Reaction Kinetics

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

GCE A-Level H2 Chemistry 9476-2027

Learning goals

Rate Equations, Orders, Rate Constant
Initial Rates Method
Concentration–Time Graphs and Half-life
Activation Energy and Boltzmann Distribution
Mechanisms and Rate-determining Step
Catalysis and Enzymes

H2 Reaction Kinetics learning outcomes

This hub builds the full chain: experiment → rate law → mechanism → explanation.

Before you start

Beginner path: follow the Lessons (Recommended Order) from top to bottom.

Revision: Jump to: Quick Reference · What You Must Memorise · Common Exam Traps · Hub Quiz.

Prerequisites: skim the Prerequisites list first if you’re rusty.

Useful links: A Level portal · Exam Skills

What You’ll Learn

Deduce orders and rate equations from data (initial rates and concentration–time).
Use rate laws to predict rate changes and units of k.
Link mechanisms and rate-determining steps to observed rate equations.

Lessons (Recommended Order)

Rate Equations, Orders, Rate Constant
Write rate laws and units of k.
Open lesson
Initial Rates Method
Compare experiments to deduce reaction orders.
Open lesson
Concentration–Time Graphs and Half-life
Interpret graphs and identify first-order behaviour.
Open lesson
Mechanisms and Rate-determining Step
Link steps to the observed rate equation.
Open lesson
Activation Energy and Boltzmann Distribution
Explain temperature effects and catalyst action.
Open lesson
Catalysis and Enzymes
Heterogeneous/homogeneous catalysts and enzyme conditions.
Open lesson

A lower-temperature curve has a taller peak at lower energy. A higher-temperature curve is lower and broader. Uncatalysed and catalysed activation-energy thresholds mark the fractions energetic enough to react. — Kinetics bridge: temperature changes the energy distribution; a catalyst lowers Ea. Both increase the fraction able to react, but neither lets you infer the rate equation from the balanced equation.

Values for Maxwell–Boltzmann energy distributions at two temperatures
Particle energy (relative units)	Lower temperature	Higher temperature
0	0	0
0.5	0.2289	0.0974
1	0.3018	0.1516
1.5	0.2984	0.1771
2	0.2623	0.1839
2.5	0.2162	0.1791
3	0.171	0.1673
3.5	0.1315	0.1521
4	0.0991	0.1353
4.5	0.0735	0.1186
5	0.0538	0.1026
5.5	0.039	0.0879
6	0.0281	0.0747
6.5	0.0201	0.063
7	0.0142	0.0528
7.5	0.0101	0.0441
8	0.0071	0.0366
8.5	0.005	0.0303
9	0.0035	0.025
9.5	0.0024	0.0205
10	0.0017	0.0168

Values for Maxwell–Boltzmann energy distributions at two temperatures
Particle energy (relative units)	Lower temperature	Higher temperature
0	0	0
0.5	0.2289	0.0974
1	0.3018	0.1516
1.5	0.2984	0.1771
2	0.2623	0.1839
2.5	0.2162	0.1791
3	0.171	0.1673
3.5	0.1315	0.1521
4	0.0991	0.1353
4.5	0.0735	0.1186
5	0.0538	0.1026
5.5	0.039	0.0879
6	0.0281	0.0747
6.5	0.0201	0.063
7	0.0142	0.0528
7.5	0.0101	0.0441
8	0.0071	0.0366
8.5	0.005	0.0303
9	0.0035	0.025
9.5	0.0024	0.0205
10	0.0017	0.0168

Quick Reference

If the question asks…	Start with…	What to show
“find the order” (initial rates)	compare experiments where only one concentration changes	show ratios, then state the order clearly
“write the rate equation”	use your orders	Rate = k[A]^m[B]ⁿ (define k)
“find units of k”	write rate units and concentration units	divide by concentration powers (use overall order)
“is it first order?”	compare successive halving intervals on concentration–time data	“constant half-life” wording
“link rate law to mechanism”	identify the rate-determining step	use the stated fast equilibrium where needed to remove an intermediate

What You Must Memorise

Rate equation form: rate = k[A]^m[B]ⁿ (orders are found experimentally).
Order meaning: doubling [A] changes rate by 2^m; zero order means no rate change when concentration changes.
Initial rates ratio (when only [A] changes): rate₂/rate₁ = ([A]₂/[A]₁)^m.
First-order signature required here: constant half-life from concentration–time data.
Boundary: integrated rate equations, logarithmic plots, the Arrhenius equation and steady-state derivations are not required by 9476.
Units anchor: rate typically in mol dm⁻³ s⁻¹; units of k depend on overall order.
Catalyst language: alternative pathway with lower Eₐ → larger fraction with E ≥ Eₐ.

Hub Quiz and Check Your Understanding

Use the quiz without notes first. Classify each error as rate-factor reasoning, graph evidence, mechanism consistency, collision theory, catalysis or enzyme conditions.

H2 Reaction Kinetics Quiz
Check orders, rate constants, graphical evidence, mechanisms, Boltzmann reasoning, catalysis and enzymes.
Start hub quiz
H2 Reaction Kinetics Knowledge Check
Find the reasoning step that needs more work, use the feedback, then try a fresh question independently.
Practice onlyCheck understanding

Practise

Work through questions with marking and feedback as you learn.

About 10 minutes

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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.

A reaction A + 2B → D occurs by Step 1 (slow): A + B → C, then Step 2 (fast): C + B → D. What is the rate equation?

rate = k[A][B]
rate = k[A][B]²
rate = k[C][B]
rate = k[A]

Practise after feedback

After a check, practise the skills it showed you need to work on.

About 10 minutes

Questions are picked at random each time you start. You'll see the answer after each question. It's for practice only 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 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.

A reaction A + 2B → D occurs by Step 1 (slow): A + B → C, then Step 2 (fast): C + B → D. What is the rate equation?

rate = k[A][B]
rate = k[A][B]²
rate = k[C][B]
rate = k[A]

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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

For 2NO₂(g) + F₂(g) → 2NO₂F(g), the rate equation is rate = k[NO₂][F₂]. Which mechanism is consistent with this?

Step 1 (slow): NO₂ + F₂ → NO₂F + F; Step 2 (fast): NO₂ + F → NO₂F
One step: 2NO₂ + F₂ → 2NO₂F
Step 1 (fast): NO₂ + F₂ → NO₂F + F; Step 2 (slow): NO₂ + F → NO₂F
Step 1 (slow): F₂ → 2F; Step 2 (fast): 2NO₂ + 2F → 2NO₂F

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Reaction Kinetics

H2 Reaction Kinetics learning outcomes

What You’ll Learn

Lessons (Recommended Order)

Rate Equations, Orders, Rate Constant

Initial Rates Method

Concentration–Time Graphs and Half-life

Mechanisms and Rate-determining Step

Activation Energy and Boltzmann Distribution

Catalysis and Enzymes

Quick Reference

What You Must Memorise

Hub Quiz and Check Your Understanding

H2 Reaction Kinetics Quiz

H2 Reaction Kinetics Knowledge Check

Practise

Practice complete

Your answers

Practise after feedback

Practice complete

Your answers

Check what I know

Starting check complete

Your answers

Check my progress

Progress check complete

Your answers

Check again

Progress check complete

Your answers

Review

Review complete

Your answers