Reaction Kinetics

Physical Chemistry Weightage: AS · Papers 1, 2 and 3 (assumed in Papers 4 and 5) Topic 8

“Welcome! Kinetics is the topic where every answer follows the same chain: a change, then the frequency of effective collisions, then the rate. Learn the chain, learn the two Boltzmann diagrams, and you will collect most of the marks. Paper 3 also loves a rate experiment, so we will practise the graph work too.”

— SCORECHEM ACADEMIC TEAM

0. What Examiners Want

How marks are earned in kinetics

Definitions: rate of reaction (change in concentration per unit time) and activation energy (the minimum energy that colliding particles need for a reaction to occur).
Explanations: a factor, then the effect on the frequency of effective collisions, then the rate. Skipping the middle link loses the mark.
Diagrams: labelled axes, E_a marked, curves the right shape, and the area beyond E_a shaded.
Graph work: a tangent drawn with a ruler, a large triangle, the gradient with units.
Paper 3: a table with headings and units, a scale that fills the grid, a best-fit line, and a conclusion that refers to the graph.

1. Rate of Reaction

Rate of reaction: the change in concentration of a reactant or product per unit time.

Units: mol dm⁻³ s⁻¹ (or mol dm⁻³ min⁻¹).
Reactant concentration falls and product concentration rises, but the rate is always quoted as a positive number.
The rate is fastest at the start (highest reactant concentration) and falls as the reactant is used up.

Average rate = change in concentration ÷ time taken. Rate at one moment = the gradient of a tangent to the concentration–time curve at that time.

Finding the rate at one instant. Draw a tangent to the curve at the chosen time, extend it to both axes, then divide the change in concentration by the change in time. The gradient of a falling reactant curve is negative; the rate is quoted as a positive number in mol dm⁻³ s⁻¹.

Following the rate in practice (choose a property that changes steadily as the reaction goes):

Method	Suits a reaction that...
Measure the volume of gas given off at time intervals (gas syringe or upturned measuring cylinder)	produces a gas
Measure the mass lost on a balance	loses a dense gas such as CO₂
Take samples, quench (stop) the reaction, then titrate	has a reactant or product that can be titrated
Measure a colour change with a colorimeter	involves a coloured substance
Measure the time for a fixed change to happen (a mark disappears, a colour appears)	shows a clear visible change

Quenching means stopping the reaction in the sample suddenly, for example by adding a large volume of cold water or removing a reactant, so that the titration result reflects that moment.

⚠️ Examiner Trap: Gradient and units. The gradient of a falling reactant curve is negative, but the rate is the positive value. Use a long tangent and a big triangle, read the values from the axes (not from the curve), and write the units: concentration unit divided by time unit. The rate of a gas-volume experiment has units such as cm³ s⁻¹, not mol dm⁻³ s⁻¹.

2. Collision Theory

For particles to react they must:

collide, and
collide with energy equal to or greater than the activation energy, E_a, and
collide in the correct orientation.

Activation energy: the minimum energy that colliding particles must have for a reaction to occur (the energy needed to break the bonds that must break).

A collision that meets both conditions is an effective collision. Only a small fraction of collisions are effective. The rate depends on the frequency of effective collisions.

According to collision theory, particles react only if they collide (1) with at least the activation energy and (2) in the correct orientation. Most collisions are ineffective. Anything that increases the frequency of effective collisions increases the rate.

⚠️ Examiner Trap: Frequency, not number. Write “more frequent effective collisions” or “greater frequency of collisions with E ≥ E_a”. A bare “more collisions” does not distinguish a faster reaction from a larger amount of reaction, and is not credited.

3. Concentration, Pressure and Surface Area

Change	Effect on particles	Effect on rate
Increase the concentration of a solution	more particles per unit volume, so they collide more often	more frequent effective collisions, so the rate increases
Increase the pressure of a gas	more molecules per unit volume (they are closer together)	more frequent effective collisions, so the rate increases
Increase the surface area of a solid (smaller pieces, powder)	more particles exposed at the surface	more frequent effective collisions, so the rate increases

In each case the proportion of particles with E ≥ E_a is unchanged; only the frequency of collisions changes.
Doubling the surface area or concentration changes the rate. It does not change the total amount of product, which is set by the amount of the limiting reactant.

The graphs below show the shape of a gas-volume experiment. A higher concentration gives a steeper start, but the same final volume when the amount of limiting reagent is the same.

Left: the same mass of solid with acid of two concentrations. The higher concentration gives a steeper initial gradient, and the same final volume if the solid is limiting. Right: in a clock experiment the relative rate (1000 ÷ t) is plotted against the volume of reactant used; with a constant total volume, this volume is proportional to concentration.

⚠️ Examiner Trap: Same final volume. When you sketch a faster experiment, draw a steeper curve that levels off at the same final volume, as long as the limiting reagent is unchanged. A curve that ends higher is a wrong answer, and so is one that ends lower.

4. Temperature and the Boltzmann Distribution

Molecules in a gas or liquid have a range of energies. The Boltzmann distribution shows the number of molecules with each energy.

The curve starts at the origin (no molecules have zero energy) and rises to a peak.
It does not touch the x-axis at high energy (there is no upper limit).
The area under the curve is the total number of molecules; it is the same at every temperature.
The area to the right of E_a is the number of molecules able to react on collision.

At a higher temperature:

the peak is lower and moves to the right (the average energy is greater);
the curve is flatter and wider, and the area under it is unchanged;
the area beyond E_a is much larger.

Boltzmann distribution at two temperatures (not to scale). At the higher temperature the peak is lower and further to the right, the total area (number of molecules) is the same, and the area beyond E_a is larger. So a greater proportion of collisions are effective, and the rate increases.

Why does the rate increase with temperature? There are two effects.

The molecules move faster, so they collide more often (a small effect).
A much greater proportion of collisions has energy ≥ E_a, so the frequency of effective collisions increases (the large effect).

A rise of about 10 °C roughly doubles the rate of many reactions. The rate rises steeply because the tail of the distribution grows quickly, not because collisions become twice as frequent.

⚠️ Examiner Trap: Temperature does not lower E_a. Heating gives molecules more energy; E_a stays the same. Do not write that “the activation energy decreases”. Say that a greater proportion of molecules has energy at least equal to E_a. Also convert temperatures if a question uses the Kelvin scale (T = θ + 273).

5. Catalysis

Catalyst: a substance that increases the rate of a reaction by providing an alternative reaction pathway with a lower activation energy. It is not used up (it is chemically unchanged at the end).

Reaction pathway for an exothermic reaction with and without a catalyst. The catalyst provides an alternative route with a lower activation energy. The reactants and products are unchanged, so ΔH is the same, and the activation energy of the reverse reaction is also lowered.

The catalysed route has a lower E_a, so at the same temperature a greater proportion of collisions is effective.
ΔH is unchanged; the reactants and products are at the same energy.
The E_a of the reverse reaction is lowered by the same amount, so equilibrium is reached faster but the position does not change.
E_a(reverse) = E_a(forward) − ΔH (an exothermic reaction has E_a(reverse) larger).

A catalyst does not change the shape of the Boltzmann curve or the temperature. It provides an alternative route with a lower activation energy, so a larger proportion of the molecules have enough energy to react and the rate increases.

Types of catalyst

Type	Meaning	Examples
Homogeneous	same physical state as the reactants	H⁺(aq) in the esterification of a carboxylic acid with an alcohol
Heterogeneous	different state from the reactants (usually a solid with gases or liquids)	Fe in the Haber process; Ni in the hydrogenation of alkenes; V₂O₅ in the Contact process; MnO₂ with H₂O₂

In heterogeneous catalysis, reactant molecules are adsorbed onto active sites on the surface, bonds weaken, products form and are desorbed.
Catalysts allow a lower temperature to be used, which saves energy and can reduce emissions.

⚠️ Examiner Trap: What a catalyst changes. A catalyst provides an alternative route with a lower E_a. It does not raise the energy of the molecules, it is not used up, it does not change ΔH, and it does not change the equilibrium yield. In a profile diagram the reactant and product levels must be the same as the uncatalysed curve.

6. Rate Experiments (Paper 3)

Sodium thiosulfate and hydrochloric acid (the “disappearing cross”):

Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + S(s) + SO₂(g) + H₂O(l)

Sulfur forms a fine yellow precipitate that makes the mixture cloudy; a cross under the flask disappears.
Measure the time, t, for the cross to be hidden.
Vary the volume of thiosulfate and add water so the total volume is constant; then the concentration of thiosulfate is proportional to its volume.
Rate is proportional to 1 ÷ t, so use relative rate = 1000 ÷ t (units s⁻¹, times 1000 to give convenient numbers).
Plot relative rate (y) against volume of thiosulfate (x). A straight line through the origin shows that the rate is proportional to concentration.

Sources of error and improvements

Judging when the cross disappears is subjective; use the same observer and the same flask, or use a light sensor.
The temperature must be kept constant (a water bath); the reaction mixture cools or warms slightly.
Start the timer as soon as the last reagent is added and mix in the same way each time.
Repeat and average; discard anomalous results.

Safety: the reaction produces SO₂, a toxic gas that can trigger asthma. Use small volumes in a well-ventilated room or a fume cupboard, and wear eye protection. Dispose of the mixture by pouring it down the sink with plenty of water (or as instructed).

Marble chips (CaCO₃) and hydrochloric acid are followed by collecting the CO₂ in a gas syringe (or by loss of mass) at regular intervals. Plot volume against time and draw a tangent to find the rate at a chosen time.

⚠️ Examiner Trap: 1 ÷ t is the rate. The time taken is inversely proportional to the rate. Plotting time against concentration gives a curve, not a straight line. In a table, give the heading “relative rate / s⁻¹” (or 1000 ÷ t), and keep the same number of decimal places in each row.

7. Quick Sheet and Checklist

Idea	Say or write
Rate	change in concentration per unit time; mol dm⁻³ s⁻¹
Rate at a moment	gradient of the tangent
E_a	minimum energy needed for a reaction to occur on collision
Effective collision	E ≥ E_a and correct orientation
Concentration or pressure	more particles per unit volume, so more frequent effective collisions
Temperature	greater proportion with E ≥ E_a (main effect); more frequent collisions (minor)
Catalyst	alternative route, lower E_a; ΔH and equilibrium position unchanged
Relative rate (Paper 3)	1000 ÷ t

Before you leave the question, check:

Each explanation uses “frequency of effective collisions” or “proportion of molecules with E ≥ E_a”.
Boltzmann sketches have the right shape, equal areas, E_a marked and the area shaded.
A rate from a graph has a tangent, a large triangle and units.
A catalyst profile keeps the same start and finish levels.
Gas-volume sketches end at the same final volume unless the amount of limiting reactant changed.