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Rates of Reaction: Question 8

Syllabus 6.2

Structured Core 6 marks

A cleaning-products factory tests whether manganese(IV) oxide can be used as a catalyst to speed up the decomposition of hydrogen peroxide solution into water and oxygen gas: 2H2O2(aq)→2H2O(l)+O2(g)2\text{H}_2\text{O}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g)

A technician sets up two trials, each using the same volume and concentration of hydrogen peroxide solution, at the same temperature:

  • Trial 1: the hydrogen peroxide solution alone, with no catalyst added.
  • Trial 2: the same hydrogen peroxide solution, with a small measured mass of powdered manganese(IV) oxide added at the start.

The volume of oxygen gas collected in a gas syringe is recorded every 20 s20\text{ s}.

Time / s 0 20 40 60 80 100 120
Volume of O2 in Trial 1 / cm3 0 4 8 11 14 16 18
Volume of O2 in Trial 2 / cm3 0 30 46 54 58 60 60

(a) State the factor being investigated by comparing Trial 1 and Trial 2. [1]

(b) Calculate the average rate of reaction in Trial 2 between t=0t=0 and t=20t=20 seconds. Give the units of your answer. [2]

(c) State what the total volume of oxygen gas collected by the end of each trial will eventually be, once both reactions are complete, and explain your answer. [2]

(d) The technician weighs the manganese(IV) oxide before adding it to Trial 2, and weighs it again (after filtering and drying it) once the reaction has finished. Suggest what this second measurement would show, and explain why this shows that manganese(IV) oxide is acting as a catalyst rather than as a reactant in this reaction. [1]

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Worked solution

Part (a): Identifying the factor investigated

Everything about the two trials is kept the same (the volume and concentration of hydrogen peroxide solution, and the temperature) except that Trial 2 has a small measured mass of powdered manganese(IV) oxide added at the start, while Trial 1 has none. The factor being investigated is therefore whether a catalyst is present.

Part (b): Calculating the average rate in Trial 2

Between t=0t=0 and t=20t=20 seconds, the volume of oxygen collected in Trial 2 rises from 00 to 30 cm330\text{ cm}^3:

average rate=change in volumechange in time=30−020−0=3020\text{average rate} = \frac{\text{change in volume}}{\text{change in time}} = \frac{30-0}{20-0} = \frac{30}{20}

average rate=1.5 cm3/s\text{average rate} = 1.5\text{ cm}^3\text{/s}

Part (c): Why both trials reach the same final volume

Trial 2’s readings have already levelled off at 60 cm360\text{ cm}^3 by t=100 st=100\text{ s}, showing its reaction is complete. Trial 1 is still producing gas at t=120 st=120\text{ s} (only 18 cm318\text{ cm}^3 so far), so it has not finished within the time shown, but it uses exactly the same volume and concentration of hydrogen peroxide as Trial 2, so it contains the same total number of moles of hydrogen peroxide to decompose.

Manganese(IV) oxide does not appear on either side of the balanced equation 2H2O2(aq)→2H2O(l)+O2(g)2\text{H}_2\text{O}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g). It is not a reactant, so it cannot add to or take away from the amount of hydrogen peroxide available. This means Trial 1 will eventually also reach a total volume of 60 cm360\text{ cm}^3 of oxygen, just far more slowly than Trial 2. The catalyst changes only the rate at which the oxygen is released, not the total amount that is ever produced.

Part (d): Evidence that manganese(IV) oxide is a catalyst

If the manganese(IV) oxide is filtered out, dried, and weighed again once the reaction is finished, its mass would be found to be unchanged from its starting mass. A true reactant would be chemically converted into products and would be used up, so its mass would decrease as the reaction proceeded. Because the manganese(IV) oxide’s mass stays the same, it has not been chemically changed by the reaction. It has simply provided a quicker pathway for the hydrogen peroxide to decompose and can, in principle, be recovered and reused. This unchanged mass is exactly what identifies it as a catalyst rather than a reactant.

Final answers

  • (a) Whether a catalyst (manganese(IV) oxide) is present.
  • (b) 1.5\boxed{1.5} cm3/s.
  • (c) Both trials eventually produce 60\boxed{60} cm3 of oxygen: the catalyst changes the rate, not the total amount, since manganese(IV) oxide is not a reactant and the same amount of hydrogen peroxide is used in both trials.
  • (d) The mass of manganese(IV) oxide is unchanged before and after. Showing it is not used up, so it is acting as a catalyst, not a reactant.