1. Recap
1. Two species, P and Q, react together according to the following equation. P + Q → R The accepted mechanism for this reaction is P + P P2 fast P2 + Q → R + P slow What is the order with respect to P and Q? P Q A. 1 B. 2 C. D.

2. Arrhenius Equation

3. Activation Energy
Activation energy is the minimum energy to colliding particles need in order to react
You can think of it as:
The energy required to begin breaking bonds
The energy that particles need to overcome the mutual repulsion of their electron shells.
Can you think of an analogy?

4. Activation Energy

5. The Arrhenius Equation
We met the rate constant, k, a couple of lessons ago
The Arrhenius Equation tells us how k is related to a variety of factors:
Where:
k is the rate constant
Ea is the activation energy
T is the temperature measured in Kelvins
R is the gas constant, J mol-1 K-1.
e is Euler’s number
A is the ‘frequency factor’

6. Rearranging Arrhenius
If we take logs of both sides, we can re-express the Arrhenius equation as follows:
This may not look like it, but is actually an equation in the form y = mx + c
Where:
‘y’ is ln k
‘m’ is -Ea/R
‘x’ is 1/T
‘c’ is ln A

7. To determine Ea Experimentally: (Assuming we know the rate equation)
Measure the rate of reaction at various different temperatures.
Keeping all concentrations the same
Calculate the rate constant, k, at each temperature.
Plot a graph of ln k (y-axis) vs 1/T (x-axis)
The gradient of this graph is equal to ‘-Ea/R’, this can be rearranged to calculate Ea.

8. Arrhenius Equation A plot of ln k vs 1/T will give a straight line
The slope of line will equal Ea/R
The activation energy may be found by multiplying the slope by “R”

9. Graphing with Arrhenius
Rate constants for the reaction
CO(g) + NO2(g)  CO2(g) + NO(g)
Were measured at four different temperatures. Plot the data to determine activation energy in kJ/mol.

10. Graphing
318
0.332
308
0.184
298
0.101
288
0.0521
T (Kelvin)
k (M1 s1)‏

11. Graphing Steps: Make a column of ln k data
Make a column of inverse temp (1/T)
Plot ln k vs 1/T
Calculate the slope
Multiply slope by R

12. Arrhenius Another Way!
Another useful arrangement of the Arrhenius equation enables calculation of: 1) Ea (with k at two different temps)
2) the rate constant at a different temperature (with Ea, k and temps)

13. Arrhenius Another Way!
Use the data to calculate activation energy of the reaction mathematically
7.0 x 101
500
6.1 x 102
450
2.9 x 103
400
k (s1)
T (Kelvin)

14. Arrhenius

15. Problems
Find the activation energy (in kJ/mol) of the reaction if the rate constant at 600K is 3.4 M-1s-1 and 31.0 at 750K.
Find the rate constant if the temperature is 289K, Activation Energy is 200kJ/mol and pre-exponential factor is 9 M-1s-1
Find the new rate constant at 310K if the rate constant is 7 M-1s-1 at 370K, Activation Energy is 900kJ/mol 
Calculate the activation energy if the pre-exponential factor is 15 M-1s-1, rate constant is 12M-1s-1and it is at 22K
Find the new temperature if the rate constant at that temperature is 15M-1s-1 while at temperature 389K the rate constant is 7M-1s1,  the Activation Energy is 600kJ/mol