Hydrogen peroxide, H2O2, decomposes to produce water and oxygen according to the following balanced equation: 2H2O2(l) → 2H2O(l) + O2(g) 5.1 The activation energy (E_A) for this reaction is 75 kJ and the heat of reaction (ΔH) is -196 kJ - NSC Physical Sciences - Question 5 - 2016 - Paper 2

The completed potential energy diagram should reflect the following:

• The reactants start at a potential energy level corresponding to the energy of H2O2.
• The peak of the graph represents the activation energy (75 kJ) at the activated complex.
• The products (water and oxygen) should be located lower on the y-axis at -196 kJ.

The reaction path with manganese dioxide added should show a lower energy activation pathway leading to the activated complex, depicted by a dotted line peaking at the same activation energy level but lowering the overall energy of the reaction.

Manganese dioxide serves as a catalyst to lower the activation energy needed for the decomposition of hydrogen peroxide. According to collision theory, this increases the frequency of successful collisions between the reacting molecules, thereby accelerating the reaction rate.

The average rate of reaction can be calculated using the formula:

$ext{Average rate} = \frac{\Delta V}{\Delta t}$

For graph A, this would be the change in volume of oxygen produced between these times divided by the time interval.

Using the volume of oxygen produced from graph A, you can calculate the moles of oxygen using the formula:

$\text{Moles of } O_2 = \frac{\text{Volume}}{24 \, ext{dm}^3/\text{mol}}$

Then, using the stoichiometry of the balanced equation, convert moles of oxygen to moles of hydrogen peroxide and finally calculate the mass using the molar mass of hydrogen peroxide (H2O2).

To determine this, compare the moles of hydrogen peroxide deduced from the respective graphs based on the oxygen output. The mass needed for graph B will depend on whether the volume of oxygen produced is more or less than that for graph A.

The curve that represents 1 mol of oxygen gas produced at 120 °C is likely to be the one that shows higher kinetic energy distribution.

The curve that represents 2 moles of oxygen gas produced at 90 °C would typically align with increased molecular distribution at that kinetic energy compared to the curve representing a single mole.