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8873 / 2027
Reaction Kinetics

Lesson 3 of 3

Enzymes

Apply specificity and condition sensitivity.

A-Level 8873, revised syllabus (2026-2027)

  • I can explain enzyme properties.Syllabus

    Syllabus 8873, 7(i). Explain enzyme properties.

An enzyme has a specific active site and suitable operating conditions

Use the lock-and-key model without confusing cooling with denaturation.

Make a guess. You are not marked.

An enzyme works well at 37 degrees C. It is boiled, then cooled back to 37 degrees C. What happens to its activity?

  1. It returns to normal once the enzyme has cooled down
  2. It is higher than before, because heat activated it
  3. It stays very low, as the active site has lost its shape
Show the answer

It stays very low, as the active site has lost its shape

Boiling denatures the enzyme: the protein unfolds and the active site loses its shape. Cooling does not restore that shape.

In this course, enzymes are proteins that act as biological catalysts. Each folded protein has an active site with a shape and chemical groups that suit particular substrates. In the lock-and-key model, a matching substrate binds to the active site. The enzyme catalyses one particular reaction, then releases the products and can be used again.

A selective catalytic cycle
  1. Matching substrate

    Its shape and interactions suit the active site.

  2. Enzyme-substrate complex

    Binding enables a lower-activation-energy pathway.

  3. Products released

    The enzyme is available for another suitable substrate.

At low temperatures, particles have less kinetic energy and there are fewer effective collisions, so the enzyme works slowly. Raising the temperature first increases the rate. At high enough temperatures, the interactions that hold the protein in shape break. The active site changes shape, and activity falls. This is denaturation. Cooling a folded enzyme only slows it down, but overheating can damage it for good.

A change in pH can change the charged groups in the protein and at its active site. So each enzyme works only over a limited pH range, and extreme pH can denature it. Specificity covers both which substrate binds and which type of reaction is catalysed. An enzyme does not catalyse every reaction that its substrate could undergo.

Check your understanding

An enzyme reaction slows in a refrigerator and speeds up again when gently warmed. Is denaturation the best explanation?

  1. Yes. Cold changes the enzyme's shape, and gentle warming then refolds it back.
  2. No. Cooling lowers particle energies; the active site stays intact, so the rate recovers.
  3. No. Cooling raises the activation energy of the reaction, so fewer collisions succeed.
Show the answer

No. Cooling lowers particle energies; the active site stays intact, so the rate recovers.

Fewer particles have enough energy at low temperature, and collisions are less frequent. Recovery on warming shows the enzyme's shape was not destroyed.

Exam-style question

Paper 2 Section B style

This question draws on Chemical Energetics: Thermochemistry, and Reaction Kinetics.

Hydrogen peroxide decomposes slowly. Manganese(IV) oxide, a black solid, and the enzyme catalase both speed up the reaction.

2H2O2(aq) -> 2H2O(l) + O2(g)

Experiment 1. A student added a little manganese(IV) oxide to 50.0 cm3 of 1.00 mol dm-3 H2O2 in a polystyrene cup. The temperature rose by 21.5 °C. Take the specific heat capacity of the solution as 4.18 J g-1 K-1 and its density as 1.00 g cm-3.

Experiment 2. In a second run, the concentration of H2O2 was measured over time at a constant temperature.

Table 1. Concentration of H2O2 in Experiment 2
time / min0102030
[H2O2] / mol dm-31.000.500.250.125

Bond energies / kJ mol-1: O-H 460; O-O 150; O=O 496.

  1. (a) Explain, in terms of bond breaking and bond making, why a reaction is exothermic. [1]
  2. (b) Use Experiment 1 to calculate the enthalpy change, in kJ per mole of H2O2. [3]
  3. (c) The accepted value is about -98 kJ mol-1. Suggest two reasons why the value from (b) is less exothermic. [2]
  4. (d) Use the bond energies to calculate the enthalpy change for 2H2O2 -> 2H2O + O2, with every species treated as a gas. [3]
  5. (e) Suggest why an enthalpy change from bond energies may differ from a measured value. [1]
  6. (f) Use Table 1 to show that the reaction is first order with respect to H2O2, and write the rate equation. [3]
  7. (g) The initial rate in Experiment 2 is 0.069 mol dm-3 min-1. Calculate the rate constant, with its units. [2]
  8. (h) Explain, with reference to the Boltzmann distribution, how manganese(IV) oxide increases the rate. [3]
  9. (i) The rate of the catalase-catalysed reaction falls sharply above about 40 °C. Explain why. [2]
 [20]

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