AQA GCSE Chemistry Chemistry Paper 2 (Higher), November 2020: Question 8

12 marks · Standard Demand difficulty · Short Answer

Investigate the rate of reaction between hydrochloric acid and calcium carbonate by plotting a graph, calculating mean rate, and determining surface area to volume ratios.

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Question

Question text

08 This question is about the rate of the reaction between hydrochloric acid and

calcium carbonate.

A student investigated the effect of changing the size of calcium carbonate lumps on

the rate of this reaction.

This is the method used.

1. Pour 40 cm3 of hydrochloric acid into a conical flask.

2. Add 10.0 g of small calcium carbonate lumps to the conical flask.

3. Attach a gas syringe to the conical flask.

4. Measure the volume of gas produced every 30 seconds for 180 seconds.

5. Repeat steps 1 to 4 using 10.0 g of large calcium carbonate lumps.

The student calculated the number of moles of gas from each volume of

gas measured.

Table 4 shows the student’s results for large calcium carbonate lumps.

Table 4

Time in seconds Number of moles of gas

0 0.0000

30 0.0011

60 0.0020

90 0.0028

120 0.0034

150 0.0038

180 0.0040

The student plotted the results for small calcium carbonate lumps on Figure 4.

08.1 Complete Figure 4.

You should:

• plot the data for large calcium carbonate lumps from Table 4

• draw a line of best fit.

[3 marks]

Figure 4

08.2 Determine the mean rate of reaction for small calcium carbonate lumps

between 20 seconds and 105 seconds.

Give the unit.

Use Figure 4.

[4 marks]

Mean rate of reaction = Unit

08.3 The student concluded that the large calcium carbonate lumps reacted more slowly

than the small calcium carbonate lumps.

How do the student’s results show that this conclusion is correct?

[1 mark]

The difference in the rates of reaction of large lumps and of small lumps of

calcium carbonate depends on the surface area to volume ratios of the lumps.

*23* Figure 5 shows a cube of calcium carbonate.

Figure 5

08.4 Calculate the surface area to volume ratio of the cube in Figure 5.

Give your answer as the simplest whole number ratio.

[3 marks]

Surface area : volume = :

08.5 A larger cube of calcium carbonate has sides of 5 cm

Describe how the surface area to volume ratio of this larger cube differs from that of

the cube shown in Figure 5.

[1 mark]

Mark scheme

Show the mark scheme

Question 8

AO /

Question Answers Extra information Mark

Spec. Ref.

08.1 all seven points plotted allow a tolerance of ±½ small 2 AO2

correctly square 4.6.1.1

allow 1 mark for five or six

points plotted correctly

line of best fit 1

08.2 0.0038 and 0.0014 1 AO2

4.6.1.1

0.0038 - 0.0014 allow correct use of incorrectly 1

105 - 20 determined mole value(s)

= 0.000028 1

= 2.8 x 10-5

mol/s allow moles per second 1

allow converse statement for

08.3 small lumps AO2

(for large lumps) a smaller 1 4.6.1.1

number of moles of gas is

collected in the same time

(for large lumps) more time is

needed to collect the same

number of moles of gas

the line (of best fit for large

lumps) is less steep

allow the line (of best fit for large

lumps) takes more time to

become horizontal

Question 8 (continued)

AO /

18 Spec. Ref.

08.4 (surface area = 6 x 0.5 x 0.5) = 1 AO2

1.5 (cm2) 4.6.1.3

(volume = 0.5 x 0.5 x 0.5) = 1

0.125 (cm3)

(surface area : volume =) allow correctly calculated ratio 1

12 : 1 using incorrectly calculated

values for surface area and/or

volume

08.5 decreases by a factor of 10 allow 10 times smaller 1 AO2

allow one tenth 4.2.4.1

allow 1/10 4.6.1.3

allow 1 : 10 (large cube to small

cube)

Total 12

How to answer it

Rates of Reaction: Marble Chips, Graphing & Surface Area

WHAT THIS QUESTION TESTS

This question assesses core experimental and mathematical chemistry skills from Topic 6: The Rate and Extent of Chemical Change:

Graph Plotting: Accurately plotting experimental data points and constructing a smooth curved line of best fit.
Rate Calculations from Curves: Calculating the mean rate of reaction across a specified time interval ( Δy / Δx ) and stating the correct derived unit.
Interpreting Rates: Explaining how graph gradients show differences in reaction speed.
Mathematical Modelling: Calculating surface area to volume ( SA : V ) ratios of geometric shapes and predicting the scaling effect when side length increases.

QUESTION 08.1 · 3 MARKS

Plotting Reaction Data & Line of Best Fit

Plotting points from Table 4 for large calcium carbonate lumps and drawing a curve of best fit

✅ Model Answer & Marking Points

[2 marks] All 7 points plotted accurately within ±½ small square:
(0, 0.0000) , (30, 0.0011) , (60, 0.0020) , (90, 0.0028) , (120, 0.0034) , (150, 0.0038) , (180, 0.0040) .
(Award 1 mark if 5 or 6 points are correctly plotted).
[1 mark] A single, smooth curved line of best fit starting at (0,0) , passing through or close to all plotted points, and levelling off below the small lumps curve.

🧠 Exam Technique: Examiner Tips

Use neat crosses (×): Dots can be lost under thick lines; a sharp pencil cross allows the examiner to check accuracy within ±½ small grid square.
One continuous curve: Do not join the dots with a ruler ("dot-to-dot"). A reaction rate decreases continuously, so the curve must be smooth.
Check scale values: On the y-axis, each large grid box is 0.0010, meaning each small square represents 0.0001 moles.

❌ Common Errors

Point at 30 s: Plotting 0.0011 as 0.0015 (misreading the grid). 0.0011 is exactly 1 small square above 0.0010.
Double/Feathered lines: Sketching or going over the line multiple times loses the line mark. Turn your paper so your wrist curves naturally to draw one continuous stroke.

QUESTION 08.2 · 4 MARKS

Calculating Mean Rate from a Curve

Determining the mean rate of reaction for small lumps between 20 s and 105 s with units

📐 Step-by-Step Calculation

Step 1: Read values from the small lumps line on Figure 4 [1 mark]
At time = 20 s : moles = 0.0014
At time = 105 s : moles = 0.0038
Step 2: Set up the mean rate equation [1 mark]
Mean rate = change in moles ÷ change in time
Mean rate = (0.0038 - 0.0014) ÷ (105 - 20)
Mean rate = 0.0024 ÷ 85
Step 3: Calculate the numerical value [1 mark]
Mean rate = 0.000028 or in standard form: 2.8 × 10⁻⁵
Step 4: Determine the correct unit [1 mark]
Moles divided by seconds = mol/s (or moles per second )

✅ Model Answer Summary

Mean rate of reaction: 0.000028 (or 2.8 × 10⁻⁵ )

Unit: mol/s

Full 4 marks require both coordinates correct, correct substitution, correct final value, and the unit.

❌ Common Traps

Wrong curve: Reading data from the newly plotted large lumps curve instead of the small lumps curve printed on the exam paper.
Dividing by 105 s: Failing to subtract the start time ( 105 - 20 = 85 s ).
Missing or incorrect unit: Writing cm³/s or g/s out of habit instead of checking the axis label ( moles ).

QUESTION 08.3 · 1 MARK

Interpreting Graphical Rate Evidence

How the results show that large calcium carbonate lumps react more slowly

✅ Acceptable Answers (Any 1 of the following)

For large lumps, a smaller number of moles of gas is collected in the same time (e.g. at 60 s, large has 0.0020 mol vs small has ~0.0030 mol).
For large lumps, more time is needed to produce the same number of moles of gas.
The line of best fit for large lumps is less steep (has a lower gradient).
The curve for large lumps takes more time to level off (reach horizontal).

1 mark total. A converse statement for small lumps is also fully accepted.

🧠 Exam Technique: Be Specific

Do not simply state: "The line is lower." State why that matters scientifically: the curve is less steep / has a lower gradient, or specify that less gas is produced in a given time.

QUESTION 08.4 · 3 MARKS

Calculating Surface Area to Volume Ratio

Finding the simplest whole number ratio for a 0.5 cm cube

📐 Step-by-Step Calculation

Step 1: Calculate Total Surface Area [1 mark]
Area of 1 face = 0.5 cm × 0.5 cm = 0.25 cm²
A cube has 6 identical faces:
Total Surface Area = 6 × 0.25 cm² = 1.5 cm²
Step 2: Calculate Volume [1 mark]
Volume = length × width × height
Volume = 0.5 cm × 0.5 cm × 0.5 cm = 0.125 cm³
Step 3: Calculate Simplest Whole Number Ratio [1 mark]
SA : Volume = 1.5 : 0.125
Divide both sides by 0.125 :
1.5 ÷ 0.125 = 12
0.125 ÷ 0.125 = 1
Ratio = 12 : 1

✅ Model Answer

Surface area : volume = 12 : 1

Error carried forward (ecf) applies if an incorrect surface area or volume is correctly simplified into a ratio.

❌ Common Errors

Forgetting 6 faces: Calculating the area of only 1 face ( 0.25 cm² ) gives a ratio of 2 : 1 (loses 1 mark).
Not simplifying fully: Leaving the answer as 1.5 : 0.125 ignores the instruction for simplest whole number ratio.
Inverting the ratio: Writing 1 : 12 instead of 12 : 1 .

QUESTION 08.5 · 1 MARK

Scale Factor Effect on SA:V Ratio

How the SA:V ratio changes when cube side length increases from 0.5 cm to 5 cm

✅ Model Answer

Decreases by a factor of 10

Also accepted by examiner:

10 times smaller
One tenth (1/10)
1 : 10 (large cube to small cube)

💡 Key Knowledge: The Cube Scaling Rule

For any cube of side length x :

SA = 6x² and Volume = x³

SA : V = (6x²) / x³ = 6 / x

For side 0.5 cm : 6 / 0.5 = 12 cm⁻¹ (Ratio 12 : 1)
For side 5.0 cm : 6 / 5.0 = 1.2 cm⁻¹ (Ratio 1.2 : 1)

When the side length increases by a factor of 10 ( 0.5 cm → 5 cm ), the SA:V ratio decreases by a factor of 10.

🧠 Exam Technique: Be Quantitative

Simply writing "it is smaller" or "it decreases" is not enough to gain the mark. The question asks how it differs; you must state the numerical factor (by a factor of 10 or one tenth).

Topics

Chemistry · C2: Bonding, Structure and the Properties of Matter · C6: The Rate and Extent of Chemical Change

Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 2 (Higher), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.