Factors Which Affect the Rates of Chemical Reactions — AQA GCSE Chemistry

MOD 6 • RATE • RP5 PART 2

Factors Affecting Reaction Rate

Five factors that change the rate. Both halves of Required Practical 5. How to sketch graphs for changed conditions without losing the plateau mark.

Why Factors Matter

Lesson 6.01 showed you how to measure rate from a graph. This lesson is about what changes the rate — and the patterns examiners reuse year after year.

Spec 4.6.1.2 — Recall Required

Students should be able to recall how changing each of the following factors affects the rate of chemical reactions:

Concentration of reactants in solution

Pressure of reacting gases

Surface area of solid reactants

Temperature

Presence of a catalyst

Where the Reasoning Lives

This lesson covers what each factor does and how to spot it on a graph. The full collision theory explanation — frequency, activation energy, why temperature is special — sits in Lesson 6.03. We'll touch on the basic mechanism here, but the deep "why" comes next.

The Five Factors

Each factor either speeds up or slows down a reaction. The direction is intuitive — the wording is what earns the marks.

Concentration

Higher concentration of a reactant in solution → faster rate.

Pressure

Higher pressure of reacting gases → faster rate. (Same idea as concentration but for gases.)

Surface Area

Larger surface area of a solid (e.g. powder) → faster rate.

Temperature

Higher temperature → faster rate. The most powerful factor.

Catalyst

Adding a catalyst → faster rate. The catalyst itself is not used up.

Concentration & Pressure

Concentration applies to solutions; pressure is its equivalent for gases. Both work the same way.

Mechanism (preview of 6.03)

Higher concentration / pressure means more reactant particles per unit volume. Particles encounter each other more often, so collisions happen more frequently — and so does reaction.

Effect on a Volume-vs-Time Graph

Steeper initial gradient — the reaction is faster at the start.

Reaches the plateau sooner — finishes faster.

Plateau height depends on which substance is limiting (next slides cover this).

Examiner Alert — P2 Jun22 Q08.3

Asked why rate decreases over time, students wrote "concentration decreases" — too vague. Mark scheme: "concentration of reactants decreases". Always specify reactants, not generic concentration.

Surface Area

For solid reactants, only the surface is in contact with the other reactant. The inside of a solid lump can't react until the surface above it has been used up.

Lump

Most atoms are buried inside. Only the outside layer can collide with the other reactant. Slow rate.

Powder

The same mass is broken into many small pieces — much more surface exposed. More collision sites. Faster rate.

Surface Area to Volume Ratio

The same mass of solid has the same volume, but breaking it into smaller pieces increases the surface area while volume stays the same. So the surface area to volume ratio (SA : V) goes up.

Higher SA : V → more particles exposed at the surface → more collisions per second → faster rate.

The spec (4.6.1.3) explicitly requires this language: "predict and explain the effects of changes in the size of pieces of a reacting solid in terms of surface area to volume ratio".

Temperature

Temperature is the most powerful rate factor. The mark scheme structure is consistent across all years: one cause + at least one consequence.

The Cause (1 mark)

Particles have more (kinetic) energy — equivalent: particles move faster. Both wordings score the same mark.

The Consequence — Either Alternative Scores (1 mark)

Frequency of collisions increases (more collisions per second) — because faster particles encounter each other more often.

A greater proportion of collisions have enough energy to react — because more particles have energy ≥ activation energy.

Both effects are real — both stem from the same cause (more KE). The mark scheme awards 1 mark for either alternative on Q09.5-style questions.

Examiner Alert — P2 Jun24 Q09.5

Three-mark question structure: M1 = "rate increases" + M2 = "particles have more energy / move faster" + M3 = "frequency increases" or "greater proportion have enough energy to react". The MS explicitly says ignore "successful" on its own — write "have enough energy to react" instead. Top-marks habit: include both consequences anyway — it doesn't cost marks and protects you against year-on-year MS variation.

Coming Up in 6.03

Lesson 6.03 (Collision Theory and Activation Energy) gives you the full mechanism — what activation energy actually is, why concentration / pressure / surface area only increase collision frequency, and why temperature uniquely also increases the energy of those collisions.

Catalysts

A catalyst speeds up a reaction without being used up. After the reaction, you can recover the catalyst unchanged — it can be used again and again.

Catalyst (Mark-Scheme Definition)

A substance that increases the rate of reaction without being used up. It works by providing an alternative pathway with lower activation energy.

Examples You Should Know

Manganese dioxide (MnO₂) — catalyses the decomposition of hydrogen peroxide (used in P2 Jun23 Q07).

Iron — used in the Haber process (covered in 6.07).

Enzymes — biological catalysts in living systems.

Coming Up in 6.04

The full catalyst mechanism — alternative pathway diagrams, activation energy reduction, and the "catalyst doesn't change yield in equilibrium" subtlety — is covered in Lesson 6.04. For now: catalyst → faster rate → not used up.

Required Practical 5 — Two Methods

RP5 investigates how concentration affects rate. AQA requires you to know both methods — gas collection AND a colour/turbidity method. Lesson 6.01 covered Method A; this lesson completes the pair with Method B.

Spec Note — RP5 Hypothesis

The spec specifies RP5 should "involve developing a hypothesis". A typical hypothesis: "Increasing the concentration of [reactant] will increase the rate of reaction, because more reactant particles per unit volume produces more frequent collisions." Be ready to write one if asked.

Method A

Gas Collection (covered in 6.01)

Reaction: Mg + 2HCl → MgCl₂ + H₂

IV: concentration of HCl. DV: volume of H₂ at fixed time intervals.

Apparatus: conical flask + gas syringe (or inverted measuring cylinder).

Method B

Disappearing Cross (this lesson)

Reaction: Na₂S₂O₃ + 2HCl → 2NaCl + H₂O + SO₂ + S

IV: concentration of Na₂S₂O₃. DV: time for cross to disappear.

Apparatus: conical flask placed on a paper cross.

Disappearing Cross — Method

Sodium thiosulfate reacts with hydrochloric acid to form a fine yellow sulfur precipitate. The mixture turns cloudy as the precipitate builds up — eventually a black cross drawn under the flask becomes invisible.

Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + SO₂(g) + S(s)

Method (5 steps)

Draw a black cross on white paper. Place a conical flask on top.

Measure a fixed volume of sodium thiosulfate into the flask.

Add hydrochloric acid; start the timer immediately.

View the cross from above. Stop the timer when the cross is no longer visible.

Repeat with different concentrations of thiosulfate (control: same volume, same temperature, same acid concentration).

Rate ∝ 1 ÷ time → shorter time = faster rate

Why Rate ∝ 1/time?

The cross disappears when the same amount of sulfur has formed each time. So if it takes half as long, twice as much sulfur formed per second — twice the rate.

Worked Example — Concentration Effect

3 MARKS

Question: A student doubled the concentration of hydrochloric acid in a reaction with magnesium ribbon. Describe and explain the effect on the rate of reaction.

M1

Describe the effect: The rate of reaction increases (the reaction is faster).

M2

Explain — particles per volume: The doubled concentration means there are more H⁺ particles per unit volume of acid.

M3

Explain — collisions: More particles per volume means more collisions per second between H⁺ ions and Mg atoms — so the rate is faster.

Examiner note: "More collisions" alone is too vague — must say "more collisions per second" or "greater frequency of collisions". This is the recurring marker phrase across multiple papers.

Sketching Curves Under Changed Conditions

"Sketch the curve if concentration is halved" is one of the most predictable Paper 2 questions — and one of the most missed. The trick is to think about both the gradient AND the plateau.

The Two-Part Rule

Initial gradient — gets less steep if concentration is halved (rate is slower).

Plateau height — depends on whether the substance whose concentration changed is the limiting reactant or in excess:

Halved substance is LIMITING

Half the moles of limiting reactant → half the maximum amount of product. Plateau is at half the original height.

Example: Jun24 Q09.4 — excess zinc, halved acid → plateau halves.

Halved substance is EXCESS

Limiting reactant is unchanged → same maximum amount of product forms. Plateau stays at the same height.

Example: Jun22 Q08.5 — thiosulfate doubled (was already excess) → plateau unchanged.

Examiner Alert — P2 Jun23 Q07.4 / P2 Jun24 Q09.4

Students drew a less-steep curve but stopped it at the same plateau even when the halved substance was the limiting reactant. Always check which reactant is in excess before deciding the plateau height.

Worked Example — Sketch a Halved-Concentration Curve

2 MARKS

Question (P2 Jun24 Q09.4 style): Zinc reacts with sulfuric acid to produce hydrogen. The original graph shows results for excess zinc with 0.40 mol/dm³ sulfuric acid; the curve plateaus at 0.0168 mol of H₂. Sketch the expected curve if 0.20 mol/dm³ sulfuric acid (excess zinc again) were used instead.

M1

Less-steep gradient: Halving the acid concentration halves the initial rate, so the curve climbs less steeply from (0, 0).

M2

Plateau at HALF height: Sulfuric acid is the limiting reactant (zinc is in excess). Halving its concentration halves the moles of acid → halves the moles of H₂ produced. So the plateau is at 0.0084 mol — half of 0.0168.

Mark scheme phrasing: "less steep than existing line" + "becomes level at 0.0084 mol" (allow tolerance ±½ small square).

Anomalies and Means

RP5 questions almost always include a repeats table with one obviously bad result. The mark for "calculate the mean" is split — and you can lose all of it by including the anomaly.

The Two-Step Mean

Identify and discard the anomaly — the result that doesn't fit the trend (visually obvious, or large percentage deviation from the others).

Calculate the mean of the remaining values only.

Examiner Alert — P2 Jun23 Q01.2

Students computed the mean of all four trials including the anomaly — even getting the arithmetic right. They lost the discard mark. The mark scheme awards one mark for discarding the anomaly and a separate mark for the correct mean — so include all values and you cap at 1 of 2.

How to Identify an Anomaly

If three trials gave 0.78, 0.81, 0.81 and one gave 0.68 — the 0.68 is your anomaly. Either say "trial 3 was anomalous and was discarded" or just calculate without it and label clearly: mean = (0.78 + 0.81 + 0.81) ÷ 3 = 0.80 cm³/s.

Method Evaluation — Gas Escape

If the gas-collection method is used (Method A), the volume collected is almost always less than the volume produced. Two-mark questions ask why — and the mark scheme is precise.

Vague — 0 marks

"Some gas is lost." / "Gas evaporates from the flask." / "There's a leak."

Mark-Scheme Pair

(1) "(some) hydrogen / gas escapes from the flask"

(2) "(because the reaction starts) before the stopper is put in" — or "stopper cannot be inserted instantly".

Examiner Alert — P2 Jun24 Q09.1 / P2 Jun23 Q07.2

"Gas evaporates" is wrong wording — gases don't evaporate (only liquids do). Use "escape" for gases. For 2-mark questions, the second mark is the mechanism: explain when or where the gas escapes (e.g. before the stopper is inserted). 1-mark fallback alternatives accepted by Q09.1 MS: "some hydrogen remains in the conical flask" or "some hydrogen remains in the delivery tube".

Method Improvements

Use a gas syringe instead of a delivery tube into water (avoids gas dissolving). Insert the stopper quickly after mixing. Use a flask with a tight-fitting bung. These are common 1-mark "suggest an improvement" answers.

Practice 1 — Temperature Explanation

3 MARKS

Question: Explain how increasing the temperature of a reaction mixture increases the rate of reaction.

M1

Rate increases: increasing the temperature increases the rate of reaction.

M2

Cause: particles have more energy (or: particles move faster — accepted alternative).

M3

Consequence (either alternative scores):(a) the frequency of collisions increases, or(b) a greater proportion of collisions have enough energy to react.

P2 Jun24 Q09.5 mark scheme exactly. M3 awards 1 mark for either (a) or (b) — not both required. Top-marks habit: include both anyway, since other temperature questions across years have asked for both effects. Note: the word "successful" alone is ignored — say "have enough energy to react" instead.

Practice 2 — Sketch the Curve

2 MARKS

Question (P2 Jun22 Q08.5 style): A reaction between 0.10 mol/dm³ sodium thiosulfate solution and hydrochloric acid was monitored using a light sensor. The thiosulfate was in excess. The graph plots % of light reaching the sensor (y-axis) against time (x-axis); the curve decreases from 95% and levels off at 24%. Describe the curve you would expect for 0.20 mol/dm³ thiosulfate — for both initial gradient and final level.

M1

Initial gradient: the curve for 0.20 mol/dm³ would be steeper than the curve for 0.10 mol/dm³ (steeper drop in % light). Higher concentration of thiosulfate → more frequent collisions → faster initial rate of sulfur formation → faster cloudiness.

M2

Final level: the curve levels off at the same 24% as the original. Thiosulfate is in excess; the limiting reactant (HCl) is unchanged. The same final mass of sulfur forms either way, blocking the same fraction of light.

Mark scheme verbatim (P2 Jun22 Q08.5): "decreasing curve starting at (0,95) and steeper initially than curve for 0.10 mol/dm³" + "levelling at 24%". Tied to Q08.4: "(hydrochloric) acid is used up" / "acid is the limiting reactant" — name the limiting reactant; "reactants used up" alone is ignored.

Lesson Recap

You should now be able to:

Name the five factors that affect rate: concentration, pressure, surface area, temperature, catalyst.

State the direction of effect for each: higher concentration / pressure / surface area / temperature → faster rate; catalyst → faster rate (without being used up).

Give the two effects of temperature : more frequent collisions AND greater proportion with enough energy to react.

Describe both RP5 methods : gas collection (Mg + HCl, gas syringe) and disappearing cross (Na₂S₂O₃ + HCl, time for cross to disappear).

Use rate ∝ 1/time for the disappearing-cross method.

Sketch a halved-concentration curve : less steep initial gradient + plateau at half height (if limiting) or same height (if excess).

Discard anomalies before calculating means in RP5.

Use precise wording : "concentration of reactants decreases", "gas escaped from the flask", "more collisions per second ".

Top-Marks Habit

Before sketching any changed-condition curve, ask: which substance is limiting? If the changed substance is limiting, plateau changes proportionally; if it's in excess, plateau is unchanged. This single check separates 2/2 from 1/2 on the most predictable Paper 2 question type.

UP NEXT • LESSON 6.03

Collision Theory and Activation Energy

The mechanism behind every rate factor — what "frequency of collisions" really means, what activation energy is, and why temperature is the only factor that affects both at once.