Reaction Kinetics

Learn and apply Reaction Kinetics in the published Chemistry course sequence.

GCE A-Level H1 Chemistry 8873-2027

Learning goals

explain and use the terms: rate of reaction; rate equation; order of reaction; rate constant; half-life of a reaction; activation energy; catalysis
construct and use rate equations of the form rate = k[A]m[B]n (limited to simple cases of single-step reactions for which m and n are 0, 1 or 2), including: — deducing the order of a reaction by the initial rates method
construct and use rate equations of the form rate = k[A]m[B]n (limited to simple cases of single-step reactions for which m and n are 0, 1 or 2), including: — justifying, for zero- and first-order reactions, the order of reaction from concentration-time graphs
construct and use rate equations of the form rate = k[A]m[B]n (limited to simple cases of single-step reactions for which m and n are 0, 1 or 2), including: — calculating an initial rate using concentration data [integrated forms of rate equations are not required]
show understanding that the half-life of a first-order reaction is independent of concentration
explain qualitatively, in terms of frequency of collisions, the effect of concentration changes on the rate of a reaction
show understanding, including reference to the Boltzmann distribution, of what is meant by the term activation energy
explain qualitatively, in terms of both the Boltzmann distribution and of collision frequency, the effect of temperature change on a rate constant (and hence, on the rate) of a reaction
— explain that, in the presence of a catalyst, a reaction follows a different pathway, i.e. one of lower activation energy, giving a larger rate constant
— interpret this catalytic effect in terms of the Boltzmann distribution
outline the mode of action of heterogeneous catalysis, as exemplified by the catalytic removal of oxides of nitrogen in the exhaust gases from car engines
describe enzymes as protein molecules that act as biological catalysts with high specificity (in the reactions that they catalyse and in their choice of substrates as exemplified by the lock-and-key model), temperature sensitivity and pH sensitivity [Knowledge of the levels of structure of proteins is not required. Details of the denaturation process will be discussed in 9 (m).]

H1 Reaction Kinetics learning outcomes

Reaction Kinetics connects measurements to rate equations, graph evidence and particle explanations. Strong H1 answers keep the 8873 boundary clear: use simple single-step equations, justify claims from data and do not import H2 mechanism demands.

Follow the five lessons in order when learning. During revision, classify an error as rate-factor reasoning, graph evidence, collision theory, catalyst pathway or enzyme conditions.

What You’ll Learn

Use rate terms and simple single-step rate equations with individual orders zero, one or two.
Deduce reaction orders by the method of initial rates and calculate an initial rate from concentration data.
Justify zero- and first-order behaviour from concentration–time evidence.
Recognise that first-order half-life is independent of concentration.
Explain concentration effects using collision frequency and temperature effects using a Boltzmann distribution.
Explain how a catalyst provides a lower-activation-energy pathway and therefore a larger rate constant.
Apply heterogeneous catalysis to catalytic NOx removal and explain enzyme specificity and sensitivity to temperature and pH.

Lessons (Recommended Order)

Build the mathematical model first, test it with initial-rate and graph evidence, then explain rate changes and catalysts at particle level.

Simple Rate Equations and Orders
Build simple rate equations, interpret orders and calculate an initial rate.
Open lesson
Method of Initial Rates
Use controlled comparisons to deduce individual reaction orders.
Open lesson
Zero- and First-order Graphs and Half-life
Distinguish zero and first order from gradients and successive half-lives.
Open lesson
Activation Energy and Boltzmann Distribution
Explain concentration, temperature and catalyst effects with collision and energy evidence.
Open lesson
Heterogeneous Catalysis and Enzymes
Apply lower-Ea reasoning to catalytic NOx removal and enzyme behaviour.
Open lesson

Quick Reference

Idea	Exam-ready reminder
Rate equation	rate = k[A]ᵐ[B]ⁿ; orders come from experiment, not equation coefficients
Individual order	doubling a concentration multiplies rate by 2ⁿ; n is restricted to 0, 1 or 2 here
Overall order	add the individual orders in the experimentally determined rate equation
Initial-rates comparison	choose a pair where only one reactant concentration changes, then compare rate and concentration factors
Zero order	a straight concentration–time line with constant negative gradient
First order	a curved concentration–time plot with a constant half-life independent of concentration
Concentration	higher concentration increases collision frequency, not the energy of each collision
Temperature	the Boltzmann distribution broadens and shifts so a larger fraction has energy at least Ea
Catalyst	an alternative lower-Ea pathway gives a larger k at the same temperature
Enzyme	an active site binds a specific substrate; excessive temperature or unsuitable pH disrupts its effective shape

Use simple single-step rate equations. Do not import integrated rate equations, mechanism or rate-determining-step deductions, homogeneous catalysis, or extra H2 catalyst examples.

Practice and Check Your Understanding

H1 Reaction Kinetics Topic Practice
Course-specific 8873 practice across rate equations, data, graph evidence, Boltzmann reasoning, catalysts and enzymes.
Start topic practice
H1 Reaction Kinetics: Check Your Understanding
Check the reasoning links, read the targeted feedback, then try a fresh question independently.
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For 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g), rate = k[NO]²[H₂]. When [NO] = 0.010 and [H₂] = 0.020 mol dm⁻³, the initial rate is 2.4 × 10⁻⁵ mol dm⁻³ s⁻¹. What is k?

12 mol⁻² dm⁶ s⁻¹
0.12 mol⁻¹ dm³ s⁻¹
12 mol dm⁻³ s⁻¹
8.3 × 10⁻² mol⁻² dm⁶ s⁻¹

Initial-rate data: Exp 1 [P] = 0.10, [Q] = 0.10, rate = 1.0 × 10⁻³; Exp 2 [P] = 0.20, [Q] = 0.10, rate = 2.0 × 10⁻³; Exp 3 [P] = 0.20, [Q] = 0.20, rate = 2.0 × 10⁻³. What is the rate equation?

rate = k[P]
rate = k[P][Q]
rate = k[P]²
rate = k[Q]

The concentration of reactant B falls in a straight line from 0.60 mol dm⁻³ at 0 s to 0.30 mol dm⁻³ at 120 s and reaches zero at 240 s. Which conclusion is correct?

Zero order: the gradient, and so the rate, stays constant as [B] falls, and successive half-lives shorten (120 s, then 60 s).
First order, because the first half-life is 120 s.
First order, because the concentration of B decreases with time.
Zero order, with a constant half-life of 120 s.

The pressure of a mixture of reacting gases is increased at constant temperature by compressing it. Why does the rate increase?

The molecules are closer together, so there are more molecules per unit volume and collisions are more frequent.
Compression gives each molecule more kinetic energy, so more collisions exceed the activation energy.
Compression lowers the activation energy.
More molecules are present in the container.

Compared with the Boltzmann distribution of a gas at T₁, how does the distribution at a higher temperature T₂ differ?

The peak is lower and at a higher energy, the total area is unchanged, and a larger area lies beyond the activation energy.
The peak is higher and at a higher energy.
The total area under the curve increases.
The activation energy moves to a lower energy.

Which statement about the energy profile and Boltzmann distribution for a catalysed reaction is correct?

The catalysed pathway has a lower activation energy and the same ΔH, so on the unchanged Boltzmann distribution more molecules have enough energy to react.
The catalysed pathway has a lower activation energy and a less negative ΔH.
The catalyst raises the energy of the reactants so they reach the same activation energy more easily.
The catalysed pathway has a higher activation energy, but molecules collide more often.

Which sequence describes heterogeneous catalysis on a solid surface?

Adsorption of reactants onto active sites → weakening of bonds and reaction → desorption of products
Desorption of reactants → reaction → adsorption of products
Absorption of reactants into the metal → reaction inside the solid → release of products
Adsorption of reactants → reaction → products remain adsorbed permanently

The rate of an enzyme-catalysed reaction rises as temperature increases to about 37 °C and then falls sharply. Which explanation of the fall is correct?

Above the optimum temperature the enzyme's active site changes shape, so the substrate no longer fits and fewer enzyme–substrate complexes form.
The enzyme is used up by the reaction at higher temperatures.
The substrate molecules are denatured by heat.
Molecules move too fast to collide above 37 °C.

Practise after feedback

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About 10 minutes

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This check needs JavaScript to record answers. The reviewed questions are listed below for study.

For 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g), rate = k[NO]²[H₂]. When [NO] = 0.010 and [H₂] = 0.020 mol dm⁻³, the initial rate is 2.4 × 10⁻⁵ mol dm⁻³ s⁻¹. What is k?

12 mol⁻² dm⁶ s⁻¹
0.12 mol⁻¹ dm³ s⁻¹
12 mol dm⁻³ s⁻¹
8.3 × 10⁻² mol⁻² dm⁶ s⁻¹

rate = k[P]
rate = k[P][Q]
rate = k[P]²
rate = k[Q]

The concentration of reactant B falls in a straight line from 0.60 mol dm⁻³ at 0 s to 0.30 mol dm⁻³ at 120 s and reaches zero at 240 s. Which conclusion is correct?

Zero order: the gradient, and so the rate, stays constant as [B] falls, and successive half-lives shorten (120 s, then 60 s).
First order, because the first half-life is 120 s.
First order, because the concentration of B decreases with time.
Zero order, with a constant half-life of 120 s.

The pressure of a mixture of reacting gases is increased at constant temperature by compressing it. Why does the rate increase?

The molecules are closer together, so there are more molecules per unit volume and collisions are more frequent.
Compression gives each molecule more kinetic energy, so more collisions exceed the activation energy.
Compression lowers the activation energy.
More molecules are present in the container.

Compared with the Boltzmann distribution of a gas at T₁, how does the distribution at a higher temperature T₂ differ?

The peak is lower and at a higher energy, the total area is unchanged, and a larger area lies beyond the activation energy.
The peak is higher and at a higher energy.
The total area under the curve increases.
The activation energy moves to a lower energy.

Which statement about the energy profile and Boltzmann distribution for a catalysed reaction is correct?

The catalysed pathway has a lower activation energy and the same ΔH, so on the unchanged Boltzmann distribution more molecules have enough energy to react.
The catalysed pathway has a lower activation energy and a less negative ΔH.
The catalyst raises the energy of the reactants so they reach the same activation energy more easily.
The catalysed pathway has a higher activation energy, but molecules collide more often.

Which sequence describes heterogeneous catalysis on a solid surface?

Adsorption of reactants onto active sites → weakening of bonds and reaction → desorption of products
Desorption of reactants → reaction → adsorption of products
Absorption of reactants into the metal → reaction inside the solid → release of products
Adsorption of reactants → reaction → products remain adsorbed permanently

The rate of an enzyme-catalysed reaction rises as temperature increases to about 37 °C and then falls sharply. Which explanation of the fall is correct?

Above the optimum temperature the enzyme's active site changes shape, so the substrate no longer fits and fewer enzyme–substrate complexes form.
The enzyme is used up by the reaction at higher temperatures.
The substrate molecules are denatured by heat.
Molecules move too fast to collide above 37 °C.

Check what I know

Start here to see which parts you already know.

About 8 minutes

Answer 8 short questions. It shows what to work on next and doesn't count towards mastery.

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This check needs JavaScript to record answers. The reviewed questions are listed below for study.

For the reaction A + 2B → products, rate = k[A][B]² with k = 2.0 × 10⁻² mol⁻² dm⁶ s⁻¹. What is the initial rate when [A] = 0.10 mol dm⁻³ and [B] = 0.20 mol dm⁻³?

8.0 × 10⁻⁵ mol dm⁻³ s⁻¹
4.0 × 10⁻⁴ mol dm⁻³ s⁻¹
8.0 × 10⁻⁴ mol dm⁻³ s⁻¹
4.0 × 10⁻⁵ mol dm⁻³ s⁻¹

Initial-rate data: Exp 1 [A] = 0.10, [B] = 0.10, rate = 2.0 × 10⁻⁴; Exp 2 [A] = 0.20, [B] = 0.10, rate = 8.0 × 10⁻⁴; Exp 3 [A] = 0.10, [B] = 0.30, rate = 6.0 × 10⁻⁴ (concentrations in mol dm⁻³, rates in mol dm⁻³ s⁻¹). What is the rate equation?

rate = k[A]²[B]
rate = k[A][B]
rate = k[A][B]²
rate = k[A]⁴[B]³

The concentration of reactant A is 0.80 mol dm⁻³ at 0 s, 0.40 at 50 s, 0.20 at 100 s and 0.10 at 150 s. Which conclusion is correct?

First order: the half-life is constant at 50 s whatever the starting concentration.
Zero order: the concentration falls by a constant amount in each 50 s interval.
First order: the half-life doubles each time, from 50 s to 100 s.
Second order, with a half-life of 50 s.

Why does increasing the concentration of hydrochloric acid increase the rate of its reaction with magnesium ribbon?

There are more acid particles per unit volume, so collisions with the magnesium surface are more frequent and more successful collisions happen per second.
The acid particles gain more kinetic energy, so more collisions exceed the activation energy.
The activation energy of the reaction is lowered.
A larger proportion of collisions have energy equal to or above the activation energy.

A 10 K rise in temperature often roughly doubles the rate of a reaction. Which explanation is correct?

The Boltzmann distribution shifts to higher energies, so the fraction of molecules with energy at least equal to the activation energy increases greatly; collisions also become slightly more frequent, so k increases.
Collisions become twice as frequent, which doubles the rate.
The activation energy decreases as the temperature rises.
The total number of molecules, the area under the Boltzmann curve, increases.

How does a catalyst increase the rate of a reaction, interpreted using the Boltzmann distribution?

It provides a pathway with a lower activation energy; the distribution is unchanged, but a larger fraction of molecules has at least the new, lower Eₐ, so k is larger.
It shifts the Boltzmann distribution to higher energies.
It increases the frequency of collisions between reactant molecules.
It makes ΔH more negative, so more energy is available for reaction.

In a car's catalytic converter, platinum and rhodium catalyse 2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g). How does this heterogeneous catalyst work?

CO and NO adsorb onto active sites on the metal surface, their bonds are weakened and they react, and the CO₂ and N₂ formed desorb, freeing the sites.
The metal dissolves in the exhaust gases and reacts with them in the gas phase.
Platinum reacts permanently with NO to form platinum oxide, removing NO.
The gases adsorb so strongly that the products stay bound to the surface.

Which description of enzymes is correct?

Protein catalysts that are highly specific for their substrates, as described by the lock-and-key model, and whose activity falls away from an optimum temperature and pH
Carbohydrate catalysts that work equally well at all temperatures
Protein catalysts whose rate keeps increasing as temperature rises
Protein catalysts that each speed up many unrelated reactions

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Reaction Kinetics

H1 Reaction Kinetics learning outcomes

What You’ll Learn

Lessons (Recommended Order)

Simple Rate Equations and Orders

Method of Initial Rates

Zero- and First-order Graphs and Half-life

Activation Energy and Boltzmann Distribution

Heterogeneous Catalysis and Enzymes

Quick Reference

Practice and Check Your Understanding

H1 Reaction Kinetics Topic Practice

H1 Reaction Kinetics: Check Your Understanding

Practise

Practice complete

Your answers

Practise after feedback

Practice complete

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Check what I know

Starting check complete

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Check my progress

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Check again

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Review

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