Reaction Kinetics — A-Level Chemistry Study Guide

For: A-Level Chemistry candidates sitting A-Level Chemistry.

Covers: Rate of reaction measurement, collision theory and activation energy, factors affecting reaction rate, rate equations/orders/half-lives, and Maxwell-Boltzmann distribution curves for kinetic analysis.

You should already know: IGCSE Chemistry, basic algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Chemistry style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Reaction Kinetics?

Reaction kinetics is the branch of physical chemistry that studies the speed of chemical reactions, the factors that alter reaction speed, and the step-by-step mechanisms by which reactions proceed. Also referred to as chemical kinetics, it is a core physical chemistry topic tested in both AS and A2 A-Level Chemistry papers, typically accounting for 8-12 marks per exam, often combined with equilibrium or organic mechanism questions in long-answer sections.

2. Rate of reaction — measurement and units

The rate of reaction is defined as the change in concentration of a reactant or product per unit time. Since reactant concentration decreases over time and product concentration increases, a negative sign is added for reactants to ensure rate is always a positive value: $$r=\frac{\Delta [P]}{\Delta t}=-\frac{\Delta [R]}{\Delta t}$$ Where $[P]$ = product concentration, $[R]$ = reactant concentration, and $\Delta t$ = time interval.

Standard units for rate are ${\text{mol dm}}^{-3}{\text{ s}}^{-1}$, though units may use minutes or hours for slow reactions.

Common measurement methods tested in A-Level exams:

1. Gas volume collection: For reactions producing gas (e.g. Mg + HCl → MgCl₂ + H₂), measure volume of gas collected at fixed time intervals.
2. Mass loss: For reactions producing gaseous products that escape an open flask, measure mass decrease over time.
3. Colorimetry: For reactions with colored reactants or products, measure absorbance of light (proportional to concentration) over time.
4. Quench titration: Stop the reaction at fixed intervals (via cooling, dilution, or removing the catalyst) then titrate to find remaining reactant concentration.

Worked Example

For the decomposition reaction $2H_2{O}_2(aq)\to 2H_{2}O(l)+O_2(g)$, the concentration of $H_2{O}_2$ falls from $2.0{\text{mol dm}}^{-3}$ to $1.6{\text{mol dm}}^{-3}$ in 10 seconds. Calculate the average reaction rate.

Solution: $$r=-\frac{\Delta [H_2{O}_2]}{\Delta t}=-\frac{1.6-2.0}{10}=0.04{\text{mol dm}}^{-3}{\text{ s}}^{-1}$$ To get the overall reaction rate (adjusted for stoichiometry), divide by the stoichiometric coefficient of $H_2{O}_2$ (2): $$\text{Overall rate}=\frac{0.04}{2}=0.02{\text{mol dm}}^{-3}{\text{ s}}^{-1}$$

Exam tip: Examiners often ask you to select the appropriate measurement method for a given reaction, so always check for gaseous products, colored species, or acid/base reactants that can be titrated.

3. Collision theory and activation energy

Collision theory is the foundational model used to explain why reaction rates change with different conditions. Its three core postulates are:

1. For a reaction to occur, reactant particles must collide with each other.
2. Collisions must have kinetic energy equal to or greater than the activation energy ($E_a$) of the reaction to be successful.
3. Collisions must have the correct spatial orientation to break existing reactant bonds and form new product bonds.

Activation energy is defined as the minimum kinetic energy colliding particles must possess to initiate a reaction. It represents the energy barrier required to break bonds in reactant molecules before products can form.

Worked Example

Explain why the highly exothermic combustion of methane ($\Delta H=-890{\text{kJ mol}}^{-1}$) does not occur spontaneously at room temperature.

Solution: Even though the overall reaction releases large amounts of energy, reactant methane and oxygen molecules at room temperature do not have enough kinetic energy to overcome the activation energy barrier for the reaction. A spark or flame is required to provide the initial energy to reach $E_a$ and start the reaction, after which the heat released sustains the process.

Exam note: Always explicitly reference "successful collisions" when explaining rate changes using collision theory: stating "more collisions" alone will lose marks, as only collisions meeting the energy and orientation requirements contribute to reaction progress.

4. Effects of temperature, concentration, surface area, catalyst

Each factor that alters reaction rate can be directly explained using collision theory:

1. Concentration (solutions) / Pressure (gases): Increasing concentration increases the number of reactant particles per unit volume, raising the frequency of collisions, and thus the frequency of successful collisions, leading to a faster reaction rate. Increasing pressure for gas reactions compresses the gas, producing the same effect as higher concentration.
2. Surface area (solid reactants): Breaking a solid into smaller fragments increases the surface area exposed to liquid or gas reactants, increasing the frequency of collisions between reactant phases, leading to more successful collisions per second.
3. Temperature: Increasing temperature raises the average kinetic energy of all particles. The primary effect is that a far larger proportion of particles have energy ≥ $E_a$; a secondary effect is a small increase in collision frequency. Combined, this leads to a dramatic increase in successful collision frequency: a 10°C temperature rise roughly doubles the rate of most reactions.
4. Catalyst: A substance that increases reaction rate without being consumed in the reaction. It works by providing an alternative reaction pathway with a lower activation energy ($E_a^{\prime }<E_a$), so a larger proportion of particles have energy ≥ the lower $E_a^{\prime }$, increasing successful collision frequency. Catalysts do not alter the enthalpy change of a reaction, or the position of equilibrium, only the speed at which equilibrium is reached.

Worked Example

Explain why powdered calcium carbonate reacts much faster with dilute hydrochloric acid than equal-mass lumps of calcium carbonate.

Solution: Powdered CaCO₃ has a far larger exposed surface area than solid lumps. More CaCO₃ particles are accessible to H⁺ ions from HCl, so collision frequency between reactants increases, leading to a higher frequency of successful collisions and a faster reaction rate.

5. Rate equation, orders, half-life

The rate equation (or rate law) is an experimentally derived expression that relates reaction rate to reactant concentrations. For a general reaction $aA+bB\to \text{products}$, the rate equation takes the form: $$r=k[A{]}^m[B{]}^n$$ Where:

• $k$ = rate constant, a temperature-dependent constant unique to the reaction
• $m$ = order of reaction with respect to A
• $n$ = order of reaction with respect to B
• Overall order of reaction = $m+n$

Critical note: The values of $m$ and $n$ are not equal to the stoichiometric coefficients $a$ and $b$ unless the reaction is a single-step elementary process. Orders must always be determined from experimental data, and correspond to the number of molecules of each species involved in the slowest (rate-determining) step of the reaction.

Half-life ($t_{1/2}$) is the time taken for the concentration of a reactant to fall to half its initial value. For first-order reactions (overall order = 1), half-life is constant, independent of initial concentration: $$t_{1/2}=\frac{\ln 2}{k}\approx \frac{0.693}{k}$$ For zero-order reactions, half-life decreases as concentration decreases; for second-order reactions, half-life increases as concentration decreases.

Units of $k$ are derived by rearranging the rate equation and substituting units for rate and concentration:

• Zero order: $k$ units = ${\text{mol dm}}^{-3}{\text{ s}}^{-1}$
• First order: $k$ units = ${\text{s}}^{-1}$
• Second order: $k$ units = ${\text{mol}}^{-1}{\text{ dm}}^3{\text{ s}}^{-1}$

Worked Example

Experimental data for the reaction $2NO(g)+O_2(g)\to 2N{O}_2(g)$ gives the rate equation $r=k[NO{]}^2[O_2]$. (a) What is the overall order? (b) What are the units of $k$? (c) How does the rate change if [NO] is doubled, with [O₂] held constant?

Solution: (a) Overall order = 2 + 1 = 3 (b) Rearrange for $k$: $k=\frac{r}{[NO{]}^2[O_2]}$. Substitute units: $$k=\frac{{\text{mol dm}}^{-3}{\text{ s}}^{-1}}{({\text{mol dm}}^{-3}{)}^2({\text{mol dm}}^{-3})}={\text{mol}}^{-2}{\text{ dm}}^6{\text{ s}}^{-1}$$ (c) If [NO] doubles, $[NO{]}^2$ increases by a factor of 4, so reaction rate increases 4 times.

6. Maxwell-Boltzmann distribution

The Maxwell-Boltzmann (MB) distribution is a probability curve that shows the distribution of kinetic energies of particles in a gas or solution at a fixed temperature. Key features:

• X-axis = kinetic energy of particles, Y-axis = number (or fraction) of particles with a given kinetic energy
• No particles have zero kinetic energy, so the curve starts at the origin
• The peak corresponds to the most probable kinetic energy (the energy of the largest number of particles)
• The curve has a long asymptotic tail that never touches the x-axis: there is no upper limit to particle kinetic energy
• The total area under the curve equals the total number of particles in the sample, so it remains constant for the same sample at different temperatures

Effect of temperature increase: The curve shifts to the right, becomes shorter and broader, and the area under the curve to the right of $E_a$ increases significantly, meaning more particles have sufficient energy for successful collisions. Effect of catalyst addition: The $E_a$ mark shifts left on the x-axis, so the area under the curve to the right of $E_a$ increases, without changing the shape of the distribution curve itself.

Exam tip: When asked to draw MB curves, never draw the tail touching the x-axis, and keep the area under the curve the same for different temperatures of the same sample, or you will lose marks.

7. Common Pitfalls (and how to avoid them)

• Wrong: Stating that higher temperature increases rate only because collision frequency increases. Why: Students forget the primary effect of temperature is increasing the proportion of particles with energy ≥ $E_a$. Correct: Always mention both effects: more particles have energy ≥ $E_a$ (primary) and collision frequency increases slightly (secondary), leading to more successful collisions.
• Wrong: Using stoichiometric coefficients from the balanced equation to write the rate equation without experimental data. Why: Students confuse rate laws with reaction stoichiometry. Correct: Reaction orders are always derived from experimental data, they only match stoichiometric coefficients for single-step elementary reactions, which are rare.
• Wrong: Claiming catalysts change the enthalpy change of a reaction or shift equilibrium position. Why: Students assume catalysts alter the reaction itself, rather than just the reaction pathway. Correct: Catalysts only lower activation energy, they do not change ΔH, do not affect reaction yield, and speed up forward and reverse reactions equally.
• Wrong: Giving incorrect units for the rate constant $k$. Why: Students memorize units instead of deriving them. Correct: Always derive $k$ units by rearranging the rate equation, substituting units for rate and concentration, and cancelling terms.
• Wrong: Drawing a Maxwell-Boltzmann curve that touches the x-axis at high kinetic energy, or changing the area under the curve when temperature changes. Why: Students forget the tail is asymptotic and area equals total particle count. Correct: Draw the curve approaching but never touching the x-axis, and keep area identical for different temperatures of the same sample.

8. Practice Questions (A-Level Chemistry Style)

Question 1

The reaction between bromate, bromide and hydrogen ions follows the equation: $$Br{O}_3^{-}(aq)+5B{r}^{-}(aq)+6H^{+}(aq)\to 3B{r}_2(aq)+3H_{2}O(l)$$ The initial rate data below was collected at 298K:

(a) Deduce the order of reaction with respect to each reactant. (b) Write the full rate equation for the reaction. (c) Calculate the value of the rate constant $k$ at 298K, including units.

Solution: (a) Compare 1 and 2: $[Br{O}_3^{-}]$ doubles, rate doubles → order 1 with respect to $Br{O}_3^{-}$. Compare 1 and 3: $[B{r}^{-}]$ triples, rate triples → order 1 with respect to $B{r}^{-}$. Compare 1 and 4: $[H^{+}]$ doubles, rate quadruples → order 2 with respect to $H^{+}$. (b) Rate equation: $r=k[Br{O}_3^{-}][B{r}^{-}][H^{+}{]}^2$ (c) Substitute values from experiment 1: $$k=\frac{1.2\times 10^{-3}}{(0.10)(0.10)(0.10{)}^2}=12{\text{mol}}^{-3}{\text{ dm}}^9{\text{ s}}^{-1}$$

Question 2

(a) Define the term activation energy of a reaction. (b) Explain, using collision theory, how a catalyst increases the rate of a reaction. (c) State two properties of a catalyst that remain unchanged at the end of a reaction.

Solution: (a) Activation energy is the minimum kinetic energy that colliding reactant particles must possess for a successful reaction to occur, i.e. to break existing bonds and form product molecules. (b) A catalyst provides an alternative reaction pathway with a lower activation energy than the uncatalysed pathway. A greater proportion of reactant particles have kinetic energy ≥ the lower activation energy, so the frequency of successful collisions increases, leading to a faster reaction rate. (c) Mass of the catalyst, chemical composition of the catalyst.

Question 3

The decomposition of hydrogen peroxide is a first-order reaction with rate constant $k=2.5\times 10^{-4}{\text{s}}^{-1}$ at 300K. (a) Calculate the half-life of the reaction. (b) If the initial concentration of $H_2{O}_2$ is $1.6{\text{mol dm}}^{-3}$, what will its concentration be after 3 half-lives? (c) Explain why the half-life of this first-order reaction is constant, regardless of initial concentration.

Solution: (a) For first-order reactions: $t_{1/2}=\frac{\ln 2}{k}=\frac{0.693}{2.5\times 10^{-4}}=2772\text{s}\approx 2800\text{s}$ (2 significant figures) (b) After 1 half-life: $0.8{\text{mol dm}}^{-3}$, after 2: $0.4{\text{mol dm}}^{-3}$, after 3: $0.2{\text{mol dm}}^{-3}$ (c) For first-order reactions, rate is proportional to reactant concentration: $r=k[A]$. As concentration decreases, the reaction rate decreases proportionally, so the time taken for concentration to halve remains constant, no matter the starting concentration.

9. Quick Reference Cheatsheet

10. What's Next

Reaction kinetics is a foundational physical chemistry topic that connects directly to multiple later A-Level Chemistry syllabus units. You will use your understanding of rate equations and reaction orders to deduce reaction mechanisms in A2 organic and physical chemistry papers, and your knowledge of activation energy and catalysts will be critical when studying chemical equilibrium, industrial processes (such as the Haber and Contact processes), and enzyme kinetics in biological chemistry applications. Kinetics questions are frequently combined with equilibrium questions in high-mark long-answer sections of A2 papers, so mastering this topic will give you a significant advantage in these sections.

If you are struggling with any part of reaction kinetics, from calculating rate constants to interpreting Maxwell-Boltzmann distribution curves, you can ask Ollie, our AI tutor, for personalized explanations, additional practice questions, or step-by-step walkthroughs of past paper problems. You can also find more A-Level Chemistry study resources on the homepage, including guides for chemical bonding, energetics, and organic reaction mechanisms to help you prepare for your exams.

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