Worked examples: Rate of Reaction

Work through each example by covering the answer, attempting it yourself, then checking your method against the solution. Every example flags the slip that most often costs marks. These are original questions in SPM style, not past-year papers.

Example 1, Average rate from data

Question. In a reaction between excess zinc and dilute hydrochloric acid, 60 cm³ of hydrogen is collected in 30 s. Calculate the average rate of reaction over this time.

Solution. Average rate = total volume of gas ÷ total time = 60 cm³ ÷ 30 s = 2 cm³ s⁻¹. Always divide the change in the measured quantity by the time, and state the unit.

Common slip: leaving off the unit, or dividing time by volume. Rate is quantity per unit time, so the time is the denominator.

Example 2, Instantaneous rate as a gradient

Question. A graph of volume of gas against time is a curve. Explain how you would find the rate of reaction at exactly 20 s.

Solution. Draw a tangent to the curve at the point where t = 20 s, then find the gradient of that tangent (the change in volume divided by the change in time along the tangent). That gradient is the instantaneous rate at 20 s.

Common slip: reading the volume at 20 s and calling it the rate. The rate is the gradient, not the height of the curve.

Example 3, Explaining concentration through collision theory

Question. When the concentration of hydrochloric acid is increased, the reaction with zinc becomes faster. Explain this using the collision theory.

Solution. When the concentration is higher, there are more acid particles in the same volume. The particles collide more frequently, so the frequency of effective collisions increases, and the rate of reaction increases.

Common slip: stopping at “the particles collide more often”. You must reach effective collisions and then the rate, that is the marked chain.

Example 4, Explaining temperature through collision theory

Question. Explain, using the collision theory, why a reaction is faster at 60 °C than at 30 °C.

Solution. At the higher temperature the particles gain more kinetic energy and move faster, so they collide more frequently. More importantly, a larger fraction of the particles now have energy equal to or greater than the activation energy. Both effects raise the frequency of effective collisions, so the rate increases.

Common slip: mentioning only “particles move faster” without the key point that more particles now exceed the activation energy.

Example 5, Comparing two curves

Question. The same mass of marble is reacted with the same acid, once as large chips and once as powder. On the same axes, both curves are drawn. Describe and explain the two differences you expect.

Solution. First, the powder curve rises more steeply at the start, because the larger surface area gives more frequent effective collisions and a higher initial rate. Second, both curves reach the same final volume, because the same mass of marble produces the same amount of carbon dioxide, surface area changes the speed, not the quantity.

Common slip: saying the powder gives more gas. It gives the same amount of gas, only faster.

Example 6, What a catalyst does

Question. Manganese(IV) oxide is added to hydrogen peroxide and the decomposition speeds up. State what a catalyst is and explain, in terms of activation energy, why the rate increases.

Solution. A catalyst is a substance that alters the rate of reaction while remaining chemically unchanged at the end. It provides an alternative path with a lower activation energy, so a larger fraction of collisions now have enough energy to be effective; the frequency of effective collisions rises and the rate increases.

Common slip: writing that the catalyst “lowers the activation energy of the reaction”. It provides an alternative path of lower activation energy, the original path is unchanged.

Example 7, The thiosulfate “disappearing cross”

Question. Sodium thiosulfate solution reacts with dilute acid to form a sulfur precipitate. A cross drawn under the flask takes 40 s to disappear at one temperature and 20 s at a higher temperature. Compare the rates and explain the result.

Solution. Rate is proportional to 1 ÷ time. At 40 s the relative rate is 1/40 = 0.025 s⁻¹; at 20 s it is 1/20 = 0.050 s⁻¹, so the reaction is twice as fast at the higher temperature. This is because the higher temperature raises the frequency of effective collisions, so the sulfur precipitate forms more quickly and hides the cross sooner.

Common slip: saying the reaction is slower because the time is shorter. A shorter time means a faster rate, remember rate ∝ 1 ÷ time.

Example 8, Predicting the maximum volume of gas

Question. 0.05 mol of calcium carbonate reacts completely with excess dilute hydrochloric acid. Calculate the maximum volume of carbon dioxide produced at room conditions.

Solution. The equation CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ gives 1 mol of CO₂ per mol of CaCO₃, so 0.05 mol of CaCO₃ makes 0.05 mol of CO₂. Volume = moles × molar volume = 0.05 × 24 dm³ = 1.2 dm³ (1200 cm³), using Molar volume of a gas is 24 dm³ mol⁻¹ at room conditions.

Common slip: forgetting the mole ratio from the equation, or mixing dm³ and cm³. Keep the units consistent throughout.

Using these examples

Notice that every calculation ends with a unit, every explanation ends with “frequency of effective collisions”, and every graph answer separates speed (gradient) from quantity (plateau), the three habits this chapter rewards. Once the method feels automatic, move on to the practice questions and mark yourself the same way. A one-to-one teacher can check that your collision-theory chains are complete, which is exactly where marks are gained or lost in the SPM Chemistry written papers.

From examples to marks

These eight examples cover the whole chapter and train the same core moves: calculate a rate with its unit, choose average or instantaneous correctly, run the full collision-theory chain for concentration, temperature, surface area and catalyst, read two curves without confusing “how fast” with “how much”, and turn moles into a gas volume. Rehearse them until the reasoning is automatic, then test yourself with the practice questions for this chapter and check each answer against the definitions in the revision notes.