Revision notes: Rate of Reaction

These notes work through the Rate of Reaction chapter one content standard at a time, so you can revise each part and check you have the definitions and the collision-theory chain the marking scheme expects. Keep them beside past-paper questions and use them to confirm your answers.

7.1 The concept of rate of reaction

The rate of reaction is the change in the amount of a reactant or a product per unit time. Because a reaction can be followed by anything that changes as it proceeds, you can measure rate in several ways: the volume of gas released (using a gas syringe), the loss of mass (using an electronic balance), the time taken for a fixed amount of precipitate to form, or a change in colour, pH or conductivity. Whatever the method, rate has units of a quantity divided by time, for example cm³ s⁻¹, g s⁻¹ or mol dm⁻³ s⁻¹.

Two rates to distinguish. The average rate over a period is the total change divided by the total time. The instantaneous rate at a particular moment is the gradient of the curve at that point. The exam asks for both, so read the question carefully.

A classic measurable reaction is sodium thiosulfate with dilute acid, where sulfur forms a cloudy precipitate; you time how long a mark under the flask takes to disappear, and the rate is proportional to 1 ÷ time.

7.2 Factors affecting the rate of reaction

Four factors are examined again and again, and you must be able to name each and predict its effect:

• Concentration of a solution, a higher concentration gives a faster rate.
• Temperature, a higher temperature gives a faster rate.
• Surface area (the total exposed area of a solid, so a smaller particle size or a powder), a larger surface area gives a faster rate.
• A catalyst, a suitable catalyst gives a faster rate.

For gases, increasing the pressure has the same kind of effect as increasing concentration. The important skill is not just listing the factors but explaining every one through the same idea, collision theory, which is the heart of this chapter.

7.3 The collision theory

Collision theory is the single explanation the exam wants for all four factors, so learn it precisely. Its claims are:

• Particles must collide before they can react.
• Only some collisions lead to reaction. A collision that does is an effective collision, the colliding particles must have energy equal to or greater than the activation energy and must collide in the correct orientation.
• The activation energy is the minimum energy that colliding particles must have for a collision to be effective.
• The rate of reaction depends on the frequency of effective collisions, how many effective collisions happen per unit time.

The chain that scores. Every explanation should end with “…so the frequency of effective collisions increases, and the rate increases.” Answers that stop at “more collisions” are incomplete.

Now apply the chain to each factor. Higher concentration: more reactant particles are present in the same volume, so collisions are more frequent, so effective collisions are more frequent, so the rate increases. Higher temperature: particles move faster and have more kinetic energy, so they collide more frequently and a larger fraction of collisions have energy above the activation energy, both raise the frequency of effective collisions, so the rate increases. Larger surface area: more particles of the solid are exposed at the surface, so collisions with the other reactant are more frequent, so effective collisions are more frequent, so the rate increases. A catalyst: it provides an alternative path with a lower activation energy, so a larger fraction of collisions is now effective, so the rate increases.

7.4 Catalysts in reactions

A catalyst is a substance that alters the rate of a chemical reaction while it remains chemically unchanged at the end. In this chapter you deal mainly with a positive catalyst, which speeds a reaction up. Remember these features:

• It works by providing an alternative reaction path with a lower activation energy; it does not change the reactants or products, and it does not make more product form.
• Its mass and chemical composition are the same at the end as at the start, so it can be recovered and reused. Its physical form may look different, but chemically it is unchanged.
• Only a small amount is usually needed, and a catalyst is often specific to a particular reaction.

Common examples to quote are manganese(IV) oxide in the decomposition of hydrogen peroxide, iron in the Haber process, vanadium(V) oxide in the Contact process, and enzymes as biological catalysts. On an energy profile, a catalyst lowers the “hill” that particles must climb, which is why more collisions clear it.

7.5 Rate calculations and graphs

Most rate graphs plot the volume of gas (or the mass of the flask) against time. The curve rises steeply at first, then bends and flattens to a plateau.

• The gradient at any point is the rate at that moment. It is steepest at the start, because the reactant concentration is highest and effective collisions are most frequent.
• The gradient decreases as reactants are used up, and becomes zero at the plateau, where the reaction has stopped.
• Average rate = total volume of gas ÷ total time taken.
• The height of the plateau shows the total amount of product, which depends on the amount of limiting reactant, not on the rate.

Reading two curves. A steeper initial gradient means a faster reaction. If two experiments use the same amount of limiting reactant, they reach the same plateau even if one is faster; a higher plateau means more reactant was used. This comparison is a very common graph question.

For a gas, you can predict the maximum volume from moles: volume = moles of gas × molar volume, using Molar volume of a gas is 24 dm³ mol⁻¹ at room conditions.

How to use these notes

Revise one content standard at a time and say the collision-theory chain aloud until it is automatic. For 7.1, list the measurable quantities and their units. For 7.2, name the four factors and their direction of effect. For 7.3, write the full chain for each factor. For 7.4, learn the catalyst definition word for word. For 7.5, sketch a volume-time curve and label where the rate is fastest, where it is slowing, and where it is zero. Then attempt the practice questions and mark yourself against these definitions.

A one-to-one teacher can check that every factor answer ends with “frequency of effective collisions”, which is the single place students most often drop marks in this chapter. Because rate ideas return in industrial processes and in Form 5 equilibrium across SPM Chemistry, securing collision theory now pays back later.

Worked detail: why powder reacts faster than lumps

Suppose the same mass of calcium carbonate reacts with dilute acid, once as a single lump and once as a fine powder. The powder has a far larger total surface area, so many more carbonate particles are exposed to the acid. More exposed particles means collisions between acid and carbonate are more frequent, so effective collisions are more frequent, so the initial rate is higher, the powder’s curve rises more steeply. Because the mass of carbonate is the same, both reach the same final volume of carbon dioxide; only the speed differs. Being able to separate “how fast” (the gradient) from “how much” (the plateau) is exactly the distinction the exam rewards.