What is a simple rate equation in the context of A-Level Chemistry?

In A-Level Chemistry, a simple rate equation expresses the rate of a chemical reaction as a function of the concentration of reactants. It is typically written in the form: rate = k[A]^m[B]^n, where k is the rate constant, [A] and [B] are the concentrations of the reactants, and m and n are the orders of reaction with respect to each reactant.

How do you determine the order of reaction in a chemical process?

The order of reaction can be determined experimentally by observing how the rate of reaction changes with varying concentrations of reactants. This is often done using the method of initial rates, where the initial rate of reaction is measured for different initial concentrations. The order with respect to each reactant is the exponent to which its concentration is raised in the rate equation.

What is the significance of the rate constant in reaction kinetics?

The rate constant, denoted as k, is a crucial parameter in reaction kinetics that provides information about the speed of a reaction. It is specific to a particular reaction at a given temperature and is independent of the concentrations of reactants. The value of k can be influenced by factors such as temperature and the presence of a catalyst.

Can you explain the difference between zero, first, and second order reactions?

In zero-order reactions, the rate is independent of the concentration of reactants, meaning the rate remains constant over time. In first-order reactions, the rate is directly proportional to the concentration of one reactant. For second-order reactions, the rate is proportional to the square of the concentration of one reactant or to the product of the concentrations of two different reactants.

Why is understanding reaction orders important in A-Level Chemistry?

Understanding reaction orders is essential in A-Level Chemistry because it helps predict how changes in concentration affect the rate of reaction, which is crucial for controlling industrial processes and laboratory experiments. It also aids in elucidating reaction mechanisms, providing insights into the steps involved in a chemical reaction.

Why is "Reading orders from the stoichiometric coefficients." a common mistake in Simple rate equations, orders of reaction and rate constants?

Why it happens: Assuming order = coefficient. How to avoid it: Orders are found ONLY by experiment (initial rates / graphs). Source: 9701 Examiner Reports 2022-2024

Why is "Quoting k without units or with the wrong units." a common mistake in Simple rate equations, orders of reaction and rate constants?

Why it happens: Memorising one set of units. How to avoid it: Derive units from the rate equation each time (they depend on overall order).

Why is "Using t½ = 0.693/k for any order." a common mistake in Simple rate equations, orders of reaction and rate constants?

Why it happens: Applying the first-order formula generally. How to avoid it: The constant-half-life relationship is for FIRST order only.

Why is "Including species that appear after the RDS in the rate equation." a common mistake in Simple rate equations, orders of reaction and rate constants?

Why it happens: Matching the rate equation to the overall equation. How to avoid it: Only species up to and including the rate-determining step appear in the rate equation.

Reaction Kinetics(A-Level Physical Chemistry)Cambridge International A Level Chemistry 9701

Simple rate equations, orders of reaction and rate constants

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Rate Equations, Orders and Rate Constants — Cambridge International AS & A Level Chemistry 9701 (2025-2027 syllabus)

Finding orders from initial-rate data and graphs, writing the rate equation, working out the rate constant with units, using first-order half-life, and linking orders to the mechanism.

What you’ll learn

Mapped to the Cambridge International A Level 9701 syllabus (2025-2027).

26.1 — Define order of reaction, rate constant and rate-determining step.
26.1 — Deduce orders and the rate equation from initial-rate or graphical data.
26.1 — Calculate the rate constant (with units) and use the half-life of a first-order reaction.
26.1 — Relate the rate equation to a reaction mechanism.

Orders and the rate equation

Orders come only from experiment; compare runs where one concentration changes.

The rate equation links the rate to reactant concentrations:

$\text{rate} = k[\text{A}]^m[\text{B}]^n$

The powers m and n are the orders with respect to A and B; their sum is the overall order. Crucially, orders are found only by experiment — they are not the coefficients in the chemical equation.

Initial-rates method. Compare experiments in which only one concentration changes:

rate doubles when [X] doubles → first order in X.
rate ×4 when [X] doubles → second order.
rate unchanged → zero order.

Worked. If doubling [A] doubles the rate, and doubling [B] quadruples it, then rate = k[A][B]² (overall third order).

rate = k[A]^m[B]^n; overall order = m + n.
Orders found by experiment, not from coefficients.
×2 rate → 1st; ×4 → 2nd; no change → 0.

See the full worked example for simple rate equations, orders of reaction and rate constants →

The rate constant and first-order half-life

Find k by substituting a data row; its units follow from the overall order. First order has a constant half-life.

Once the orders are known, find k by substituting one experiment's values into the rate equation. Its units depend on the overall order:

Overall order	Units of k
0	mol dm⁻³ s⁻¹
1	s⁻¹
2	mol⁻¹ dm³ s⁻¹
3	mol⁻² dm⁶ s⁻¹

Half-life. For a first-order reaction the half-life is constant (independent of concentration):

$t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$

So if a concentration–time graph shows the time to halve is always the same, the reaction is first order. (A reaction whose half-life increases as it proceeds is second order; a linear concentration–time graph is zero order.)

k from substituting a data row; quote units.
k units depend on overall order.
First order: constant half-life, t½ = 0.693/k.

See the full worked example for simple rate equations, orders of reaction and rate constants →

Orders, graphs and the mechanism

Graph shapes confirm the order; the rate equation reveals the rate-determining step.

Graphical diagnostics:

Concentration–time: zero order = straight line; first order = constant half-life; second order = increasing half-life.
Rate–concentration: zero order = horizontal; first order = straight line through the origin; second order = upward curve.

Linking to the mechanism. The rate equation tells you which species (and how many) are in the rate-determining step (the slowest step). Only species involved up to and including the RDS appear in the rate equation.

Worked. A reaction A + 2B → products with rate = k[A][B] must have a rate-determining step containing one A and one B; the second B reacts in a later, fast step (so it is absent from the rate equation). This is why orders need not match the overall equation — they probe the mechanism.

Graph shapes confirm 0/1st/2nd order.
Rate equation = species up to and including the RDS.
Species after the RDS don't appear in the rate equation.

See the full worked example for simple rate equations, orders of reaction and rate constants →

How it’s examined

A core Paper 4 question: deduce orders from an initial-rates table, write the rate equation, calculate k with units, and often link to a mechanism or a half-life. Examiner reports flag reading orders off the equation, missing units on k, and using the first-order half-life formula for other orders.

Sources: Cambridge International AS & A Level Chemistry 9701 syllabus (2025-2027), section 26.1; 9701 Examiner Reports 2022-2024; 9701/41 & 9701/42 May/Jun 2024 question papers and mark schemes. Last reviewed 2026-05-20.

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Step-by-step worked examples — Simple rate equations, orders of reaction and rate constants

Step-by-step solutions to past-paper-style questions on simple rate equations, orders of reaction and rate constants, written exactly the way a tutor would explain them at the board.

1Order from doubling a concentration (2 marks)
Getting started• order
Question
Doubling [A] (with [B] constant) doubles the rate; doubling [B] (with [A] constant) leaves the rate unchanged. State the order with respect to A and B.
Step-by-step solution
1. Step 1
  A: rate ×2 when [A] ×2 → rate ∝ [A]¹ → first order in A.
2. Step 2
  B: rate unchanged when [B] ×2 → rate ∝ [B]⁰ → zero order in B.
Answer
First order in A; zero order in B.
Examiner tip
×2 rate → order 1; ×4 → order 2; no change → order 0. Compare experiments where only ONE concentration changes.
2Units of the rate constant (2 marks)
Getting started• rate constant
Question
A reaction is overall second order with rate = k[A][B]. State the units of k. (rate in mol dm⁻³ s⁻¹)
Step-by-step solution
1. Step 1
  Rearrange k = rate/([A][B]).
  $k = \frac{\text{mol dm}^{-3}\,\text{s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})}$
2. Step 2
  Cancel.
  $k:\ \text{mol}^{-1}\,\text{dm}^{3}\,\text{s}^{-1}$
Answer
mol⁻¹ dm³ s⁻¹.
Examiner tip
Work units out from the rate equation each time: 0 order mol dm⁻³ s⁻¹; 1st s⁻¹; 2nd mol⁻¹ dm³ s⁻¹.
3Rate equation from initial-rates data (4 marks)
Building confidence• order, rate constant
Question
For 3 experiments: (1) [A]=0.10, [B]=0.10, rate=2.0×10⁻³; (2) [A]=0.20, [B]=0.10, rate=4.0×10⁻³; (3) [A]=0.10, [B]=0.20, rate=8.0×10⁻³ (mol dm⁻³ s⁻¹). Find the rate equation and k.
Step-by-step solution
1. Step 1
  1→2: [A]×2, rate×2 → first order in A.
2. Step 2
  1→3: [B]×2, rate×4 → second order in B.
  $\text{rate} = k[\text{A}][\text{B}]^2$
3. Step 3
  Substitute experiment 1 to find k.
  $k = \frac{2.0\times10^{-3}}{(0.10)(0.10)^2} = 2.0\,\text{mol}^{-2}\,\text{dm}^{6}\,\text{s}^{-1}$
Answer
rate = k[A][B]²; k = 2.0 mol⁻² dm⁶ s⁻¹ (overall 3rd order).
Examiner tip
Pick pairs of experiments where only one concentration changes; then substitute any row to find k (with units).
4First-order half-life (3 marks)
Building confidence• half-life
Question
A first-order reaction has a constant half-life of 30 s. Calculate the rate constant, and state how you would recognise first order from a concentration–time graph.
Step-by-step solution
1. Step 1
  For first order, t½ is constant: k = ln2/t½.
  $k = \frac{0.693}{30} = 0.0231\,\text{s}^{-1}$
2. Step 2
  Recognise first order: the half-life is constant (independent of concentration) on a concentration–time graph.
Answer
k = 0.0231 s⁻¹; constant half-life = first order.
Examiner tip
A constant half-life is the signature of first order; t½ = ln2/k = 0.693/k.
5Reading order from graphs (4 marks)
Stretch• graphs
Question
Describe how to identify zero, first and second order from (a) a concentration–time graph and (b) a rate–concentration graph.
Step-by-step solution
1. Step 1
  Zero order: concentration–time is a straight line of constant gradient (rate independent of concentration); rate–concentration is horizontal.
2. Step 2
  First order: concentration–time has a constant half-life; rate–concentration is a straight line through the origin.
3. Step 3
  Second order: concentration–time falls steeply then levels with an increasing half-life; rate–concentration is an upward curve (rate ∝ [X]²).
Answer
Zero: linear conc–t / horizontal rate–conc. First: constant t½ / straight rate–conc line. Second: increasing t½ / curved rate–conc.
Examiner tip
The two most reliable diagnostics: constant half-life (1st order) and a straight rate-vs-concentration line through the origin (1st order).
6Order and the rate-determining step (4 marks)
Stretch• mechanism
Question
A reaction A + 2B → products has rate = k[A][B]. A proposed mechanism has a slow step followed by a fast step. Deduce what the slow (rate-determining) step contains, and comment.
Step-by-step solution
1. Step 1
  Orders give the species (and counts) in the RDS: first order in A and first order in B.
2. Step 2
  So the rate-determining step involves one A and one B (e.g. A + B → intermediate, slow).
3. Step 3
  The second B reacts in a later fast step, so it does not appear in the rate equation.
Answer
RDS contains 1 A + 1 B (matching rate = k[A][B]); the extra B reacts after the RDS.
Examiner tip
Species that appear AFTER the rate-determining step do not appear in the rate equation — orders ≠ stoichiometric coefficients.

Key Formulae — Simple rate equations, orders of reaction and rate constants

The formulae you need to memorise for simple rate equations, orders of reaction and rate constants on the Cambridge International A Level 9701 paper, with every variable defined in plain English and a note on when to use it.

Rate equation
$\text{rate} = k[\text{A}]^m[\text{B}]^n$
$k$
rate constant
$m, n$
orders with respect to A and B (found by experiment)
When to use
Once orders are known; overall order = m + n. Orders are NOT the equation coefficients.
First-order half-life
$t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$
$t_{1/2}$
half-life (constant for first order)
When to use
First-order reactions only; a constant half-life confirms first order.
Units of the rate constant
$k\ \text{units} = \text{mol}^{(1-N)}\,\text{dm}^{3(N-1)}\,\text{s}^{-1}$
$N$
overall order of the reaction
When to use
Deriving k's units (0 order mol dm⁻³ s⁻¹; 1st s⁻¹; 2nd mol⁻¹ dm³ s⁻¹).

Key Definitions and Keywords — Simple rate equations, orders of reaction and rate constants

Definitions to memorise and the exact keywords mark schemes credit for simple rate equations, orders of reaction and rate constants answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9701 sitting.

Order of reaction
Examiner keyword
The power to which a reactant's concentration is raised in the rate equation; the overall order is the sum of the individual orders. Found only by experiment.
Rate constant (k)
Examiner keyword
The constant in the rate equation; it increases with temperature and its units depend on the overall order.
Half-life (t½)
Examiner keyword
The time for the concentration of a reactant to halve; for a first-order reaction it is constant.
Rate-determining step (RDS)
Examiner keyword
The slowest step in a reaction mechanism; the species in the rate equation are those involved up to and including this step.

Common Mistakes and Misconceptions — Simple rate equations, orders of reaction and rate constants

The traps other students keep falling into on simple rate equations, orders of reaction and rate constants questions — taken from recent Cambridge International A Level 9701 examiner reports and mark schemes — and how to avoid them.

✕Reading orders from the stoichiometric coefficients.
9701 Examiner Reports 2022-2024
Why it happens
Assuming order = coefficient.
How to avoid it
Orders are found ONLY by experiment (initial rates / graphs).
✕Quoting k without units or with the wrong units.
Why it happens
Memorising one set of units.
How to avoid it
Derive units from the rate equation each time (they depend on overall order).
✕Using t½ = 0.693/k for any order.
Why it happens
Applying the first-order formula generally.
How to avoid it
The constant-half-life relationship is for FIRST order only.
✕Including species that appear after the RDS in the rate equation.
Why it happens
Matching the rate equation to the overall equation.
How to avoid it
Only species up to and including the rate-determining step appear in the rate equation.

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