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T-3
For a given reaction at constant temperature, k is constant even though the concentrations of reactants
change. However, k increases if the temperature increases (as in the H2 and O2 reaction mentioned in
experiment S). At higher temperatures the molecules move faster resulting in a greater number of
collisions and a greater proportion of molecules will collide with energy equal to or greater than E
a
and
will be able to cross over the energy barrier to form products. The relationship of the rate constant of a
chemical reaction to the temperature (in Kelvin ) can be expressed by the Arrhenius equation:
k = Ae
-E
a
/RT
(1)
In equation (1), A is the proportionality constant, R is the Universal Gas Constant using the value of
8.314 J mol
-1
K
-1
and e is the base of natural logarithms (e = 2.718...). By taking the natural logarithm
of equation (1), it is converted to:
ln k = -
E
R
a
1
T
+ ln A
(2)
in which “ln” refers to natural logarithm of base “e ”. Comparing equation (2) to the equation of a
straight-line, y = mx + b, indicates that if we plot ln k versus 1/T, we would obtain a straight line with
slope equal to: -
E
R
a
where R = 8.314 J mol
-1
K
-1
. From this graphical approach, we may determine the activation energy,
E
a,
for the reaction.
Every chemical reaction has a specific activation energy but it is possible to change the reaction rate by
use of a catalyst. A catalyst provides an alternate pathway with a lower activation energy for the
reaction. A good catalyst will provide a new reaction mechanism with a much lower activation energy.
As a result, the rate of reaction (at a given temperature) is faster with a catalyst than without.
EXPERIMENTAL METHOD
For the oxidation-reduction reaction:
2 I
-
(aq)
+ S2O
8
2-
(aq)
2 SO
4
2-
(aq)
+ I
2(aq)
(3)
it was found in Experiment S, that the relationship between the rate of reaction and the concentration of
reactants is:
