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What controls how fast a chemical reaction goes?

Use collision theory to explain how concentration, temperature, surface area and pressure change reaction rate.

Collision theory, activation energy, the Maxwell-Boltzmann distribution, and how concentration, temperature, surface area and pressure change the rate of reaction, with worked TASC-style examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

TASC expects you to explain rate in terms of collisions between particles, use the Maxwell-Boltzmann distribution, and predict how each experimental variable changes the rate.

What rate means

The rate of reaction is how quickly reactants are used up or products are formed per unit time. You can follow it by measuring a change such as gas volume produced, mass lost, colour intensity or conductivity, then plotting that quantity against time. The gradient of the curve gives the rate, which is usually fastest at the start and slows as reactants are consumed.

Collision theory

For a reaction to occur, particles must collide, but not every collision leads to reaction. A successful (effective) collision must satisfy two conditions:

the particles collide with energy equal to or greater than the activation energy;
the particles collide in the correct orientation so bonds can break and form.

The Maxwell-Boltzmann distribution

At any temperature the particles in a sample have a spread of kinetic energies, shown by the Maxwell-Boltzmann distribution. Only the particles to the right of $E_a$ on this curve can react. Raising the temperature shifts the whole distribution to higher energies and, crucially, greatly increases the fraction of particles above $E_a$ . This is why a modest temperature rise produces a large rate increase.

How each factor changes the rate

Concentration: more particles in the same volume means more frequent collisions, so the rate rises.
Pressure (gases): higher pressure squeezes gas particles closer together, equivalent to raising concentration, so collision frequency and rate increase.
Surface area: powdering a solid exposes far more particles at the surface, so the rate rises; a powder reacts much faster than a single lump of the same mass.
Temperature: faster-moving particles collide more often and, far more importantly, more energetically, so the fraction of collisions exceeding $E_a$ rises sharply.

Explaining a surface-area effect

Set up the comparison

Marble chips ( $\text{CaCO}_3$ ) react with hydrochloric acid to give carbon dioxide: $\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$ . The same mass of marble is used as a single lump in one flask and as a powder in another, with identical acid.

Isolate the variable

The chemical reaction and the concentrations are identical, so the difference must be physical: the surface area exposed to the acid.

Apply collision theory

Powder has a far greater total surface area, exposing many more carbonate particles to acid particles, so collisions between acid and solid are more frequent per second. The powdered marble therefore reacts faster and releases gas more quickly.

Measuring rate from a graph

When you plot a measurable quantity (such as gas volume or mass) against time, the gradient of the tangent at any point gives the instantaneous rate at that moment. The steepest gradient is at the start, where reactant concentrations are highest, and the gradient falls to zero when the curve flattens and the reaction is complete. Comparing two runs is straightforward: a steeper initial gradient means a faster reaction, and a curve that levels off at the same final value but sooner indicates a faster rate with the same total amount of product. This lets you distinguish a change in rate (different gradient) from a change in amount of product (different final height).

In the exam, link every rate change back to either the frequency of collisions or the proportion that exceeds the activation energy, and use the Maxwell-Boltzmann curve to justify temperature effects.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20233 marksIn an experiment, excess hydrochloric acid reacts with magnesium and the volume of hydrogen produced is graphed against time. The reaction uses

2\ \text{g}

of magnesium powder and

0.5\ \text{mol L}^{-1}

hydrochloric acid:

\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)

. Use collision theory to explain why the reaction rate is fastest at the start and falls to zero over time.

Show worked answer →

The rate is greatest at the start and gradually slows until the curve flattens. (1 mark)

At the beginning the concentration of $\text{HCl}$ and the amount of magnesium surface are greatest, so the frequency of successful collisions between $\text{H}^+$ ions and the magnesium surface is highest, giving the steepest part of the curve. (1 mark)

As the reaction proceeds the magnesium is consumed and $[\text{HCl}]$ falls, so collisions become less frequent and the rate decreases. When magnesium runs out the rate becomes zero and the graph levels off. (1 mark)

TCE 20213 marksA graph shows the distribution of molecular energies of a gas mixture at

250\ ^\circ\text{C}

, with the activation energy

E_a

marked. With reference to the distribution of molecular energies, explain how a small increase in temperature causes a large increase in reaction rate.

Show worked answer →

Raising the temperature shifts the Maxwell-Boltzmann distribution so its peak moves to higher energy and the curve broadens and flattens, increasing the average kinetic energy of the molecules. (1 mark)

Only molecules with energy greater than or equal to $E_a$ can react. Because the distribution falls away steeply at high energy, even a small temperature rise moves a disproportionately large fraction of molecules past $E_a$ . (1 mark)

That large increase in the proportion of sufficiently energetic (successful) collisions, together with a modest rise in collision frequency, produces a large increase in reaction rate. (1 mark)