ChemistryCBSE Class 12 · Chemical Kinetics

Rate Law, Order and Molecularity: Class 12 Chemical Kinetics

Updated 15 October 2026 · Fact-checked

A rate law expresses the rate as $k[A]^x[B]^y$, where the powers $x$ and $y$ are found by experiment. Their sum is the order of the reaction, which can be zero, fractional or whole. Molecularity is the number of particles colliding in one elementary step: always a small whole number, never zero.

Understand Rate Law, Order and Molecularity

Rate of a reaction. The rate is the change in concentration of a reactant or product per unit time. For $R \to P$ , $rate = - \frac{Δ [ R ]}{Δ t} = + \frac{Δ [ P ]}{Δ t}$ ; the minus sign makes the rate positive because $[R]$ falls. Over a finite interval this is the average rate; as $Δ t \to 0$ it becomes the instantaneous rate, $- \frac{d [ R ]}{d t}$ , which is the slope of the tangent on a concentration–time graph. When coefficients are not 1, divide each rate by its coefficient so that all give the same value: for $2 H I \to H_{2} + I_{2}$ , $rate = - \frac{1}{2} \frac{d [ H I ]}{d t} = \frac{d [ H _{2} ]}{d t} = \frac{d [ I _{2} ]}{d t}$ . Units are mol L⁻¹ s⁻¹ (or atm s⁻¹ for gases).

Rate law. At a fixed temperature the rate depends on concentrations. For $a A + b B \to c C + d D$ the experimental rate law is $rate = k [A]^{x} [B]^{y}$ . The constant $k$ is the rate constant (rate when every concentration is 1 mol L⁻¹). The exponents $x$ and $y$ need not equal $a$ and $b$ : they must be found by experiment. For example, $2 N O + O_{2} \to 2 N O_{2}$ has rate $= k [N O]^{2} [O_{2}]$ , while $C H C l_{3} + C l_{2} \to C C l_{4} + H C l$ has rate $= k [C H C l_{3}] [C l_{2}]^{1/2}$ .

Order. The sum $x + y$ is the overall order; $x$ alone is the order with respect to A. Order can be 0, 1, 2, 3 or a fraction. Because rate has units mol L⁻¹ s⁻¹, the units of $k$ depend on the order: $(mol L^{- 1})^{1 - n} s^{- 1}$ . Zero order: mol L⁻¹ s⁻¹; first: s⁻¹; second: L mol⁻¹ s⁻¹.

Order from initial rates. Run the reaction several times, changing one concentration at a time, and compare initial rates. If doubling [A] doubles the rate, the order in A is 1; if it quadruples, 2; if nothing changes, 0. In general $\frac{r _{2}}{r _{1}} = (\frac{[ A ] _{2}}{[ A ] _{1}})^{x}$ .

Elementary and complex reactions. An elementary reaction happens in a single step. A complex reaction is a sequence of elementary steps, its mechanism. The slowest step controls the overall rate and is the rate-determining step. NCERT's example: $2 H_{2} O_{2} I^{-} 2 H_{2} O + O_{2}$ has rate $= k [H_{2} O_{2}] [I^{-}]$ , explained by a slow step $H_{2} O_{2} + I^{-} \to H_{2} O + I O^{-}$ followed by a fast step $H_{2} O_{2} + I O^{-} \to H_{2} O + I^{-} + O_{2}$ .

Molecularity. The number of reacting species (atoms, ions or molecules) that must collide at the same time in an elementary step is its molecularity: unimolecular ( $N H_{4} N O_{2} \to N_{2} + 2 H_{2} O$ ), bimolecular ( $2 H I \to H_{2} + I_{2}$ ) or termolecular ( $2 N O + O_{2} \to 2 N O_{2}$ ). Three-body collisions are rare, so molecularity above 3 is not found. Molecularity applies only to elementary steps; for a complex reaction it has no meaning, and the order is decided by the slowest step.

Pseudo first order. If one reactant is present in large excess, its concentration hardly changes and drops out of the rate law. Hydrolysis of ethyl acetate in a large excess of water, and inversion of cane sugar in acid, are bimolecular but follow first-order kinetics: these are pseudo first order reactions.

Rate Law, Order and Molecularity: formulas and key facts

Rate in terms of each species: $a A + b B \to c C + d D : rate = - \frac{1}{a} \frac{d [ A ]}{d t} = - \frac{1}{b} \frac{d [ B ]}{d t} = \frac{1}{c} \frac{d [ C ]}{d t} = \frac{1}{d} \frac{d [ D ]}{d t}$; Gives one value of rate whichever species you follow.
Rate law: $rate = k [A]^{x} [B]^{y}, order = x + y$; $x$ , $y$ from experiment, not from the balanced equation.
Units of k for order n: $(mol L^{- 1})^{1 - n} s^{- 1}$; n = 0: mol L⁻¹ s⁻¹; n = 1: s⁻¹; n = 2: L mol⁻¹ s⁻¹; n = 3: L² mol⁻² s⁻¹.
Order from initial rates: $\frac{r _{2}}{r _{1}} = (\frac{[ A ] _{2}}{[ A ] _{1}})^{x} (other concentrations fixed)$; Take logs if the ratio is not an obvious power.
Average rate: $r_{a v} = - \frac{Δ [ R ]}{Δ t}$; Instantaneous rate is the slope of the tangent, $- \frac{d [ R ]}{d t}$ .

How to solve Rate Law, Order and Molecularity questions

Three question types dominate: relating rates of different species, finding order and $k$ from data, and predicting how rate changes when concentrations or volume change. Theory questions ask you to contrast order with molecularity.

1For rate relations, divide each $d [] / d t$ by its coefficient and put a minus sign on reactants; set them all equal.
2For initial-rate tables, pick two experiments in which only one concentration changes. Write $\frac{r _{2}}{r _{1}} = (\frac{c _{2}}{c _{1}})^{x}$ and solve for $x$ . Repeat for each reactant.
3Write the rate law with the orders, add them for the overall order, then substitute one experiment to get $k$ with its units.
4For 'how does the rate change' questions, multiply the factors: $(factor in [A])^{x} \times (factor in [B])^{y}$ .
5If the volume of a gas vessel is reduced to $1/ f$ , every concentration rises by $f$ , so the rate rises by $f^{order}$ .
6Give the units of $k$ from the overall order, $(mol L^{- 1})^{1 - n} s^{- 1}$ , or work back from $k$ 's units to find the order.
7For mechanism questions, the rate law follows the slow step; a species that appears only after the slow step does not appear in the rate law.

Rate Law, Order and Molecularity in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Order tells you how sensitive the rate is to concentration

Suppose a pharmaceutical lab finds that when the concentration of a drug in a syrup is doubled, the rate at which it decomposes becomes four times as fast. The lab plans to market a diluted version with one third of the original concentration.

What is the order of decomposition with respect to the drug? By what factor will the initial decomposition rate change in the diluted syrup?

Try it first, then show the solutionSolution

$\frac{r _{2}}{r _{1}} = 4 = 2^{x}$ , so $x = 2$ : second order in the drug.
Rate law: rate $= k [drug]^{2}$ .
For one third the concentration: rate factor $= (\frac{1}{3})^{2} = \frac{1}{9}$ .

Answer: Second order; the initial rate falls to one ninth.

Memory hook: The order is the power: double the dose, square of two, four times the rate.

Tricks to remember Rate Law, Order and Molecularity

Short memory aids. Say each one aloud once, then use it on a question the same day.

Order is Observed, Molecularity is Mechanism
O for Order and Observed (from experiment, any value); M for Molecularity and Mechanism (from a single step, whole number 1-3).
Use it when: Every 'distinguish' question.
Units of k: 1 minus n
$k$ has units $(mol L^{- 1})^{1 - n} s^{- 1}$ : zero order keeps mol L⁻¹, first order loses it, second order flips it.
Use it when: Finding units or order instantly.
Slowest step sets the pace
Like a queue at a single ticket window: the slowest step controls the whole line.
Use it when: Mechanism-based questions.

Rate Law, Order and Molecularity: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Units of k · 1 mark · MCQBoard level

The unit of the rate constant of a second-order reaction is:
(a) s⁻¹ (b) mol L⁻¹ s⁻¹ (c) L mol⁻¹ s⁻¹ (d) L² mol⁻² s⁻¹

Try it first, then show the solutionSolution

$k = \frac{rate}{[ A ] ^{2}} = \frac{mol L ^{- 1} s ^{- 1}}{( mol L ^{- 1} ) ^{2}}$ .

Answer: (c) L mol⁻¹ s⁻¹.

Example 2: Relating rates of different species · 2 marks · Very short answerBoard level

For $2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)$ , the rate of formation of $N O_{2}$ at an instant is $2.0 \times 1 0^{- 4}$ mol L⁻¹ s⁻¹. Calculate (a) the rate of disappearance of $N_{2} O_{5}$ , (b) the rate of formation of $O_{2}$ and (c) the rate of the reaction.

Try it first, then show the solutionSolution

Rate $= - \frac{1}{2} \frac{d [ N _{2} O _{5} ]}{d t} = \frac{1}{4} \frac{d [ N O _{2} ]}{d t} = \frac{d [ O _{2} ]}{d t}$ (½ mark).
(c) Rate $= \frac{2.0 \times 1 0 ^{- 4}}{4} = 5.0 \times 1 0^{- 5}$ mol L⁻¹ s⁻¹ (½ mark).
(a) $- \frac{d [ N _{2} O _{5} ]}{d t} = 2 \times 5.0 \times 1 0^{- 5} = 1.0 \times 1 0^{- 4}$ mol L⁻¹ s⁻¹ (½ mark).
(b) $\frac{d [ O _{2} ]}{d t} = 5.0 \times 1 0^{- 5}$ mol L⁻¹ s⁻¹ (½ mark).

Answer: (a) $1.0 \times 1 0^{- 4}$ , (b) $5.0 \times 1 0^{- 5}$ , (c) $5.0 \times 1 0^{- 5}$ mol L⁻¹ s⁻¹.

Example 3: Rate law from initial rates · 3 marks · Short answerHarder board level

For $A + B \to$ products, the following initial rates were measured at 300 K:
Expt 1: [A] = 0.10 M, [B] = 0.10 M, rate = $2.0 \times 1 0^{- 3}$ mol L⁻¹ s⁻¹
Expt 2: [A] = 0.20 M, [B] = 0.10 M, rate = $8.0 \times 1 0^{- 3}$ mol L⁻¹ s⁻¹
Expt 3: [A] = 0.10 M, [B] = 0.20 M, rate = $4.0 \times 1 0^{- 3}$ mol L⁻¹ s⁻¹
Find the order with respect to A and B, write the rate law, and calculate $k$ with units.

Try it first, then show the solutionSolution

Expts 1 and 2 ([B] fixed): $\frac{8.0}{2.0} = 4 = 2^{x}$ , so $x = 2$ (½ mark).
Expts 1 and 3 ([A] fixed): $\frac{4.0}{2.0} = 2 = 2^{y}$ , so $y = 1$ (½ mark).
Rate $= k [A]^{2} [B]$ ; overall order 3 (1 mark).
$k = \frac{2.0 \times 1 0 ^{- 3}}{( 0.10 ) ^{2} ( 0.10 )} = 2.0$ L² mol⁻² s⁻¹ (1 mark).

Answer: Second order in A, first order in B; rate $= k [A]^{2} [B]$ ; $k = 2.0$ L² mol⁻² s⁻¹.

Example 4: Changing concentrations and volume · 3 marks · Short answerHarder board level

The rate law for a gas-phase reaction is rate $= k [A] [B]^{2}$ . How will the rate change if (a) [B] is tripled and [A] is halved, (b) the volume of the reaction vessel is halved at constant temperature?

Try it first, then show the solutionSolution

(a) New rate $= k (\frac{[ A ]}{2}) (3 [B])^{2} = \frac{9}{2} k [A] [B]^{2}$ (1 mark).
So the rate becomes 4.5 times (½ mark).
(b) Halving the volume doubles every concentration: rate $= k (2 [A]) (2 [B])^{2} = 8 k [A] [B]^{2}$ (1 mark).
The rate becomes 8 times, i.e. $2^{order}$ with order 3 (½ mark).

Answer: (a) 4.5 times. (b) 8 times.

Example 5: Checking a mechanism against the rate law · 3 marks · Short answerHOTS / competency

For $N O_{2} (g) + C O (g) \to N O (g) + C O_{2} (g)$ below 500 K, the rate law found by experiment is rate $= k [N O_{2}]^{2}$ . A proposed mechanism is:
Step 1 (slow): $N O_{2} + N O_{2} \to N O_{3} + N O$
Step 2 (fast): $N O_{3} + C O \to N O_{2} + C O_{2}$
(a) Show that the steps add up to the overall reaction. (b) Is the mechanism consistent with the rate law? (c) What are the molecularity of step 1 and the order of the overall reaction? (d) Why does [CO] not appear in the rate law?

Try it first, then show the solutionSolution

(a) Adding: $2 N O_{2} + N O_{3} + C O \to N O_{3} + N O + N O_{2} + C O_{2}$ ; cancelling $N O_{3}$ and one $N O_{2}$ gives $N O_{2} + C O \to N O + C O_{2}$ (½ mark).
(b) The slow step decides the rate: rate $= k [N O_{2}]^{2}$ , matching the experiment (1 mark).
(c) Step 1 is bimolecular (two $N O_{2}$ collide); overall order = 2 (½ mark).
(d) CO reacts only in the fast step after the rate-determining step, so its concentration does not affect the overall rate (1 mark).

Answer: (a) Steps add to the overall equation. (b) Yes. (c) Molecularity 2; order 2. (d) CO enters only after the slow step.

Where students lose marks in Rate Law, Order and Molecularity

The slips examiners see most often, and what to write instead.

Mistake: Taking orders from the balanced equation
Coefficients look like the obvious powers.
Wrong working
$C H C l_{3} + C l_{2} \to C C l_{4} + H C l$ : rate $= k [C H C l_{3}] [C l_{2}]$
Right working
Rate $= k [C H C l_{3}] [C l_{2}]^{1/2}$ (found by experiment)
Do this: Use coefficients only for elementary steps; otherwise orders come from data.
Usually costs: 1 mark.
Mistake: Forgetting to divide by the stoichiometric coefficient
Wrong working
Rate of reaction $= \frac{d [ N O _{2} ]}{d t} = 2.0 \times 1 0^{- 4}$
Right working
Rate $= \frac{1}{4} \frac{d [ N O _{2} ]}{d t} = 5.0 \times 1 0^{- 5}$ mol L⁻¹ s⁻¹
Do this: Write the full rate expression with fractions before substituting.
Usually costs: 1 mark.
Mistake: Wrong units of k
Students write mol L⁻¹ s⁻¹ for every order.
Do this: Use $(mol L^{- 1})^{1 - n} s^{- 1}$ with the overall order.
Usually costs: ½ mark.
Mistake: Saying molecularity can be zero or fractional
Order and molecularity are mixed up.
Do this: Zero and fractional values belong only to order.
Usually costs: 1 mark in MCQs and distinguish questions.

Exam tips

In initial-rate tables, state which two experiments you compare for each reactant.
Always give units of $k$ in the final answer.
For distinguish questions, write points in a two-column format (order | molecularity).
Quote NCERT's pseudo first order examples: hydrolysis of ethyl acetate and inversion of cane sugar.
Remember that rate law and order are experimental quantities: write 'determined experimentally' in definitions.

Practice questions on Rate Law, Order and Molecularity

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which statement about molecularity is correct?
(a) It can be zero.
(b) It can be fractional.
(c) It applies only to elementary reactions.
(d) It is found only by experiment.
Hint
Molecularity counts colliding particles in one step.
Answer
(c).
Q2 · 2 marks · Very short answerBoard level
The reaction $2 N O (g) + O_{2} (g) \to 2 N O_{2} (g)$ is believed to occur in a single elementary step. Write its rate law and state its order and molecularity. Why can the order of a complex reaction be fractional while molecularity cannot?
Hint
For an elementary step, the coefficients are the powers.
Answer
Rate $= k [N O]^{2} [O_{2}]$ ; order 3 and molecularity 3 (termolecular). Order is found from experiment and, for a complex reaction, is set by its mechanism, so it can be fractional (e.g. $C H C l_{3} + C l_{2}$ : order 3/2). Molecularity counts the whole particles colliding in one step, so it is always 1, 2 or 3.
Q3 · 2 marks · Very short answerHarder board level
Identify the order of the reaction from the units of its rate constant: (a) $k = 4.5 \times 1 0^{- 3}$ mol L⁻¹ s⁻¹, (b) $k = 1.6 \times 1 0^{- 2}$ L² mol⁻² s⁻¹. Also write the units of $k$ for a reaction of order 3/2.
Hint
Units are $(mol L^{- 1})^{1 - n} s^{- 1}$ .
Answer
(a) Zero order. (b) Third order. For $n = 3/2$ : $(mol L^{- 1})^{- 1/2} s^{- 1} = mol^{- 1/2} L^{1/2} s^{- 1}$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Hydrolysis of ethyl acetate in a large excess of water follows first-order kinetics.
Reason (R): The concentration of water hardly changes during the reaction.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Pseudo first order.
Answer
(a). It is a pseudo first order reaction.
Q5 · 2 marks · Very short answerHarder board level
For $A \to 2 B$ , the concentration of A falls from 0.80 mol L⁻¹ to 0.50 mol L⁻¹ in 15 minutes. Calculate the average rate of the reaction in mol L⁻¹ min⁻¹ and mol L⁻¹ s⁻¹, and the average rate of formation of B.
Hint
Rate $= - \frac{Δ [ A ]}{Δ t} = \frac{1}{2} \frac{Δ [ B ]}{Δ t}$ .
Answer
Rate $= \frac{0.30}{15} = 0.020$ mol L⁻¹ min⁻¹ $= 3.3 \times 1 0^{- 4}$ mol L⁻¹ s⁻¹. Rate of formation of B $= 0.040$ mol L⁻¹ min⁻¹.
Q6 · 3 marks · Short answerHarder board level
Explain with NCERT's example how a complex reaction's rate law is decided by its slowest step. Why can molecularity not be assigned to the overall reaction $2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$ (catalysed by $I^{-}$ )?
Hint
Two steps, one slow.
Answer
Mechanism: $H_{2} O_{2} + I^{-} \to H_{2} O + I O^{-}$ (slow) then $H_{2} O_{2} + I O^{-} \to H_{2} O + I^{-} + O_{2}$ (fast). The slow step controls the rate, so rate $= k [H_{2} O_{2}] [I^{-}]$ , first order in each, second order overall. The overall reaction is complex (two steps), and molecularity is defined only for individual elementary steps; each step here is bimolecular.
Q7 · 3 marks · Short answerHOTS / competency
For the reaction $X + Y \to Z$ , initial rates were:
Expt 1: [X] = 0.010 M, [Y] = 0.020 M, rate = $3.0 \times 1 0^{- 4}$ mol L⁻¹ s⁻¹
Expt 2: [X] = 0.020 M, [Y] = 0.020 M, rate = $6.0 \times 1 0^{- 4}$ mol L⁻¹ s⁻¹
Expt 3: [X] = 0.020 M, [Y] = 0.040 M, rate = $6.0 \times 1 0^{- 4}$ mol L⁻¹ s⁻¹
(a) Write the rate law and find $k$ . (b) Predict the rate when [X] = 0.040 M and [Y] = 0.010 M. (c) What does the result for Y suggest about the mechanism?
Hint
Changing [Y] does not change the rate.
Answer
(a) Order 1 in X (rate doubles when [X] doubles), 0 in Y; rate $= k [X]$ ; $k = \frac{3.0 \times 1 0 ^{- 4}}{0.010} = 0.030$ s⁻¹. (b) Rate $= 0.030 \times 0.040 = 1.2 \times 1 0^{- 3}$ mol L⁻¹ s⁻¹. (c) Y is not involved in the slow step; it reacts in a fast step after the rate-determining step, so the reaction is complex.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose an industrial lab studies the decomposition of ammonia on a hot platinum surface at high pressure: $2 N H_{3} (g) P t N_{2} (g) + 3 H_{2} (g)$ . They find that the rate does not change when the pressure of ammonia is changed, and $k = 1.6 \times 1 0^{- 4}$ mol L⁻¹ s⁻¹.
(a) What is the order of the reaction? Give a reason for this behaviour. (1 mark)
(b) Calculate the rates of formation of $N_{2}$ and $H_{2}$ and the rate of disappearance of $N H_{3}$ . (2 marks)
(c) If the platinum surface area were doubled, would you expect $k$ to change? Explain. (1 mark)
Hint
Rate $= - \frac{1}{2} \frac{d [ N H _{3} ]}{d t} = \frac{d [ N _{2} ]}{d t} = \frac{1}{3} \frac{d [ H _{2} ]}{d t} = k$ .
Answer
(a) Zero order. At high pressure the metal surface is fully covered (saturated) with ammonia, so adding more gas does not increase the rate. (b) $\frac{d [ N _{2} ]}{d t} = 1.6 \times 1 0^{- 4}$ , $\frac{d [ H _{2} ]}{d t} = 4.8 \times 1 0^{- 4}$ , $- \frac{d [ N H _{3} ]}{d t} = 3.2 \times 1 0^{- 4}$ mol L⁻¹ s⁻¹. (c) Yes; the rate is limited by the available surface, so more surface means a larger observed rate constant.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Units of k for a zero order reaction?Show answer
mol L⁻¹ s⁻¹.
2.Can order be fractional?Show answer
Yes.
3.Maximum molecularity normally observed?Show answer
3 (termolecular).
4.What is a rate-determining step?Show answer
The slowest elementary step of a mechanism; it controls the overall rate.
5.Example of a pseudo first order reaction?Show answer
Acid hydrolysis of ethyl acetate in excess water (or inversion of cane sugar).
6.Rate change if all concentrations double for a third order reaction?Show answer
8 times.

Rate Law, Order and Molecularity: frequently asked questions

What is a rate law in Class 12 Chemistry?

It is the experimentally found expression of the reaction rate in terms of the concentrations of reactants, each raised to some power: rate = k[A]^x[B]^y.

What is the difference between order and molecularity?

Order is the sum of powers in the experimental rate law and can be zero or fractional. Molecularity is the number of species that collide in one elementary step and is always 1, 2 or 3.

How do I find the order of a reaction from initial-rate data?

Compare two experiments in which only one concentration changes: if the concentration is multiplied by f and the rate by f^x, the order in that reactant is x.

What is a pseudo first order reaction?

A reaction that involves two reactants but behaves as first order because one reactant is in such large excess that its concentration is effectively constant.

Why can't molecularity be determined for a complex reaction?

A complex reaction occurs in several elementary steps with different molecularities; molecularity is defined only for a single elementary step.