Hydrogen peroxide solution is an oxidising reagent - Leaving Cert Chemistry - Question 3 - 2008

The balanced equation for the decomposition of hydrogen peroxide is: $2H_2O_2 (aq) \rightarrow 2H_2O (l) + O_2 (g)$

A suitable apparatus for this experiment includes a conical flask containing hydrogen peroxide, a delivery tube leading to a water-filled graduated cylinder to measure the gas produced, and a balance for weighing the manganese(IV) oxide catalyst.

To plot the graph, use the values from the table. The x-axis will represent time (in minutes), and the y-axis will represent volume (in cm³). Each point can be plotted according to the data provided, connecting the points smoothly to represent gas production over time.

The graph is steepest at the beginning because the reaction starts with a higher concentration of hydrogen peroxide, leading to a faster rate of reaction and gas production. As the reaction proceeds, the concentration of hydrogen peroxide decreases, resulting in a slower rate of gas production.

To find the instantaneous rate of gas production at 5 minutes, calculate the slope of the tangent at that point. Between the values at 5 minutes and the previous time (2.0 minutes), the rate is: $\text{Rate} = \frac{\text{Gas Volume at } 5.0\, \text{min} - \text{Gas Volume at } 2.0\, \text{min}}{5.0 - 2.0} = \frac{50.5 - 20.0}{5.0 - 2.0} = \frac{30.5}{3.0} \approx 10.17 \text{ cm}^3/\text{min}$

To calculate the total mass of gas produced, we need to convert the total volume of oxygen gas produced at 12 minutes into mass:

• Volume at 12 minutes = 78.0 cm³.
• Using the ideal gas law under standard conditions, we find the mass: 1 mole of gas (O2) at STP = 22,400 cm³. Therefore: $\text{Mass of gas} = \text{Volume} \times \left(\frac{\text{Molar Mass of O}_2}{\text{Molar Volume}}\right)$ Where the Molar Mass of O2 = 32 g/mol. Hence, $\text{Mass} = 78.0 \times \frac{32}{22400} \approx 0.111 \text{ g}$