always numerical, and among the most formulaic marks in the paper once the integrated laws are secure

Chemical Kinetics

Thermodynamics says whether a reaction can happen; kinetics says how fast. The rate law cannot be read off the balanced equation — it must be measured — and temperature works through the exponentially small fraction of molecules that clear the activation barrier.

⏱ 25 min readMedium difficulty✓ Free · exam-pattern aligned

Timed practiceMCQs from this chapter with instant scoring

FlashcardsFormulas & traps on spaced repetition

JEE Main PYQsSolved previous-year questions

By the end of this chapter you'll be able to…

1Define average, instantaneous and initial rate with stoichiometric coefficients divided out, and list what changes a rate
2Distinguish order from molecularity, explain why order is experimental, and interpret fractional and negative orders
3Deduce a rate law from initial rate data, read the order off the units of $k$ , and test a proposed mechanism against an observed rate law
4Apply the zero- and first-order integrated laws and their half-lives, and identify order from successive half-lives or from which plot is linear
5Recognise pseudo-order conditions and recover the true rate constant from an observed one
6Apply the Arrhenius equation in its two-point form, explain temperature dependence through the Maxwell-Boltzmann tail, and quantify the effect of a catalyst

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Why this chapter matters in JEE Main

Thermodynamics says whether a reaction can happen; kinetics says how fast. Two facts run the chapter and both are easy to state and easy to get wrong. The rate law is measured and cannot be read off the balanced equation, which is why order and molecularity are different things and why order can be zero, fractional or negative. And temperature acts through the exponentially small fraction of molecules above the activation barrier, not through mean speed, which is why a 10 K rise can double a rate and why a catalyst that removes 67 kJ per mole speeds a reaction by a factor of ten to the eleven. A half-life is a constant only for first order, and three reactions sharing a starting concentration and a first half-life are in three different places 30 minutes later.

Before you start — revise these

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Logarithms and exponentials, and reading a straight-line plot

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Equilibrium and the reaction quotient, for fast pre-equilibrium mechanisms

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Enthalpy and Gibbs energy from Chemical Thermodynamics, to see what a catalyst leaves untouched

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Concentration units and stoichiometry from the mole concept

Chemical Kinetics

Three reactions. Each starts at exactly 1.00 M. Each has a half-life of exactly 10 minutes.

Where is each one after 30 minutes?

The obvious answer is that they are all in the same place: three half-lives, so 0.125 M each.

That is right for exactly one of them.

Order	After 10 min	After 20 min	After 30 min
Zero	0.500 M	0 M, finished	already over
First	0.500 M	0.250 M	0.125 M
Second	0.500 M	0.250 M	0.250 M

Identical starting point, identical first half-life, and by thirty minutes one reaction has been finished for ten minutes, one has twice as much left as another, and only the first-order one behaved the way "three half-lives" suggests.

The reason is that a half-life is only a constant for a first-order reaction.

$t_{1/2} = \frac{[ A ] _{0}}{2 k} (zero), \frac{0.693}{k} (first), \frac{1}{k [ A ] _{0}} (second)$

Zero order has half-lives that halve each time, so the reaction ends abruptly. Second order has half-lives that double each time, so it drags on for ever. Only first order repeats.

That single comparison sets up the whole chapter, because it shows that the order is not an accounting label. It is a statement about the shape of the future, and it can only be found by measurement.

The two facts	What they explain
The rate law is measured, never read off the balanced equation	Why order and molecularity are different things
Temperature acts through the exponentially small fraction above the barrier	Why 10 K can double a rate, and why a catalyst matters so much

Thermodynamics told you whether a reaction can go. It said nothing about when: hydrogen and oxygen at room temperature have an enormously negative $Δ G$ for forming water and sit unchanged for centuries.

1. Rate of Reaction

Because stoichiometry makes different species change at different speeds, rate is defined with the coefficients divided out.

For $a A + b B \to c C + d D$ ,

$rate = - \frac{1}{a} \frac{d [ A ]}{d t} = - \frac{1}{b} \frac{d [ B ]}{d t} = + \frac{1}{c} \frac{d [ C ]}{d t}$

The minus signs make the rate positive for reactants, whose concentrations fall. Defined this way a single number describes the reaction whichever species you watched.

Average rate is measured over a finite interval. Instantaneous rate is the limit as that interval shrinks, the slope of the tangent to the concentration-time curve. Initial rate, taken at the moment of mixing, is the most useful in practice, because no products have yet accumulated to complicate matters.

What changes the rate

Concentration, temperature, catalyst, surface area and, for gases, pressure. Light matters for photochemical reactions, and the nature of the reactants matters most: ionic reactions in solution are essentially instantaneous, while covalent rearrangements can take hours.

Illustration 1

In $N_{2} + 3 H_{2} \to 2N H_{3}$ , hydrogen is consumed at $0.060 mol L^{- 1} s^{- 1}$ . Find the rate of reaction and the rate of change of each of the other two species.

The rate of reaction is defined so that it comes out the same whichever species is watched, which means dividing each rate of change by that species' coefficient:

$rate = - \frac{d [ N _{2} ]}{d t} = - \frac{1}{3} \frac{d [ H _{2} ]}{d t} = + \frac{1}{2} \frac{d [ N H _{3} ]}{d t}$

$rate = \frac{0.060}{3} = 0.020 mol L^{- 1} s^{- 1}$

From that single number the others follow:

$- \frac{d [ N _{2} ]}{d t} = 0.020, + \frac{d [ N H _{3} ]}{d t} = 2 (0.020) = 0.040 mol L^{- 1} s^{- 1}$

Three different numbers, one rate. Reporting 0.060 as "the rate of the reaction" is the standard error, and it is off by exactly the coefficient of whichever species happened to be measured. Ammonia appears twice as fast as nitrogen disappears, which is stoichiometry showing up in the timing.

2. Rate Law, Order and Molecularity

$rate = k [A]^{x} [B]^{y}$

The exponents are found by experiment and have no necessary connection to the stoichiometric coefficients. This is the single most important sentence in the chapter.

The order is $x + y$ . It may be zero, fractional or negative, none of which a coefficient could ever be.

Order against molecularity

Order	Molecularity
Experimental	Theoretical
Applies to overall reactions	Applies to elementary steps only
Can be zero, fractional or negative	Always a positive whole number
Determined from the rate law	Determined by counting colliding species

For an elementary reaction, happening in one step, order and molecularity coincide. For a complex reaction they need not, because of the rate-determining step: a multi-step reaction is no faster than its slowest step, so the observed rate law reflects that step and whatever precedes it, not the overall equation.

Molecularity above three is never observed, because four particles colliding simultaneously with the right energy and orientation is vanishingly improbable.

Illustration 2

The reaction $2NO + O_{2} \to 2N O_{2}$ is found to be third order overall, rate $= k [NO]^{2} [O_{2}]$ . Here the exponents happen to match the coefficients. Does that make it termolecular?

It does not, and assuming so would be the trap.

A termolecular elementary step requires three molecules to meet at one instant, which is rare enough to be almost never the answer. The accepted mechanism is two bimolecular steps.

$NO + NO ⇌ N_{2} O_{2} (fast equilibrium)$

$N_{2} O_{2} + O_{2} \to 2N O_{2} (slow)$

The slow step fixes the rate, so rate $= k_{2} [N_{2} O_{2}] [O_{2}]$ . The fast equilibrium gives $[N_{2} O_{2}] = K [NO]^{2}$ , and substituting yields

$rate = k_{2} K [NO]^{2} [O_{2}]$

which is exactly the observed third-order law, with $k = k_{2} K$ .

So the rate law matched the stoichiometry by coincidence, through a two-step mechanism with no termolecular step anywhere. A rate law is consistent with a mechanism; it never proves one.

Illustration 3

Two measured rate laws:

$rate = k [C H_{3} CHO]^{3/2} and rate = k \frac{[ O _{3} ] ^{2}}{[ O _{2} ]}$

What do the exponents $3/2$ and $- 1$ mean physically?

Neither could ever be a stoichiometric coefficient, and that is the point.

The first is acetaldehyde decomposition, a chain reaction. Its order of $3/2$ comes from combining the initiation, propagation and termination steps, whose individual rate constants appear inside a square root. No single step is one-and-a-half molecular; the fraction is an artefact of the algebra.

The second is ozone decomposition, first order negative in oxygen. Its mechanism starts with a reversible dissociation, $O_{3} ⇌ O_{2} + O$ , so adding oxygen pushes that equilibrium back and starves the slow step of oxygen atoms. A product appears in the rate law and slows the reaction down.

Both cases make the same point from opposite directions. Order is a number extracted from data, and its job is to constrain mechanisms, not to describe the balanced equation.

Units of the rate constant

$units of k = mo l^{1 - n} L^{n - 1} s^{- 1}$

Order	Units of $k$
Zero	mol L $^{- 1}$ s $^{- 1}$
First	s $^{- 1}$
Second	L mol $^{- 1}$ s $^{- 1}$

Trap. If a question gives you the units of $k$ , it has given you the order. This is free information and it is routinely left on the table.

Illustration 4

A rate constant is quoted with units $mo l^{- 1} L s^{- 1}$ . What is the overall order?

Rate always carries units of $mol L^{- 1} s^{- 1}$ , and the rate law sets $k$ against $n$ powers of concentration, so

$[k] = \frac{mol L ^{- 1} s ^{- 1}}{( mol L ^{- 1} ) ^{n}} = (mol L^{- 1})^{1 - n} s^{- 1}$

Running $n$ through the small cases:

Order	Units of $k$
0	$mol L^{- 1} s^{- 1}$
1	$s^{- 1}$
2	$mo l^{- 1} L s^{- 1}$
3	$mo l^{- 2} L^{2} s^{- 1}$

So the quoted units belong to a second order reaction.

The units alone fix the overall order, which is why a question supplying nothing but units is asking for exactly that. Notice also that first order is the only case where $k$ carries no concentration unit at all — which is the same fact as a first-order half-life not depending on how much you started with.

Four ways to find the order

Method	What you do
Initial rate	Change one concentration at a time and watch the initial rate respond
Integrated	Substitute data into each integrated law and see which returns a constant $k$
Graphical	Plot the candidate functions and see which is straight
Half-life	Use $t_{1/2} \propto [A]_{0}^{1 - n}$

Illustration 5

Initial rates are measured for $A + B \to products$ . Find the rate law and the value of $k$ .

$[A]_{0}$	$[B]_{0}$	Initial rate ( $mol L^{- 1} s^{- 1}$ )
0.10	0.10	$2.0 \times 1 0^{- 3}$
0.20	0.10	$8.0 \times 1 0^{- 3}$
0.20	0.20	$1.6 \times 1 0^{- 2}$

Compare rows where only one concentration moves — which is why the table is built this way.

Rows 1 to 2: $[A]$ doubles at fixed $[B]$ and the rate quadruples, so $2^{x} = 4$ and the order in A is 2.

Rows 2 to 3: $[B]$ doubles at fixed $[A]$ and the rate doubles, so the order in B is 1.

$rate = k [A]^{2} [B], overall order 3$

Substituting the first row:

$k = \frac{2.0 \times 1 0 ^{- 3}}{( 0.10 ) ^{2} ( 0.10 )} = 2.0 mo l^{- 2} L^{2} s^{- 1}$

The units land on $mo l^{- 2} L^{2} s^{- 1}$ , which the table above assigns to third order — an independent confirmation that the exponents were read correctly, and worth doing every time.

3. Integrated Rate Laws

Zero order

$[A] = [A]_{0} - k t, t_{1/2} = \frac{[ A ] _{0}}{2 k}$

A plot of concentration against time is straight with slope $- k$ , and the half-life is proportional to the initial concentration.

Zero order happens when something other than concentration is limiting. Ammonia decomposing on hot platinum is zero order because the surface is saturated: adding more ammonia cannot help when every active site is already occupied. Photochemical reactions are often zero order for the same reason, limited by light intensity.

Illustration 6

A zero-order reaction has $k = 2.0 \times 1 0^{- 2} mol L^{- 1} s^{- 1}$ and $[A]_{0} = 0.50 mol L^{- 1}$ . Find the half-life and the time at which the reactant is completely consumed.

Concentration falls in a straight line, $[A] = [A]_{0} - k t$ , so both answers are read straight off it:

$t_{1/2} = \frac{[ A ] _{0}}{2 k} = \frac{0.50}{2 ( 2.0 \times 1 0 ^{- 2} )} = 12.5 s$

$t_{co m pl e t e} = \frac{[ A ] _{0}}{k} = 25 s$

Two things here have no counterpart in first order. A zero-order reaction genuinely finishes, at a definite time, whereas a first-order concentration only approaches zero and never arrives.

And the half-life depends on the starting concentration, $t_{1/2} \propto [A]_{0}$ , so doubling the initial amount doubles the half-life. First order is the sole case where the half-life is independent of concentration, and second order runs the other way with $t_{1/2} \propto 1/ [A]_{0}$ . That triple contrast is what half-life questions are usually built on.

First order

$k = \frac{2.303}{t} lo g \frac{[ A ] _{0}}{[ A ]}, t_{1/2} = \frac{0.693}{k}$

A plot of $lo g [A]$ against time is straight with slope $- k /2.303$ .

The half-life is independent of concentration. That is its defining signature and the fastest way to identify first-order behaviour from raw data: if successive half-lives are equal, the reaction is first order.

All radioactive decay is first order, which is why a half-life is a property of an isotope rather than of a sample.

Illustration 7

A wooden artefact gives 30.0 per cent of the carbon-14 activity of living wood. Carbon-14 has a half-life of 5730 years. How old is it?

Radioactive decay is strictly first order, so the ordinary integrated law applies with activity standing in for concentration.

$k = \frac{0.693}{5730} = 1.21 \times 1 0^{- 4} y r^{- 1}$

$t = \frac{2.303}{k} lo g \frac{100}{30.0} = \frac{2.303}{1.21 \times 1 0 ^{- 4}} \times 0.523 = 9950 years$

Note what made this possible. Because the half-life of a first-order process does not depend on how much material is present, a sample that has lost 70 per cent of its carbon-14 dates the same whether it weighs a gram or a tonne.

Try the same trick on a second-order reaction and it fails: its half-life depends on the starting concentration, so "how far through" tells you nothing until you know where it started. First-order kinetics is what makes dating possible at all.

Reading the graphs

Plot	Zero order	First order
Concentration against time	Straight, slope $- k$	Exponential decay
Log of concentration against time	Curved	Straight, slope $- k /2.303$
Rate against concentration	Horizontal line	Straight through the origin
Half-life against initial concentration	Straight through the origin	Horizontal line

The last row is the most useful in practice, because it separates the orders from raw data with no plotting at all.

Illustration 8

A reaction's successive half-lives are measured as 10, 20 and 40 minutes. Identify the order and find $k$ , given $[A]_{0} = 1.00$ M.

Equal successive half-lives would mean first order. These double each time, which is the signature of second order, where $t_{1/2} = 1/ (k [A]_{0})$ : as $[A]_{0}$ halves, the half-life doubles.

Check it against the general relation $t_{1/2} \propto [A]_{0}^{1 - n}$ . For $n = 2$ the exponent is $- 1$ , so halving the concentration doubles the half-life. It fits.

$k = \frac{1}{t _{1/2} [ A ] _{0}} = \frac{1}{10 \times 1.00} = 0.100 L mo l^{- 1} mi n^{- 1}$

The units confirm second order independently, which is the check worth doing every time.

Note the practical consequence. After three half-lives a first-order reaction has taken 30 minutes to reach 0.125 M. This one has taken 70 minutes to reach the same place, and the gap widens without limit. A second-order reaction never really finishes.

Pseudo-first order

A genuinely second-order reaction behaves as first order when one reactant is in large excess, because its concentration barely changes and gets absorbed into the rate constant.

Ester hydrolysis in dilute aqueous solution is the standard case: water is both reactant and solvent, so its concentration is effectively constant at about 55 M. Cane sugar inversion is the other classic.

Illustration 9

Ester hydrolysis in water gives an observed first-order constant of $k_{o b s} = 1.8 \times 1 0^{- 5} s^{- 1}$ . Find the true second-order rate constant.

The real rate law is second order overall.

$rate = k_{2} [ester] [H_{2} O]$

Water at about 55.5 M is in such excess that its concentration is unchanged over the whole reaction, so it folds into the constant.

$rate = (k_{2} \times 55.5) [ester] = k_{o b s} [ester]$

$k_{2} = \frac{1.8 \times 1 0 ^{- 5}}{55.5} = 3.2 \times 1 0^{- 7} L mo l^{- 1} s^{- 1}$

The units are the tell. $k_{o b s}$ carries s $^{- 1}$ , which looks first order, while $k_{2}$ carries L mol $^{- 1}$ s $^{- 1}$ , which is honestly second order. Nothing about the chemistry changed; the excess reactant simply hid inside the constant.

4. Temperature and the Arrhenius Equation

As a rough guide, a 10 K rise doubles or triples many reaction rates near room temperature. That ratio is the temperature coefficient.

Faster collisions cannot explain it. Going from 300 K to 310 K raises the mean molecular speed by under 2 per cent, which could not possibly double anything.

The explanation is the tail of the energy distribution. Only molecules with at least the activation energy react, and that fraction is exponentially sensitive to temperature even when the average barely moves.

Two terms are worth separating. Threshold energy is the minimum total energy colliding molecules must possess. Activation energy is the extra they need above their average, so it is the threshold minus the average energy of the reactants.

$k = A e^{- E_{a} / R T}$

with $A$ the frequency factor, related to how often collisions occur with the right orientation.

Using it

$lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$

A plot of $lo g k$ against $1/ T$ is straight with slope $- E_{a} /2.303 R$ , which is the standard experimental route to $E_{a}$ .

Illustration 10

The rule of thumb says a 10 K rise doubles the rate. Test it for $E_{a} = 50$ kJ mol $^{- 1}$ , first from 300 to 310 K and then from 600 to 610 K.

$ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$

From 300 to 310 K:

$ln \frac{k _{2}}{k _{1}} = \frac{50000}{8.314} (\frac{1}{300} - \frac{1}{310}) = 6014 \times 1.075 \times 1 0^{- 4} = 0.647 \Rightarrow \frac{k _{2}}{k _{1}} = 1.91$

From 600 to 610 K:

$ln \frac{k _{2}}{k _{1}} = 6014 \times (2.73 \times 1 0^{- 5}) = 0.164 \Rightarrow \frac{k _{2}}{k _{1}} = 1.18$

The same 10 K nearly doubles the rate at room temperature and adds barely 18 per cent at 600 K.

The reason is in the bracket. What matters is not $Δ T$ but the change in $1/ T$ , and $1/ T$ flattens out as $T$ grows, so the same ten degrees buys progressively less. The rule of thumb is a room-temperature accident, not a law, and a question that quotes it at 800 K is testing whether you noticed.

A larger $E_{a}$ also means a more temperature-sensitive reaction, because the exponent is larger. Reactions with small barriers speed up much less on heating.

Illustration 11

A reaction's rate constant exactly doubles when the temperature rises from 300 K to 310 K. Find the activation energy.

The two-temperature form of the Arrhenius equation removes the pre-exponential factor, which is never known:

$ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$

$\frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{( 300 ) ( 310 )} = 1.075 \times 1 0^{- 4} K^{- 1}$

$E_{a} = \frac{R l n 2}{1.075 \times 1 0 ^{- 4}} = \frac{( 8.314 ) ( 0.693 )}{1.075 \times 1 0 ^{- 4}} \Rightarrow E_{a} \approx 53.6 kJ mo l^{- 1}$

This is the previous illustration run backwards, and the two agree. The rule of thumb that a 10 K rise doubles the rate is not a general law; it is the statement that near room temperature a barrier of roughly 50 kJ mol⁻¹ behaves that way, and 53.6 is how close the rule actually sits to the round number.

Reactions with much smaller barriers are far less temperature-sensitive, and those with much larger ones far more so — which is why the rule is safe for a rough estimate and unsafe as an answer.

5. Collision Theory

$rate = Z \times e^{- E_{a} / R T} \times P$

The first factor is the collision frequency, the second the fraction of collisions with enough energy, the third the steric or probability factor, being the fraction with the right orientation.

Most collisions achieve nothing. Each molecule in a typical gas undergoes billions of collisions per second, and yet reactions take seconds or hours, because the energy fraction is exponentially small.

The orientation requirement is the theory's second insight. Two molecules can collide hard enough and still bounce apart if the reacting groups were not facing each other, which is why $P$ for reactions between large molecules can be below $1 0^{- 5}$ .

Collision theory works well for simple gas-phase reactions and poorly in solution and for complex molecules, where $P$ becomes an empirical fudge rather than a prediction.

Catalysis

A catalyst provides an alternative route with a lower activation energy.

It does not change $Δ H$ , $Δ G$ or $K$ . The initial and final states are untouched, so the thermodynamics is untouched; only the barrier between them is lowered. Because it lowers the barrier for both directions equally, it speeds forward and reverse by the same factor, which is exactly why it cannot shift an equilibrium.

Homogeneous catalysis has catalyst and reactants in the same phase, as when an acid catalyses ester hydrolysis. Heterogeneous catalysis has them in different phases, as when iron catalyses ammonia synthesis at a gas-solid interface.

Enzymes are biological catalysts of extraordinary specificity. Their power comes from lowering activation energy, exactly as for any catalyst, and their selectivity from the shape of the active site.

Illustration 12

Hydrogen peroxide decomposition has $E_{a} = 75$ kJ mol $^{- 1}$ uncatalysed. The enzyme catalase reduces it to about 8 kJ mol $^{- 1}$ . By what factor is the rate increased at 300 K?

Only the exponential term changes, so the frequency factors cancel to a good approximation.

$\frac{k _{c a t}}{k _{u n c a t}} = \frac{e ^{- 8000/ R T}}{e ^{- 75000/ R T}} = e^{67000/ (8.314 \times 300)} = e^{26.9}$

$\frac{k _{c a t}}{k _{u n c a t}} \approx 5 \times 1 0^{11}$

Five hundred billion times faster, from removing 67 kJ mol $^{- 1}$ from a barrier.

Put that in human terms. A reaction that would take a century uncatalysed finishes in under a hundredth of a second. This is why a cut fizzes the instant hydrogen peroxide touches it: your cells are full of catalase, and every one of them is running the same reaction that a bottle of peroxide performs imperceptibly slowly on a shelf.

And notice what has not changed. $Δ H$ for the decomposition is identical, $Δ G$ is identical, and $K$ is identical. The peroxide was always going to decompose. The enzyme only decided when.

Summary

A half-life is a constant only for a first-order reaction. Zero-order half-lives halve each time so the reaction ends abruptly, second-order half-lives double each time so it never really finishes, and three reactions with the same starting concentration and the same first half-life are in three different places 30 minutes later.

The rate law is measured and cannot be read off the balanced equation. Order can be zero, fractional or negative, and a product can appear in it with a negative exponent, as oxygen does for ozone decomposition.

Order and molecularity coincide only for elementary steps. A rate law that happens to match the stoichiometry, as for $2NO + O_{2}$ , still need not imply a termolecular step, and a rate law is consistent with a mechanism rather than proof of one.

The units of $k$ are $mo l^{1 - n} L^{n - 1} s^{- 1}$ , so quoting them gives away the order, and checking them catches most arithmetic errors for free.

Zero order gives $[A] = [A]_{0} - k t$ with $t_{1/2} = [A]_{0} /2 k$ ; first order gives $k = (2.303/ t) lo g ([A]_{0} / [A])$ with $t_{1/2} = 0.693/ k$ . Equal successive half-lives identify first order, doubling ones identify second order.

Radiocarbon dating works only because first-order half-lives are independent of amount, so a fraction remaining fixes an age without knowing the original mass.

A reaction in large excess of one reactant is pseudo-first order: the excess concentration hides inside $k_{o b s}$ , and dividing by 55.5 M recovers the honest second-order constant with honest units.

Temperature acts through the tail of the Maxwell-Boltzmann distribution, not through mean speed. The 10 K doubling rule is a room-temperature accident: for $E_{a} = 50$ kJ mol $^{- 1}$ it gives a factor of 1.91 from 300 K and only 1.18 from 600 K, because what matters is the change in $1/ T$ .

Collision theory writes the rate as $Z \times e^{- E_{a} / R T} \times P$ , and a catalyst changes only the exponent. Catalase drops the peroxide barrier by 67 kJ mol $^{- 1}$ and speeds the reaction by about $5 \times 1 0^{11}$ , while leaving $Δ H$ , $Δ G$ and $K$ exactly where they were.

Key formulas & results

Everything to memorise for the exam hall, in one card. Screenshot this for revision.

The organising principle

the rate law is measured, never read off the equation; temperature acts on the tail, not the average

The first fact separates order from molecularity. The second explains why 10 K can double a rate and why a small reduction in the barrier changes everything.

Rate of reaction

$\text{rate} = -\dfrac{1}{a}\dfrac{d[A]}{dt} = +\dfrac{1}{c}\dfrac{d[C]}{dt}$

Coefficients divided out so that one number describes the reaction whichever species you watched. Initial rate is the most useful in practice, since no products have accumulated to complicate matters.

Rate law and order

$\text{rate} = k[A]^x[B]^y$, order $= x + y$

Order may be zero, fractional or negative, none of which a coefficient could be. Acetaldehyde decomposition is order $3/2$ from a chain mechanism, and ozone decomposition is order $-1$ in oxygen because a product starves the slow step.

Order against molecularity

order is experimental and unbounded; molecularity is theoretical, whole and never above three

They coincide only for an elementary step. $\mathrm{2NO + O_2}$ is third order and still not termolecular: a fast $\mathrm{NO + NO \rightleftharpoons N_2O_2}$ pre-equilibrium feeding a slow step reproduces the law exactly.

Units of the rate constant

$\mathrm{mol^{1-n}\,L^{n-1}\,s^{-1}}$

Zero order mol L$^{-1}$ s$^{-1}$, first order s$^{-1}$, second order L mol$^{-1}$ s$^{-1}$. If a question quotes the units it has quoted the order, and checking them catches most arithmetic errors free.

Zero order

$[A] = [A]_0 - kt, \qquad t_{1/2} = \dfrac{[A]_0}{2k}$

The only case that genuinely finishes, at $t = [A]_0/k$. Occurs when something other than concentration limits: a saturated catalyst surface, or light intensity in a photochemical reaction.

First order

$k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}, \qquad t_{1/2} = \dfrac{0.693}{k}$

The half-life is independent of concentration, which is its signature and what makes radiocarbon dating possible: a fraction remaining fixes an age without knowing the original mass.

Identifying the order

$t_{1/2} \propto [A]_0^{\,1-n}$

Successive half-lives that are **equal** mean first order, that **halve** mean zero order, and that **double** mean second order. Equivalently, whichever of $[A]$, $\ln[A]$ or $1/[A]$ plots straight against $t$ names the order.

Pseudo-order

$k_{obs} = k_2[\mathrm{H_2O}]$, with water at about 55.5 M

A genuinely second-order reaction reads as first order when one reactant is in large excess. The units give it away: $k_{obs}$ in s$^{-1}$ against a true $k_2$ in L mol$^{-1}$ s$^{-1}$.

Arrhenius equation

$k = Ae^{-E_a/RT}, \qquad \log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{T_2 - T_1}{T_1T_2}\right)$

A plot of $\log k$ against $1/T$ is straight with slope $-E_a/2.303R$. What matters is the change in $1/T$, not in $T$, which is why the 10 K doubling rule is a room-temperature accident.

Collision theory

$\text{rate} = Z \times e^{-E_a/RT} \times P$

Collision frequency, energy fraction and steric factor. Most collisions achieve nothing because the exponential is tiny, and $P$ can fall below $10^{-5}$ for large molecules. Threshold energy is the total needed; $E_a$ is the excess above the average.

Catalysis

$\dfrac{k_{cat}}{k_{uncat}} = e^{\Delta E_a / RT}$, with $\Delta H$, $\Delta G$ and $K$ unchanged

Catalase lowers the peroxide barrier from 75 to 8 kJ mol$^{-1}$, a factor of about $5\times10^{11}$ at 300 K. The barrier falls equally in both directions, which is exactly why an equilibrium cannot shift.

How half-life depends on concentration

t_{1/2} \propto [A]_{0} (n = 0), t_{1/2} = \frac{0.693}{k} (n = 1), t_{1/2} \propto \frac{1}{[ A ] _{0}} (n = 2)

First order is the only case where the half-life ignores how much you started with, which is the same fact as its rate constant carrying no concentration unit. Zero order also genuinely reaches completion, at $t = [\mathrm{A}]_0/k$, where a first-order concentration only ever approaches zero.

JEE Main test series

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Chemical Kinetics shows up in every JEE Main paper — sit a timed one and find out where you actually stand.

~1Est. Qs

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Traps JEE Main sets — and how to dodge them

These are the exact option-traps and misreads that cost marks under negative marking.

WATCH OUT

✗ Reading the exponents of the rate law off the balanced equation

✓ Order is measured.

2N O_{2} + F_{2}

is first order in

N O_{2}

, not second, and orders of

3/2

and

- 1

occur where no coefficient could. Even when a rate law does match the stoichiometry, as for

2NO + O_{2}

, it need not imply an elementary step: a fast pre-equilibrium feeding a slow bimolecular step reproduces the same third-order law with no termolecular collision anywhere.

Why it happens: The two coincide often enough, and the equation is the only thing written down.

WATCH OUT

✗ Treating half-life as a constant of every reaction

✓

t_{1/2} \propto [A]_{0}^{1 - n}

. For zero order the half-life is proportional to concentration, so successive ones halve and the reaction ends abruptly; for second order it is inversely proportional, so successive ones double and the reaction never really finishes. Three reactions starting at 1.00 M with the same first half-life of 10 minutes are at 0 M, 0.125 M and 0.250 M after 30 minutes.

Why it happens: It is introduced through first-order and radioactive decay, where it genuinely is one.

WATCH OUT

✗ Ignoring the units of

k

✓ The units encode the order through

mo l^{1 - n} L^{n - 1} s^{- 1}

, so a

k

quoted in L mol

^{- 1}

^{- 1}

has already told you the reaction is second order. Used the other way they are a free check: if your working produced a first-order constant with second-order units, an exponent went astray. This is the cheapest error-detector in the chapter.

Why it happens: They look like a formality attached after the number.

WATCH OUT

✗ Explaining temperature dependence by faster collisions

✓ From 300 K to 310 K the mean speed rises by under 2 per cent, which cannot double anything. What doubles is the fraction of molecules beyond the activation energy, which sits far out in the tail of the Maxwell-Boltzmann distribution where the curve is steepest. The dependence is exponential through

e^{- E_{a} / R T}

, so a shift too small to matter for the average matters enormously for the reactive minority.

Why it happens: Molecules do move faster when heated, and collisions do become more frequent.

WATCH OUT

✗ Applying the 10 K doubling rule at any temperature

✓ The Arrhenius bracket contains the change in

1/ T

, not in

T

, and

1/ T

flattens as

T

rises. For

E_{a} = 50

kJ mol

^{- 1}

, 300 to 310 K gives a factor of 1.91 while 600 to 610 K gives only 1.18. The rule is a room-temperature accident, and a question quoting it at high temperature is testing whether you noticed.

Why it happens: It is quoted as a general rule of thumb and it works well in the laboratory.

WATCH OUT

✗ Reporting a pseudo-first-order constant as the true rate constant

✓ The excess reactant has been absorbed into the constant, so

k_{o b s} = k_{2} [H_{2} O]

with water at about 55.5 M. Dividing recovers the honest second-order constant, and the units confirm it:

1.8 \times 1 0^{- 5}

^{- 1}

becomes

3.2 \times 1 0^{- 7}

L mol

^{- 1}

^{- 1}

. Nothing about the chemistry changed; the excess simply hid inside the number.

Why it happens: The data fit a first-order plot perfectly, so the reaction looks first order.

Exam-pattern practice

PYQ-style questions with full solutions. Work through them as a readiness check — mark yourself honestly and get your gap report at the end.

Readiness check

Are you exam-ready for Chemical Kinetics?

12 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

12 questions~8 min worth ~8 marks in JEE Main exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Rate is defined with coefficients divided out; initial rate is the cleanest measurement
•The rate law is measured, never read off the equation; order can be zero, fractional or negative
•Order and molecularity coincide only for elementary steps; molecularity never exceeds three
•A matching rate law does not prove an elementary step: $2NO + O_{2}$ is third order via a fast pre-equilibrium
•Units of $k$ are $mo l^{1 - n} L^{n - 1} s^{- 1}$ , so quoted units give away the order
•Zero: $[A] = [A]_{0} - k t$ , $t_{1/2} = [A]_{0} /2 k$ , and it genuinely finishes at $[A]_{0} / k$
•First: $k = (2.303/ t) lo g ([A]_{0} / [A])$ , $t_{1/2} = 0.693/ k$ , independent of concentration
•Successive half-lives equal means first, halving means zero, doubling means second order
•First-order half-life independence is what makes radiocarbon dating work at all
•Pseudo-order: $k_{o b s} = k_{2} \times 55.5$ for water; divide back out and check the units
•Temperature acts on the Maxwell-Boltzmann tail; the 10 K rule gives 1.91 at 300 K and only 1.18 at 600 K
• $rate = Z e^{- E_{a} / R T} P$ ; a catalyst changes only the exponent and leaves $Δ H$ , $Δ G$ and $K$ alone

JEE Main question blueprint

How this topic is asked, tier by tier — so you can prep to the pattern.

Typical weightage: ~2 questions (8 marks) of the 100-mark Chemistry section

Question style	Marks each	Typical count	What it tests
Integrated rate laws and half-life	1	1	Zero and first order integrated forms and their straight-line plots, half-life scaling with concentration across the three orders, and identifying order from successive half-lives
Rate law, order and molecularity	1	1	Relating the single reaction rate to each species' rate of change through its coefficient, deducing order by the initial-rate method, reading overall order straight off the units of $k$, and separating order from molecularity
Temperature, Arrhenius and collision theory	1	1	$\ln(k_2/k_1)$ between two temperatures to extract $E_a$, the limits of the ten-degree rule of thumb, and the orientation and energy requirements of a successful collision
Catalysis and pseudo-order	1	1	Why a catalyst lowers $E_a$ without shifting equilibrium, enzyme efficiency, and recognising a pseudo-first-order reaction where one reactant is in large excess

Exam-hall strategy

Battle-tested tips from mentors and toppers for this topic under the sectional clock.

Read the units of $k$ first, in either direction. Given units, you have the order; having computed an order, check the units match before writing anything down.
For order from data, change one concentration at a time and take ratios of rates. Never attempt to infer exponents from the balanced equation, even when they would happen to agree.
Identify order from successive half-lives before reaching for a formula: equal means first, halving means zero, doubling means second.
In Arrhenius problems, work with $1/ T_{1} - 1/ T_{2}$ rather than $Δ T$ , and keep $E_{a}$ in joules to match $R = 8.314$ . Mixing kilojoules into that expression is the most common numerical slip in the chapter.
For catalyst and enzyme questions only the exponential term changes, so cancel the frequency factors immediately and compute $e^{Δ E_{a} / R T}$ in one step.

Beyond the exam

Where this skill shows up in the job you're competing for — and in life.

Radiocarbon dating works only because first-order half-li…

Radiocarbon dating works only because first-order half-lives are independent of amount, so 30 per cent of living carbon-14 activity dates a sample at about 9950 years whether it weighs a gram or a tonne

Catalase in your cells drops the activation energy for hy…

Catalase in your cells drops the activation energy for hydrogen peroxide decomposition from 75 to about 8 kJ per mole, a rate increase near $5 \times 1 0^{11}$ , which is why a cut fizzes instantly while the bottle on the shelf decomposes imperceptibly

Food is refrigerated rather than sterilised because spoil…

Food is refrigerated rather than sterilised because spoilage reactions obey Arrhenius: dropping from 298 K to 277 K cuts rates several-fold through the exponential term, buying days without changing the chemistry at all

Where else this topic is tested

Prepare once, score in every exam that asks it.

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Questions aspirants ask

Pulled from the Q&A community and mentor sessions.

Because order is a number fitted to data, not a count of anything. Acetaldehyde decomposition comes out at order $3/2$ because it is a chain reaction: combining the initiation, propagation and termination steps puts a ratio of rate constants under a square root, and the fractional exponent is an artefact of that algebra rather than a claim that one-and-a-half molecules collide. Ozone decomposition is order $- 1$ in oxygen because the mechanism opens with a reversible dissociation, so adding oxygen pushes that equilibrium backwards and starves the slow step of oxygen atoms. A product appearing in the rate law with a negative exponent is a strong hint of a pre-equilibrium, which is exactly the kind of structural information order is for.

Not necessarily, and assuming so is a standard trap. $2NO + O_{2} \to 2N O_{2}$ is third order overall, rate $= k [NO]^{2} [O_{2}]$ , which looks exactly like a termolecular step. Termolecular steps are rare enough to be almost never the answer, and the accepted mechanism is two bimolecular steps: a fast equilibrium $NO + NO ⇌ N_{2} O_{2}$ followed by a slow $N_{2} O_{2} + O_{2}$ . Substituting $[N_{2} O_{2}] = K [NO]^{2}$ into the slow step reproduces the observed law exactly. A rate law is consistent with a mechanism and never proof of one, which is why questions ask what a rate law reveals rather than what it establishes.

Because its half-life doubles each time. $t_{1/2} = 1/ (k [A]_{0})$ , so as the concentration halves the next half-life takes twice as long: 10 minutes, then 20, then 40, then 80. The reactant approaches zero but the time to remove each remaining half keeps growing, so completion is asymptotic. First order behaves better, repeating the same half-life for ever, which still never quite reaches zero but at least at a predictable pace. Zero order is the only one that genuinely ends, at $t = [A]_{0} / k$ , because the rate does not fall as the reactant is consumed. This is why the phrase 'the reaction is complete' means something quite different for the three orders.

Threshold energy is the total energy that colliding molecules must possess between them for reaction to be possible. Activation energy is the extra energy needed above what the reactants already have on average, so $E_{a}$ is the threshold energy minus the average energy of the reactants at that temperature. The distinction matters because the Arrhenius factor $e^{- E_{a} / R T}$ measures the fraction of molecules that must be found in the tail, which is about the excess rather than the absolute. In most JEE problems only $E_{a}$ appears and the distinction is definitional, but questions do occasionally supply a threshold energy and expect you to subtract.

Both, and it does it by changing the pathway rather than the reaction. A catalyst offers an alternative route with a lower activation energy, so along that route the rate constant is genuinely larger: $k_{c a t} / k_{u n c a t} = e^{Δ E_{a} / R T}$ , which for catalase and hydrogen peroxide is about $5 \times 1 0^{11}$ at 300 K. What it cannot change is anything about the reactants or the products, since it touches neither, so $Δ H$ , $Δ G$ and $K$ are untouched. And because the new route is available in both directions, both rate constants rise by the same factor and their ratio, the equilibrium constant, is unmoved. A catalyst decides when a reaction happens, never whether.