Calculating Rates of Reactions

MOD 6 • RATE • RP5 PART 1

Mean rate, instantaneous rate, and the tangent technique. The most heavily examined graph skill on Paper 2 — and the one students lose the most marks on.

What is Rate of Reaction?

Rate measures how quickly a reaction proceeds. It's a measure of change per unit time — and the units depend on what you're measuring.

Rate of Reaction

The change in the amount of reactant or product per unit time.

Common Units

cm³/s — when measuring volume of gas produced

g/s — when measuring mass change (gas escaping)

mol/s — when measuring moles of product or reactant HT only

Always state the unit clearly — it must match the y-axis of the graph.

Two Ways to Follow a Reaction

You can track a reaction's progress by measuring either side of the equation:

Method A

Measure PRODUCT formed

Volume of gas collected, mass of precipitate formed, depth of colour appearing.

Graph shape: starts at 0, climbs steeply, then levels off as reaction completes.

Method B

Measure REACTANT used

Mass loss when gas escapes, time for a cross to disappear (turbidity).

Graph shape: starts at maximum, falls steeply, then levels off.

Examiner Insight

Most exam graphs show volume of gas vs time — Method A. The curve increases over time, but the rate still decreases — read carefully.

Mean Rate

Mean rate is the average rate over a defined time interval. Use it when the question asks "between t = 0 and t = 60 s".

Mean Rate

Total change in amount of reactant or product, divided by the total time taken.

Mean rate = (amount of product formed) ÷ (time taken)

Mean rate = (amount of reactant used up) ÷ (time taken)

From a Graph

Read the y-value at the start and end of the interval. Subtract to get Δy. The time interval is Δx. Divide.

Mean rate = Δy ÷ Δx, where the two points are the endpoints of the interval — not a tangent.

Worked Example — Mean Rate

2 MARKS

In a reaction between magnesium and dilute hydrochloric acid, 48 cm³ of hydrogen gas was collected after 60 seconds. Calculate the mean rate of reaction over this interval.

1

Identify the change in amount of product: Δy = 48 cm³.

2

Identify the time interval: Δx = 60 s.

3

Apply: Mean rate = Δy ÷ Δx = 48 ÷ 60 = 0.80 cm³/s.

Answer: 0.80 cm³/s — units must be stated for the final mark.

Instantaneous Rate

Instantaneous rate is the rate at one specific moment in time — for example, "the rate at t = 30 s". Because the curve is curved, you can't just use endpoints — you need a tangent.

Instantaneous Rate

The rate at a specific moment. Found by drawing a tangent to the curve at that point and using its slope as a measure of the rate.

Tier Split — Spec 4.6.1.1

Foundation tier: draw the tangent and use its slope qualitatively — e.g. "tangent A is steeper than tangent B, so the reaction is faster at A".

Higher tier: calculate the actual numerical gradient of the tangent (Δy ÷ Δx) — and may use mol/s units. The next four slides cover this HT calculation skill.

Instantaneous rate = gradient of tangent = Δy ÷ Δx HT

Drawing a Tangent — Technique

Tangent quality is examined directly. Mark schemes give a separate mark for "tangent drawn at correct time" — but only if your line is a good tangent.

Three-Step Technique

Touch: place your ruler so it just touches the curve at the specified time — don't cross through.

Extend: draw the tangent well beyond the touch point in both directions — across several major gridlines.

Read: pick two points far apart on your tangent, read their actual axis values, and calculate Δy ÷ Δx.

Worked Example — Instantaneous Rate

4 MARKS

HT only

A tangent has been drawn at t = 30 s on a volume vs time graph. The tangent passes through (10 s, 18 cm³) and (50 s, 66 cm³). Calculate the instantaneous rate of reaction at t = 30 s. Give your answer to 2 significant figures.

1

Read y-step from tangent endpoints: Δy = 66 − 18 = 48 cm³.

2

Read x-step from tangent endpoints: Δx = 50 − 10 = 40 s.

3

Calculate gradient: rate = Δy ÷ Δx = 48 ÷ 40 = 1.2 cm³/s.

4

Round to 2 sig figs: 1.2 cm³/s.

Answer: 1.2 cm³/s. Mark scheme structure (P2 Jun23 Q07.3): 1 mark for x and y step values, 1 mark for substitution, 1 mark for calculation, 1 mark for sig figs.

The #1 Mistake — Vertical Line ≠ Tangent

This single error appeared in P2 Jun23 Q07.3 and again in P2 Jun24 Q09.3. Students who made it lost up to 4 of the 5 available marks.

Wrong

This calculates mean rate from t=0 — not instantaneous. Loses 3 of 4 marks.

Right

Tangent touches the curve at t; gradient = instantaneous rate. Full marks available.

Examiner Alert — P2 Jun23 Q07.3 / P2 Jun24 Q09.3

Without a tangent, the only mark available is the final sig figs / standard form mark. Always read the question carefully: "instantaneous rate" → tangent, not vertical drop.

Read Axis VALUES — Not Grid Squares

A subtle error that costs marks every year. Students count grid squares (e.g. "tangent rises 5 squares over 4 squares = 1.25") instead of reading the actual numbers on the axes.

Examiner Alert — P2 Jun22 Q08.2

Students counted squares on axes instead of reading actual values — lost MP1 (the y-step / x-step value mark). Each axis has a scale; one square may equal 5 cm³ or 0.001 mol — you must convert square count into axis value.

Two-Part Check

Read axis labels first: what's the scale? "Each major gridline = 10 cm³"? "Each small square = 0.001 mol"?

Then read the value at each tangent endpoint by tracing horizontally and vertically to the axes — write down the actual number, not the square count.

Tangent Length Tip

Extend your tangent across several major gridlines. A tangent only 2 squares long introduces large rounding errors — examiners explicitly flagged this in P2 Jun22 Q08.2.

Why Rate Decreases Over Time

Every rate graph starts steep and gradually flattens. Students often describe the shape — but lose the explanation mark by being too vague.

The Mark-Scheme Answer

As the reaction proceeds, the concentration of REACTANTS decreases. Fewer reactant particles per unit volume → fewer collisions per second → fewer successful collisions → rate decreases.

Examiner Alert — P2 Jun22 Q08.3

Students who wrote "concentration decreases" without specifying reactants did not gain the mark. Be explicit: concentration of reactants decreases.

Full 3-Mark Answer Template

Concentration of reactants decreases over time.

So there are fewer collisions per second between reactant particles.

Therefore the rate of reaction decreases.

When the Reaction Is Complete

The graph levels off — becomes a horizontal line. The gradient is zero, so the rate is zero.

Why Does It Level Off?

One of the reactants — the limiting reactant — has been used up entirely. With no limiting reactant left, no more product can form, so the volume of gas stops changing.

RP5 — Gas Collection Method

Required Practical 5 investigates how concentration affects rate using two methods. Method A — gas collection — is the focus here. Full method detail appears in lesson 6.02.

Standard Reaction

Magnesium ribbon + dilute hydrochloric acid → magnesium chloride + hydrogen gas.

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

VariableWhat it is

IV (independent)Concentration of HCl (e.g. 0.5, 1.0, 2.0 mol/dm³)

DV (dependent)Volume of H₂ collected at fixed time intervals (gas syringe)

ControlsSame volume of acid; same mass and shape of Mg; same temperature

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

"Gas evaporates" is wrong (gases don't evaporate). "Gas is lost" is too vague. The mark scheme wants: "gas escaped from the flask". For 2-mark gas-collection error questions, also state the mechanism — e.g. "the reaction starts before the stopper is put in" — so some hydrogen escapes before being trapped.

Practice 1 — Mean Rate

2 MARKS

A student measured 30 cm³ of carbon dioxide gas in 20 seconds when calcium carbonate reacted with dilute hydrochloric acid.

Calculate the mean rate of reaction. Give the unit.

1

Mean rate = Δy ÷ Δx = 30 ÷ 20.

2

= 1.5 cm³/s (unit required for the second mark).

Practice 2 — Instantaneous Rate

4 MARKS

HT only

A tangent drawn at t = 45 s on a moles vs time graph passes through (20 s, 0.0040 mol) and (70 s, 0.0090 mol).

Calculate the instantaneous rate of reaction at t = 45 s. Give your answer in standard form, with units.

(Modeled on P2 Jun24 Q09.3 — 5-mark structure)

1

y-step from tangent: Δy = 0.0090 − 0.0040 = 0.0050 mol.

2

x-step from tangent: Δx = 70 − 20 = 50 s.

3

Rate = Δy ÷ Δx = 0.0050 ÷ 50 = 0.0001 mol/s.

4

In standard form: 1.0 × 10⁻⁴ mol/s.

Common Errors — Quick Recap

Every error below was directly cited in examiner reports for AQA GCSE Chemistry Paper 2.

Students wroteMark scheme wantedSource

Vertical drop to axis from a curve pointTangent at that time, then gradientP2 Jun23 Q07.3

Counted grid squaresRead actual axis values at tangent endpointsP2 Jun22 Q08.2

"Rate decreases because concentration decreases"Concentration of reactants decreasesP2 Jun22 Q08.3

"Gas evaporates from the flask"Gas escaped from the flaskP2 Jun23 Q07.2

Calculated Δx ÷ ΔyΔy ÷ Δx (= gradient)P2 Jun24 Q09.3

No units on the final ratecm³/s, g/s, or mol/s — match y-axisP2 Jun23 Q07.3

Lesson Recap — You Should Now Be Able To...

Define rate of reaction as the change in amount of reactant or product per unit time.

Calculate mean rate by reading two endpoints from a graph and applying Δy ÷ Δx — universal across both tiers.

Draw a tangent at a specified time and use its slope to compare rates qualitatively — universal across both tiers.

Calculate the gradient of a tangent as a numerical instantaneous rate HT only

Distinguish mean rate (vertical line drop) from instantaneous rate (tangent) — and never confuse them.

Read actual axis values rather than counting grid squares.

Explain why rate decreases over time — the concentration of reactants decreases.

State the units of rate to match the y-axis (cm³/s, g/s — and mol/s for HT).

Identify when a reaction is complete from the graph (curve levels off; rate = 0).

Top Marks Habit

When the question says "instantaneous", draw the tangent before reaching for a calculator. The tangent is worth a separate mark — and unlocks every other mark in the question.

UP NEXT • LESSON 6.02

Factors Affecting Rate

Concentration, surface area, temperature, catalysts. The full RP5 method — gas collection AND the disappearing cross. How to sketch graphs for changed conditions without losing the plateau mark.