The decomposition of hydrogen peroxide in the presence of a manganese(IV) oxide catalyst produces water and oxygen gas.
2H2O2(aq) → 2H2O(l) + O2(g)
0.2 g of manganese(IV) oxide granules are added to 50 cm3 of 0.1 mol dm–3 hydrogen peroxide at 20 °C. The volume of gas collected is shown on the graph as curve X.
A second experiment is carried out at 20 °C using the same mass of manganese(IV) oxide. The volume of gas collected is shown as curve Y. Which of the following conditions could result in curve Y?

Answer:

14) A

We know that the total volume of gas produced is lower in curve Y and the reaction takes place faster (initial line is very steep).
The steep line in curve Y shows that the reaction happened very quickly (the rate of reaction is higher than curve X). This indicates that the manganese (IV) oxide particle size would have been smaller. This is because the smaller the particle size, the more surface area is present and hence faster reaction (provided that mass of manganese (IV) oxide is same, which it is). The answer is either A, B or C.

We also know less gas is produced. This means there would be lesser number of moles of hydrogen peroxide. We use the formula, moles = volume x concentration (remember, volume should be in dm-3). Moles in curve X = 0.05 x 0.1 = 0.005 mol
Moles of hydrogen peroxide in option A = 0.02 x 0.2 = 0.004 mol
Moles of hydrogen peroxide in option B = 0.025 x 0.2 = 0.005 mol
Moles of hydrogen peroxide in option C = 0.05 x 0.1 = 0.005mol

Therefore, A is the answer.