Rates of Reaction — Measuring and Calculating

Understanding rates of reaction is a core part of the AQA GCSE Combined Science Trilogy (8464) chemistry specification. In this lesson you will learn how to measure the rate of a chemical reaction using different experimental methods, how to calculate the mean rate, and how to interpret graphical data including drawing tangents for instantaneous rate.

---

What Is the Rate of Reaction?

The rate of reaction measures how quickly reactants are used up or how quickly products are formed during a chemical reaction.

• A fast reaction (such as an explosion) has a high rate.
• A slow reaction (such as rusting) has a low rate.

Reaction Example	Approximate Timescale	Rate Description
Explosion of hydrogen and oxygen	Fractions of a second	Very fast
Burning magnesium ribbon	A few seconds	Fast
Marble chips reacting with dilute acid	A few minutes	Moderate
Rusting of iron	Days to months	Slow
Weathering of limestone	Years to centuries	Very slow

Exam Tip (AQA 8464): The specification uses the term "rate of reaction" rather than "speed of reaction." Always use the correct terminology in your written answers.

---

Calculating Mean Rate of Reaction

The mean rate of reaction is calculated using:

$\text{mean rate of reaction} = \frac{\text{quantity of reactant used or product formed}}{\text{time taken}}$mean rate of reaction=time takenquantity of reactant used or product formed​

The quantity measured depends on the experimental method used:

Measurement	Units	Method
Volume of gas produced	cm³	Gas syringe or collection over water
Mass lost	g	Measuring decrease in mass on a balance
Time for a precipitate to obscure a mark	s	Disappearing cross experiment
Colour change	s	Timing a colour change (colorimeter)

The units of rate depend on the quantities involved:

Quantity Unit	Time Unit	Rate Unit
cm³	s	cm³/s
g	s	g/s
mol	s	mol/s
cm³	min	cm³/min

Exam Tip: Always include units with your calculated rate. A numerical answer without units will lose marks.

---

Worked Example 1: Mean Rate from Volume

Question: In an experiment, 60 cm³ of gas was collected in 150 seconds. Calculate the mean rate of reaction.

Solution:

$\text{mean rate} = \frac{\text{volume of gas}}{\text{time taken}}$mean rate=time takenvolume of gas​

$\text{mean rate} = \frac{60\text{ cm}^3}{150\text{ s}}$mean rate=150 s60 cm3​

$\text{mean rate} = 0.40\text{ cm}^3\text{/s}$mean rate=0.40 cm3/s

---

Worked Example 2: Mean Rate from Mass Loss

Question: A reaction caused the mass of a flask to decrease by 0.90 g over 3 minutes. Calculate the mean rate of reaction in g/s.

Solution:

First convert minutes to seconds: 3 min = 180 s

$\text{mean rate} = \frac{0.90\text{ g}}{180\text{ s}} = 0.005\text{ g/s}$mean rate=180 s0.90 g​=0.005 g/s

Common Mistake: Forgetting to convert minutes to seconds when the question asks for a rate in g/s or cm³/s. Always check the units required in the question.

---

Methods for Measuring Rate

1. Gas Syringe Method

A gas syringe collects the gas produced and measures its volume at regular intervals. This is accurate for gases that do not dissolve in water.

2. Mass Loss Method

Place the reaction flask on a balance. As gas escapes, the mass decreases. Record mass at regular time intervals.

3. Disappearing Cross Method

A cross is drawn on paper beneath a conical flask. When the reaction produces a precipitate, the solution becomes opaque. The time taken for the cross to disappear is recorded.

Method	Advantages	Disadvantages
Gas syringe	Accurate volume readings; easy to plot	Gas can leak if syringe sticks
Mass loss	Continuous data; simple equipment	Small mass changes hard to detect; drafts affect readings
Disappearing cross	Quick and simple	Subjective endpoint; only gives one time reading

---

Interpreting Rate Graphs

A typical graph of product formed (y-axis) against time (x-axis) shows:

graph TD
    A["Start: reactant concentration highest"] --> B["Steep curve — fast rate"]
    B --> C["Curve becomes less steep — rate slows"]
    C --> D["Curve levels off — reaction complete"]
    D --> E["Flat line — no more product formed"]

Feature	Meaning
Steep gradient at start	Rate is fastest when reactant concentration is highest
Decreasing gradient	Rate slows as reactants are used up
Horizontal line	Reaction has finished — limiting reactant used up
Final y-value	Total amount of product formed

---

Finding Instantaneous Rate: Drawing a Tangent

The mean rate gives the average over the whole reaction. The instantaneous rate at a particular time is found by drawing a tangent to the curve at that point.

Steps to Draw a Tangent

1. Identify the time point on the x-axis.
2. Place a ruler so it just touches the curve at the corresponding point without cutting through it.
3. Draw a straight line — this is the tangent.
4. Choose two points on the tangent that are far apart.
5. Calculate the gradient:

$\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$gradient=ΔxΔy​=x2​−x1​y2​−y1​​

6. The gradient equals the instantaneous rate at that time.

Worked Example: Tangent Method

A graph shows volume of gas (cm³) against time (s). A tangent drawn at t = 40 s passes through (10, 8) and (70, 56).

$\text{gradient} = \frac{56 - 8}{70 - 10} = \frac{48}{60} = 0.80\text{ cm}^3\text{/s}$gradient=70−1056−8​=6048​=0.80 cm3/s

The instantaneous rate at 40 seconds is 0.80 cm³/s.

Exam Tip: When drawing a tangent in an exam, make it as long as possible across the graph. Longer lines give more accurate gradient calculations because small errors in reading coordinates have less impact.

---

Summary

• Rate of reaction = quantity of reactant used or product formed ÷ time taken
• Common methods: gas syringe, mass loss, disappearing cross
• Graphs of product vs time show rate changes: steep = fast, flat = finished
• Instantaneous rate is found by drawing a tangent and calculating its gradient
• Always include units and watch for unit conversions (minutes to seconds)

Exam Tip: Rate calculation questions appear frequently on AQA 8464 papers. Practise with different unit combinations and always show your working clearly.

---

Extended Worked Examples and Exam Technique

Worked Example 3: Comparing Mean Rates Across Two Intervals

Question: A student reacted calcium carbonate with dilute hydrochloric acid and recorded the total volume of carbon dioxide produced at regular intervals:

Time (s)	Volume of CO₂ (cm³)
0	0
30	24
60	40
120	56
180	60
240	60

(a) Calculate the mean rate of reaction over the first 30 seconds. (b) Calculate the mean rate between 60 s and 180 s. (c) Explain why the rate changed.

Solution:

(a) $\text{mean rate}_{0-30} = \frac{24 - 0}{30 - 0} = 0.80\text{ cm}^3\text{/s}$mean rate0−30​=30−024−0​=0.80 cm3/s

(b) $\text{mean rate}_{60-180} = \frac{60 - 40}{180 - 60} = \frac{20}{120} = 0.17\text{ cm}^3\text{/s}$mean rate60−180​=180−6060−40​=12020​=0.17 cm3/s

(c) In the first interval the concentration of hydrochloric acid was at its highest, so the frequency of successful collisions was greatest and the rate was fastest. Between 60 s and 180 s the acid was partly used up, the concentration of reactant particles had fallen, collisions became less frequent, so the rate decreased. After 180 s the volume is constant at 60 cm³, meaning the limiting reactant has been fully consumed and the reaction has finished.

Worked Example 4: Instantaneous Rate by Tangent

A smooth curve of volume against time for the reaction of magnesium with hydrochloric acid passed through the points (20 s, 18 cm³) and (60 s, 54 cm³) when a tangent was drawn at t = 40 s.

$\text{instantaneous rate at 40 s} = \frac{54 - 18}{60 - 20} = \frac{36}{40} = 0.90\text{ cm}^3\text{/s}$instantaneous rate at 40 s=60−2054−18​=4036​=0.90 cm3/s

Common Mistake: Students often read the two endpoints of the curve, not the tangent. Always use the straight-line tangent coordinates, ideally spread across the full length of the line, to minimise reading error.

Comparison Table: Methods for Measuring Rate

Method	What Is Measured	Suitable For	Main Limitation
Gas syringe	Volume of gas	Reactions producing any gas	Syringe friction; gas must not dissolve significantly
Balance / mass loss	Mass lost as gas escapes	Reactions where the gas is dense (e.g. CO₂)	H₂ is too light to measure mass change accurately
Disappearing cross	Time for solution to obscure a cross	Sodium thiosulfate + HCl reaction	Subjective judgement of when cross disappears
Colorimetry	Intensity of colour change	Reactions where a coloured species forms or disappears	Requires a colorimeter; not usually Combined Science equipment

Mermaid — The Shape of a Typical Rate Graph

graph LR
    A["t = 0: steep gradient (fast)"] --> B["Curve begins to flatten (rate decreasing)"]
    B --> C["Plateau reached (reaction complete)"]

The gradient at any point equals the instantaneous rate. The steeper the curve, the faster the reaction at that moment. A flat section means the limiting reactant has been used up.

Exam Tip: Reading Rates From a Graph

When asked to calculate the mean rate between two times, always quote the coordinates you have used. If you simply write down an answer with no coordinates, you cannot be awarded a "working" mark on a two- or three-mark question.

Common Mistake: Students often confuse mean rate (use two points far apart on the curve) with instantaneous rate (draw a tangent at one point, then use two points on the tangent). These produce different numbers.

Answering at different grade levels

• Grades 1–3 answer: "The rate tells you how fast a reaction happens. You divide how much gas is made by the time."
• Grades 4–5 answer: "The rate of reaction is the quantity of product formed (or reactant used) per unit time. The equation is: rate = change in quantity ÷ time. Units are usually cm³/s or g/s. A reaction is fastest at the start when the concentration of reactants is highest."
• Grades 6–7 answer: "The mean rate is calculated over an interval using total change ÷ total time, whereas the instantaneous rate at a specific time is found from the gradient of a tangent to the curve. The rate decreases during a reaction because the concentration of reactants falls, reducing the frequency of successful collisions above the activation energy."
• Grades 8–9 answer: "Quantitative rate data can be used to probe reaction kinetics. A linear portion of a volume–time graph indicates a zero-order or pseudo-zero-order dependence on one reactant over that interval. The gradient at t = 0 corresponds to the initial rate, when reactant concentrations are known precisely and the rate is least affected by competing reverse processes. Accuracy improves when tangents are drawn over long intervals and when repeats are used to calculate a mean rate with outliers rejected."

---

Further practice questions

1. In a reaction, 3.6 g of magnesium was consumed in 2 minutes. Calculate the mean rate in g/s, giving your answer to two significant figures.
2. A volume–time graph for a reaction has a gradient of 1.2 cm³/s at t = 10 s and a gradient of 0.3 cm³/s at t = 60 s. Explain, in terms of collision theory, why the gradient has decreased.
3. A student writes: "The reaction stopped after 90 seconds because all the particles ran out of energy." Rewrite this sentence correctly using the term limiting reactant.

Attempt all three before moving on. Model answers are available in your learning path for cross-checking.

---

AQA alignment: This content is aligned with AQA GCSE Combined Science: Trilogy (8464) specification — specifically 5.6 Rates and extent of chemical change (5.6.1 Rate of reaction, 5.6.1.1 Calculating rates of reactions). Assessed on Chemistry Paper 2. Note that Trilogy excludes the deeper collision-theory and Le Chatelier extensions (5.6.1.3, 5.6.1.4 and 5.6.2.4) that appear only in the separate Chemistry (8462) specification.