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

The balanced equation for the decomposition of hydrogen peroxide can be represented as:

$2 H_2O_2 \rightarrow 2 H_2O + O_2$

This shows that two molecules of hydrogen peroxide decompose to produce two molecules of water and one molecule of oxygen gas.

A suitable apparatus for this experiment would comprise a conical flask containing the hydrogen peroxide solution and a filter funnel with manganese(IV) oxide catalyst. The setup would include a delivery tube leading to an inverted graduated cylinder to collect the gas produced. The diagram should clearly label each component, including the flask, delivery tube, graduated cylinder, and catalyst.

When plotting the graph of gas volume against time, the x-axis represents time in minutes, while the y-axis represents the volume of gas produced in cm³. Mark the data from the table accurately and draw a smooth curve through the points, ensuring that the graph starts at the origin (0,0).

The graph is steepest at the beginning because the rate of reaction is highest shortly after the catalyst is added and the hydrogen peroxide begins to decompose rapidly. This is due to the high concentration of the reactants leading to more frequent collisions that result in product formation.

To find the instantaneous rate at 5 minutes, calculate the slope of the tangent of the graph at that point. If the volumes at 4 and 6 minutes are 65.5 cm³ and 73.0 cm³ respectively, the rate can be determined using:

$\text{Rate} = \frac{V_{6min} - V_{4min}}{6 - 4} = \frac{73.0 - 65.5}{2} = 3.75 \text{ cm}^3 \text{ min}^{-1}$

The total mass of gas produced can be determined from the final volume of gas collected at the end of the experiment. At 12 minutes, the volume of gas produced is 78.0 cm³. The mass of oxygen can be calculated using the molar volume of gas at room temperature and pressure, considering the ideal gas law. Likewise, if using the ideal gas approximation, we can calculate:

Using the molar mass of O₂ (32 g/mol):

$\text{Mass} = \frac{V}{1000} \times \text{Molar Mass}$

Substituting for volume:

$\text{Mass} = \frac{78.0}{1000} \times 32 = 2.496 \text{ grams}$