a common rule of thumb is that the rate of reaction doubles for each 10c rise in temperature. What activation energy would this suggest at a temperature of 25c
Answers
The Arrhenius equation will tell us that the activation energy in this case must be 52.4 kJ/mol.
Explanation:
The Arrhenius equation says that the dependence of the rate constant
k
on temperature reads
k
=
A
e
−
E
a
R
T
For the sake of this problem, let us say that
k
1
is the initial value of the constant at a temperature
T
1
, and
k
2
is the value at temperature
T
2
To double the reaction rate, we must double the value of k, i.e. the ratio
k
2
k
1
must equal 2.
This will require the right hand side of the Arrhenius equation to change to
A
e
−
E
a
R
T
2
from
A
e
−
E
a
R
T
1
all of which can be written in ratio forms as
k
2
k
1
=
e
(
−
E
a
R
)
(
1
T
2
−
1
T
1
)
Taking the natural log of each side:
ln
2
=
−
(
E
a
R
)
(
1
T
2
−
1
T
1
)
or
E
a
=
ln
2
(
R
)
1
T
2
−
1
T
1
Using 308K for
T
2
and 298K for
T
1
E
a
=
0.693
(
8.314
)
1
308
−
1
298
E
a
=
52.4
k
J
m
o
l