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A series of experiments is carried out with compounds C and D - AQA - A-Level Chemistry - Question 3 - 2017 - Paper 2

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A series of experiments is carried out with compounds C and D. Using the data obtained, the rate equation for the reaction between the two compounds is deduced to be... show full transcript

Worked Solution & Example Answer:A series of experiments is carried out with compounds C and D - AQA - A-Level Chemistry - Question 3 - 2017 - Paper 2

Step 1

3.1 Calculate a value for the rate constant at this temperature and give its units.

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Answer

To calculate the rate constant, we use the rate equation:

k=rate[C][D]k = \frac{rate}{[C][D]}
Substituting the values:

  • Rate = 3.1×10−33.1 \times 10^{-3} dm³ s⁻¹
  • [C] = 0.48 mol dm⁻³
  • [D] = 0.23 mol dm⁻³

Thus,

k=3.1×10−3 dm3 s−10.48 mol dm−3×0.23 mol dm−3k = \frac{3.1 \times 10^{-3} \text{ dm}^3 \text{ s}^{-1}}{0.48 \text{ mol dm}^{-3} \times 0.23 \text{ mol dm}^{-3}}

Calculating the denominator:
0.48×0.23=0.1104 mol2 dm−60.48 \times 0.23 = 0.1104 \text{ mol}^2 \text{ dm}^{-6}

Now substituting back, we get:

k=3.1×10−30.1104=2.81×10−2 dm3 mol−1 s−1k = \frac{3.1 \times 10^{-3}}{0.1104} = 2.81 \times 10^{-2} \text{ dm}^3 \text{ mol}^{-1} \text{ s}^{-1}

Thus, the rate constant is approximately 2.81 x 10^-2 dm³ mol⁻¹ s⁻¹.

The units of k are:

  • dm³ mol⁻¹ s⁻¹.

Step 2

3.2 Use this equation and your answer from Question 3.1 to calculate a value, in kJ mol^-1, for the activation energy of this reaction at 25 °C.

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Answer

From Question 3.1, we found that the rate constant k is approximately 2.81×10−22.81 \times 10^{-2} dm³ mol⁻¹ s⁻¹.
Using the equation:

ln⁡k=−EaRT+ln⁡A\ln k = -\frac{E_a}{RT} + \ln A

Rearranging to find EaE_a:

Ea=−RT(ln⁡k−ln⁡A)E_a = -RT(\ln k - \ln A)

Substituting the known values:

  • R = 8.31 J K⁻¹ mol⁻¹ = 0.00831 kJ K⁻¹ mol⁻¹
  • T = 25 °C = 298 K
  • A = e^{16.9} (to find ln A)
  • k = 2.81×10−22.81 \times 10^{-2}

Calculating ln⁡k\ln k:
ln⁡(2.81×10−2)≈−3.58\ln(2.81 \times 10^{-2}) \approx -3.58

Now substituting these values into the activation energy equation:

Ea=−0.00831×298(−3.58−16.9)E_a = -0.00831 \times 298 (-3.58 - 16.9)

Calculating the value:
Ea=−0.00831×298×(−20.48)≈51.1 kJ mol−1E_a = -0.00831 \times 298 \times (-20.48) \approx 51.1 \text{ kJ mol}^{-1}

Thus, the activation energy is approximately 51.1 kJ mol⁻¹.

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