16.3 ACTIVATION ENERGY (and The ARRHENIUS EQUATION)
In 1889, Svante Arrhenius demonstrated that the rate constants of many chemical reactions vary
with ___________________ in accordance with the following equation. As the temperature
increases, the value of the rate constant increases _____________________.
Equ’n 1:
OR Taking natural log of both side gives Equ’n 2:
*Gas constant (R) = __________ J mol-1 K-1
*A = _______________ constant (related to the _________________ of the reactants at the
point of collision)
*Note that equation 2 is in the form y =_____________ and will therefore graph a
_______________ line.
That is:
y = ln k
; m = slope = -
Ea
R
; x=
1
T
*To determine Ea graphically:
1) determine slope of line: m =
2) -Ea = (R) • (slope)
*Units of activation energy = kJ mol-1
or
; b = y-intercept = ln A
 ln k
1

T
slope =
 Ea
R
(i.e. “
y
”)
x
Try the Following Activity: Determining Ea Graphically
Problem: The following data were collected for the reaction: 2NO2(g)  2NO(g) + O2(g)
Rate Constant (mol dm-3 s-1)
7.8
Temperature (ºC)
400
10
410
14
420
18
430
24
440
Temperature (K)
Determine the activation energy for the reaction in kilojoules per mole.
Start by completing the following:
(*Remember ln k = ____ and 1/T = _____)
ln k
ln(7.8) = 2.05
1/T (K-1)
1/673 = 1.486 x 10-3
Next:
 plot points on graph paper provided (use ENTIRE graph!)
 draw straight line of best fit
 calculate slope of the line(show slope calculation on graph)
 calculate the activation energy using:
 Extension: What is the approximate value of ln A?
Answers:
*Slope should be around: -1.4 x 104 K-1
*Ea should be around:
1.2 x 102 kJ mol-1