|
THE ARRHENIUS
EQUATION
The increase in the rate constant (k) with increase in temperature is found to obey an equation suggested by SVANTE ARRHENIUS in 1889.
This
proposes that the rate of a chemical reaction is a product of three factors.
Frequency
of collisions. Fraction
of collisions with
correct orientation. Fraction
of molecules with enough energy to
react (i.e.
Energy = EA
or greater).
You
are now ready to do the ILPAC practical A42 (and if required, the planning
exercise A43).
to as
Next, enter the values of
ln k and 1/8.31T for
two different sets of reults.
As for the graphical method, it is critical to take the 1/T
value to as many places of decimal as possible.
SOLUTION TO
Q1.
A. By simultaneous equations.
(1) (a)
ln 1.78
x 10-5
=
-EA x 0.00019
+ ln A
(b)
So -10.9363
= -0.000019 EA
+ ln A
(2) (a)
ln 1.07
x 10-4
= -EA
x 0.00018 EA
+ ln A
(b)
So
-9.1427 = -0.00018 EA
+ ln A
Multiply
1(b) by (-1)
(3)
10.9363 =
0.00019 EA -
lnA
Now
add the bottom two equations i.e. 2(b) + 3
1.7936 =
0.00001 EA +
ln A -
ln A
1.7936 = EA
0.00001
EA
= 179
360 J mol-1
= 179.36
kJ mol-1
B.
By
gradient.
Gradient
= -EA
= -
difference in ln k values
R
difference
in 1/T values
Multiply across by (-1) we can write
EA =
difference in ln k values x
R J mol-1
difference in 1/T values
EAc =
(10.9363 -
9.1427) x
8.31
(1.58
- 1.50)
x 10-3
EAc =
1.7936 x
8.31
0.00008
EA
= 186 310
J mol –1
= 186.3 kJ
mol -1
SOLUTION
TO Q2.
1nk
= -EA
1 +
lnA
R
T
Two
sets of k and T that we have are:
|
k |
lnk |
T |
1/T |
1/RT |
|
8.13
x 10-3 |
-4.8122 |
60°C 333K |
0.00300
= 3.00
x 10-3 |
0.000361 |
|
50
x 10-3 = 5.00
x 10-2 =
0.05 |
-2.9957 |
80°C 353K |
0.00283
= 2.83
x 10-3 |
0.000340 |
So
– 4.812
= -0.000361
EA +
ln A
and
-2.996
= -0.000340 EA
+ ln A
Multiply
first of these by (-1)
4.812
= 0.00036EA
- ln A
Now
add the bottom two together
1.816
= 0.000029 EA
- ln A
+ ln A
EA =
1.8165 J mol-1
0.000021
= 90800
J mol-1
=
90.8 kJ mol-1
B. By
gradient.
Gradient
= -EA
= -
difference in lnk values
R
difference
in 1/T values
Multiply across by (-1) we can write
EA =
difference in lnk values
x R J mol-1
difference in 1/T values
EA
=
(4.812 -
2.996) x
8.31 =
1.816 x
8.31
(3.003
- 2.832)
x 10-3 =
1.71 x 10 –3
EA
=
88 250 J mol –1
= 88.25 kJ
mol -1
Some
problems to try.
Q1. When gaseous hydrogen iodide decomposes
in accordance with the equation:
2HI(g)
D H2(g)
+ I2(g)
the reaction is found to be second order to hydrogen iodide. In a series of experiments, the rate constant for this reaction was determined at several different temperatures. The results obtained are shown in the table below.
|
1/T |
Temperature(T) / K |
Rate
constant (k)/dm3 mol-1
s-1 |
1nK |
|
1.58
x 10-3 |
633 |
1.78
x 10-5 |
-10.9360 |
|
1.50
x 10-3 |
666 |
1.07
x 10-4 |
-9.1427 |
|
1.43
x 10-3 |
697 |
5.01
x 10-4 |
-7.5989 |
|
1.40
x 10-3 |
715 |
1.05
x 10-3 |
-6.8590 |
|
1.28
x 10-3 |
781 |
1.51
x 10-2 |
-4.1931 |
Work out the
Activation Energy EA for this reaction. Also – EQUATIONS.
SOLUTION BY
GRADIENT.
SOLUTION BY
SIMULTANEOUS EQUATION.
This is shown on the next page (See Graph One).
The EA values obtained by the two methods are in reasonable agreement here
(i.e. about 186 kJ mol-1 from the graph
and 180 kJ mol by simultaneous equations).
SOLUTION BY
GRADIENT.
SOLUTION BY
SIMULTANEOUS EQUATION.
The graphical solution is shown on the next page (See
Graph Two). In this equation the EA
values obtained by the two methods are again in reasonable agreement (i.e. about
88.5kJ mol-1 by graphical means and 91 kJ mol by solving simultaneous
equations.
As for the graphical method, it is critical to take the 1/T
value to as many places of decimal as possible.
Q3.
Na2O5
(in CCl4)
N2O4 (in
CCl4) +
1/2O2(g)
t, °C
|
T,
K |
1/T, K-1 |
k,
s-1 |
ln k |
|
0 25 35 45 55 65 |
273 298 308 318 328 338 |
3.66
x 10-3 3.36
x 10-3 3.25
x 10-3 3.66
x 10-3 3.66
x 10-3 3.66
x 10-3 |
7.87
x 10-7 3.46
x 10-5 1.35
x 10-4 4.98
x 10-4 1.50
x 10-3 4.87
x 10-3 |
-14.055 -10.272 -8.910 -7.605 -6.502 -5.325 |
[ANSWER:
100 kJ mol-1]
Q4.
The
second-order rate constant for the reaction between ethyl bromide and hydroxide
ions in water,
C2H5Br(aq)
+ OH-(aq)
C2H5OH(aq) +
Br (aq)
was measured at several temperatures, with the
following results:
|
Temperature, °C |
25 |
30 |
35 |
40 |
45 |
50 |
|
K, mol dm-3s-1 |
8.8
x 10-5 |
1.6
x 10-4 |
2.8
x 10-4 |
5.0
x 10-4 |
8.5
x 10-4 |
1.40x10-3 |
[ANSWER:
89 kJ mol-1]
Q5.
HO(g)
+ H2(g)
H2O(g) +
H(g)
|
Temperature, °C |
100 |
200 |
300 |
400 |
|
k, mol dm-3s-1 |
1.1
x 10-9 |
1.8
x 10-8 |
1.2
x 10-7 |
4.4
x 10-7 |
[ANSWER:
42 kJ mol-1]
