Three experiments, A, B and C, are carried out to investigate some of the factors that affect the rate of decomposition of hydrogen peroxide, H₂O₂(t) - NSC Physical Sciences - Question 5 - 2022 - Paper 2

Curve Y represents experiment C. This is because it shows a higher kinetic energy distribution corresponding to the increased temperature of 35 °C in experiment C compared to the other experiments.

The volume of O₂(g) collected in the syringe is 320 cm³.

To calculate the mass of water produced, first find the moles of O₂ produced:

Using the molar volume, $n(O₂) = \frac{V}{V_m} = \frac{320 \text{ cm}^3}{24000 \text{ cm}^3 \cdot \text{mol}^{-1}} = 0.01333 \text{ mol}$

From the balanced equation, 2 moles of H₂O are produced for every mole of O₂. Therefore, $n(H₂O) = 2 \times n(O₂) = 2 \times 0.01333 = 0.02667 \text{ mol}$

Now, calculate the mass of H₂O: $m = n \times M = 0.02667 \text{ mol} \times 18 \text{ g/mol} = 0.480 g.$

The rate of production of oxygen gas for the interval 30 s to 36 s is calculated from the graph. The mass changes from 0.8 g to 0.9 g during this period. So, the rate is: $Rate = \frac{\Delta m}{\Delta t} = \frac{0.9 g - 0.8 g}{6 s} = 0.01667 g/s.$

Converting this to moles, using the molar mass of O₂ (32 g/mol), gives: $Rate = \frac{0.01667 g/s}{32 g/mol} = 0.00052 mol/s.$

The rate of the reaction in the interval 3 to 9 s will be GREATER THAN the rate in the interval 9 to 20 s. This is because the reaction is expected to slow down as it progresses due to the depletion of reactants.

The average rate of decomposition of hydrogen peroxide is provided as 2.1 x 10⁻³ mol·s⁻¹. Using this information, and the earlier calculated values, we can determine the time t as follows:

From the graph, if the mass of oxygen at time t can be read from the graph, we can calculate the time corresponding to that average rate:

This gives approximately t = 26.67 seconds.