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

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

Temperature / K		Time, t / s	Rate constant, k / s-1	ln k
310	3.23 x 10-3	57		–9.2
335		31	3.01 x 10-4	–8.1
360	2.78 x 10-3	19	5.37 x 10-4	–7.5
385	2.60 x 10-3	7	9.12 x 10-4

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:

Temperature / K		Time, t / s	Rate constant, k / s-1	ln k
310	3.23 x 10-3	57	1.01 x 10-4	–9.2
335	2.99 x 10-3	31	3.01 x 10-4	–8.1
360	2.78 x 10-3	19	5.37 x 10-4	–7.5
385	2.60 x 10-3	7	9.12 x 10-4	–7.0

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

ln k	=			+	ln A
y	=	m	x	+	c

• 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