JEE Mains Chemistry · Chemical Kinetics

Rate Law, Order and Molecularity

Rate = k[A]ᵐ[B]ⁿ, with the exponents found by experiment; their sum is the order, which fixes how the rate responds to a change in concentration and the unit of k.

All Chemical Kinetics notes JEE Mains Chemistry strategy

Why this matters

Fourteen PYQs, six of them multiple choice, and two from 2026. Five ask how the rate changes when concentrations, partial pressures or the volume change; five find the order from a table of initial rates and use it to fill a missing entry; four read the order from the unit of k or test statements about order and molecularity. Three ideas cover the page.

Concept 1 of 3: How the rate changes when concentrations change

The rate law

r = k[A]^m[B]^n

says how strongly the rate feels each concentration. Multiply

[A]

p

and the rate is multiplied by

p^m

. For a single-step (elementary) reaction the exponents are the coefficients. The rate depends on concentration, not on how much solution you take.

Definition

$\dfrac{r_2}{r_1} = p^m q^n$ when $[A]$ is multiplied by $p$ and $[B]$ by $q$ .
Elementary (single-step) reaction: the exponents equal the stoichiometric coefficients. Otherwise they come only from experiment.
Squeezing a gas mixture to $\dfrac{1}{f}$ of its volume multiplies every concentration by $f$ , so the rate by $f^{m+n}$ .
Gas reactions can use partial pressures. After some reaction, find each new partial pressure from the stoichiometry first.
The rate is intensive: more solution at the same concentration gives the same rate; adding water dilutes it and lowers the rate.

Ratio of rates

\frac{r_2}{r_1} = \left(\frac{[A]_2}{[A]_1}\right)^{m}\left(\frac{[B]_2}{[B]_1}\right)^{n}

Worked example

The single-step gas reaction

A(g) + 2B(g) \rightarrow C(g)

starts with

p_A = 0.8

atm and

p_B = 1.0

atm. Find the ratio of the initial rate to the rate when

p_C = 0.3

atm.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 5 Apr 2024 · Q147Moderate

Example 1 · Chemical Kinetics · Rate Law, Order and Molecularity

Consider the following single step reaction in gas phase at constant temperature.

2{\text{ }A}_{(g)}+B_{(g)}\rightarrow C_{(g)}

The initial rate of the reaction is recorded as

r_{1}

when the reaction starts with

1.5\text{ }atm

pressure of

A

and 0.7 atm pressure of B. After some time, the rate

r_{2}

is recorded when the pressure of

C

becomes

0.5\text{ }atm

. The ratio

r_{1}:r_{2}

is _____

\times10^{- 1}

. (Nearest integer)

Order read from the balanced equation

Coefficients give the exponents only for an elementary (single-step) reaction. Otherwise the order comes from experiment:

\mathrm{2N_2O_5 \rightarrow 4NO_2 + O_2}

is first order, not second.

More solution taken as a faster reaction

The rate depends on concentration. Doubling the volume of the same solution leaves the rate unchanged; adding the same volume of water halves the concentration and lowers the rate.

Drill 4 more on how the rate changes when concentrations change

Concept 2 of 3: Order from a table of initial rates

Pick two runs in which only one concentration changes. The rate ratio is then that concentration's ratio raised to its order. Do the same for the other reactant, then find

k

from any run and use it for a missing entry.

Definition

Runs with $[B]$ fixed: $\dfrac{r_2}{r_1} = \left(\dfrac{[A]_2}{[A]_1}\right)^m$ , so $m = \dfrac{\log(r_2/r_1)}{\log([A]_2/[A]_1)}$ .
Repeat with $[A]$ fixed to get $n$ .
$k = \dfrac{r}{[A]^m[B]^n}$ from any run; then fill the missing rate or concentration.
A reactant of zero order does not affect the rate, whatever its concentration.
If the table gives the rate of formation of one product, the ratios, and so the orders, are the same.

Order in A from two runs

m = \frac{\log(r_2/r_1)}{\log([A]_2/[A]_1)} \quad ([B]\ \text{fixed})

Worked example

For

A + B \rightarrow

products: run 1,

[A] = 0.10

[B] = 0.10

M, rate

2.0 \times 10^{-3}

M s

^{-1}

; run 2,

[A] = 0.20

[B] = 0.10

M, rate

8.0 \times 10^{-3}

; run 3,

[A] = 0.10

[B] = 0.30

M, rate

6.0 \times 10^{-3}

. Find the orders and

k

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 11 April 2023 · Q125Moderate

Example 2 · Chemical Kinetics · Rate Law, Order and Molecularity

For a chemical reaction

A + B \rightarrow

Product, the order is 1 with respect to

A

and

B

. The measured data are: Rate

= 0.10\ mol{\text{ }L}^{- 1}{\text{ }s}^{- 1}

when

\lbrack A\rbrack = 20\ mol{\text{ }L}^{- 1}

and

\lbrack B\rbrack = 0.5\ mol{\text{ }L}^{- 1}

; Rate

= 0.40\ mol{\text{ }L}^{- 1}{\text{ }s}^{- 1}

when

\lbrack A\rbrack = x\ mol{\text{ }L}^{- 1}

and

\lbrack B\rbrack = 0.5\ mol{\text{ }L}^{- 1}

; Rate

= 0.80\ mol{\text{ }L}^{- 1}{\text{ }s}^{- 1}

when

\lbrack A\rbrack = 40\ mol{\text{ }L}^{- 1}

and

\lbrack B\rbrack = y\ mol{\text{ }L}^{- 1}

. What is the value of

x

and

y

Two runs where both concentrations changed

[A]

and

[B]

both change between two runs, the rate ratio mixes both orders. Choose a pair where only one changes, or divide out the factor from the order you already know.

Drill 4 more on order from a table of initial rates

Concept 3 of 3: Unit of the rate constant for each order, and order versus molecularity

The rate always has the unit mol L

^{-1}

^{-1}

, so

k

must absorb whatever the concentration terms bring. That makes the unit of

k

a label for the order. Order is measured; molecularity is a count of the particles that collide in one elementary step.

Definition

Order: the sum of the exponents in the experimental rate law. It can be 0, a fraction or a whole number.
Molecularity: the number of species that collide in one elementary step. Always 1, 2 or 3; never 0 or a fraction; defined only for an elementary step.
A reactant in the equation may not appear in the rate law at all (zero order in it).
Unit of $k$ for order $n$ : $(\text{mol L}^{-1})^{1-n}\,\text{s}^{-1}$ . Only for zero order do the rate and $k$ share a unit.
The decomposition of $\mathrm{N_2O_5}$ is first order, although the equation has $\mathrm{2N_2O_5}$ .

Order	Rate law	Unit of k	Half-life
0	$r = k$	mol L $^{-1}$ s $^{-1}$	$\dfrac{[A]_0}{2k}$ , proportional to $[A]_0$
1	$r = k[A]$	s $^{-1}$	$\dfrac{0.693}{k}$ , independent of $[A]_0$
2	$r = k[A]^2$	L mol $^{-1}$ s $^{-1}$	$\dfrac{1}{k[A]_0}$ , inversely proportional to $[A]_0$
3	$r = k[A]^3$	L $^2$ mol $^{-2}$ s $^{-1}$	Proportional to $\dfrac{1}{[A]_0^2}$
$n$	$r = k[A]^n$	$(\text{mol L}^{-1})^{1-n}$ s $^{-1}$	Proportional to $[A]_0^{\,1-n}$

The unit of k names the order; the half-life column is used again on the zero-order page.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q150Moderate

Example 3 · Chemical Kinetics · Rate Law, Order and Molecularity

The number of incorrect statement/s from the following is___ A. The successive half-lives of zero order reactions decreases with time. B. A substance appearing as reactant in the chemical equation may not affect the rate of reaction C. Order and molecularity of a chemical reaction can be a fractional number D. The rate constant units of zero and second order reaction are

molL^{- 1}{\text{ }s}^{- 1}

and

mol^{- 1}Ls^{- 1}

respectively

Rate and k given the same unit

Only for a zero-order reaction. For a first-order reaction the rate is in mol L

^{-1}

^{-1}

but

k

is in s

^{-1}

A fractional molecularity

Order can be zero or a fraction, because it is measured. Molecularity counts colliding particles in one elementary step, so it is always 1, 2 or 3.

Drill 3 more on unit of the rate constant for each order, and order versus molecularity

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

How the rate changes when concentrations change
Ratio of rates
$\frac{r_2}{r_1} = \left(\frac{[A]_2}{[A]_1}\right)^{m}\left(\frac{[B]_2}{[B]_1}\right)^{n}$
Order from a table of initial rates
Order in A from two runs
$m = \frac{\log(r_2/r_1)}{\log([A]_2/[A]_1)} \quad ([B]\ \text{fixed})$

Reference tables (1)

Unit of the rate constant for each order, and order versus molecularity5 rows

Order	Rate law	Unit of k	Half-life
0	$r = k$	mol L $^{-1}$ s $^{-1}$	$\dfrac{[A]_0}{2k}$ , proportional to $[A]_0$
1	$r = k[A]$	s $^{-1}$	$\dfrac{0.693}{k}$ , independent of $[A]_0$
2	$r = k[A]^2$	L mol $^{-1}$ s $^{-1}$	$\dfrac{1}{k[A]_0}$ , inversely proportional to $[A]_0$
3	$r = k[A]^3$	L $^2$ mol $^{-2}$ s $^{-1}$	Proportional to $\dfrac{1}{[A]_0^2}$
$n$	$r = k[A]^n$	$(\text{mol L}^{-1})^{1-n}$ s $^{-1}$	Proportional to $[A]_0^{\,1-n}$

The unit of k names the order; the half-life column is used again on the zero-order page.

Watch out for (5)

Order read from the balanced equation
→ How the rate changes when concentrations change
More solution taken as a faster reaction
→ How the rate changes when concentrations change
Two runs where both concentrations changed
→ Order from a table of initial rates
Rate and k given the same unit
→ Unit of the rate constant for each order, and order versus molecularity
A fractional molecularity
→ Unit of the rate constant for each order, and order versus molecularity

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