Rates of reaction

Chemistry 1110 · CCEA A Level · 10 min read · Updated: Aug 31, 2026

Welcome to A2 Chemical Kinetics: Mastering Rates of Reaction

Welcome to one of the most exciting and scoring topics in CCEA A2 1 Chemistry (Topic 4.3)! At AS Level, you learned how factors like temperature, concentration, surface area, and catalysts alter the speed of a reaction. Now, in A2, we take this a step further by mathematically measuring, expressing, and predicting exactly how fast reactions happen and uncovering the step-by-step pathways (mechanisms) molecules take when they transform.

Don't worry if mathematical expressions in chemistry seem intimidating at first! We will break down every equation, graph, and concept step-by-step so you can easily master both the written exam (ACH12) and practical paper (Unit A2 3).

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1. Core Definitions and the Rate Equation

Let's begin with the foundational language used in kinetics. Examiners frequently test these definitions word-for-word in structured questions.

Rate of Reaction: The change in concentration of a reactant or a product per unit time. Its standard units are \(\text{mol dm}^{-3}\text{ s}^{-1}\).

The Rate Equation (Rate Expression): An experimentally determined equation showing how the rate depends on the concentrations of reactants:

\(\text{Rate} = k[\text{A}]^m[\text{B}]^n\)

Here is what each term means:

• \([\text{A}]\) and \([\text{B}]\) represent the molar concentrations of reactants \(\text{A}\) and \(\text{B}\) (in \(\text{mol dm}^{-3}\)).
• \(m\) and \(n\) are the Orders of Reaction with respect to each reactant (at A-Level, these are integers: \(0\), \(1\), or \(2\)).
• \(k\) is the Rate Constant.
• Overall Order of Reaction: The sum of individual orders (\(m + n\)).

Order of Reaction (with respect to a reactant): The power to which the concentration of that reactant is raised in the rate equation.
• Zero Order (\(m = 0\)): Changing the concentration has no effect on the rate (\([\text{A}]^0 = 1\)).
• First Order (\(m = 1\)): The rate is directly proportional to the concentration. Doubling \([\text{A}]\) doubles the rate (\(2^1 = 2\)).
• Second Order (\(m = 2\)): The rate is proportional to the square of the concentration. Doubling \([\text{A}]\) quadruples the rate (\(2^2 = 4\)).

The Rate Constant (\(k\)): The proportionality constant linking rate to concentration. For a specific reaction, \(k\) remains constant at a fixed temperature, but its value changes when the temperature changes or when a catalyst is added.

Half-Life (\(t_{1/2}\)): The time taken for the concentration of a reactant to decrease to half of its initial value.

How to Deduce Units for the Rate Constant (\(k\))

One of the easiest ways to pick up marks (or lose them!) in CCEA exams is calculating the units of \(k\). Never guess them—always rearrange the rate equation:

\(k = \frac{\text{Rate}}{[\text{A}]^m[\text{B}]^n}\)

Let's work out the units for different overall orders:

• Zero Order Overall (\(\text{Rate} = k\)):
\(\text{Units} = \text{mol dm}^{-3}\text{ s}^{-1}\)

• First Order Overall (\(\text{Rate} = k[\text{A}]\)):
\(k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{\text{mol dm}^{-3}} = \text{s}^{-1}\)

• Second Order Overall (\(\text{Rate} = k[\text{A}]^2\) or \(\text{Rate} = k[\text{A}][\text{B}]\)):
\(k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})} = \frac{\text{s}^{-1}}{\text{mol dm}^{-3}} = \text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}\)

• Third Order Overall (\(\text{Rate} = k[\text{A}]^2[\text{B}]\)):
\(k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\)

Key Takeaway: Rate equations can only be found through experimental data, never by simply looking at the big stoichiometric balancing numbers in an ordinary chemical equation.

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2. Determining Orders from Experimental Data and Graphs

Method A: The Initial Rates Method (Table Interpretation)

In this method, scientists perform several runs of a reaction at constant temperature, varying initial concentrations and measuring the initial rate for each run.

Example Walkthrough:
• Experiment 1: \([\text{A}] = 0.10\text{ mol dm}^{-3}\), \([\text{B}] = 0.10\text{ mol dm}^{-3}\), \(\text{Initial Rate} = 2.0 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\)
• Experiment 2: \([\text{A}] = 0.20\text{ mol dm}^{-3}\), \([\text{B}] = 0.10\text{ mol dm}^{-3}\), \(\text{Initial Rate} = 4.0 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\)
• Experiment 3: \([\text{A}] = 0.10\text{ mol dm}^{-3}\), \([\text{B}] = 0.20\text{ mol dm}^{-3}\), \(\text{Initial Rate} = 8.0 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\)

Step 1: Find the order with respect to \(\text{A}\):
Compare Exp 1 and Exp 2 (where \([\text{B}]\) is kept constant): \([\text{A}]\) doubles (\(\times 2\)) and the rate doubles (\(\times 2\)). Since \(2^1 = 2\), the reaction is first order with respect to \(\text{A}\).

Step 2: Find the order with respect to \(\text{B}\):
Compare Exp 1 and Exp 3 (where \([\text{A}]\) is kept constant): \([\text{B}]\) doubles (\(\times 2\)) and the rate quadruples (\(\times 4\)). Since \(2^2 = 4\), the reaction is second order with respect to \(\text{B}\).

Step 3: Write the Rate Equation:
\(\text{Rate} = k[\text{A}][\text{B}]^2\) (Overall order \(= 1 + 2 = 3\)).

Method B: Concentration–Time Graphs (\([\text{A}]\) vs \(t\))

By plotting reactant concentration against time, the shape of the decay curve reveals the order:

• Zero Order: A straight line with a constant negative gradient. The rate does not slow down as reactant is used up.
• First Order: An exponential decay curve with a constant half-life (\(t_{1/2}\)). Whether you start at \(1.0\text{ mol dm}^{-3}\) or \(0.5\text{ mol dm}^{-3}\), the time taken to halve remains identical.
• Second Order: A steep curve where successive half-lives double (\(t_{1/2}\) increases as concentration decreases).

Calculating instantaneous rate: To find the rate at any time \(t\), draw a sharp tangent to the curve at that exact point and calculate its gradient: \(\text{Gradient} = \frac{\Delta y}{\Delta x}\).

Method C: Rate–Concentration Graphs (\(\text{Rate}\) vs \([\text{A}]\))

• Zero Order (\(m = 0\)): A horizontal straight line. Changing concentration has no effect on rate.
• First Order (\(m = 1\)): A straight line passing directly through the origin (\(\text{Rate} \propto [\text{A}]\)).
• Second Order (\(m = 2\)): An upward-curving parabola (\(\text{Rate} \propto [\text{A}]^2\)).

Key Takeaway: Always check your graph axes carefully! A horizontal line on a Rate vs Concentration graph means zero order, but a straight diagonal line on a Concentration vs Time graph also means zero order.

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3. Experimental Techniques for Monitoring Reaction Rates

CCEA requires you to select and explain suitable experimental techniques for measuring reaction rates continuously or via initial rates.

1. Colorimetry:
Used when a reactant or product is coloured (e.g., tracking the loss of brown/orange colour as \(\text{Br}_2\) or \(\text{I}_2\) reacts). A colorimeter measures light absorbance over time; absorbance is directly proportional to concentration.

2. Sampling, Quenching, and Titration:
Small portions (aliquots) are removed at regular time intervals and immediately quenched to stop the reaction. Quenching is achieved by rapid cooling on ice, large dilution with cold water, or neutralizing an acid/base catalyst. The remaining reactant or formed product is then determined by volumetric titration (e.g., titrating liberated iodine with standard sodium thiosulfate using a starch indicator).

3. Gas Collection or Mass Loss:
Used when a reaction produces a gas. We can measure the volume of gas produced over time using a sealed gas syringe, or monitor mass loss over time using a precision balance (effective for dense gases like \(\text{CO}_2\)).

4. Electrical Conductivity (Conductimetry):
Used when there is a change in the total number or type of ions in solution as the reaction proceeds.

5. Clock Reactions (e.g., The Iodine Clock):
An initial rate method where a small, fixed amount of an added reagent produces a sharp, visible colour change (such as the sudden appearance of the blue-black starch-iodine complex). The time taken (\(t\)) to reach this visual endpoint is recorded. Because the amount of product formed is fixed and small:

\(\text{Initial Rate} \propto \frac{1}{t}\)

Key Takeaway: Choose the method that matches the physical property changing during the reaction (colour \(\rightarrow\) colorimetry; gas \(\rightarrow\) gas syringe; ions \(\rightarrow\) conductivity).

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4. Reaction Mechanisms and the Rate-Determining Step (RDS)

Most chemical reactions do not happen in one single collision; they occur via a sequence of elementary steps called a reaction mechanism.

The Rate-Determining Step (RDS): The slowest single step in a multi-step reaction. Just like traffic moving along a motorway is limited by a single bottleneck lane closure, the overall speed of a chemical reaction is entirely limited by its slowest step.

Rules connecting Mechanisms to Rate Equations:
1. Any reactant (or catalyst) that appears in the rate equation must be present in the rate-determining step (or in an equilibrium step before the RDS).
2. Species that react only after the RDS do not affect the rate and will be zero order (they do not appear in the rate equation).

Crucial Organic Chemistry Link: Nucleophilic Substitution Mechanisms

CCEA directly links kinetics to the mechanisms of halogenoalkanes:

1. \(\text{S}_\text{N}1\) Mechanism (Tertiary Halogenoalkanes):
• Step 1 (Slow / RDS): The carbon–halogen (\(\text{C–X}\)) bond breaks heterolytically to form a stable tertiary carbocation intermediate.
• Step 2 (Fast): The nucleophile (\(\text{OH}^-\)) rapidly attacks the carbocation.
Because only the halogenoalkane participates in the slow step, the rate equation is first order overall:

\(\text{Rate} = k[\text{tertiary halogenoalkane}]\)

The concentration of hydroxide ions has zero effect on the rate (\([\text{OH}^-]^0\)).

2. \(\text{S}_\text{N}2\) Mechanism (Primary Halogenoalkanes):
• Occurs in a single, concerted step where the nucleophile attacks the primary carbon at the same time the halide leaving group departs, passing through a bimolecular transition state.
Because both molecules collide in this single step, the rate equation is second order overall:

\(\text{Rate} = k[\text{primary halogenoalkane}][\text{OH}^-]\)

Key Takeaway: Kinetics provides physical proof for organic mechanisms: \(\text{S}_\text{N}1\) is 1st order overall (unimolecular RDS), while \(\text{S}_\text{N}2\) is 2nd order overall (bimolecular RDS).

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5. Temperature and the Arrhenius Equation

Why does increasing temperature drastically increase reaction rates? Raising the temperature gives particles more kinetic energy, meaning a significantly larger fraction of collisions have energy greater than or equal to the activation energy (\(E \ge E_a\)). This causes the rate constant (\(k\)) to increase exponentially.

The mathematical relationship is given by the Arrhenius Equation:

\(k = A e^{-\frac{E_a}{RT}}\)

Taking the natural logarithm (\(\ln\)) of both sides gives the linear form:

\(\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\)

Let's define the terms:
• \(k\) = Rate constant
• \(A\) = Arrhenius pre-exponential factor (frequency factor related to collision frequency and orientation)
• \(E_a\) = Activation energy in \(\text{J mol}^{-1}\)
• \(R\) = Gas constant (\(8.31\text{ J K}^{-1}\text{ mol}^{-1}\))
• \(T\) = Absolute temperature in Kelvin (\(\text{K}\))

Arrhenius Graphs: Finding \(E_a\) and \(A\)

The linear Arrhenius equation matches the equation of a straight line (\(y = mx + c\)):

• \(y = \ln k\)
• \(x = \frac{1}{T}\)
• \(\text{Gradient } (m) = -\frac{E_a}{R}\)
• \(y\text{-intercept } (c) = \ln A\)

When you plot \(\ln k\) against \(\frac{1}{T}\), you obtain a straight line with a negative gradient.

Step-by-step determination of \(E_a\):
1. Calculate the gradient from the graph: \(\text{Gradient} = \frac{\Delta y}{\Delta x}\) (this will be a negative number).
2. Rearrange the gradient formula: \(E_a = -\text{gradient} \times R\).
3. Multiply by \(8.31\text{ J K}^{-1}\text{ mol}^{-1}\) to get \(E_a\) in \(\text{J mol}^{-1}\).
4. Crucial step: Divide by \(1000\) to convert your final answer into \(\text{kJ mol}^{-1}\) if requested by the question!

Key Takeaway: On an Arrhenius plot, the steeper the downward slope, the higher the activation energy of the reaction.

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6. Top CCEA Exam Pitfalls to Avoid

• Stoichiometry Trap: Never write a rate equation from the balancing numbers of an overall chemical equation. Orders must be determined experimentally.
• Temperature Units: Always convert temperatures from Celsius to Kelvin by adding \(273\) (\(T(\text{K}) = \theta(^\circ\text{C}) + 273\)) before calculating \(\frac{1}{T}\).
• Arrhenius Units: Remember that \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\) uses Joules, not kilojoules. Your direct calculation gives \(E_a\) in \(\text{J mol}^{-1}\). Convert to \(\text{kJ mol}^{-1}\) only at the end by dividing by \(1000\).
• Rate Constant Units: Always re-evaluate the units of \(k\) from scratch for each specific rate equation. They change depending on the overall order!
• Drawing Tangents: When finding the initial rate at \(t = 0\text{ s}\), ensure your ruler forms a clean tangent touching the origin without cutting across the curve.

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Quick Summary Checklist

• Rate: \(\text{mol dm}^{-3}\text{ s}^{-1}\)
• Zero order: Rate is independent of concentration; \([\text{A}]\) vs \(t\) is a straight downward line.
• First order: Constant half-life (\(t_{1/2}\)); \(\text{Rate} \propto [\text{A}]\); units of \(k = \text{s}^{-1}\).
• Second order: Successive half-lives double; \(\text{Rate} \propto [\text{A}]^2\); units of \(k = \text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}\).
• Rate-Determining Step: The slowest step controlling overall rate; dictates which species appear in the rate equation.
• Arrhenius Plot: Plotting \(\ln k\) vs \(\frac{1}{T}\) gives \(\text{gradient} = -\frac{E_a}{R}\) and \(\text{intercept} = \ln A\).

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the definition and standard unit for the rate of reaction?

Rate of reaction is the change in concentration of a reactant or product per unit time, with standard units of \(\text{mol dm}^{-3}\text{ s}^{-1}\).

What is meant by the order of reaction with respect to a given reactant?

The order of reaction is the power to which the concentration of that reactant is raised in the experimentally determined rate equation.

How do you deduce the units of the rate constant (\(k\)) for a given rate equation?

Rearrange the rate equation to make \(k\) the subject (\(k = \frac{\text{Rate}}{[\text{Reactants}]^n}\)) and substitute units of \(\text{mol dm}^{-3}\text{ s}^{-1}\) for rate and \(\text{mol dm}^{-3}\) for concentration.

How can the order of a reaction be determined using the initial rates method?

By comparing experiments where one reactant's concentration changes while all others remain constant, and observing the proportional change in initial rate.

How can concentration–time graphs be used to distinguish between zero-, first-, and second-order reactions?

Zero order gives a straight line with constant negative slope; first order gives an exponential decay with constant half-life (\(t_{1/2}\)); second order gives a steep decay curve where successive half-lives double.

What shapes do rate–concentration graphs have for zero-, first-, and second-order reactions?

Zero order produces a horizontal line, first order produces a straight line through the origin, and second order produces an upward curve (parabola).

How is the initial rate calculated in a clock reaction?

Initial rate is inversely proportional to the time taken (\(t\)) to reach a sharp, visible endpoint: \(\text{Initial Rate} \propto \frac{1}{t}\).

What is the rate-determining step (RDS) in a reaction mechanism, and how does it relate to the rate equation?

The RDS is the slowest step in a multi-step reaction; only species involved in or prior to the RDS appear in the rate equation.

How do the kinetics of \(\text{S}_\text{N}1\) and \(\text{S}_\text{N}2\) nucleophilic substitution reactions differ?

\(\text{S}_\text{N}1\) is first order overall (\(\text{Rate} = k[\text{halogenoalkane}]\)), whereas \(\text{S}_\text{N}2\) is second order overall (\(\text{Rate} = k[\text{halogenoalkane}][\text{Nu}^-]\)).

What are the Arrhenius equation and its logarithmic linear form?

The exponential form is \(k = A e^{-\frac{E_a}{RT}}\), and its linear form is \(\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\).

How is activation energy (\(E_a\)) calculated from an Arrhenius plot of \(\ln k\) against \(\frac{1}{T}\)?

Calculate the negative gradient of the line (\(m\)) and multiply by \(-R\): \(E_a = -\text{gradient} \times 8.31\text{ J K}^{-1}\text{ mol}^{-1}\), then divide by 1000 for \(\text{kJ mol}^{-1}\).

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