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Determining Speed of a Chemical Reaction

Measuring rate of reaction: time taken, gas volume and mass loss methods, plus interpreting graphs (gradient/tangent) in exam questions.

  • SEC G3 Pure Chemistry 2027
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Learning objectives

  • suggest a suitable method for investigating the effect of a given variable on the rate of a reaction
  • interpret data obtained from experiments concerned with rate of reaction.

1. Definition

The rate (speed) of reaction is the change in amount of reactant used up or product formed per unit time.

To understand what “rate” looks like on graphs (gradient/tangent), revise: Speed of Reaction.

2. Key Ideas

  • You must measure a quantity that changes with time (e.g. volume of gas, mass, time for a fixed change).
  • Rate = gradient of the relevant graph:
    • Volume of gas vs time → gradient is rate of gas production.
    • Mass of mixture vs time (gas escapes) → gradient is rate of mass loss.
  • A reaction is usually fastest at the start because reactant concentrations are greatest and any solid surface has not yet been depleted, so effective collisions are most frequent.
  • When a graph becomes horizontal, the rate is zero (reaction has stopped / limiting reactant used up).

3. Detailed Explanations

Quick Recall (methods + graphs)
  • Rate is “change per time”: decide what changes (volume, mass, concentration, time for a fixed change).
  • On a graph, rate = gradient; on a curve, rate at a specific time = tangent gradient.
  • A horizontal graph means rate is zero (reaction stopped / limiting reactant used up).
  • Fair test: change one factor only and keep the rest constant.

A. Method 1: Time taken for a fixed change

If you keep the amount of reactant the same each time (same mass/length/surface area), you can compare reactions by timing how long they take to reach a fixed end-point (e.g. “magnesium completely dissolves”).

Example reaction (magnesium + dilute sulphuric acid):

Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g)

Do not overclaim: 1/time is a comparison tool

Using 1/time gives a relative rate only when you are comparing similar experiments (same amount of reactant, same end-point). It is not the definition of rate.

At a fixed end-point, a shorter time means a faster reaction. Do not compare bubble size or make a rate conclusion from a single observation.

B. Method 2: Volume of gas produced vs time (gas syringe)

Use this method if the reaction produces a gas (e.g. marble + dilute hydrochloric acid).

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

Gas-syringe method for measuring reaction rateA conical flask containing marble chips and dilute hydrochloric acid is sealed with a bung and connected by a delivery tube to a graduated gas syringe. Carbon dioxide collected is recorded at regular time intervals.marble chips + dilute HClgraduated gas syringerecord CO₂ volume / cm³ at regular timesairtight delivery tube
Gas-syringe method: record the volume of carbon dioxide at regular time intervals. Keep every variable except the one being investigated constant.

Plot a graph of volume of gas (e.g. CO₂) against time.

Volume of Gas vs Time

Example volume-of-gas vs time curve for a reaction that slows down as reactants are used up; gradient represents rate.

Scroll across the graph to read all labels.

Example volume-of-gas vs time curve for a reaction that slows down as reactants are used up; gradient represents rate.Example volume-of-gas vs time curve for a reaction that slows down as reactants are used up; gradient represents rate.
Typical shape: steep at the start (fast rate), then levels off as a reactant is used up. The gradient is the rate.
Open full-size graph
View figure data
Values for Volume of Gas vs Time
Time (s)CO2 produced
00
1010
2018
3025
4030
5034
6037
7039
8040
Instantaneous vs average rate

If the graph is curved, the rate at a particular time is found by drawing a tangent and calculating its gradient. Do not use two random points on the curve unless the question asks for an average rate over that time interval.

C. Method 3: Mass loss vs time (gas escapes)

Use this method when a gas escapes from the reaction mixture. As gas leaves, the mass decreases.

Mass-loss method for measuring reaction rateA conical flask containing marble chips and dilute hydrochloric acid stands on an electronic balance. A loose cotton-wool plug reduces spray while allowing carbon dioxide to escape, so the decreasing mass is recorded at regular time intervals.marble chips + dilute HClloose cotton-wool plugreduces spray; does not seal flaskCO₂ escapesmass / gelectronic balancerecord decreasing mass at regular times
Mass-loss method: record the mass at regular time intervals as carbon dioxide escapes. Cotton wool reduces spray but must not seal the flask.

Plot a graph of mass against time.

Mass of Mixture vs Time (Gas Escapes)

Example mass vs time curve showing mass decreases as gas escapes; the magnitude of the negative gradient indicates rate.

Scroll across the graph to read all labels.

Example mass vs time curve showing mass decreases as gas escapes; the magnitude of the negative gradient indicates rate.Example mass vs time curve showing mass decreases as gas escapes; the magnitude of the negative gradient indicates rate.
Mass decreases as gas escapes. The gradient is negative; the magnitude of the gradient shows the rate of gas production.
Open full-size graph
View figure data
Values for Mass of Mixture vs Time (Gas Escapes)
Time (s)Mass
0120
10119.3
20118.8
30118.4
40118.1
50117.9
60117.8
Safety and good practical technique

Use dilute acids as instructed and keep your face away from the flask. Do not block the flask neck: gas must escape, or pressure can build up.

D. Choosing the best method (quick guide)

Reaction featureBest measurementWhat you plotWhat gives the rate
Gas is produced and you can collect itVolume of gasVolume vs timeGradient (or tangent)
Gas is produced but collection is awkwardMass lossMass vs timeGradient (magnitude)
Reaction has a clear end-point (e.g. disappears, precipitate appears)Time takenN/A1/time for comparisons only
Macroscopic → particle → symbolic
  • Macroscopic observation: gas volume rises quickly at first, then levels off.
  • Particle explanation: reactant concentration falls as particles are used up, so effective collisions become less frequent.
  • Symbolic representation: the gradient, Δ V/Δ t, decreases and finally becomes zero.

4. Common Mistakes

  • Writing “faster reaction makes more gas” (wrong): the total gas depends on starting amounts; rate is about how fast gas is produced.
  • Reading a curved graph by “joining two points on the curve” when the question wants the rate at a specific time (needs a tangent).
  • Comparing times without controlling the amount/surface area of reactant (unfair test).
  • Forgetting units (e.g. writing “0.8” instead of “0.8 cm³/s”).

5. Exam Tips

Mark-scheme phrases

Use these exact ideas:

  • “Rate is the gradient of the graph.”
  • “The reaction is fastest at the start because there are more reactant particles, so there are more frequent effective collisions.”
  • “When the graph becomes horizontal, the rate is zero (reaction has stopped).”

6. Worked Examples

Modelled example 1

Comparing rates using time

Core

Problem

Two identical pieces of magnesium ribbon, with the same length and thickness, are added to two acids. Magnesium disappears in 25 s in Acid A and 35 s in Acid B. Which reaction is faster?
Study the worked solution

  1. Check that the endpoint is comparable

    Method

    Use complete disappearance of identical magnesium pieces as the same fixed change.

    Reason

    Time alone compares rate only when the experiments reach an equivalent endpoint with the same reactant amount.

    Working

    Both trials use identical ribbon and the same endpoint.

  2. Compare the times

    Method

    Choose Acid A because it reaches the endpoint sooner.

    Reason

    For the same change, a shorter time means a greater change per unit time.

    Working

    25 s < 35 s, so A is faster.

  3. Confirm with relative rate

    Method

    Compare reciprocal times.

    Reason

    1/t is a valid relative-rate tool for this controlled fixed endpoint.

    Working

    A: 1/25 = 0.040 s⁻¹; B: 1/35 = 0.029 s⁻¹.

Guided practice 2

Average rate from data

About 5 min

Problem

A reaction produces 48 cm³ of CO₂ in 60 s. Calculate the average rate in cm³/s.

Calculate change per elapsed time

Unit

Hints

Hint 1: average-rate relationship
Divide the total gas-volume change by the complete 60 s interval.
View solution step by step

  1. Divide change by interval

    Method

    Divide the 48 cm³ product increase by 60 s.

    Reason

    An average rate spreads the measured change across the full stated interval.

    Working

    48/60 = 0.80.

  2. Report the measured rate

    Method

    Give the gas-volume-per-time unit and two significant figures.

    Reason

    The unit follows the axes and 48 cm³ has two significant figures.

    Working

    0.80 cm³ s⁻¹.

Common misconception 3

Interpreting a gas-volume graph

Find and correct the mistake

Learner claim

A CO₂-volume graph is steep at first and then becomes horizontal. A student says, “The horizontal section means the reaction has reached its fastest rate because the gas volume is greatest.” Explain the mistake and correct the interpretation.

Read gradient separately from amount

Steep initial section
Horizontal section

View solution step by step

  1. Interpret the initial gradient

    Method

    Identify the start as the fastest section.

    Reason

    Reactant concentrations and available solid surface are greatest initially, so effective collisions are most frequent.

    Working

    Steep gradient ⇒ high gas-production rate.

  2. Explain the decreasing gradient

    Method

    State that rate falls as reactants are used up.

    Reason

    Fewer available reactant particles produce fewer effective collisions per unit time.

    Working

    The curve progressively becomes shallower.

  3. Separate final amount from rate

    Method

    Assign zero rate to the horizontal section.

    Reason

    The graph may show the greatest accumulated gas volume while its gradient is zero because the limiting reactant is used up.

    Working

    Horizontal line ⇒ no new gas ⇒ rate = 0.

Challenge 4

Instantaneous rate from a tangent

Minimal support

Graph-gradient transfer

A tangent drawn to a gas-volume curve at 30 s passes through (10 s,18 cm³) and (50 s,34 cm³). Calculate the instantaneous rate at 30 s.

Use two tangent points

Hints

Hint 1: point selection
Use the two stated points on the tangent, even though they need not be experimental data points on the curve.
Hint 2: gradient
Calculate (34-18)/(50-10) with volume on the numerator.
View solution step by step

  1. Find the tangent changes

    Method

    Subtract coordinates in the same order.

    Reason

    The tangent gradient is change in gas volume divided by change in time.

    Working

    Δ V = 34-18 = 16 cm³; Δ t = 50-10 = 40 s.

  2. Calculate the gradient

    Method

    Divide the volume change by the time change.

    Reason

    The tangent represents the curve’s instantaneous slope at 30 s.

    Working

    rate at 30 s = 16/40 = 0.40 cm³ s⁻¹.

7. Mind Stretchers

Mind stretcher 1: Tangent trapExtension

Question: A student finds the rate at 2 minutes by using the points (0 min, 0 cm³) and (4 min, 40 cm³) on a curved volume–time graph. Explain what this actually gives, and what they should do instead.

Show Answer

Answer:

  • Using two points far apart gives an average rate over 0–4 minutes.
  • To find the rate at 2 minutes, they must draw a tangent at 2 minutes and calculate the gradient of the tangent.

8. Quiz

Practise this lesson

The shared K324 / 6092 practice includes method selection, fair tests, graph gradients, tangents and rate calculations.

K324 / 6092 Practice