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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

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{\text{rate}}{[C][D]}

Substituting in the known values:

k=3.1×10−3 mol dm−3 s−1(0.48 mol dm−3)(0.23 mol dm−3)k = \frac{3.1 \times 10^{-3} \text{ mol dm}^{-3} \text{ s}^{-1}}{(0.48 \text{ mol dm}^{-3})(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 this back into the equation for k:

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:

Rate constant = 2.81 x 10^-2 dm³ mol^-1 s^-1

Units = dm³ mol^-1 s^-1

Step 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

First, calculate ln k using the rate constant found in Question 3.1:

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

Calculating:

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

Now, we will use the Arrhenius equation:

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

Rearranging gives:

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

Substituting the values:

  • Gas constant, R = 8.31 J K^-1 mol^-1
  • Temperature, T = 298 K (25 °C)
  • ln A = 16.9

So,

Ea=−(8.31)(298)(−3.58−16.9)E_a = - (8.31)(298)(-3.58 - 16.9)

Calculating:

Ea=(8.31)(298)(−20.48) J mol−1E_a = (8.31)(298)(-20.48)\text{ J mol}^{-1}

Converting J to kJ:

Ea=(8.31)(298)(−20.48)/1000E_a = (8.31)(298)(-20.48) / 1000

Hence:

Ea=51.1 kJ mol−1E_a = 51.1 \text{ kJ mol}^{-1}

Thus, the activation energy is approximately:

Activation energy = 51.1 kJ mol^-1

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