Mini Chemistry
Part of Mini EducationSwitch subject

Initial Rates Method

Learn and apply Initial Rates Method in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
On this page

Method of Initial Rates: Orientation

Initial rates questions look scary because they’re a table of numbers, but the logic is always the same: compare two experiments to isolate one reactant’s effect, deduce the order, then write the full rate equation and solve for k.

Link this page to Rate Equations, Orders, and Rate Constant and the Reaction Kinetics hub to keep mechanism and data interpretation consistent.

Definitions (Must Know)

A. Initial rate

The initial rate is the rate measured at (or very near) time t = 0, before concentrations change significantly.

B. Method of initial rates

The method of initial rates deduces reaction orders by comparing initial rates from experiments with different starting concentrations.

C. Order with respect to a reactant

The order with respect to A is the power of [A] in the rate equation.

D. Rate equation (rate law)

rate = k[A]^m[B]ⁿ

Quick visual (example where B is constant):

Initial Rate vs Concentration of A (Example Pattern)

Initial Rate vs Concentration of A (Example Pattern). Example data plotted as Initial rate against Concentration of A.

Scroll across the graph to read all labels.

Initial Rate vs Concentration of A (Example Pattern). Example data plotted as Initial rate against Concentration of A.Initial Rate vs Concentration of A (Example Pattern). Example data plotted as Initial rate against Concentration of A.
Example pattern: when A doubles, the initial rate increases by a factor of 4 (consistent with rate proportional to [A]^2, with other conditions constant).
Open full-size graph
View figure data
Values for Initial Rate vs Concentration of A (Example Pattern)
Concentration of A (mol dm^-3)Example data
0.11
0.24
0.39

Detailed Explanations

A. Workflow (always works)

  1. Write the general form: rate = k[A]^m[B]ⁿ.
  2. Pick two experiments where only [A] changes: rate₂/rate₁ = ([A]₂/[A]₁)^m Solve for m.
  3. Pick two experiments where only [B] changes and solve for n.
  4. Write the full rate equation with your m and n.
  5. Substitute one experiment to find k (and units).

B. Full example (two reactants)

Data:

Experiment[A] / mol dm⁻³[B] / mol dm⁻³Initial rate / mol dm⁻³ s⁻¹
10.100.102.0 × 10⁻³
20.200.108.0 × 10⁻³
30.100.306.0 × 10⁻³

Find order in A (compare 1 → 2; B constant):

(8.0 × 10⁻³)/(2.0 × 10⁻³) = (0.20/0.10)^m

So 4 = 2^m ⇒ m = 2.

Find order in B (compare 1 → 3; A constant):

(6.0 × 10⁻³)/(2.0 × 10⁻³) = (0.30/0.10)ⁿ

So 3 = 3ⁿ ⇒ n = 1.

Rate equation: rate = k[A]²[B]

Find k (use experiment 1):

k = rate/[A]²[B] = (2.0 × 10⁻³)/(0.10)²(0.10) = 2.0

Overall order is 3, so k units are dm⁶ mol⁻²s⁻¹.

C. How initial rates are measured

Choose a measurable property that changes as the reaction proceeds, such as gas volume, mass, colour intensity, pH or concentration. Keep temperature and every starting concentration except the one being investigated constant.

  1. Prepare the reactant mixtures and bring them to the chosen temperature.
  2. Start the reaction in a repeatable way, mix promptly and start timing immediately.
  3. Record the chosen property at short, regular intervals near t = 0.
  4. Plot the measured quantity against time. Draw a tangent at t = 0; its gradient gives the initial rate after applying any required stoichiometric conversion.
  5. Repeat the experiment and compare the initial gradients. Investigate an anomalous run instead of silently discarding it.

For a clock reaction, a fixed visual endpoint can compare relative initial rates when only a small, constant amount has reacted. Then initial rate ∝ 1/t. This shortcut is only valid when the same endpoint and total conditions are used each time.

Choosing a method

Match the method to the chemistry. A gas syringe is useful only when a gas is formed and retained; mass loss works only when material leaves the apparatus; colorimetry needs a coloured species and a suitable calibration or absorbance relationship.

Worked Examples

Modelled example 1

Deduce an order from factor data

Core

Problem

When [A] doubles while other conditions stay constant, the initial rate increases by a factor of 8. Find the order with respect to A.
Study the worked solution

  1. Write the factor equation

    Method

    Raise the concentration factor 2 to the unknown order m.

    Reason

    Only A changes, so its rate-law term accounts for the full rate factor.

    Working

    2^m = 8.

  2. Match powers

    Method

    Express 8 as a power of 2.

    Reason

    8 = 2³.

    Working

    m = 3: the reaction is third order in A.

Guided practice 2

Deduce a fractional order

About 6 min

Problem

In an experiment, [B] is quadrupled while other conditions remain constant and the initial rate doubles. Find the order with respect to B.

Try this before viewing the solution

Hints

Hint 1: translate the factors
Write 4ⁿ = 2.
Hint 2: use a common base
Express 4 as 2², then compare the power with 2¹.
View solution step by step

  1. Set up the relationship

    Method

    Equate B’s concentration factor raised to n with the rate factor.

    Reason

    Only B changes.

    Working

    4ⁿ = 2.

  2. Solve

    Method

    Rewrite 4ⁿ as 2²ⁿ.

    Reason

    Equal bases allow their powers to be equated.

    Working

    2n = 1, so n = 1/2.

Common misconception 3

Correct an uncontrolled row comparison

Find and correct the mistake

Learner method

Between two runs, both [A] and [B] double and the rate increases eight-fold. A learner concludes that A is third order. Explain why that conclusion is unsupported and what comparison is needed.

Assess the evidence

This comparison establishes

View solution step by step

  1. Expose the confounding

    Method

    Write the rate factor as 2^m2ⁿ.

    Reason

    Both concentration changes contribute to the observed factor 8.

    Working

    2^(m + n) = 8, so only m + n = 3 follows.

  2. Request isolating evidence

    Method

    Compare a pair where A changes while B is constant, or vice versa.

    Reason

    Holding one variable fixed allows the other order to be assigned independently.

    Working

    The individual order in A is not determined by the mixed comparison alone.

Examiner practice 4

Deduce a rate equation from initial-rate data

5 marks

Problem

Using the data below, determine both orders and write the rate equation. [5 marks]

Try this before viewing the solution

View solution step by step

  1. Choose A comparison

    1 mark

    Method

    Compare experiments 1 and 2, where B is constant.

    Reason

    Only A’s concentration term can explain the rate change.

    Working

    A doubles while rate becomes four times larger.

  2. Deduce A order

    1 mark

    Method

    Solve 2^m = 4.

    Reason

    The concentration factor raised to the order equals the rate factor.

    Working

    m = 2.

  3. Choose B comparison

    1 mark

    Method

    Compare experiments 1 and 3, where A is constant.

    Reason

    This isolates B’s effect.

    Working

    B doubles while rate doubles.

  4. Deduce B order

    1 mark

    Method

    Solve 2ⁿ = 2.

    Reason

    The observed factors match directly.

    Working

    n = 1.

  5. Write the equation

    1 mark

    Method

    Insert both experimentally deduced powers.

    Reason

    A complete rate equation includes k.

    Working

    rate = k[A]²[B].

Challenge 5

Predict a new initial rate

Minimal support

Problem

For a reaction with rate = k[A]²[B], an experiment at [A] = 0.100 and [B] = 0.200 mol dm⁻³ gives an initial rate of 1.20 × 10⁻³ mol dm⁻³s⁻¹. Determine k with units, then predict the initial rate at [A] = 0.150 and [B] = 0.300 mol dm⁻³.

Try this before viewing the solution

Hints

Hint 1: calibrate k
Use the first experiment in k = rate/([A]²[B]).
Hint 2: apply the calibrated law
Substitute the second pair of concentrations into k[A]²[B] without rounding k early.
View solution step by step

  1. Calculate k

    Method

    Rearrange and substitute the calibration experiment.

    Reason

    The established orders make k the only unknown.

    Working

    k = (1.20 × 10⁻³)/(0.100)²(0.200) = 0.600 dm⁶ mol⁻²s⁻¹.

  2. Predict the new rate

    Method

    Insert the new concentrations into the same rate law.

    Reason

    k is unchanged at the same temperature.

    Working

    rate = 0.600(0.150)²(0.300) = 4.05 × 10⁻³ mol dm⁻³s⁻¹.

Mind Stretchers

Mind stretcher 1Extension

The rate equation has the form rate = k[A]^m[B]ⁿ. Between two experiments, [A] increases by a factor of 2.0 and [B] decreases by a factor of 2.0, but the rate stays the same. Deduce the relationship between m and n.

Show Hint

Use pairs that isolate one changing concentration; do not compare a row where two reactants change at once.

Show Answer

Mark scheme:

  • Rate ratio is 1: 1 = ([A]₂/[A]₁)^m([B]₂/[B]₁)ⁿ = (2)^m(1/2)ⁿ
  • So 2^m = 2ⁿ ⇒ m = n.

Mind stretcher 2: Resolving a mixed comparisonExtension

Question. From experiment 1 to 2, [A] doubles at fixed [B] and rate quadruples. From 2 to 3, [A] halves and [B] triples while rate changes by a factor 3/4. Deduce both orders.

Show Hint

Find the order in A first, then remove its factor of (1/2)² from the second comparison.

Show Answer

A is second order because doubling it quadruples rate. In the second comparison A contributes 1/4; the observed factor is 3/4, so B contributes 3 when tripled and is first order.