The volume of oxygen released by hydrogen peroxide solution, in the presence of granular manganese(IV) oxide (MnO2), was measured at one minute intervals until the reaction was complete - Leaving Cert Chemistry - Question 3 - 2009

To carry out this experiment, a diagram should include a reaction vessel (beaker), a delivery tube connected to the vessel, and a graduated collection vessel or gas syringe to measure the volume of gas released. The apparatus should be set up to ensure that the oxygen produced is directed into the collection vessel without any loss.

For this step, plot the volume of oxygen produced (in cm³) on the y-axis against time (in minutes) on the x-axis. The points to plot based on the data are (0, 0), (1, 38), (2, 56), (3, 67), (4, 74), (5, 78), (6, 80), (7, 80), (8, 80). Ensure axes are properly labeled and the graph is scaled to fit all points.

To find the average rate of oxygen production during the first three minutes, we first determine the change in volume over the time period. The volume at 3 minutes is 67 cm³ and at 0 minutes is 0 cm³. Thus, the change in volume is: 67 cm³ - 0 cm³ = 67 cm³. The time interval is 3 minutes. Therefore, the average rate of production is calculated as:

$ext{Average Rate} = \frac{\text{Change in Volume}}{\text{Time Interval}} = \frac{67 \text{ cm}^3}{3 \text{ min}} \approx 22.33 \text{ cm}^3\text{ per minute}$.

(i) If the reaction had been carried out at a higher temperature, the reaction rate would likely be faster. This is because increased temperature generally provides more kinetic energy to the reactant molecules, leading to more frequent and effective collisions, which can accelerate the reaction rate.

(ii) If the granules of manganese(IV) oxide were ground to a fine powder, the reaction rate would increase. A fine powder has a larger surface area compared to larger granules, allowing more collisions between the reactants and the catalyst, thus speeding up the reaction.