Hydrogen peroxide solution decomposes to form water and oxygen - AQA - A-Level Chemistry - Question 4 - 2022 - Paper 3

A suitable tangent should be drawn touching the curve at the point where [H2O2] is 0.05 mol dm−3. This tangent can be drawn by ensuring it touches the curve at the designated concentration, and it should be able to extend to both the left and the right without crossing the curve.

The gradient of the tangent can be calculated using the formula:

$Gradient = \frac{change \ in \ concentration}{change \ in \ time}$

Using the values obtained from the tangent points, for example: if the tangent rises by 0.0015 mol dm−3 for a time change of 10 seconds, the gradient would be 0.0015 / 10 = 0.00015 mol dm−3 s−1.

From the experiment, we have:

• Vt = 20 cm3
• Vmax = 100 cm3 To find [H2O2], we first need the concentration at that volume:

$[H2O2] = [H2O2]_{initial} \times (\frac{Vt}{Vmax})$ If [H2O2]_{initial} was 0.08 mol dm−3, substituting gives: $[H2O2] = 0.08 \times (\frac{20}{100}) = 0.016 \ mol \ dm^{-3}.$

To plot the data from Table 5, firstly, identify the concentration values listed in the table (for instance, 0.02, 0.03, 0.05, etc.) and their corresponding rates. Each concentration should be placed along the x-axis while the rates are plotted along the y-axis. Ensure to accurately reflect the relationship shown in the table.

The order of reaction can be determined by analyzing the shape of the line drawn in Figure 2. If the line is straight and passes through the origin, this indicates a first-order reaction. If it curves upwards or downwards, it may suggest a second-order or zero-order reaction respectively. A straight line through the origin demonstrates that the rate is directly proportional to the concentration of [H2O2].