Determining Activation Energy & the Arrhenius Factor (HL) (DP IB Chemistry)

The data in the table below was collected for a reaction.

1. Complete the table

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

3. Use this to calculate:

   1. the activation energy, E_a, in kJ mol^-1

   2. the Arrhenius factor, A, of the reaction

Answer 1:

Answer 2:

• Choose a suitable scale for the axes

  • The scale does not need to start from (0,0)

  • The plotted points should fill as much of the graph provided as possible

Answer 3a:

• To calculate E_a:

  • Gradient = = –3666.6

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

• Convert to kJ mol^-1:

  • E_a = 30.5 kJ mol^-1

Answer 3b:

• Choose a point on the graph:

  • (2.60 x 10^-3, –7)

• Use the logarithmic form of the Arrhenius equation

• From the point chosen from the graph:

  • ln k = –7.0

  • = 2.60 x 10^-3

  • = = –3666.6

• Substituting these values:

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

  • –7.0 = –9.53 + ln A

• Rearranging gives:

  • ln A = –7.0 + 9.53 = 2.53

  • A = e^2.53

  • A = 12.55

• You are not required to learn these equations as they are given in the Data Booklet

• However, you do need to be able to rearrange them, and knowing them is helpful in understanding the effects of temperature on the rate constant.