Chemistry 12 (HL)
Unit 2 / IB Topic 16.3
Kinetics 5
Activation Energy and the Arrhenius Theory
RATE AND TEMPERATURE
Rate is dependent on temperature. This may be explained qualitatively using the collision theory:
At higher temperatures, there are more molecules with sufficient kinetic energy to overcome
the activation energy, as shown by the
Therefore, there are
shaded area (red +blue) under the curve
more successful collisions per unit time, and rate increases.
QUANTITATIVE RELATIONSHIP BETWEEN TEMPERATURE AND RATE?
One “rule of thumb” for the effect of temperature on rate is that rate DOUBLES
for every 10ºC increase in temperature.
Consider the rate expression for a reaction:
rate = k [A]m [B]n
You might also say that the value of the rate constant
reactant concentrations.
e.g.
k at 25ºC = 0.035 s
-1
k at 35ºC = 0.070 s
-1
k at 15ºC = 0.018 s
-1
increases in the same way, for any given
p. 1
Chemistry 12 (HL)
Unit 2 / IB Topic 16.3
THE ARRHENIUS EQUATION
You must be asking “Is there a more precise mathematical relationship”? Of course!
The ARRHENIUS EQUATION gives the relationship between k, temperature AND the activation
energy for a reaction.
k=
−Ea
Ae RT
This equation is
in Table 1 in the
Chemistry Data
Booklet.
EXAMPLE:
http://www.chemguide.co.uk/physical/basicrates/arrhenius.html
A more useful form of the Arrhenius equation is:
This equation is
in Table 1 in the
Chemistry Data
Booklet.
E
lnk = − a + ln A
RT
or
! –E $ ! 1 $
ln k = ## a && # & + lnA
" R % "T%
y = mx + b
p. 2
Chemistry 12 (HL)
Unit 2 / IB Topic 16.3
Graphical analysis using the Arrhenius equation:
Example:
-1
Given this data, find the activation energy for the reaction in kJ mol using a graphical method.
-1
rate constant (s )
temperature (ºC)
47
2.88 x 10
-4
4.87 x 10
-4
67
7.96 x 10
-4
87
1.26 x 10
-3
107
1.94 x 10
-3
127
temperature (K)
1/T
ln k
Draw an Arrhenius plot. Use Excel or your GC.
What do you need to do with the temperature first?
What do you need to do with the k values?
Find the gradient (slope) = -3050.2
Recall that the gradient = - Ea / R
4
Calculate Ea: Ea = -slope x R = -(-3050.2)(8.31) = 2.53 x 10 J mol
-1
= 25.3 kJ mol
-1
p. 3
Chemistry 12 (HL)
Unit 2 / IB Topic 16.3
PRACTICE PROBLEMS
1.
The rate constant, k, was determined for the reaction decomposition of hydrogen peroxide at
different temperatures. The results are given below.
temperature (K)
550
600
650
700
750
a)
b)
c)
2.
ln k
-15.6
-12.2
-9.4
-7.0
-4.9
-1
Plot a graph of lnk vs T .
Calculate the gradient, and use it to determine the value of the activation energy for
this reaction.
Without obtaining the actual value, state how you could determine the value of the
Arrhenius constant.
The rate constant for the decomposition of ethanol were measured at five different
temperatures.
CH3CHO (g) à CH4 (g) + CO (g)
rate constant k
-1
3 -1
(mol dm s )
0.011
0.035
0.105
0.343
0.789
temperature
(K)
700
730
760
790
810
Calculate the value for the energy of activation for this reaction.
p. 4