How to Study Chemical Reactions (Rates and Equilibrium) for IGCSE

The chemical reactions topic in 0620 covers two related but distinct areas: rates of reaction (how fast reactions happen) and reversible reactions/equilibrium (whether reactions go to completion). Students often mix up the explanations for these two areas, which costs marks. Study them as separate blocks, then learn how they connect.

Block 1: Rates of reaction

The four factors

The rate of a reaction is affected by four factors. For each one, you need to state the effect and explain it using collision theory.

Temperature: Higher temperature increases the rate. Explanation: particles have more kinetic energy, move faster, collide more frequently, and importantly, a greater proportion of collisions exceed the activation energy.

The second part — activation energy — is the mark that distinguishes a good answer from a mediocre one. “Particles move faster and collide more often” is worth one mark. Adding “a greater proportion of collisions have energy equal to or greater than the activation energy” earns the second mark. Many students miss this.

Concentration (for solutions) / Pressure (for gases): Higher concentration or pressure increases the rate. Explanation: more particles per unit volume, so collisions are more frequent.

Surface area: Smaller pieces (or powder) have a larger surface area exposed to the other reactant, so more collisions occur per second. The total mass of solid has not changed; only the exposed surface has.

Catalyst: Increases the rate without being consumed. Explanation: provides an alternative reaction pathway with lower activation energy, so more collisions are successful at the same temperature.

Collision theory: the unifying explanation

All four factors are explained by the same theory: for a reaction to occur, particles must collide with sufficient energy (the activation energy) and correct orientation. Anything that increases the frequency of collisions or the proportion of sufficiently energetic collisions increases the rate.

Study method: write a paragraph explaining why the rate increases for each factor, using collision theory. Then close your notes and rewrite each explanation from memory. Check for precision — “particles collide more” is vague; “particles collide more frequently per unit time” is precise. See active recall techniques for this type of practice.

Rate graphs

The exam frequently presents graphs of gas volume collected (or mass lost) against time. Know how to read them:

• A steeper initial gradient means a faster initial rate.
• The curve levels off when the reaction is complete (one reactant is used up).
• The final volume (or mass change) tells you the total amount of product formed.
• Comparing two curves on the same axes lets you identify which conditions gave a faster rate and whether the same total product was formed.

Example: if you increase concentration, the curve is steeper initially (faster rate) but levels off at the same final volume (same total product because the same amount of limiting reactant was used). If you increase temperature, the curve is steeper AND may level off at the same final volume (same total product) or slightly different if temperature affects the equilibrium of a reversible reaction.

Practise sketching graphs for each factor. The interpreting graphs guide covers how to read and describe graph features for exam answers.

Block 2: Reversible reactions and equilibrium

What is a reversible reaction?

A reversible reaction proceeds in both the forward and backward directions. The symbol is a double-headed arrow. Example: the Haber process (N2 + 3H2 reversible 2NH3) and the hydration of copper(II) sulfate (CuSO4.5H2O reversible CuSO4 + 5H2O).

Dynamic equilibrium

In a closed system, a reversible reaction reaches dynamic equilibrium: the forward and reverse rates are equal, and the concentrations of reactants and products remain constant (but not necessarily equal). Dynamic means both reactions continue; equilibrium means the rates balance.

Effect of changing conditions

Temperature:

• Increasing temperature shifts equilibrium in the endothermic direction.
• If the forward reaction is exothermic, increasing temperature shifts equilibrium backwards (less product).

Pressure (gas reactions only):

• Increasing pressure shifts equilibrium towards the side with fewer gas moles.
• In the Haber process (1 + 3 = 4 moles on the left, 2 moles on the right), high pressure favours the forward reaction (more NH3).

Concentration:

• Adding more reactant shifts equilibrium forward (more product).
• Removing product shifts equilibrium forward (more product to replace it).

Catalyst:

• Does not change the equilibrium position. It speeds up both the forward and reverse reactions equally, so equilibrium is reached faster, but the proportions of reactant and product at equilibrium are the same.

The Haber process as the key example

Almost every equilibrium question on 0620 uses the Haber process. Learn it thoroughly:

• Equation: N2 + 3H2 ⇌ 2NH3 (forward reaction is exothermic)
• Conditions: 450 degrees C, 200 atm, iron catalyst
• Why 450 degrees C and not lower? Lower temperature favours more NH3 (exothermic direction) but the rate is too slow to be economical. 450 degrees C is a compromise between yield and rate.
• Why 200 atm? High pressure favours fewer gas moles (forward direction, more NH3). Very high pressures are expensive and dangerous, so 200 atm is a compromise.
• Why iron catalyst? Increases the rate of reaching equilibrium, making the process economical, without changing the equilibrium position.

The “compromise” reasoning is what the examiner wants for extended answers. State the effect on equilibrium, state the effect on rate, and explain why the actual conditions are a compromise.

Connecting rates and equilibrium

The link between the two blocks: changing temperature affects both the rate of reaction (collision theory) and the equilibrium position (favouring exothermic or endothermic direction). The exam may ask about both in the same question, so keep your explanations distinct. A rate answer uses collision theory. An equilibrium answer uses the effect on forward and reverse reaction directions.

Study plan

Week 1: Rates

• Days 1-2: Learn the four factors and their collision theory explanations. Write explanations from memory.
• Days 3-4: Practise rate graphs. Sketch curves for different conditions and describe what each shows.
• Day 5: Past paper questions on rates from the chemical reactions exam guide.

Week 2: Equilibrium

• Days 1-2: Learn the definitions (reversible reaction, dynamic equilibrium). Study the effects of temperature, pressure, concentration, and catalyst on equilibrium position.
• Days 3-4: Master the Haber process example. Write the conditions and explain each compromise.
• Day 5: Past paper questions on equilibrium.

Week 3: Integration

• Mixed past paper questions covering both rates and equilibrium.
• Practise distinguishing which type of answer each question needs.