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*Revision Materials and Past Papers* 1 Atomic Structure and the Periodic Table 2 Bonding and Structure 3 Redox I 4 Inorganic Chemistry and the Periodic Table 5 Formulae, Equations and Amounts of Substance 6 Organic Chemistry I 7 Modern Analytical Techniques I 8 Energetics I 9 Kinetics I 10 Equilibrium I 11 Equilibrium II 12 Acid-base Equilibria 13 Energetics II 14 Redox II 15 Transition Metals 16 Kinetics II 17 Organic Chemistry II 18 Organic Chemistry III 19 Modern Analytical Techniques II RP Required Practicals

16 Kinetics II

16 Kinetics II

Kinetics II

Specification Reference Topic 16, points 1–12 (Edexcel A-Level Chemistry)

Quick Notes

  • Rate of reaction: how quickly a chemical change takes place, units mol dm−3 s−1.
  • Rate equation: rate = k[A]m[B]n — shows how rate depends on concentration.
  • Order: Tells us how changes in concentration affect the rate.
  • Overall order: Add up the powers (orders) in the rate equation.
  • Methods to determine orders of reaction:
    • Initial Rate Method – Measures how rate changes with concentration.
    • Edexcel A-Level Chemistry plots showing initial rate comparisons to deduce reaction order.
    • Continuous Monitoring – Plots concentration vs. time to determine order graphically.
    • Edexcel A-Level Chemistry concentration–time profiles for zero, first and second order reactions.
  • Half-life: Time it takes for a reactant’s concentration to halve.
  • Rate-determining step (RDS): The slowest step in a multistep reaction and limits the overall rate.
  • Activation energy (Ea) is the minimum energy particles need to react.
  • Catalysts provide alternative pathways with lower activation energy.
  • The Arrhenius equation links rate constant with temperature and activation energy.
  • Edexcel A-Level Chemistry Arrhenius equation linking k to A, Ea, R and T.

Full Notes

The rate of a reaction measures how quickly a chemical change takes place. It’s usually expressed as the change in concentration of a reactant (or product) per unit time, e.g. mol dm−3 s−1.

The rate equation is a mathematical expression linking rate to concentrations of reactants:

rate = k[A]m[B]n

k is the rate constant, which varies with temperature.
m and n are the orders of reaction with respect to each reactant.

Orders of a reaction

Order of reaction tells us how the rate responds to changes in concentration of a reactant:

Orders of reactions can only be determined experimentally.

The overall order of a reaction is the sum of all the individual orders.

Measuring Reaction Rates – Example Techniques

You should be comfortable with the following practical methods to measure rate:

Experimental Approaches

There are different techniques we can use to determine orders experimentally, depending on the type of reaction.

Initial Rate Method

We can measure how the initial reaction rate of a reaction changes when reactant concentrations are varied.

Steps:

Worked Example: Determining a Rate Equation from Initial Rates

Reaction: A + B → Products

exp [A] (mol dm−3) [B] (mol dm−3) Initial Rate (mol dm−3 s−1)
1 0.10 0.10 0.02
2 0.20 0.10 0.04
3 0.20 0.20 0.16

[A] has doubled from exp. 1 to exp. 2 and [B] is constant. Rate has doubled (0.02 → 0.04).
This means the reaction is first order with respect to A.

[B] has doubled from exp. 1 to exp. 3 and [A] is constant. Rate has quadrupled (×4), 0.04 → 0.16.
This means the reaction is second order with respect to B.

Thus, the rate equation is:
Rate = k [A]1 [B]2

You can also plot rate against concentration and use the shape of the graph to identify whether the reactant is 0, 1st or 2nd order.

Edexcel A-Level Chemistry plots showing initial rate comparisons to deduce reaction order.

Continuous monitoring

For some reactions, it is possible to measure how concentration changes over time.

This allows us to produce a graph of concentration vs time. From the graph, you can calculate the rate at any point (e.g. by drawing a tangent).

Edexcel A-Level Chemistry tangent on concentration–time curve showing instantaneous rate from gradient.

The shape of the curve can also helps deduce the order:

Edexcel A-Level Chemistry concentration–time profiles for zero, first and second order reactions.

Using Rate Equations to Understand Mechanisms

Most reactions do not happen in one simple step, but in a series of steps.

These steps do not always occur at the same rate.

When looking at how fast a reaction can happen, it's the slowest step that determines the overall rate.

Example Reaction: NO2 + CO → NO + CO2

The rate equation is Rate = k [NO2]2, because CO isn’t in the rate equation, we know it isn’t involved in the rate determining step (RDS).

Possible Mechanism:
NO2 + NO2 → NO3 + NO (slow)
NO3 + CO → NO2 + CO2 (fast)

Step 1 is the slow step (RDS) and doesn’t include CO.

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Matt’s exam tip

Remember that mechanisms are proposed using rate equations. They are only predictions, there may be more than one possible mechanism for a given rate equation.

Kinetics of Acid-Catalysed Iodination of Propanone

The acid-catalysed iodination of propanone is often used as an example to determine a rate equation and propose a mechanism.

Obtaining Data to Determine Reaction Orders

The reaction:
CH3COCH3 + I2 → CH2ICOCH3 + HI (acid-catalysed)

Typical experimental result:
Rate = k[CH3COCH3][H+]
i.e., first order in propanone and H+, zero order in iodine.

Using Rate Data to Predict Species in the Rate-Determining Step

Deducing a Possible Mechanism

Proposed mechanism:

  1. CH3COCH3 + H+ → Enol (slow)
  2. Enol + I2 → CH2ICOCH3 + HI (fast)

Hydrolysis of Halogenoalkanes: Rate Equations as Evidence for SN1 or SN2

The rate equation for the hydrolysis of a halogenoalkane gives information and evidence about the mechanism that it follows.

Primary Halogenoalkanes - SN2

For the hydrolysis of primary halogenoalkanes, the rate equation is:

Rate = k[halogenoalkane][OH]

This shows both reactants are involved in the rate-determining step, providing evidence for a one-step mechanism (SN2).

Edexcel A-Level Chemistry SN2 mechanism diagram showing one-step bimolecular substitution.

Primary Halogenoalkanes - SN1

For the hydrolysis of tertiary halogenoalkanes, the rate equation is:

Rate = k[halogenoalkane]

Since OH doesn't appear in the rate equation, it suggests a two-step mechanism (SN1), with carbocation formation as the slow, rate-determining step.

Edexcel A-Level Chemistry SN1 mechanism diagram showing carbocation intermediate and two-step pathway.

Activation Energy and the Arrhenius Equation

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.

Edexcel A-Level Chemistry Arrhenius equation linking k to A, Ea, R and T.

where:

This also shows why k increases with temperature - the e−Ea/RT part of the expression gets bigger as 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:

Edexcel A-Level Chemistry natural log linear form of Arrhenius equation for plotting.

This allows us to plot ln k vs. 1/T (called an Arrhenius plot). The gradient of the line = −Ea / R, allowing activation energy calculation.

Edexcel A-Level Chemistry Arrhenius plot with gradient equal to −Ea/R.

−Ea÷R = m
Ea = − m × R

Performing Arrhenius Equation Calculations

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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).

1. Calculate Activation Energy (Ea) Using Two Rate Constants

Worked Example: Calculate Activation Energy (Eₐ)

Determine the activation energy (Eₐ) when the following rate constants are measured at two different temperatures:

  • At 300 K, k₁ = 2.5 × 10⁻³ s⁻¹
  • At 350 K, k₂ = 5.0 × 10⁻³ s⁻¹

Arrhenius equation (two-point form):

ln (k₂ / k₁) = (–Eₐ / R) × (1/T₂ – 1/T₁)

Step 1: Substitute known values

ln (5.0 × 10⁻³ / 2.5 × 10⁻³) = (–Eₐ / 8.31) × (1/350 – 1/300)

Step 2: Simplify

ln 2 = (–Eₐ / 8.31) × (–4.76 × 10⁻⁴)

Step 3: Rearrange for Eₐ

Eₐ = (ln 2 × 8.31) / (4.76 × 10⁻⁴)

Final Answer:

Eₐ = 12 110 J mol⁻¹ = 12.1 kJ mol⁻¹

2. Using 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.

Summary