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*Revision Materials and Past Papers* 1 Atomic Structure 2 Amounts of Substance 3 Bonding 4 Energetics 5 Kinetics 6 Chemical Equilibria & Kc 7 Redox Equations 8 Thermodynamics 9 Rate Equations 10 Kp (Equilibrium Constant) 11 Electrode Potentials & Cells 12 Acids and Bases 13 Periodicity 14 Group 2: Alkaline Earth Metals 15 Group 7: The Halogens 16 Period 3 Elements & Oxides 17 Transition Metals 18 Reactions of Ions in Aqueous Solution 19 Intro to Organic Chemistry 20 Alkanes 21 Halogenoalkanes 22 Alkenes 23 Alcohols 24 Organic Analysis 25 Optical Isomerism 26 Aldehydes & Ketones 27 Carboxylic Acids & Derivatives 28 Aromatic Chemistry 29 Amines 30 Polymers 31 Amino Acids, Proteins & DNA 32 Organic Synthesis 33 NMR Spectroscopy 34 Chromatography RP1–RP12 Required Practicals

1.9 Rate Equations (A-level only)

1.9.1 Rate Equations 1.9.2 Determination of Rate Equation

Rate Equations and the Arrhenius Equation

Specification Reference Physical Chemistry, Rate equations 3.1.9.1

Quick Notes

  • Rate equations describe how a reactions rate depends on reactant concentrations.
  • General form of the rate equation:
    AQA A‑Level Chemistry general rate equation showing Rate equals k times concentrations of reactants raised to their orders
    k = Rate constant (varies with temperature)
    [A], [B] = Reactant concentrations (mol dm−3)
    m, n = Orders of reaction (must be determined experimentally)
  • Overall order of reaction = sum of individual orders (m + n).
  • The rate constant (k) changes with temperature, following the Arrhenius equation:
    AQA A‑Level Chemistry Arrhenius equation k equals A e to the power of negative Ea over RT
    where:
    A = Arrhenius constant (pre‑exponential factor)
    Ea = Activation energy (J mol−1)
    R = Gas constant (8.31 J mol−1 K−1)
    T = Temperature (K)
  • The equation can be rearranged into a linear form:
    AQA A‑Level Chemistry linear form of the Arrhenius equation ln k equals ln A minus Ea over R times 1 over T
    This allows us to find activation energy (Ea) and A by plotting a ln k vs. 1/T graph, with slope = −Ea/R. AQA A‑Level Chemistry Arrhenius plot example showing straight line of ln k against 1 over T with negative slope

Full Notes

Rate Equation and Definitions

The rate of a reaction is a measure of how fast a reaction is occurring.

AQA A‑Level Chemistry rate law diagram Rate equals k times concentrations to powers m and n

Rate equations shows how the rate of a reaction depends on reactant concentrations and the rate constant (k). where

Order of reaction:
Orders of a reaction link changes in concentrations of reactants to changes in the rate of a reaction. The order of a reaction ‘with respect to…’ just means how changing the concentration of a particular reactant affects the rate of the reaction (independent of other reactants).

There are three types of orders you need to know for A‑level Chemistry – zero order, first order and second order. Graph shapes and determining orders are covered here.

Orders of reactions can only be determined experimentally and the overall order of a reaction is the sum of all orders for each reactant.

For Example:
For the reaction
A + B → Products
the rate equation is:
Rate = k [A]1 [B]2
Order with respect to A = 1 (First Order).
Order with respect to B = 2 (Second Order).
Overall order = 1 + 2 = 3 (Third Order).

Rate constant (k):
Temperature, activation energy and the proportion of collisions with correct orientation also determine a reactions rate. These are accounted for in a rate equation by the rate constant ‘k’.

k varies with temperature.

Rate Equation Calculations

Rate constants can be calculated using experimental data and a given rate equation. The units of k can vary for different reactions and depend on the overall order of reaction.

Worked Example

A reaction with the following rate equation and reactant concentrations has a reaction rate of 0.05 mol dm−3 s−1. Calculate the rate constant, k, for the reaction and give its units.

  • Rate = 0.05 mol dm−3 s−1
  • [A] = 0.10 mol dm−3
  • [B] = 0.20 mol dm−3
  • Rate equation: Rate = k [A]1 [B]2

Solve for k:
k = Rate ÷ ([A]1 [B]2)
k = (0.05) ÷ (0.10 × 0.202)
k = 0.05 ÷ (0.10 × 0.04)
k = 12.5 mol−2 dm6 s−1
Units of k depend on the overall order of reaction.

The Arrhenius Equation and Arrhenius Plots

The Arrhenius Equation, how to rearrange it and using Arrhenius Plots has been covered in much more detail here. What is on this page is just the essentials you need to know for AQA A‑level chemistry.

The Arrhenius equation shows how activation energy (Ea), temperature (T) and the proportion of collisions with correct orientation (A) can be linked together by the rate constant, k.

AQA A‑Level Chemistry Arrhenius equation k equals A e to the minus Ea over RT

The equation shows why k increases with temperature – the e−Ea/RT part of the expression gets bigger as temperature (T) increases.

The Arrhenius equation is an exponential equation because it contains ‘e’. To make it easier to work with, we can rearrange it to a straight line form by multiplying both sides by ln:

AQA A‑Level Chemistry natural‑log form of the Arrhenius equation ln k equals ln A minus Ea over R times 1 over T

This can be rearranged a little to follow the y = mx + c function of a straight line. Where y = ln k and x = 1/T.

AQA A‑Level Chemistry mapping ln k vs 1 over T to y equals mx plus c form with slope minus Ea over R

This allows us to plot ln k vs. 1/T (called an Arrhenius plot).

AQA A‑Level Chemistry Arrhenius plot example showing straight line of ln k against 1 over T with negative slope

The gradient of the line = −Ea/R, allowing activation energy calculation.
−Ea ÷ R = m   →   Ea = − m × R

Arrhenius Equation Calculations

WORKED EXAMPLE Calculate Activation Energy (Ea) Using Two Rate Constants

A reaction has the following rate constants, k, at 300 K and 350 K. Determine the activation energy, Ea, for the reaction.

  • At 300 K, k = 2.5 × 10−3 s−1
  • At 350 K, k = 5.0 × 10−3 s−1

Use:
ln (k2 / k1) = (−Ea / R) × (1/T2 − 1/T1)
ln (5.0 × 10−3 / 2.5 × 10−3) = (−Ea / 8.31) × (1/350 − 1/300)
ln 2 = (−Ea / 8.31) × (−4.76 × 10−4)
Ea = (ln 2 × 8.31) / (4.76 × 10−4)
Ea = 12 110 J mol−1 = 12.1 kJ mol−1


Using an Arrhenius Graph to Find Activation Energy
The gradient of an Arrhenius plot is −Ea/R. By plotting ln k vs. 1/T, we can use the gradient of the straight line to find Ea.
Plot ln k vs. 1/T.
The gradient = −Ea / R.
Multiply the gradient by −8.31 to find Ea.

Photo of Matt
Matt’s exam tip

When using the Arrhenius equation in calculations, don’t forget units of activation energy (Ea) are kJ mol−1! The gas constant, R, has units of J K−1 mol−1, meaning you must convert any calculated Ea value to kJ (divide by 1000).


Summary

Concept Definition / Relationship Notes
Rate equation Rate = k [A]m[B]n m, n determined experimentally
Order (with respect to a reactant) Power to which its concentration is raised Zero: no effect; First: factor-for-factor; Second: factor squared
Overall order m + n Determines the units of k
Rate constant, k Proportionality constant in rate law Depends on temperature; units vary with overall order
Arrhenius equation k = A e−Ea/RT k increases with T; linear form: ln k = ln A − (Ea/R)(1/T)
Arrhenius plot ln k vs 1/T Slope = −Ea/R; intercept = ln A