Module A11 · CSEC Chemistry
Rates ofReaction

Chemical kinetics, collision theory, rate curves, and four key factors — understand why some reactions are fast as lightning while others take centuries.

Section 1 — What is the Rate of Reaction?

DefinitionThe rate of reaction measures how quickly reactants are converted into products — the change in concentration of a reactant or product per unit time.
Rate = Δ[concentration] ÷ Δt

Units: mol dm⁻³ s⁻¹  |  cm³ s⁻¹  |  g s⁻¹

How Do We Measure Rate? (Measurable Properties)

MethodWhat's MeasuredExample Reaction
Gas syringe / graduated cylinderVolume of gas produced over timeMg + HCl → H₂ gas collected
Mass balanceDecrease in mass as gas escapesCaCO₃ + HCl → CO₂ escapes, mass falls
Cloudiness / turbidityTime for solution to turn cloudy (cross experiment)Na₂S₂O₃ + HCl → S precipitate forms
Colour intensityChange in colour of a coloured reactant/productBleaching of dye by KMnO₄

Section 2 — Collision Theory

For a reaction to occur, bonds must break in reactants and new bonds must form in products. But not every collision causes a reaction!

Three Conditions for an Effective Collision

  • 1Collision — Reactant particles MUST collide with each other. No collision = no reaction.
  • 2Activation Energy — Particles must collide with enough energy (≥ activation energy, Ea) to break existing bonds.
  • 3Correct Orientation — Particles must be aligned correctly so energy can be transferred to the bonds that need to break.
Effective CollisionA collision that results in a chemical reaction. Only effective collisions produce products. The RATE of reaction depends on the FREQUENCY of effective collisions per second.
💡 AnalogyThink of LEGO bricks: (1) bring them close enough to touch, (2) push with enough force to click, (3) align them in the right direction. Only when ALL THREE conditions are met do they connect — just like reacting particles!

Section 3 — Rate Curves

A rate curve plots a measurable property (e.g. volume of gas, mass of flask) against time. All rate curves share the same characteristic shape.

Time → Volume of gas / cm³ Faster (↑ concentration / temp / surface area / catalyst) Slower Same final level Steep gradient (fast at start) Gradient decreasing Flat = reaction complete
Rate Curve FeatureWhat It Means
Steep gradient at startReaction is FASTEST — reactant concentration is highest, most collisions per second
Gradient becomes shallowerReaction SLOWING DOWN — reactants being used up, fewer collisions
Curve becomes flat (horizontal)Reaction has STOPPED — limiting reactant fully used up
Same final level for faster/slower variantsSame total product made — limiting reactant amount is unchanged

Worked Example — Calculating Average Rate

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) First 60 s: 28 cm³ H₂ collected Average rate (0–60 s) = 28 ÷ 60 = 0.47 cm³ s⁻¹ Next 60 s (60–120 s): 10 cm³ H₂ collected Average rate (60–120 s) = 10 ÷ 60 = 0.17 cm³ s⁻¹ Rate is decreasing — reactants are being used up!

Section 4 — Factors Affecting Rate of Reaction

💡 Golden Exam RuleAlways explain using collision theory. Never just say "rate increases" — say WHY in terms of collision frequency, activation energy, and effective collisions.
⬆ Concentration

Effect: ↑ concentration → ↑ rate

Collision theory: More particles per unit volume → collisions happen more frequently → more effective collisions per second.

🌍 Bleach works faster when concentrated — same principle!

🌡️ Temperature

Effect: ↑ temperature → ↑ rate. Every +10°C roughly doubles the rate!

Collision theory: Particles gain kinetic energy → move faster (more frequent collisions) AND hit harder (more collisions exceed activation energy). Both effects increase effective collisions.

🌍 Refrigerators slow food spoilage by lowering reaction rates. Pressure cookers cook faster by raising temperature above 100°C.

🪨 Surface Area / Particle Size

Effect: Smaller particles (greater surface area) → ↑ rate

Collision theory: Reactions occur AT THE SURFACE of solids. Grinding solid into smaller pieces exposes MORE surface to the other reactant → more collisions per second.

⚠️ DANGER: Finely divided flour or coal dust in air can cause devastating explosions — enormous surface area means incredibly fast reaction with O₂!

⚗️ Catalysts

Effect: Catalyst → ↑ rate without being permanently changed

Collision theory: Provides an alternative reaction pathway with lower activation energy. More collisions have enough energy to react — without needing higher temperature.

Example: MnO₂ catalyses decomposition of H₂O₂:

2H₂O₂(aq) →[MnO₂] 2H₂O(l) + O₂(g)

🌍 Catalytic converters use Pt/Pd to convert toxic exhaust gases. Enzymes are biological catalysts — without them, your body's reactions would be too slow to sustain life!

Inhibitors (Negative Catalysts)Some substances slow reactions by providing a pathway with HIGHER activation energy. They reduce the frequency of effective collisions. Example: preservatives in food inhibit spoilage reactions.

Section 5 — Effect on Rate Curves

When Rate INCREASES

  • Steeper gradient at the start
  • Horizontal sooner — completes faster
  • Same final level — same total product made

When Rate DECREASES

  • Shallower gradient at the start
  • Horizontal later — takes longer to complete
  • Same final level — if limiting reactant unchanged
🧠 Key Point — Changing Concentration of Excess ReactantIf you use MORE of the excess reactant (e.g. more acid when marble is limiting), the rate increases BUT the final level stays the same — because the limiting reactant amount hasn't changed.

Section 6 — Resources & Simulations

Section 7 — CSEC Practice Questions

Question 1 — Rate Calculation
CaCO₃ crystals react with excess HCl. CO₂ collected: t=0 → 0 cm³, t=30s → 54 cm³, t=60s → 88 cm³, t=90s → 106 cm³, t=120s → 118 cm³. (a) Average rate in first 30 s. (b) Average rate between 60–90 s. (c) Why is rate higher at 30 s than 90 s?
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a Rate (0–30 s) = ΔV ÷ Δt = (54 − 0) ÷ 30 = 1.8 cm³ s⁻¹

b Rate (60–90 s) = (106 − 88) ÷ 30 = 18 ÷ 30 = 0.60 cm³ s⁻¹

c At 30 s, the concentration of HCl is much higher (reactant barely used up). More HCl particles per unit volume → collisions are more frequent → more effective collisions per second → higher rate. By 90 s, HCl has been largely consumed → lower concentration → fewer collisions → lower rate.

✅ (a) 1.8 cm³ s⁻¹   (b) 0.60 cm³ s⁻¹   (c) Higher HCl concentration at 30 s → more frequent collisions → faster rate.
Question 2 — Collision Theory
Explain, using collision theory, why increasing temperature increases the rate of a reaction. Your answer should mention at least TWO distinct effects.
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1 Effect 1 — Increased collision frequency: At higher temperatures, particles have more kinetic energy and therefore move faster. Faster-moving particles collide with each other more frequently per second.

2 Effect 2 — More collisions exceed activation energy: Because particles move faster, they also collide with greater energy. A larger proportion of these collisions now have energy equal to or greater than the activation energy (Ea). More effective collisions per second → faster rate.

✅ Two effects: (1) Higher collision frequency (particles move faster). (2) Greater proportion of collisions exceed activation energy. Both increase the frequency of effective collisions, raising the rate.
Question 3 — Catalysts
Hydrogen peroxide solution decomposes slowly at room temperature. Adding manganese(IV) oxide powder causes rapid gas evolution. (a) What is the role of MnO₂? (b) Write the equation. (c) How would you prove MnO₂ was not permanently changed? (d) How does MnO₂ change the activation energy?
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a MnO₂ acts as a catalyst — it increases the rate of decomposition of H₂O₂ without itself undergoing any permanent chemical change.

b 2H₂O₂(aq) → 2H₂O(l) + O₂(g)   [gas produced = oxygen, test: glowing splint relights]

c Filter off the MnO₂ at the end of the reaction. Wash, dry and weigh it — the mass will be the same as at the start. You could also re-add it to fresh H₂O₂ solution to confirm it still catalyses the reaction.

d MnO₂ provides an alternative reaction pathway with a lower activation energy. More collisions now have enough energy to result in a reaction, so the rate increases without raising the temperature.

✅ Catalyst. Lowers activation energy. Proved unchanged by same mass before/after or repeat activity.
Question 4 — Rate Curve Interpretation
The experiment in Q1 is repeated with calcium carbonate POWDER (same mass, same acid). Describe how the rate curve would differ from the original. Give a full explanation using collision theory.
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1 The new curve would have a steeper gradient at the start — the reaction proceeds faster.

2 The curve would become horizontal sooner — the reaction would be complete in less time.

3 The final level would be the same — the same mass of CaCO₃ is used, so the same total volume of CO₂ is produced. Only the rate has changed, not the amount of limiting reactant.

Explanation (collision theory): Powder has a much larger surface area than crystals. More CaCO₃ surface is exposed to the HCl acid. More acid particles can simultaneously collide with CaCO₃ particles → frequency of collisions increases → more effective collisions per second → higher rate.

✅ Steeper curve, reaches same final level sooner. Same total gas produced. Larger surface area → more frequent collisions → faster rate.
Question 5 — Multiple Choice
Which of the following would NOT increase the rate of a reaction between marble chips (CaCO₃) and hydrochloric acid?
(A) Increasing the temperature   (B) Using marble powder instead of chips   (C) Using a larger volume of the same concentration of acid   (D) Increasing the concentration of the acid
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A Would INCREASE rate — higher temperature gives particles more KE, more frequent and more energetic collisions.

B Would INCREASE rate — powder has greater surface area than chips, more frequent collisions.

C Using MORE volume of the SAME concentration means the same number of particles per cm³ but a larger total amount. The concentration (particles per cm³) is unchanged, so collision frequency per unit volume is unchanged. Rate does NOT increase (though more product will eventually be made).

D Would INCREASE rate — more acid particles per cm³ → more frequent collisions.

✅ Answer: (C) — using a larger volume of the SAME concentration does not increase the rate. Only concentration per unit volume affects collision frequency.