IB DP Chemistry Reactivity 2.2: How fast? Rates and collision theory

This guide teaches the first part of Reactivity 2.2, How fast? The rate of chemical change, based on the IB Diploma Programme Chemistry guide (February 2023, first assessment 2025). Its scope is understandings Reactivity 2.2.1 to 2.2.5, which are studied at both SL and HL. The additional higher level understandings (Reactivity 2.2.6 to 2.2.13: mechanisms, rate equations and the Arrhenius equation) are covered in the HL rates guide.

For the whole course, see the IB DP chemistry hub or print the checklist.

What this unit covers

Understanding	Skills you need	SL/HL
Reactivity 2.2.1	Express rate as how fast the concentration of one named reactant or product changes with time; work out rates, including from tangents to graphs of concentration, volume or mass against time	SL and HL
Reactivity 2.2.2	Explain that reaction needs collisions with enough energy and the right geometry; link average kinetic energy to temperature in kelvin	SL and HL
Reactivity 2.2.3	Predict and explain how pressure, concentration, surface area, temperature and a catalyst change the rate	SL and HL
Reactivity 2.2.4	Define activation energy, Ea; use Maxwell–Boltzmann curves to explain the effect of temperature	SL and HL
Reactivity 2.2.5	Sketch energy profiles for exothermic and endothermic reactions, with and without a catalyst, and explain them; use Maxwell–Boltzmann curves to show the effect of a lower Ea; know that enzymes are biological catalysts	SL and HL

Rate of reaction (2.2.1)

The rate of reaction tells you how quickly the concentration of one named reactant or product changes, per unit time. For a reactant the concentration falls, so you take the size of the change; rate is always quoted as a positive number.

rate = Δ[concentration] / Δt unit: mol dm⁻³ s⁻¹

You cannot usually watch a concentration directly. Instead you follow a property that changes as the reaction goes:

Property followed	Use it when	Typical apparatus
Volume of gas	A gas is made	Gas syringe or inverted measuring cylinder
Mass	A gas escapes from an open flask	Balance, with a cotton wool plug in the flask neck
Colour intensity	One species is coloured	Colorimeter (absorbance against time)
Conductivity or pH	The number or type of ions changes	Conductivity probe or pH probe
Time for a visible change	A colour appears or vanishes, or a precipitate forms	Stopwatch

In the first four methods time is the independent variable: you choose the times and read the property. In the last method time is the dependent variable: you measure how long a fixed amount of change takes, and the rate is proportional to 1/time. Rates from volume or mass data come out in cm³ s⁻¹ or g s⁻¹; convert to moles if the question asks for mol s⁻¹.

Average rate and instantaneous rate

• Average rate over an interval = total change ÷ time taken.
• Instantaneous rate at a moment = gradient of a tangent drawn to the curve at that moment. The initial rate is the gradient of the tangent at t = 0.

Draw the tangent long, touching the curve at one point only. Pick two points on it far apart: gradient = (y₂ − y₁)/(x₂ − x₁).

Worked example 1 (colorimetry). Crystal violet is a purple dye that hydroxide ions turn colourless:

CV⁺(aq) + OH⁻(aq) → CVOH(aq)

Colorimeter absorbance readings are converted into [CV⁺]. At t = 0, [CV⁺] = 2.40 × 10⁻⁵ mol dm⁻³. The tangent at t = 0 runs from (0 s, 2.40 × 10⁻⁵ mol dm⁻³) to (300 s, 0). At t = 300 s, [CV⁺] = 0.90 × 10⁻⁵ mol dm⁻³, and the tangent there passes through (0 s, 1.80 × 10⁻⁵ mol dm⁻³) and (600 s, 0).

1. Initial rate = (2.40 × 10⁻⁵ − 0)/(300 − 0) = 8.0 × 10⁻⁸ mol dm⁻³ s⁻¹ (rate of loss of CV⁺).
2. Rate at 300 s = 1.80 × 10⁻⁵/600 = 3.0 × 10⁻⁸ mol dm⁻³ s⁻¹, less than half the starting rate, because fewer CV⁺ ions remain to collide with OH⁻.
3. Average rate over the first 300 s = (2.40 − 0.90) × 10⁻⁵/300 = 5.0 × 10⁻⁸ mol dm⁻³ s⁻¹, between the two instantaneous values.
4. The equation is 1 : 1, so OH⁻ is used up at the same rate as CV⁺. With a 1 : 2 ratio the product would form twice as fast, so always say which species a rate refers to.

Worked example 2 (mass loss). Citric acid solution is added to sodium hydrogencarbonate in an open flask on a balance. The reading falls from 212.47 g to 212.03 g during the first 40 s, because CO₂ escapes.

1. Mass of CO₂ lost = 212.47 − 212.03 = 0.44 g.
2. Average rate = 0.44/40 = 0.011 g s⁻¹.
3. In moles: 0.44/44.01 = 0.0100 mol CO₂, so the rate = 0.0100/40 = 2.5 × 10⁻⁴ mol s⁻¹.

Collision theory (2.2.2)

Particles react only when they collide, and a collision leads to reaction only if both of these are true:

1. The particles collide with at least the activation energy, Ea.
2. They collide with the proper orientation (collision geometry), so the atoms that must bond actually meet.

Collision theory builds on the kinetic molecular theory from Structure 1.1. The average kinetic energy of the particles is directly proportional to the absolute temperature, in kelvin (K = °C + 273.15). Doubling the kelvin temperature doubles the average kinetic energy; doubling the Celsius temperature does not.

Geometry matters most for larger molecules. Hydroxide ions react with bromomethane only when OH⁻ approaches the carbon atom on the side opposite the bromine; a hit elsewhere, however energetic, gives no reaction.

Worked example 3. A reaction is run at 17 °C and again at 47 °C. By what factor does the average kinetic energy of the particles change?

1. Convert: 17 °C = 290.15 K and 47 °C = 320.15 K.
2. Ratio = 320.15/290.15 = 1.10, an increase of only about 10%.

Yet a 30 K rise often makes a reaction several times faster. The Maxwell–Boltzmann section below explains why.

Factors that affect the rate (2.2.3)

Each factor works by changing the frequency of collisions, the fraction of collisions with E ≥ Ea, or both.

Change	Effect on rate	Explanation
Higher concentration of a solution	Faster	More particles in each dm³, so collisions happen more often
Higher pressure of a gas (same temperature)	Faster	Particles are closer together, so they collide more often
Larger surface area of a solid (smaller pieces, powder)	Faster	More particles exposed, so surface collisions happen more often
Higher temperature	Faster	Particles move faster, so more frequent collisions, and a much larger fraction of collisions have E ≥ Ea (the main effect)
Adding a catalyst	Faster	Alternative pathway with lower Ea, so a larger fraction of collisions are successful

Write “more frequent collisions”, not just “more collisions”.

Controlling variables. To study one factor, keep the others fixed: temperature (water bath), the volume and concentration of each solution, the mass and particle size of any solid, the shape and size of the flask, and whether the mixture is stirred.

Errors on graphs. Random errors scatter points on both sides of the best-fit line. A systematic error shifts every point the same way; for example, a stopwatch always started 2 s late, or a line that should pass through the origin but cuts an axis. Repeating readings reduces the effect of random error but does nothing for systematic error.

Activation energy and the Maxwell–Boltzmann distribution (2.2.4)

Activation energy, Ea, means the smallest amount of energy a collision must bring for the colliding particles to react.

A Maxwell–Boltzmann energy distribution curve shows how kinetic energy is spread among the particles of a gas at a fixed temperature.

How to sketch it

• Axes: x-axis “kinetic energy” (or “energy”); y-axis “number of particles” (or “fraction of particles”).
• The curve starts at the origin: no particle has zero energy.
• It climbs to a single maximum (the most probable energy) and then drops away.
• The high-energy tail approaches the x-axis but never meets it.
• Draw a vertical Ea line on the high-energy side of the maximum. The area beyond that line represents the particles with enough energy to react.

At a higher temperature the curve is lower and wider, and its peak moves to the right. The area enclosed is unchanged: there are still as many particles. Far more area now lies beyond Ea.

That is the answer to worked example 3. A modest rise in average kinetic energy moves a much larger share of particles past Ea, so the fraction of successful collisions rises steeply. This effect matters far more than the small increase in collision frequency.

Linking question: fuels. Methane and air can be mixed safely at room temperature even though combustion is strongly exothermic. Very few collisions have energy above the high Ea until a spark or flame supplies energy. A high Ea is useful in a fuel: it can be stored safely and burns only when ignited. Alkanes are described as kinetically stable (high Ea, so slow) but thermodynamically unstable (their combustion products are lower in energy). Bond strength matters too. The very strong triple bond in N₂ makes reactions that must break it slow, which helps explain why nitrogen gas is so unreactive (a link to bond strength in Structure 2.2).

Catalysts and energy profiles (2.2.5)

A catalyst increases the rate by providing an alternative reaction pathway with a lower activation energy, and is chemically unchanged at the end. Enzymes are biological catalysts. (The different mechanisms of homogeneous and heterogeneous catalysts are not assessed.)

Energy profiles

An energy profile plots potential energy (y-axis) against the reaction pathway or reaction coordinate (x-axis).

• Exothermic: products below reactants; ΔH negative.
• Endothermic: products above reactants; ΔH positive.
• Ea is the height from the reactants to the highest point of the profile.
• With a catalyst, draw a lower hump between the same start and finish levels. ΔH does not change.

Worked example 4. For an exothermic reaction, Ea = 96 kJ mol⁻¹ and ΔH = −58 kJ mol⁻¹. With a catalyst, Ea = 61 kJ mol⁻¹.

1. Activation energy of the reverse reaction (uncatalysed) = 96 + 58 = 154 kJ mol⁻¹: it starts from the lower product level.
2. Reverse reaction with the catalyst = 61 + 58 = 119 kJ mol⁻¹.
3. The catalyst lowers both barriers by the same amount, 35 kJ mol⁻¹. That is why a catalyst speeds up the forward and backward reactions together and does not change the position of equilibrium (Reactivity 2.3).

For an endothermic reaction, the reverse barrier is Ea(forward) − ΔH. For example, with Ea = 118 kJ mol⁻¹ and ΔH = +46 kJ mol⁻¹, the reverse barrier is 72 kJ mol⁻¹.

Catalysts on the Maxwell–Boltzmann curve

Draw one curve: the temperature has not changed. Mark Ea (uncatalysed) and, further left, Ea (catalysed). The area beyond the catalysed Ea is larger, so a greater fraction of collisions have enough energy. Do not draw a second curve or move the peak.

HL link. Transition elements make useful catalysts; their features are covered in AHL Structure 3.1.

Common errors

• Reading a rate off the curve at a point instead of drawing a tangent and finding its gradient.
• Saying higher temperature works “because there are more collisions” and missing the larger fraction with E ≥ Ea.
• Using Celsius when comparing average kinetic energies.
• Drawing a Maxwell–Boltzmann curve that touches the x-axis at high energy, or a hotter curve with a larger area.
• Showing ΔH changing when a catalyst is added.
• Writing that a catalyst “gives particles more energy”.

Where to go next

Next, try the practice questions, and use the revision notes for quick review. The particle model behind collision theory is in the Structure 1 study guide. Our free diagnostics show which topics need more work.

Official syllabus

Aligned with the IB’s DP Chemistry guide (International Baccalaureate; issued February 2023; first assessment 2025). This page follows Reactivity 2.2, How fast? The rate of chemical change, understandings 2.2.1 to 2.2.5, which are common to SL and HL.