Finding Activation Energy (AQA A Level Chemistry): Revision Note

Calculate the activation energy of a reaction which takes place at 400 K, where the rate constant of the reaction is 6.25 x 10^-4 s^-1.

A = 4.6 x 10¹³ and R = 8.31 J K^-1 mol^-1.

Answer:

$\mathbf{ln}\mathit{k}\mathbf{=}\mathbf{ln}\mathit{A}\mathbf{-}\frac{{\mathit{E}}_{\mathit{a}}}{\mathit{R}\mathit{T}}$

• Rearrange the equation for E_a:

  • $\frac{{\mathit{E}}_{\mathit{a}}}{\mathit{R}\mathit{T}}\mathbf{+}\mathbf{ln}\mathit{k}\mathbf{=}\mathbf{ln}\mathit{A}$

  • $\frac{{\mathit{E}}_{\mathit{a}}}{\mathit{R}\mathit{T}}\mathbf{=}\mathbf{ln}\mathit{A}\mathbf{-}\mathbf{ln}\mathit{k}$

  • ${\mathit{E}}_{\mathbf{a}}\mathbf{=}\mathbf{(}\mathbf{ln}\mathbf{A}\mathbf{-}\mathbf{ln}\mathbf{k}\mathbf{)}\mathbf{\times }\mathit{R}\mathit{T}$

• Insert the values from the question:

  • ${\mathit{E}}_{\mathbf{a}}\mathbf{=}\mathbf{(}\mathbf{ln}\mathbf{4}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{13}}\mathbf{-}\mathbf{ln}\mathbf{6}\mathbf{.}\mathbf{25}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{)}\mathbf{\times }\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{31}\mathbf{\times }\mathbf{400}\mathbf{)}$

  • ${\mathit{E}}_{\mathbf{a}}\mathbf{=}\mathbf{(}\mathbf{31}\mathbf{.}\mathbf{4597}\mathbf{-}\left(-7.3778\right)\mathbf{)}\mathbf{\times }\mathbf{3324}$

  • E_a = 129095.85 J

• Convert from J to kJ:

  • E_a = 129 kJ

1. Complete the following table.

2. Plot a graph of ln k against 1/T.

3. Using the graph and the following equation, calculate the activation energy, Ea, for this reaction.

$\mathbf{ln}\mathit{k}\mathbf{=}\frac{\mathbf{-}{\mathit{E}}_{\mathbf{a}}}{\mathit{R}}\frac{\mathbf{1}}{\mathit{T}}\mathbf{+}\mathbf{ln}\mathit{A}$

4. Calculate the Arrhenius constant, A, for this reaction.

Answers:

1. The completed table is:

2. The graph of ln k against 1/T is:

3. To calculate the activation energy, E_a:

   

   • Gradient = -E_a / R = -3666.6

   • E_a = -(-3666.6 x 8.31) = 30, 469 J mol^-1

   • E_a = -(-3666.6 x 8.31) = 30.5 kJ mol^-1

4. To calculate the Arrhenius constant, A:

   • Choose a point on the graph, e.g (2.60 x 10^-3, -7.0)

   

   • Substitute values into the equation $\mathbf{ln}\mathit{k}\mathbf{=}\frac{\mathbf{-}{\mathit{E}}_{\mathbf{a}}}{\mathit{R}}\frac{\mathbf{1}}{\mathit{T}}\mathbf{+}\mathbf{ln}\mathit{A}$

   • -7.0 = (-3666.6 x 2.60 x 10^-3) + ln A

   • -7.0 = -9.53 + ln A

   • ln A = 2.53

   • A = $e^{2.53}$

   • A = 12.55

You are not required to learn these equations. However, you do need to be able to rearrange them, and knowing them helps you understand the effects of temperature on the rate constant.