Rates of reaction

1 Plan and risk assessment

Particles react only when they collide with enough energy. A more concentrated solution has more particles in the same volume, so collisions are more frequent and the rate is higher. Mean rate = quantity of product ÷ time; the rate at one moment is the gradient of a tangent to the curve.

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) and Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + S(s) + SO₂(g) + H₂O(l)

Prediction

2 Apparatus

• ✓100 cm³ gas syringe (resolution 1 cm³) Collects and measures the hydrogen.
• ✓Bung with delivery tube Seals the flask and leads the gas to the syringe.
• ✓Clamp stand, boss and clamp Holds the syringe level and safe.
• ✓100 cm³ conical flask (gas) Where the magnesium reacts with the acid.
• ✓Magnesium ribbon (4.0 cm pieces) and emery paper The metal; emery paper cleans off the oxide.
• ✓Hydrochloric acid, 1.0 and 2.0 mol/dm³ Reacts with the magnesium and with the thiosulfate.
• ✓Sodium thiosulfate solution, 40 g/dm³ Reacts with acid to make the sulfur that hides the cross.
• ✓Distilled water (wash bottle) Dilutes the thiosulfate; rinses the flask.
• ✓Measuring cylinders: 10 cm³ and 100 cm³ (resolution 0.2 cm³ and 1 cm³) Measure the acid, the thiosulfate and the water.
• ✓100 cm³ conical flask (cross) Where the sulfur forms.
• ✓Printed black cross on white paper What you watch disappear.
• ✓Stopclock (resolution 1 s (m:ss display)) Times each run.
• ✓Thermometer (resolution 1 °C) Room temperature, a control variable.
• ✓Safety glasses Protect your eyes from the acid.

3 Method: your choices

• Which version are you doing?: AQA required practical 5: concentration, by gas volume and the disappearing crossMagnesium ribbon in two acids with a gas syringe, then the cross at five thiosulfate concentrations.
• Check the gas syringe: DoneA plunger that sticks lags behind the gas, then jumps, so some readings come out too low.
• Check the bung and tubing are airtight: DoneGas that leaks out round a loose bung or a split tube never reaches the syringe.
• Ventilate the roomSafety: DoneThe thiosulfate reaction gives off sulfur dioxide, which can bring on an asthma attack.
• Clean the magnesium ribbon: DoneMagnesium is coated with a dull layer of magnesium oxide. It reacts with the acid without giving any hydrogen, so it slows the start.
• Put the bung in straight away: DoneAny hydrogen made before the bung is in escapes into the room and is never measured.
• When to start the stopclock: Start it as the acid goes in, swirling at the same timeThe clock times the whole reaction, as AQA’s method says.
• Who watches the cross?: The same person watches every runOne person’s judgement stays about the same, so repeats agree well (within about 5 %).
• Rinse the flask between runs: DoneSulfur left on the glass makes the next mixture cloudy before the reaction has made its own sulfur.

• 0.0 Struck through: the anomaly, retaken; the new reading is beside it.

Anomaly: Time for the cross to disappear, 40 g/dm³, trial 2, retaken. Likely cause: that one time the flask was not rinsed out properly: a film of sulfur from the run before made the mixture look cloudy sooner.

6 Analysis

1. Mean rate, 1.0 mol/dm³ acid

   mean rate of reaction = quantity of product formed ÷ time taken

   Calculate the mean rate of reaction over the first 20 s with the 1.0 mol/dm³ acid. Give your answer to 2 significant figures.

   0.75 cm³/s Right first time

2. Mean rate, 2.0 mol/dm³ acid

   mean rate of reaction = quantity of product formed ÷ time taken

   Calculate the mean rate of reaction over the first 20 s with the 2.0 mol/dm³ acid. Give your answer to 2 significant figures.

   1.3 cm³/s Right first time

3. Shape of the graph

   Describe the shape of your curve with the 1.0 mol/dm³ acid.

   Steep at first, then less steep, then level: the rate falls as the magnesium is used up. Right first time

4. Rate of reaction at 30 s

   The gradient of the tangent is the rate at that moment: change in volume ÷ change in time.

   0.550 cm³/s Right first time

5. Why the stronger acid is faster

   Your mean rate with 2.0 mol/dm³ acid was 1.3 cm³/s, about 1.7 times the rate with 1.0 mol/dm³ acid. Why is the reaction faster with the more concentrated acid?

   There are more acid particles in the same volume, so they collide with the magnesium more frequently. Right first time

6. Shape of the graph

   Describe how the time changes as the concentration goes up.

   It falls steeply at first, then more slowly: a curve. Right first time

7. Rate as 1 ÷ time

   rate = 1 ÷ mean time

   The time for the cross to disappear is shorter when the rate is higher, so 1 ÷ time can stand for the rate. Calculate 1 ÷ mean time for 40 g/dm³. Give your answer to 2 significant figures.

   0.045 /s Right first time

8. Shape of the graph

   What does this graph show?

   A straight line through (or very close to) the origin: the rate is directly proportional to the concentration. Right first time

7 Conclusion and evaluation

Conclusion

Sources of error

Improvement

Repeatable or reproducible

9 Exam questions on your data

On your own readings, self-marked against the mark scheme, with error carried forward where it applies. Total 30/30.

1. 01.1[2 marks]

   You put the bung in as soon as the magnesium was in the acid.

   Another student dropped the magnesium into the acid, watched it fizz, and then put the bung in. Their final volume of hydrogen was less than expected.

   Explain why the final volume of hydrogen was less than expected.

   Your answer

   Some hydrogen escaped before the bung was put in, so not all of the hydrogen was collected in the gas syringe.

   2/2 self-marked

   • Some hydrogen (gas) escaped before the bung (stopper) was put in
   • So not all the hydrogen made was collected in the syringe (allow: the bung cannot be put in instantly)

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: Some of the hydrogen escaped from the flask before the bung was put in, so not all the hydrogen made was collected in the gas syringe.

2. 01.2[3 marks]

   Table 1 shows your volumes of hydrogen with 1.0 mol/dm³ hydrochloric acid.

   Table 1

   Time in s	Volume of hydrogen in cm³
   0	0
   10	8
   20	15
   30	21
   40	26
   50	31
   60	34
   70	36
   80	38
   90	39

   Calculate the mean rate of the reaction in the first 20 seconds.

   Give your answer to 2 significant figures.

   Your answer

   0.75

   Working: mean rate = 15 ÷ 20 = 0.75 cm³/s

   3/3

   • 15 (cm³) and 20 (s) used
   • mean rate = 15 ÷ 20 = 0.75 (cm³/s)
   • 0.75 cm³/s to 2 significant figures

   Full marks. Well done.

   Model answer: mean rate = 15 ÷ 20 = 0.75 cm³/s

3. 01.3[2 marks]

   Look again at Table 1. The magnesium is the limiting reactant: the acid is in excess. The volume stops increasing at about 90 s.

   Describe and explain how the rate of reaction changes during the reaction.

   Your answer

   The rate decreases as time goes on and the reaction slows down. It becomes zero because the magnesium is used up.

   2/2 self-marked

   • The rate decreases (the reaction slows down) as time goes on
   • It becomes zero (stops) because the magnesium is used up

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: The rate decreases as the reaction goes on: it is fastest at the start, then less gas is made in each later 10 s. It becomes zero at about 90 s because the magnesium (the limiting reactant) has all been used up.

4. 01.4[4 marks]

   Another student plotted your results in Table 1 and drew a curve of best fit. They drew a tangent to the curve at 30 s. Table 2 gives two points on their tangent, read from the axes.

   Table 2: two points on the tangent

   Point	Time in s	Volume of hydrogen in cm³
   A	10	10
   B	50	32

   Determine the rate of reaction at 30 s.

   Give your answer to 2 significant figures.

   Your answer

   0.55

   Working: change in volume = 32 − 10 = 22 cm³ change in time = 50 − 10 = 40 s rate = 22 ÷ 40 = 0.55 cm³/s

   4/4

   • y-step (change in volume) = 22 cm³
   • x-step (change in time) = 40 s
   • rate = 22 ÷ 40 = 0.55 (cm³/s)
   • 0.55 cm³/s to 2 significant figures

   Full marks. Well done.

   Model answer: change in volume = 32 − 10 = 22 cm³ change in time = 50 − 10 = 40 s rate = 22 ÷ 40 = 0.55 cm³/s (Your own tangent in the analysis gave 0.55 cm³/s.)

5. 01.5[2 marks]

   The student repeated the gas experiment with the same length of magnesium ribbon and 0.5 mol/dm³ hydrochloric acid. The acid was still in excess.

   How would the curve of volume of hydrogen against time be different from the curve for 1.0 mol/dm³ acid?

   Tick (✓) two boxes.

   • ✓ It starts at the origin and is less steep at the start. (your choice)
   • ✓ It levels off at the same volume. (your choice)
   • It levels off at half the volume.
   • It is steeper at the start.
   • It levels off at a higher volume.

   Your answer

   It starts at the origin and is less steep at the start.; It levels off at the same volume.

   2/2

   • It starts at the origin and is less steep at the start.
   • It levels off at the same volume.

   Full marks. Well done.

   Model answer: It starts at the origin and is less steep at the start; it levels off at the same volume, because the same mass of magnesium (the limiting reactant) makes the same amount of hydrogen.

6. 02.1[1 mark]

   The student changed the concentration of the sodium thiosulfate solution and timed how long the cross took to disappear.

   Give one variable the student should control.

   Your answer

   the volume of hydrochloric acid

   1/1 self-marked

   • Any one of: volume of (hydrochloric) acid; concentration of acid; total volume (of the mixture); temperature; the same cross (or flask, or person judging)

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: The volume of hydrochloric acid (or its concentration, the total volume, the temperature, the same cross).

7. 02.2[1 mark]

   Table 3 shows your three times for 40 g/dm³ sodium thiosulfate.

   Which trial is anomalous?

   Select one row.

   Table 3

   Trial	Time for the cross to disappear in s
   1	23
   2	14
   3	23

   Your answer

   trial 2

   1/1

   • Trial 2: much shorter than the other two, which agree

   Full marks. Clear and precise. Trial 2 (14 s) is far from the other two, which agree with each other.

   Model answer: Trial 2. It is much shorter than the other two, which agree with each other.

8. 02.3[3 marks]

   The rate of the reaction can be found using: rate = 1 ÷ mean time

   Calculate the rate of the reaction from the results in Table 3.

   Do not use the anomalous result.

   Your answer

   0.0435

   Working: mean time = (23 + 23) ÷ 2 = 23 s rate = 1 ÷ 23 = 0.0435 /s

   3/3

   • mean time = (23 + 23) ÷ 2 = 23 s (anomaly left out)
   • rate = 1 ÷ their mean time
   • rate = 0.0435 /s (allow an answer consistent with their mean)

   Full marks. Well done.

   Model answer: mean time = (23 + 23) ÷ 2 = 23 s rate = 1 ÷ 23 = 0.0435 /s

9. 02.4[2 marks]

   The higher the concentration of sodium thiosulfate, the shorter the time for the cross to disappear.

   Explain why increasing the concentration increases the rate of the reaction.

   Your answer

   There are more particles in the same volume so there are more frequent collisions.

   2/2 self-marked

   • More (reactant) particles in the same volume
   • So collisions are more frequent (more collisions per second)

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: There are more particles in the same volume, so the particles collide more frequently (more collisions per second), so the rate increases.

10. 02.5[1 mark]

    Another student’s stopclock shows times to 0.01 s. They recorded each time to the nearest second.

    Suggest why recording the times to 0.01 s would not be sensible.

    Your answer

    Human reaction time is much more than 0.01 s, so the extra digits mean nothing.

    1/1 self-marked

    • Human reaction time (about 0.2–0.4 s) / judging when the cross disappears is much less precise than 0.01 s

    Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

    Model answer: The student’s reaction time in starting and stopping the clock (about 0.2–0.4 s), and judging when the cross has gone, are much bigger than 0.01 s.

11. 02.6[1 mark]

    In the standard method the whole set of five concentrations is done three times.

    Give one reason for repeating the investigation.

    Your answer

    So that a mean can be calculated, which reduces the effect of random errors.

    1/1 self-marked

    • To calculate a mean, which reduces the effect of random errors; or to identify anomalies; or to check the results are repeatable

    Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

    Model answer: So a mean can be calculated, which reduces the effect of random errors (or to spot anomalous results).

12. 02.7[2 marks]

    The reaction between sodium thiosulfate and hydrochloric acid produces sulfur dioxide gas, which is toxic.

    Give one harm sulfur dioxide could cause and one way to reduce the risk.

    Your answer

    It can cause breathing difficulties or an asthma attack. Use a well ventilated room and pour the mixture away straight after timing.

    2/2 self-marked

    • Harm: breathing difficulties / irritates the lungs or airways / can trigger an asthma attack (ignore "toxic" or "harmful" on its own)
    • Control: well-ventilated room / open windows / fume cupboard / pour the mixture away (with plenty of water) straight after timing / do not lean over the flask

    Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

    Model answer: It can cause breathing difficulties and trigger an asthma attack. Work in a well-ventilated room and pour each mixture away with plenty of water as soon as it has been timed.

13. 03[6 marks]

    A student wants to investigate how the concentration of sodium thiosulfate solution affects the rate of its reaction with hydrochloric acid.

    The student has: sodium thiosulfate solution, water, hydrochloric acid, measuring cylinders, a conical flask, a piece of paper with a black cross on it and a stopclock.

    Plan an investigation to find out how the concentration of sodium thiosulfate affects the rate of this reaction.

    Your answer

    1. Measure 10 cm³ of sodium thiosulfate solution and 40 cm³ of water with a measuring cylinder and pour them into a conical flask. 2. Put the flask on a piece of paper with a black cross drawn on it. 3. Measure 10 cm³ of hydrochloric acid in a 10 cm³ measuring cylinder. 4. Add the acid to the flask, swirl, and start the stopclock at the same time. 5. Look down through the top of the flask and stop the clock when the cross can no longer be seen. Record the time. 6. Repeat with different concentrations of sodium thiosulfate: 20 cm³ thiosulfate + 30 cm³ water, and so on, keeping the total volume at 50 cm³. 7. Keep the volume and concentration of the acid and the temperature the same. 8. Repeat each concentration three times and calculate a mean time; rate = 1 ÷ mean time.

    6/6 self-marked

    • Measure a (known) volume of sodium thiosulfate solution with a measuring cylinder
    • Put the thiosulfate in a conical flask
    • Put the flask on the paper with the black cross
    • Add (a measured volume of) hydrochloric acid and start the stopclock
    • Measure the time for the cross to become no longer visible (look down through the top)
    • Repeat for different concentrations of sodium thiosulfate (dilute with water, the same total volume)
    • Repeat each concentration and find a mean
    • Control variables: the volume and concentration of the acid; the total volume; the temperature; the same cross
    • Rate = 1 ÷ time, or plot a graph of time (or 1 ÷ time) against concentration

    Self-marked: Level 3, 6 of 6 marks. You ticked 9 of 9 indicative points. This is your own judgement against the mark scheme.

    Model answer: 1. Measure 10 cm³ of sodium thiosulfate solution and 40 cm³ of water with a measuring cylinder and pour them into a conical flask. 2. Put the flask on a piece of paper with a black cross drawn on it. 3. Measure 10 cm³ of hydrochloric acid in a 10 cm³ measuring cylinder. 4. Add the acid to the flask, swirl, and start the stopclock at the same time. 5. Look down through the top of the flask and stop the clock when the cross can no longer be seen. Record the time. 6. Repeat with different concentrations of sodium thiosulfate: 20 cm³ thiosulfate + 30 cm³ water, and so on, keeping the total volume at 50 cm³. 7. Keep the volume and concentration of the acid and the temperature the same. 8. Repeat each concentration three times and calculate a mean time; rate = 1 ÷ mean time.

Unseen data: the same skills, new apparatus

1. 01[1 mark]

   A student added manganese(IV) oxide to hydrogen peroxide solutions of different concentrations. They timed how long it took to collect 20 cm³ of oxygen.

   Table 1

   Concentration of hydrogen peroxide in mol/dm³	Time to collect 20 cm³ of oxygen in s
   0.25	168
   0.5	109
   1	81
   2	66

   The student says: “The rate of reaction is directly proportional to the concentration of hydrogen peroxide.”

   Do the results in Table 1 support this?

   Tick (✓) one box.

   • Yes: each time the concentration doubles, the time halves, so the rate doubles.
   • Yes: as the concentration increases, the time decreases.
   • ✓ No: when the concentration doubles, the time does not halve. (your choice)
   • No: the more concentrated solutions took longer.

   Your answer

   No: when the concentration doubles, the time does not halve.

   1/1

   • No: when the concentration doubles, the time does not halve.

   Full marks. Clear and precise.

   Model answer: No: when the concentration doubles, the time does not halve. Each doubling of the concentration divides the time by 1.54, 1.35, 1.23: well under 2, so the rate less than doubles.

2. 02[4 marks]

   A student put a flask of hydrochloric acid and marble chips on a balance. They recorded the mass of carbon dioxide lost every 30 s.

   Figure 2 shows their results, a curve of best fit and a tangent (dashed) drawn at 60 s.

   The tangent passes through the two marked points: (10 s, 0.36 g) and (110 s, 0.88 g).

   Determine the rate of reaction at 60 s.

   Give your answer to 2 significant figures and give the unit.

   Your answer

   0.0052 g/s

   Working: (0.88 - 0.36) / (110 - 10)

   4/4

   • Change in mass ÷ change in time between two points on the tangent
   • Rate = 0.0052
   • Given to 2 significant figures
   • Unit: g/s

   Full marks. That is exactly what the mark scheme is looking for.

   Model answer: From the tangent: (110 s, 0.88 g) and (10 s, 0.36 g). rate = (0.88 − 0.36) ÷ (110 − 10) = 0.0052 g/s

3. 03[2 marks]

   A student timed how long it took to collect 50 cm³ of hydrogen from zinc granules and sulfuric acid. They did the experiment four times.

   Table 3

   Repeat	Time in s
   1	34
   2	52
   3	49
   4	52

   Calculate the mean time.

   Do not include the anomalous result.

   Your answer

   51

   Working: 52 + 49 + 52 = 153 153 / 3 = 51

   2/2

   • Leaves out repeat 1 and adds the other three
   • Mean = 51 s

   Full marks. Your answer meets all the points for this question.

   Model answer: Repeat 1 (34 s) is anomalous. mean = (52 + 49 + 52) ÷ 3 = 51 s