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This question is about rates of reaction - AQA - A-Level Chemistry - Question 10 - 2021 - Paper 2

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This question is about rates of reaction. Iodine and propanone react together in an acid-catalysed reaction CH₃C(O)CH₃(aq) + I₂(aq) → CH₃C(O)I(aq) + HI(aq) A stude... show full transcript

Worked Solution & Example Answer:This question is about rates of reaction - AQA - A-Level Chemistry - Question 10 - 2021 - Paper 2

Step 1

Suggest why the 1.0 cm³ portions of the reaction mixture are added to an excess of Na₂S₂O₃ solution.

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Answer

The sodium hydrogencarbonate solution neutralises the acid catalyst, so it stops the reaction.

Step 2

Suggest why the order of this reaction with respect to propanone can be ignored in this experiment.

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Answer

The concentration/amount of propanone is much larger than 200 times larger than the concentration of iodine. Therefore, the change in concentration in propanone is negligible.

Step 3

Use the results in Table 5 to draw a graph of volume of sodium thiosulfate solution against time.

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Answer

Plot the data points from Table 5 onto a graph with the y-axis representing the volume of sodium thiosulfate (in cm³) and the x-axis representing time (in minutes). Draw a line of best fit that accurately represents the data, ensuring that the line captures the general trend without being skewed by any anomalous points.

Step 4

Explain how the graph shows that the reaction is zero-order with respect to iodine.

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Answer

The graph is a straight line with a constant gradient. This indicates that the rate of reaction does not change as the concentration of iodine changes, suggesting that the reaction is zero-order concerning iodine.

Step 5

Use Figure 8 to calculate a value for the activation energy (Ea) in kJ mol⁻¹.

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Answer

To calculate the activation energy, we can use the slope from the graph. The gradient is given by:

extslope=EaR ext{slope} = -\frac{E_a}{R}

Using R = 8.31 J K⁻¹ mol⁻¹, we can find Ea in joules and then convert it to kJ. The calculated slope from the graph is approximately −2412.3. Thus,

Ea=slope×R=(2412.3)×8.31E_a = -\text{slope} \times R = -(-2412.3) \times 8.31

Calculating this gives us:

Ea=20000.41extJmol1=20.00extkJmol1E_a = 20000.41 ext{ J mol}^{-1} = 20.00 ext{ kJ mol}^{-1}.

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