ChemistryISC Class 12 · Chemical Kinetics

Rate Law, Order and Molecularity: ISC Class 12 Chemical Kinetics

Updated 15 October 2026 · Fact-checked

A rate law, rate = k[A]ˣ[B]ʸ, is found by experiment; the order is x + y and can be zero, fractional or even negative. Molecularity is the number of particles colliding in one elementary step and is always a small whole number. Order comes from data; molecularity comes from the mechanism.

Understand Rate Law, Order and Molecularity

Rate of reaction. For $a A + b B \to c C + d D$ , the rate is the change in concentration per unit time, divided by the coefficient so that it is the same whichever substance you follow: $rate = - \frac{1}{a} \frac{d [ A ]}{d t} = - \frac{1}{b} \frac{d [ B ]}{d t} = \frac{1}{c} \frac{d [ C ]}{d t} = \frac{1}{d} \frac{d [ D ]}{d t}$ . The minus sign makes the rate positive for reactants. Units: mol L⁻¹ s⁻¹ (or min⁻¹). The average rate is $Δ [] /Δ t$ over an interval; the instantaneous rate is the slope of the tangent to the concentration-time curve at one instant.

Law of mass action and the rate law. Guldberg and Waage's law of mass action says the rate is proportional to the product of the active masses (molar concentrations) of the reactants. For real reactions the powers cannot be read from the equation; they are found by experiment. The rate law is $rate = k [A]^{x} [B]^{y}$ , where $k$ is the rate constant (specific reaction rate: the rate when every concentration is 1 mol L⁻¹). $k$ depends on temperature and catalyst, not on concentration.

Order. $x$ is the order with respect to A, $y$ with respect to B, and $x + y$ is the overall order. Order is an experimental quantity. It may be 0, 1, 2, 3, a fraction ( $\frac{3}{2}$ for $C H_{3} C H O \to C H_{4} + C O$ ) or even negative (a product that slows the reaction). The units of k depend on order: $(mol L^{- 1})^{1 - n} s^{- 1}$ , i.e. mol L⁻¹ s⁻¹ (zero), s⁻¹ (first), L mol⁻¹ s⁻¹ (second), L² mol⁻² s⁻¹ (third). So a value of $k$ in s⁻¹ tells you at once that the reaction is first order.

Order from initial-rate data. Change one concentration at a time and compare initial rates. If doubling [A] doubles the rate, the order in A is 1; if it quadruples the rate, 2; if the rate is unchanged, 0. In general $\frac{r _{2}}{r _{1}} = (\frac{[ A ] _{2}}{[ A ] _{1}})^{x}$ . Then put one set of data into the rate law to find $k$ with its units.

Molecularity. An elementary reaction happens in a single step. Its molecularity is the number of reacting species (atoms, ions or molecules) that must collide at the same time: unimolecular ( $N H_{4} N O_{2} \to N_{2} + 2 H_{2} O$ ), bimolecular ( $2 H I \to H_{2} + I_{2}$ ), termolecular ( $2 N O + O_{2} \to 2 N O_{2}$ ). It is always a whole number, never zero, and rarely above 3 because a collision of more than three particles at once is very unlikely. Molecularity has meaning only for elementary steps.

Complex reactions and the slow step. Most reactions go through several elementary steps. The slowest step controls the overall rate and is called the rate-determining step; the rate law of the overall reaction is that of the slow step (written in terms of reactants). For example, $2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$ in the presence of iodide has rate $= k [H_{2} O_{2}] [I^{-}]$ , first order in each, because the slow step is $H_{2} O_{2} + I^{-} \to H_{2} O + I O^{-}$ , followed by the fast step $H_{2} O_{2} + I O^{-} \to H_{2} O + I^{-} + O_{2}$ . IO⁻ is an intermediate and I⁻ a catalyst.

Pseudo first order. When one reactant is in large excess its concentration hardly changes, so it is absorbed into the constant. Acid hydrolysis of ethyl acetate, $C H_{3} C O O C_{2} H_{5} + H_{2} O \to C H_{3} C O O H + C_{2} H_{5} O H$ , has rate $= k^{'} [C H_{3} C O O C_{2} H_{5}]$ with $k^{'} = k [H_{2} O]$ . Its molecularity is 2 but its order is 1. Inversion of cane sugar in acid is another example.

Order versus molecularity. Order: experimental, can be zero or fractional, applies to the overall reaction. Molecularity: theoretical, a positive integer, applies only to an elementary step. For an elementary step the two are equal.

Rate Law, Order and Molecularity: formulas and key facts

Rate in terms of each species: $rate = - \frac{1}{a} \frac{d [ A ]}{d t} = - \frac{1}{b} \frac{d [ B ]}{d t} = \frac{1}{c} \frac{d [ C ]}{d t} = \frac{1}{d} \frac{d [ D ]}{d t}$; For $a A + b B \to c C + d D$ .
Rate law: $rate = k [A]^{x} [B]^{y}, order = x + y$; $x$ and $y$ come from experiment, not from the coefficients.
Order from two experiments: $\frac{r _{2}}{r _{1}} = (\frac{[ A ] _{2}}{[ A ] _{1}})^{x} ([B] constant)$; Take logs if the ratio is not an obvious power.
Units of the rate constant: $k : (mol L^{- 1})^{1 - n} s^{- 1}$; Zero: mol L⁻¹ s⁻¹; first: s⁻¹; second: L mol⁻¹ s⁻¹; third: L² mol⁻² s⁻¹.
Pseudo first order: $rate = k [A] [B] \approx k^{'} [A], k^{'} = k [B]_{e x cess}$; Order 1, molecularity 2.

How to solve Rate Law, Order and Molecularity questions

ISC sets three kinds of questions here: rate expressions from an equation, order and k from tables of initial rates, and reasoning on order, molecularity and mechanisms.

1For rate expressions, divide each $\frac{d [ ]}{d t}$ by its coefficient and put a minus sign before reactants; then scale by coefficients to convert one rate into another.
2For data tables, pick two experiments in which only one concentration changes; write the ratio of rates as (ratio of concentrations)ˣ and solve for $x$ .
3Repeat for each reactant, then write the full rate law with $k$ in it, and add the powers for the overall order.
4Substitute one experiment's values to find $k$ ; check with a second experiment; give the units from the overall order.
5To predict a new rate, multiply by (factor)ˣ for each reactant; halving the volume of a gas doubles every concentration.
6For a mechanism, the rate law follows the slow step; a species formed and then used up is an intermediate; one used and regenerated is a catalyst.

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.

A reactant in large excess gives pseudo first order kinetics

Suppose a water-treatment engineer removes a trace pollutant P from water with chlorine, which is kept at a steady 2.0 × 10⁻³ mol L⁻¹, far more than the pollutant. Assume the rate law is rate = k[P][Cl₂] with k = 0.50 L mol⁻¹ s⁻¹.

Find the pseudo first order rate constant and the half-life of P. What happens to the half-life if the chlorine level is doubled?

Try it first, then show the solutionSolution

Chlorine is in large excess, so [Cl₂] is almost constant: rate $= k^{'} [P]$ with $k^{'} = k [C l_{2}] = 0.50 \times 2.0 \times 1 0^{- 3} = 1.0 \times 1 0^{- 3}$ s⁻¹.
For first order kinetics, $t_{1/2} = \frac{0.693}{k ^{'}} = 693$ s, about 11.5 minutes.
Doubling [Cl₂] doubles $k^{'}$ to $2.0 \times 1 0^{- 3}$ s⁻¹.
The half-life halves to about 347 s.

Answer: $k^{'} = 1.0 \times 1 0^{- 3}$ s⁻¹; $t_{1/2} \approx 693$ s. Doubling the chlorine halves the half-life to about 347 s.

Memory hook: A reactant in huge excess hides inside the rate constant: the reaction looks first order.

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 Modelled
Order: Observed in the lab, can be 0 or ½. Molecularity: Modelled from one step, always 1, 2 or 3.
Use it when: Two-mark 'difference between' questions.
Units ladder
Each step up in order adds 'L mol⁻¹' to k: mol L⁻¹ s⁻¹ → s⁻¹ → L mol⁻¹ s⁻¹ → L² mol⁻² s⁻¹.
Use it when: Identifying order from units of k.
Double and watch
Double a concentration: rate ×1 means zero order, ×2 first, ×4 second, ×8 third.
Use it when: Reading initial-rate tables quickly.

Rate Law, Order and Molecularity: solved examples

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

Example 1: Molecularity of an elementary step · 1 mark · MCQBoard level

If the reaction $2 N O (g) + O_{2} (g) \to 2 N O_{2} (g)$ takes place in a single elementary step, its molecularity is:
(a) 1
(b) 2
(c) 3
(d) 4

Try it first, then show the solutionSolution

Molecularity of an elementary step = number of species colliding at once.
Two NO molecules and one O₂ molecule: 3.

Answer: (c) 3 (termolecular).

Example 2: Converting between rates · 2 marks · Short answerBoard level

For $4 N H_{3} (g) + 5 O_{2} (g) \to 4 N O (g) + 6 H_{2} O (g)$ , NO is formed at $3.6 \times 1 0^{- 3}$ mol L⁻¹ s⁻¹. Calculate (a) the rate of disappearance of O₂, (b) the rate of formation of H₂O and (c) the rate of reaction.

Try it first, then show the solutionSolution

Rate $= \frac{1}{4} \frac{d [ N O ]}{d t} = \frac{3.6 \times 1 0 ^{- 3}}{4} = 9.0 \times 1 0^{- 4}$ mol L⁻¹ s⁻¹ (½ mark).
(a) $- \frac{d [ O _{2} ]}{d t} = 5 \times 9.0 \times 1 0^{- 4} = 4.5 \times 1 0^{- 3}$ mol L⁻¹ s⁻¹ (½ mark).
(b) $\frac{d [ H _{2} O ]}{d t} = 6 \times 9.0 \times 1 0^{- 4} = 5.4 \times 1 0^{- 3}$ mol L⁻¹ s⁻¹ (½ mark).
(c) Rate of reaction $= 9.0 \times 1 0^{- 4}$ mol L⁻¹ s⁻¹ (½ mark).

Answer: (a) $4.5 \times 1 0^{- 3}$ ; (b) $5.4 \times 1 0^{- 3}$ ; (c) $9.0 \times 1 0^{- 4}$ mol L⁻¹ s⁻¹.

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

For the reaction $A + 2 B \to$ products, the following initial rates were measured at 300 K:
Exp 1: [A] = 0.10, [B] = 0.20 mol L⁻¹, rate = $1.5 \times 1 0^{- 3}$ mol L⁻¹ s⁻¹
Exp 2: [A] = 0.30, [B] = 0.20 mol L⁻¹, rate = $4.5 \times 1 0^{- 3}$ mol L⁻¹ s⁻¹
Exp 3: [A] = 0.10, [B] = 0.40 mol L⁻¹, rate = $6.0 \times 1 0^{- 3}$ mol L⁻¹ s⁻¹
Find the order with respect to A and B, write the rate law and calculate k with its units.

Try it first, then show the solutionSolution

Exp 2 ÷ Exp 1: [A] trebled, rate trebled ( $3^{x} = 3$ ), so first order in A (½ mark).
Exp 3 ÷ Exp 1: [B] doubled, rate ×4 ( $2^{y} = 4$ ), so second order in B (½ mark).
Rate $= k [A] [B]^{2}$ ; overall order 3 (1 mark).
$k = \frac{1.5 \times 1 0 ^{- 3}}{0.10 \times ( 0.20 ) ^{2}} = 0.375$ L² mol⁻² s⁻¹ (1 mark).

Answer: First order in A, second order in B; rate $= k [A] [B]^{2}$ ; $k = 0.375$ L² mol⁻² s⁻¹.

Example 4: Pseudo first order hydrolysis · 3 marks · Short answerHarder board level

The acid-catalysed hydrolysis of ethyl acetate in dilute aqueous solution, $C H_{3} C O O C_{2} H_{5} + H_{2} O \to C H_{3} C O O H + C_{2} H_{5} O H$ , follows first order kinetics with an observed rate constant of $4.44 \times 1 0^{- 5}$ s⁻¹. (a) Explain why it is first order although two molecules react. (b) Taking [H₂O] = 55.5 mol L⁻¹, calculate the true (second order) rate constant. (c) State its order and molecularity.

Try it first, then show the solutionSolution

(a) Water is the solvent and is in huge excess; its concentration hardly changes, so rate $= k [es t er] [H_{2} O] = k^{'} [es t er]$ (1 mark).
(b) $k = \frac{k ^{'}}{[ H _{2} O ]} = \frac{4.44 \times 1 0 ^{- 5}}{55.5} = 8.0 \times 1 0^{- 7}$ L mol⁻¹ s⁻¹ (1 mark).
(c) Order 1 (pseudo first order); molecularity 2 (1 mark).

Answer: (a) Water is in large excess. (b) $k = 8.0 \times 1 0^{- 7}$ L mol⁻¹ s⁻¹. (c) Order 1, molecularity 2.

Example 5: From a mechanism to a rate law · 5 marks · Long answerHOTS / competency

The decomposition $2 N O_{2} C l (g) \to 2 N O_{2} (g) + C l_{2} (g)$ is thought to occur in two steps:
Step 1: $N O_{2} C l \to N O_{2} + C l$ (slow)
Step 2: $N O_{2} C l + C l \to N O_{2} + C l_{2}$ (fast)
(a) Show that the two steps add up to the overall equation and name the intermediate. [1]
(b) Write the rate law and give the overall order. [1]
(c) Give the molecularity of each step. Why is it wrong to say the overall reaction is bimolecular? [2]
(d) How does the rate change if the volume of the container is halved at constant temperature? [1]

Try it first, then show the solutionSolution

(a) Adding: $2 N O_{2} C l + C l \to 2 N O_{2} + C l_{2} + C l$ ; Cl cancels, giving the overall equation. Cl is formed and then used up: the intermediate (1 mark).
(b) The slow step decides the rate: rate $= k [N O_{2} C l]$ , first order (1 mark).
(c) Step 1 is unimolecular, step 2 bimolecular (1 mark). Molecularity applies only to elementary steps; the overall reaction is complex, so it has an order (1) but no single molecularity (1 mark).
(d) Halving the volume doubles [NO₂Cl]; for a first order reaction the rate doubles (1 mark).

Answer: (a) Cl is the intermediate. (b) Rate = k[NO₂Cl], first order. (c) Steps are uni- and bimolecular; the overall reaction has no single molecularity. (d) The rate doubles.

Where students lose marks in Rate Law, Order and Molecularity

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

Mistake: Leaving k out of the rate law
ISC examiners reported rate laws written as R = [A]²[B] (2024).
Wrong working
Rate $= [A]^{2} [B]$
Right working
Rate $= k [A]^{2} [B]$
Do this: A rate law always has a rate constant in front.
Usually costs: 1 mark, and the later value of k.
Mistake: Wrong overall order
Examiners found 'second order' written for a reaction second order in A and first order in B, and wrong orders after an error in one reactant (2024, 2025).
Do this: Overall order is the sum of all the powers in the rate law: $2 + 1 = 3$ .
Usually costs: 1 mark.
Mistake: Not linking the units of k to the order
Examiners noted 'zero' written for a reaction with k in s⁻¹, and second order chosen for k in L² mol⁻² s⁻¹ (2024, 2025).
Do this: Learn the units ladder: s⁻¹ means first order; L² mol⁻² s⁻¹ means third order.
Usually costs: 1 mark per blank.
Mistake: Missing the pseudo-order when one reactant is in excess
In 2025 many candidates could not give the order when [NO] was in large excess.
Do this: A reactant in large excess drops out of the rate law; the order is the sum of the remaining powers.
Usually costs: 1 mark.
Mistake: Giving molecularity for a complex reaction
Examiners noted that candidates found the order correctly but could not state the molecularity of the given elementary step.
Do this: Molecularity belongs to one elementary step: count the species on its left side.
Usually costs: 1 mark.

Exam tips

In data questions, state which two experiments you compare and show the ratio; examiners give a mark for each order.
Always give units with k; work them out from the overall order.
When the equation is given, do not assume the orders equal the coefficients unless the reaction is stated to be elementary.
For 'how will the rate change' questions, write the new rate law with the factors substituted, then simplify.
Quote a standard pseudo first order example (acid hydrolysis of ethyl acetate or inversion of cane sugar) in theory answers.

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
The unit of the rate of a reaction is:
(a) s⁻¹
(b) mol L⁻¹ s⁻¹
(c) L mol⁻¹ s⁻¹
(d) mol⁻¹ L s
Hint
Rate is concentration change per unit time.
Answer
(b) mol L⁻¹ s⁻¹.
Q2 · 2 marks · Short answerBoard level
Give two differences between order and molecularity of a reaction.
Hint
Think about how each is found and what values each can take.
Answer
(i) Order is found by experiment; molecularity is a theoretical count of species colliding in an elementary step. (ii) Order can be zero, fractional or negative; molecularity is always a positive whole number (1, 2 or 3). Also, order applies to the overall reaction, molecularity only to elementary steps.
Q3 · 2 marks · Short answerHarder board level
For a reaction with rate $= k [A]^{2} [B]$ , by what factor does the rate change if [A] is tripled and [B] is halved at the same time?
Hint
Multiply the factor for each reactant raised to its order.
Answer
New rate $= k (3 [A])^{2} (\frac{1}{2} [B]) = 9 \times \frac{1}{2} = 4.5$ times the original rate.
Q4 · 3 marks · Short answerHarder board level
For $X + Y \to$ products at 310 K:
Exp 1: [X] = 0.010, [Y] = 0.010 mol L⁻¹, rate = $2.4 \times 1 0^{- 5}$ mol L⁻¹ s⁻¹
Exp 2: [X] = 0.020, [Y] = 0.010 mol L⁻¹, rate = $4.8 \times 1 0^{- 5}$ mol L⁻¹ s⁻¹
Exp 3: [X] = 0.020, [Y] = 0.030 mol L⁻¹, rate = $4.8 \times 1 0^{- 5}$ mol L⁻¹ s⁻¹
Determine the rate law and the value and unit of k.
Hint
Compare 2 with 1 for X, and 3 with 2 for Y.
Answer
Exp 2/1: [X] doubled, rate doubled: first order in X. Exp 3/2: [Y] trebled, rate unchanged: zero order in Y. Rate $= k [X]$ ; $k = \frac{2.4 \times 1 0 ^{- 5}}{0.010} = 2.4 \times 1 0^{- 3}$ s⁻¹.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): For an elementary reaction, the order is equal to the molecularity.
Reason (R): In an elementary step, the rate depends on the concentration of each species that collides in that step.
(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
For a single step, the rate law can be written from the equation.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
(a) Write the units of the rate constant for zero, first, second and third order reactions, with concentration in mol L⁻¹ and time in seconds. (b) A rate constant is $3.2 \times 1 0^{- 2}$ L mol⁻¹ min⁻¹. State the order and express k in L mol⁻¹ s⁻¹.
Hint
Use $(mol L^{- 1})^{1 - n}$ s⁻¹.
Answer
(a) Zero: mol L⁻¹ s⁻¹; first: s⁻¹; second: L mol⁻¹ s⁻¹; third: L² mol⁻² s⁻¹. (b) Second order; $k = \frac{3.2 \times 1 0 ^{- 2}}{60} = 5.3 \times 1 0^{- 4}$ L mol⁻¹ s⁻¹.
Q7 · 3 marks · Short answerHOTS / competency
For a gas-phase reaction $2 A \to$ products, the initial rate is 0.40 mm Hg s⁻¹ when the pressure of A is 100 mm Hg, and 0.10 mm Hg s⁻¹ when it is 50 mm Hg, at the same temperature. (a) Find the order of the reaction. (b) Predict the effect on the initial rate of halving the volume of the vessel. (c) Can you deduce the molecularity from these data?
Hint
Partial pressure is proportional to concentration.
Answer
(a) Halving the pressure divides the rate by 4, so $2^{n} = 4$ : second order. (b) Halving the volume doubles the pressure (concentration) of A, so the rate becomes 4 times. (c) No. Order is experimental; molecularity can be stated only if the reaction is known to be a single elementary step.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a chemist studies the decomposition of ozone, $2 O_{3} \to 3 O_{2}$ , and finds the rate law rate $= k \frac{[ O _{3} ] ^{2}}{[ O _{2} ]}$ . The proposed mechanism is: $O_{3} ⇌ O_{2} + O$ (fast, equilibrium) followed by $O + O_{3} \to 2 O_{2}$ (slow).
(i) What is the overall order? [1]
(ii) What does the negative power of [O₂] tell you about the effect of the product on the rate? [1]
(iii) Name the intermediate and state the molecularity of the slow step. [1]
Hint
Add the powers, including the negative one.
Answer
(i) $2 + (- 1) = 1$ . (ii) O₂ slows the reaction: as O₂ builds up, the rate falls (it shifts the fast equilibrium back and lowers [O]). (iii) The O atom is the intermediate; the slow step is bimolecular.

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.Define order of a reaction.Show answer
The sum of the powers of the concentration terms in the experimentally determined rate law.
2.Unit of k for a second order reaction?Show answer
L mol⁻¹ s⁻¹.
3.Can molecularity be zero?Show answer
No; it is always a positive whole number.
4.What is the rate-determining step?Show answer
The slowest elementary step in a mechanism, which controls the overall rate.
5.Give an example of a reaction with fractional order.Show answer
$C H_{3} C H O \to C H_{4} + C O$ , order $\frac{3}{2}$ .
6.Order and molecularity of acid hydrolysis of ethyl acetate?Show answer
Order 1 (pseudo first order), molecularity 2.

Rate Law, Order and Molecularity: frequently asked questions

What is the difference between rate and rate constant?

Rate is the speed of the reaction at given concentrations and changes as the reaction proceeds. The rate constant is the rate when all concentrations are 1 mol L⁻¹; it changes only with temperature or catalyst.

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

Compare two experiments in which only one concentration changes: the ratio of rates equals the ratio of concentrations raised to the order.

Why is molecularity rarely more than three?

A reaction step needs the reacting species to collide at the same instant; collisions of more than three particles at once are extremely unlikely.

Can the order of a reaction be negative?

Yes. If a substance slows the reaction, its power in the rate law is negative; for example, O₂ in the decomposition of ozone.

What is a pseudo first order reaction?

A reaction of higher molecularity that behaves as first order because the other reactant is in large excess, so its concentration stays almost constant.