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Rate Equations, Orders, Rate Constant

Learn and apply Rate Equations, Orders, Rate Constant in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Rate Equations, Orders and Rate Constant: Orientation

Rate laws are the bridge between experiment and explanation: data → orders → rate equation → units of k → mechanism check. This lesson builds the definitions and “rate-change” reflexes you’ll reuse across all kinetics questions.

Link this page to Activation Energy and Boltzmann Distribution and the Reaction Kinetics hub to keep mechanism and data interpretation consistent.

Definitions (Must Know)

A. Rate of reaction

The rate of reaction is the change in concentration per unit time.

  • for a reactant A: rate = -Δ[A]/(Δ t)
  • for a product P: rate = Δ[P]/(Δ t)

B. Rate equation (rate law)

A rate equation relates rate to reactant concentrations: rate = k[A]^m[B]ⁿ where m and n are the orders with respect to A and B.

C. Order of reaction (with respect to a reactant)

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

D. Overall order

The overall order is the sum of the powers in the rate equation (e.g. m + n for two reactants).

E. Rate constant, k

The rate constant, k, is a constant for a given reaction at a fixed temperature. Its units depend on the overall order.

Detailed Explanations

A. How orders control rate changes

If rate = k[A]^m (other factors constant), then multiplying [A] by a factor f multiplies the rate by f^m.

Mini example:

  • if m = 2 and [A] doubles (f = 2), the rate increases by 2² = 4.

Quick visual (how order changes the rate–concentration shape):

Rate vs Concentration for Different Orders (Schematic)

Rate vs Concentration for Different Orders (Schematic). 0th order in A, 1st order in A, 2nd order in A plotted as Rate against Concentration of A.

Scroll across the graph to read all labels.

Rate vs Concentration for Different Orders (Schematic). 0th order in A, 1st order in A, 2nd order in A plotted as Rate against Concentration of A.Rate vs Concentration for Different Orders (Schematic). 0th order in A, 1st order in A, 2nd order in A plotted as Rate against Concentration of A.
Schematic with k = 1 and other reactants held constant: 0th order is flat, 1st order is a straight line, and 2nd order curves upward.
Open full-size graph
View figure data
Values and uncertainty for Rate vs Concentration for Different Orders (Schematic)
SeriesConcentration of A (mol dm^-3)Concentration of A uncertaintyRate (relative)Rate uncertainty
0th order in A0.11
0th order in A0.31
0th order in A0.51
0th order in A0.71
0th order in A0.91
1st order in A00
1st order in A0.20.2
1st order in A0.40.4
1st order in A0.60.6
1st order in A0.80.8
1st order in A11
2nd order in A00
2nd order in A0.20.04
2nd order in A0.40.16
2nd order in A0.60.36
2nd order in A0.80.64
2nd order in A11

B. Units of k (safe method)

  1. Take the rate units from the question (commonly mol dm⁻³s⁻¹).
  2. Take concentration units (commonly mol dm⁻³).
  3. Rearrange: k = rate/[A]^m[B]ⁿ
  4. Substitute units and simplify.

Common results (if rate is in mol dm⁻³s⁻¹):

Overall orderTypical rate lawUnits of k
0rate = kmol dm⁻³s⁻¹
1rate = k[A]s⁻¹
2rate = k[A]²dm³ mol⁻¹s⁻¹
3rate = k[A]³dm⁶ mol⁻²s⁻¹

C. What changes (and what doesn’t)

  • Changing concentration changes the rate.
  • Changing temperature (or adding a catalyst) changes k.
  • The balanced equation coefficients do not determine the orders unless the reaction is stated to be an elementary step.

Worked Examples

Modelled example 1

Read orders from a rate equation

Core

Problem

For rate = k[A]²[B], state the order with respect to A, the order with respect to B and the overall order.
Study the worked solution

  1. Read individual powers

    Method

    Use each concentration exponent as that reactant’s order.

    Reason

    Reaction order is defined by the experimentally established power in the rate equation.

    Working

    Order in A = 2; order in B = 1.

  2. Find overall order

    Method

    Add the individual orders.

    Reason

    Overall order is the sum of all concentration powers.

    Working

    2 + 1 = 3: the reaction is third order overall.

Guided practice 2

Combine concentration-change factors

About 6 min

Problem

For rate = k[A]⁰[B]² at constant temperature, what happens to the rate if [A] doubles and [B] halves?

Try this before viewing the solution

Hints

Hint 1: treat each reactant separately
Raise A’s factor 2 to power 0 and B’s factor 1/2 to power 2.
Hint 2: combine multiplicatively
Multiply the two independent rate factors; do not add the orders to the concentration factors.
View solution step by step

  1. Apply A's order

    Method

    Raise the factor 2 to power 0.

    Reason

    A is zero order, so its concentration change contributes no rate change.

    Working

    2⁰ = 1.

  2. Apply B's order

    Method

    Square the factor 1/2.

    Reason

    B is second order.

    Working

    (1/2)² = 1/4.

  3. Combine

    Method

    Multiply the two rate factors.

    Reason

    Both concentration effects act in the same rate-law product.

    Working

    1 × 1/4 = 1/4: the rate becomes one quarter of its original value.

Common misconception 3

Correct a coefficient-derived rate law

Find and correct the mistake

Learner claim

For the overall equation 2A + B → products, a learner writes rate = k[A]²[B] solely from the balanced coefficients. Explain why this is not justified and what evidence is needed.

Identify the evidence

Orders for an unstated non-elementary reaction come from

View solution step by step

  1. Separate stoichiometry from kinetics

    Method

    State that the balanced equation records overall chemical amounts.

    Reason

    It need not describe the elementary molecular events controlling the observed rate.

    Working

    Coefficients 2 and 1 do not automatically become reaction orders.

  2. Name valid evidence

    Method

    Require experimentally measured rate changes as concentrations vary, unless the reaction is explicitly stated to be elementary.

    Reason

    Those data determine the powers in the empirical rate equation.

    Working

    The proposed law remains unproven without kinetic evidence.

Examiner practice 4

Calculate k and its units

4 marks

Problem

For rate = k[A]²[B], the rate is 1.60 × 10⁻³ mol dm⁻³s⁻¹ when [A] = 0.200 mol dm⁻³ and [B] = 0.0500 mol dm⁻³. Calculate k and state its units. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Rearrange

    1 mark

    Method

    Make k the subject.

    Reason

    Both concentration terms divide the measured rate.

    Working

    k = rate/[A]²[B].

  2. Substitute

    1 mark

    Method

    Insert the rate and equilibrium concentrations with A squared.

    Reason

    The rate-law powers must be preserved numerically.

    Working

    k = (1.60 × 10⁻³)/(0.200)²(0.0500) = (1.60 × 10⁻³)/0.00200.

  3. Evaluate

    1 mark

    Method

    Complete the division.

    Reason

    The data support three significant figures.

    Working

    k = 0.800.

  4. Derive units

    1 mark

    Method

    Divide the rate unit by three concentration factors.

    Reason

    The reaction is third order overall.

    Working

    k = 0.800 dm⁶ mol⁻²s⁻¹.

Challenge 5

Derive k units dimensionally

Minimal support

Problem

For rate = k[A]², rate is measured in mol dm⁻³s⁻¹ and concentration in mol dm⁻³. Derive the units of k rather than recalling them from a table.

Try this before viewing the solution

Hints

Hint 1: rearrange symbols first
Write k = rate/[A]² before inserting units.
Hint 2: subtract unit powers
The numerator contains one mol dm⁻³ factor; the denominator contains two.
View solution step by step

  1. Rearrange

    Method

    Make k the subject of the second-order rate law.

    Reason

    Dimensional cancellation must follow the algebraic relationship.

    Working

    k = rate/[A]².

  2. Insert and simplify units

    Method

    Divide the rate unit by two concentration units.

    Reason

    One concentration factor cancels, leaving inverse concentration multiplied by inverse time.

    Working

    [k] = (mol dm⁻³s⁻¹)/((mol dm⁻³)²) = dm³ mol⁻¹s⁻¹.

Mind Stretchers

Mind stretcher 1Extension

A reaction has k units of dm⁶ mol⁻²s⁻¹ (assume rate is in mol dm⁻³s⁻¹ and concentrations are in mol dm⁻³). Doubling [A] increases the rate by a factor of 4. Deduce the orders in A and B for a rate law of the form rate = k[A]^m[B]ⁿ.

Show Hint

Separate the concentration factor from the rate factor before deciding the order.

Show Answer

Mark scheme:

  • From the units of k: overall order is 3 (the 3rd-order unit pattern is dm⁶ mol⁻²s⁻¹).
  • Doubling [A] gives rate ×4, so 2^m = 4 → m = 2.
  • Overall order m + n = 3, so n = 1.
  • Rate law: rate = k[A]²[B].

Mind stretcher 2: A dimensional auditExtension

Question. For rate = k[A]²[B], state the overall order and derive the units of k when rate is measured in mol dm⁻³ s⁻¹.

Show Hint

Divide the rate unit by three concentration factors.

Show Answer

The overall order is 3. Thus [k] = (mol dm⁻³ s⁻¹)/(mol dm⁻³)³ = dm⁶ mol⁻² s⁻¹.