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O-Level Chemistry · Secondary 3 and 4

Rate of Reaction

Rate of reaction measures how fast reactants are used up or products are formed. In O-Level Chemistry you learn how to measure rates in the laboratory, how to read rate graphs, and how to explain the effects of concentration, pressure, surface area, temperature and catalysts using collision theory. The topic is very practical, and questions often combine a graph with a mole calculation, so it ties together energy changes and the mole concept.

What the syllabus expects

  • Describe how to measure rate by collecting a gas, by measuring loss in mass or by timing the appearance of a precipitate
  • Interpret graphs of volume or mass against time, and compare rates using gradients
  • Explain the effects of concentration, pressure, surface area and temperature on rate using collision theory
  • Describe how a catalyst increases rate by providing a pathway with a lower activation energy
  • Describe enzymes as biological catalysts
  • Explain how the final amount of product depends on the limiting reactant

What rate means and how to measure it

Rate is the change in the amount of a reactant or product per unit time. For a reaction that gives off a gas, such as marble chips with hydrochloric acid, you can collect the gas in a gas syringe and record its volume at regular intervals, or place the flask on a balance and record the loss in mass as carbon dioxide escapes. For a reaction that forms a precipitate, such as sodium thiosulfate with hydrochloric acid, you can time how long it takes for a cross under the flask to disappear. A shorter time means a faster rate, so rate is proportional to 1 ÷ time. Average rate over a period = change in amount ÷ time taken, for example 240 cm³ in 60 s is 4.0 cm³/s.

Reading rate graphs

On a graph of gas volume against time, the rate at any moment is the gradient of the curve. The curve is steepest at the start, when the reactants are most concentrated, and becomes less steep as they are used up. When the curve becomes horizontal, the reaction has stopped because the limiting reactant has run out. A faster reaction gives a steeper initial gradient and reaches the horizontal part sooner. The final volume depends only on the amount of limiting reactant, not on how fast the reaction is. So changing temperature, particle size or adding a catalyst changes the steepness but not the final volume, while changing the amount of the limiting reactant changes the final volume.

Collision theory

Particles react only when they collide with energy equal to or greater than the activation energy. These are effective collisions, and the rate depends on how frequently they happen. Increasing the concentration of a solution, or the pressure of a gas, puts more particles in each unit volume, so they collide more frequently. Increasing the surface area, for example by using powder instead of lumps, exposes more particles to collision, so collisions happen more frequently. Increasing temperature makes particles move faster, so they collide more frequently, but more importantly a greater proportion of the particles now have energy equal to or greater than the activation energy. So the frequency of effective collisions rises sharply, which is why a small temperature rise can have a large effect.

Catalysts and enzymes

A catalyst speeds up a reaction without being chemically changed at the end. It provides an alternative pathway with a lower activation energy, so a greater proportion of collisions are effective. Its mass is the same at the end of the reaction, which you can check by filtering, drying and weighing it. Manganese(IV) oxide catalyses the decomposition of hydrogen peroxide, 2H₂O₂ gives 2H₂O + O₂, and iron is the catalyst in the Haber process. Enzymes are biological catalysts made of protein. Each enzyme is specific to one reaction, works best at an optimum temperature (around body temperature for human enzymes) and is denatured at high temperatures, after which it no longer works.

Worked example

Excess marble chips (calcium carbonate) are added to 50.0 cm³ of 1.00 mol/dm³ hydrochloric acid. CaCO₃ + 2HCl gives CaCl₂ + H₂O + CO₂. After 60 s, 240 cm³ of gas has been collected at r.t.p. (a) Calculate the final volume of carbon dioxide. (b) Calculate the average rate over the first 60 s. (c) Describe how the final volume and initial rate change if 50.0 cm³ of 0.500 mol/dm³ acid is used instead.

  1. Moles of HCl = 1.00 × 50.0/1000 = 0.0500 mol, and the acid is limiting because the marble is in excess.
  2. Moles of CO₂ = 0.0500 ÷ 2 = 0.0250 mol, from the 2 : 1 ratio.
  3. Final volume = 0.0250 × 24 = 0.600 dm³ = 600 cm³.
  4. Average rate = 240 cm³ ÷ 60 s = 4.0 cm³/s.
  5. With 0.500 mol/dm³ acid, moles of HCl = 0.0250 mol, so the final volume halves to 300 cm³.
  6. The acid is less concentrated, so there are fewer acid particles per unit volume, collisions are less frequent and the initial rate is lower.

Answer: (a) 600 cm³. (b) 4.0 cm³/s. (c) The final volume halves to 300 cm³ and the initial gradient is less steep, because the less concentrated acid gives less frequent effective collisions.

Common mistakes

  • Saying particles 'collide more' without saying 'more frequently' or 'per unit time'. The rate depends on the frequency of effective collisions.
  • Explaining temperature only by more frequent collisions. The main reason is that a greater proportion of particles have energy at least equal to the activation energy.
  • Saying a catalyst gives particles more energy. It lowers the activation energy by providing an alternative pathway.
  • Expecting a faster reaction to give more product. The final amount depends only on the limiting reactant.
  • Judging rate from the height of a graph instead of its gradient. A higher point on the curve means more product so far, not a faster rate.
  • Saying a catalyst does not take part in the reaction. It does take part, but it is regenerated, so it is chemically unchanged at the end.

Exam tips

  • Use the phrase 'frequency of effective collisions' in every rate explanation, and link it to the specific factor in the question.
  • When sketching a second curve on the same axes, decide two things separately: is the initial gradient steeper or less steep, and is the final volume the same, higher or lower.
  • If a question gives amounts of both reactants, work out the limiting reactant before predicting the final volume of gas.

Practice questions

1. Easy: Explain why powdered zinc reacts faster with dilute hydrochloric acid than the same mass of zinc granules.

Show answer

The powder has a larger surface area, so more zinc particles are exposed to the acid. Acid particles collide with zinc particles more frequently, so the frequency of effective collisions and the rate increase.

2. Medium: 50.0 cm³ of 0.400 mol/dm³ hydrogen peroxide decomposes completely with a manganese(IV) oxide catalyst. 2H₂O₂ gives 2H₂O + O₂. Calculate the volume of oxygen at r.t.p. and state how you could show the catalyst is unchanged.

Show answer

240 cm³. Moles of H₂O₂ = 0.400 × 50.0/1000 = 0.0200 mol. The ratio is 2 : 1, so moles of O₂ = 0.0100 mol, and 0.0100 × 24 = 0.240 dm³ = 240 cm³. Weigh the manganese(IV) oxide before, then filter, dry and reweigh it afterwards. The mass is unchanged.

3. Harder: In a sodium thiosulfate and hydrochloric acid experiment, the cross disappears after 80 s at 20 °C and after 40 s at 30 °C. Calculate the relative rate (1 ÷ time) at each temperature and explain the change using collision theory.

Show answer

At 20 °C, 1 ÷ 80 = 0.0125 s⁻¹. At 30 °C, 1 ÷ 40 = 0.025 s⁻¹. The rate doubles. At the higher temperature particles move faster and collide more frequently, and a greater proportion have energy equal to or greater than the activation energy, so the frequency of effective collisions increases.

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