State that a catalyst increases the rate of a reaction and is unchanged at the end of a reaction

Chemical Reactions – Rate of Reaction

Learning Objective (AO1)

State that a catalyst increases the rate of a reaction and is unchanged at the end of the reaction.

Core Statement (AO1 – tick‑off for revision)

Assessment Objective 2 Prompt (AO2)

Explain, using the Arrhenius equation, how a catalyst speeds up a reaction. Include a short numerical illustration.

Key Concepts

• A catalyst provides an alternative pathway with a lower activation energy (E_a).
• Because E_a is lower, a larger fraction of collisions have enough energy, so the reaction rate increases.
• The catalyst participates in intermediate steps but is regenerated; therefore it is unchanged after the reaction.
• It does **not** affect the position of equilibrium for a reversible reaction (the equilibrium constant K remains the same).

How a Catalyst Works

Activation energy and the Arrhenius equation

The rate constant k is related to the activation energy by

$$k = A\,e^{-E_a/(RT)}$$

Lowering E_a makes the exponent less negative, increasing k and thus the reaction rate.

Quantitative illustration (AO2)

At 298 K a reaction has E_a = 80 kJ mol⁻¹. A catalyst reduces E_a by 20 kJ mol⁻¹ (to 60 kJ mol⁻¹). The ratio of the new to the original rate constant is

$$\frac{k'}{k}= \frac{e^{-60\,000/(RT)}}{e^{-80\,000/(RT)}} = e^{20\,000/(RT)}$$

With R = 8.314 J mol⁻¹ K⁻¹ and T = 298 K,

$$\frac{k'}{k}= e^{20\,000/(8.314\times298)} = e^{8.07}\approx 3.2\times10^{3}$$

The reaction proceeds roughly 3 000 times faster.

Collision‑theory link (Supplement)

• Only collisions with energy ≥ E_a are effective.
• By lowering E_a, a catalyst raises the proportion of effective collisions, increasing the observed rate.

Mechanistic view (Supplement)

In a mechanism the catalyst appears in intermediate steps but cancels out in the net equation.

Step 1:  H₂O₂ + Cat → Cat‑HO₂·   (fast)
Step 2:  Cat‑HO₂· → Cat + H₂O + O₂   (fast)
Overall: 2 H₂O₂ → 2 H₂O + O₂   (Cat omitted)

The catalyst is regenerated, demonstrating that it is unchanged.

Types of Catalysts

• Heterogeneous catalyst: solid phase different from reactants (e.g. MnO₂ in H₂O₂ decomposition).
• Homogeneous catalyst: same phase as reactants (e.g. H⁺/H₂SO₄ catalysing alcohol dehydration).

Practical Investigation of Rate (Core – 6.2)

Standard experiment (MnO₂ & H₂O₂)

1. Label two identical test tubes; add 20 mL of 3 % H₂O₂ to each.
2. To tube A add a measured amount of MnO₂ (catalyst); leave tube B uncatalysed.
3. Collect the evolved O₂ in a gas‑syringe or over water and record the time taken to produce a fixed volume (e.g. 20 mL).
4. Weigh the MnO₂ before and after the reaction to show it is unchanged.

Alternative practical methods (Core – 6.2)

Method	What is measured	Typical apparatus
Gas‑syringe (or eudiometer)	Volume of gas evolved vs. time	Gas syringe, rubber tubing, water trough
Mass‑loss method	Decrease in mass of solid reactant (e.g. magnesium ribbon) vs. time	Analytical balance, timer
Colour‑change (spectrophotometric) method	Decrease in absorbance of coloured reactant (e.g. iodine) vs. time	UV‑Vis spectrophotometer, cuvettes

Data‑recording & error‑analysis checklist (AO2)

• Record initial masses/volumes, temperature, and catalyst amount.
• Take at least three repeat measurements for each condition.
• Plot rate (ΔV/Δt or Δm/Δt) against catalyst amount or concentration.
• Identify systematic errors (e.g., gas leakage, incomplete mixing) and random errors (stop‑watch reaction time).
• Discuss how the catalyst influences the slope of the rate plot.

Industrial Applications (Supplement)

• Haber process: Fe (solid) catalyses N₂ + 3 H₂ ⇌ 2 NH₃ (heterogeneous, recovered unchanged).
• Contact process: V₂O₅ (solid) catalyses 2 SO₂ + O₂ ⇌ 2 SO₃ (heterogeneous, unchanged).
• Catalytic converters: Pt, Pd, Rh (solid) accelerate oxidation of CO, HC and reduction of NOₓ in exhaust gases.

Effect on Equilibrium (Core – 6.3)

• A catalyst speeds up the attainment of equilibrium but does not change the position of equilibrium; the equilibrium constant K remains the same.

Why the Catalyst Is Unchanged

During the reaction the catalyst forms transient intermediates (e.g., Cat‑HO₂·). In the final step these intermediates decompose, regenerating the original catalyst. Consequently its chemical identity and mass are unchanged, allowing it to be recovered and reused.

Summary Checklist (AO1)

• Define a catalyst and differentiate heterogeneous vs. homogeneous types.
• Explain how a catalyst lowers activation energy and link this to the Arrhenius equation (include a short numerical example).
• Show a step‑wise mechanism where the catalyst appears and then cancels out in the net equation.
• State explicitly that the catalyst does **not** appear in the overall equation and does not affect equilibrium.
• Provide at least one laboratory and one industrial example.
• Recall the core statement: “A catalyst speeds up a reaction and is recovered unchanged.”

Practice Exam Questions

1. Explain, with reference to the Arrhenius equation, why a catalyst increases the rate of a reaction. Include a brief numerical illustration.
2. In an experiment, 0.5 g of MnO₂ is added to 20 mL of 3 % H₂O₂. After the reaction the MnO₂ is dried and still weighs 0.5 g. What does this observation tell you about the role of MnO₂?
3. Write the balanced net equation for the decomposition of hydrogen peroxide and indicate whether the catalyst appears in this equation.
4. State how a catalyst affects the position of equilibrium for a reversible reaction and give a short justification.
5. Give one industrial example of a catalyst and state whether it is heterogeneous or homogeneous.
6. Describe two alternative practical methods (other than the gas‑syringe) for investigating the rate of a reaction and the type of data each yields.