EXPERIMENT 4 - EFFECT OF TEMPERATURE ON RATE PROCESSES
(ACTIVATION ENERGY)
OBJECTIVE
Part A - Energy Barrier in Liquid Flow
To illustrate the concept of the energy barrier in fluid flow
processes.
Part B - Effect of Temperature on Rate Constant of Reaction
To determine the activation energy for a chemical process
and to observe the effect of temperature on the rate
constant of the reaction.
THEORY
Part A - Energy Barrier in Liquid Flow
Viscosity is the resistance offered to a flowing liquid. Energy is required to maintain a steady
flow of liquid through a pipe because of the energy barrier presented to the moving molecules by
the stationary molecules on the walls of the pipe. The energy barrier is pictured as a simple
repulsion force encountered by moving molecules squeezing past the stationary molecules.
The rate of flow of the liquid through the pipe is greater when the energy barrier is low. Fluidity,
, is defined to be the reciprocal of viscosity. Thus, if the resistance encountered by moving
molecules is low, (ie - low viscosity), the fluidity is high. Fluidity, , follows an exponential
relationship similar to that for the rate constant, k.
Rate constant, k:
k =Ae
Fluidity, :

Ae
-
-
E act
RT
E vis
RT
eqn (4-1)
eqn (4-2)
Fluidity is high if Evis is low. Fluidity increases with temperature. At higher temperatures, a
greater portion of molecules have sufficient kinetic energy to surmount the energy barrier. Evis is
determined by dropping steel balls of uniform radius through the liquid at different temperatures
and recording the velocities at the various temperatures.
According to Stoke's law, a ball of radius 'r' moving through a fluid with viscosity coefficient, ,
experiences a viscous drag force given by the expression
f = 6rv
eqn (4-3)
or
4-1
 =
f
6 rv
eqn (4-4)
where f , the frictional resistance to the falling sphere, is constant at constant temperature. In a
viscous fluid, f of gravity is almost immediately balanced by f of viscous drag and, when
balanced, the velocity, v, is constant.
Since 1/ = , and  = (6r / f ) v,
 = K v,
eqn (4-5)
where K is a constant.
Equating eqn (4-2) and eqn (4-5),
E  1
ln v = -  vis  + c'
 R T
 E vis  1
log v = - 
 + c'
 2.303  R  T
( y =
m x +b )
eqn (4-6)
eqn (4-7)
Evis is to be found from the measurement of the rate of fall of a sphere through a viscous liquid
at various temperatures. Evis is determined from the slope of the line obtained by plotting
1
"ln v versus " in the same way as the energy of activation is determined in reaction rate
T
problems. Evis is analogous to activation energy. The activation energy for a chemical reaction
is analogous to energy barrier for viscous flow.
Part B - Effect of Temperature on Rate Constant of Reaction
In the second part of the experiment, the variation of rate constant for a chemical reaction is
determined by first determining Eact for the reaction. The reaction to be investigated is as
follows:
BrO3- + 5 Br- + 6 H+  3 Br2 + 3 H2O
This reaction follows a fourth-order rate law,
d


rate =  [BrO 3 ] = k [BrO 3 ][ Br - ][ H + ] 2
dt
rxn (4-1)
eqn (4-8)
The initial concentrations of BrO3-, Br-, and H+ are the same in each trial. Therefore,
rate = k 
eqn (4-9)
4-2
where  = [BrO3-] [Br-] [H+]2 and is a constant. Substituting eqn (4--1) for k, the rate
of the reaction is then
rate =  A e
-
E act
RT
eqn (4-10)
or
ln (rate) = -
E act  1 
 +
R T
E act  1 
 +
2.303  R  T 
=
m x +b)
log (rate) = -
(
y
eqn (4-11)
eqn (4-12)
where  = log (A), a constant since  and A are constants.
To measure the time required for rxn (4-1) to have reacted, an arbitrary, but fixed amount of
phenol and some methyl orange are added to the reaction mixture. The bromine produced reacts
immediately with phenol (to form tribromophenol) and after all the phenol has been consumed,
any further bromine bleaches the methyl orange indicator. The time for loss of colour will be
measured at various temperatures and, from this data, the activation energy will be found.
1
1
) versus " should be a straight line when the rate
t
T
constant, k, obeys Arrhenius equation. Note that eqn (4-11) can also be written as
A plot of "log rate (or log
k 
E act  1 1 
 - 
log  2  = 2.303  R  T2 T1 
 k1 
eqn (4-13)
where k2 is the rate constant of the reaction carried out at temperature T2 and k1
is the rate constant of the reaction carried out at temperature T1.
The mathematical similarity of Evis to Eact is apparent in that
1
" and,
T
1
1
Eact is the slope of the plot of "log versus ".
t
T
Evis is the slope of the plot of "log v versus
4-3
PROCEDURE
Part A - Energy Barrier in Liquid Flow
1.
Four graduated cylinders containing a viscous liquid (ie - STP oil) will be provided.
Each graduated cylinder is marked with two horizontal marks, one near the bottom and
the other near the top of the cylinder.
2.
Measure accurately the distance between the marks in centimeters as this is the distance
that the metal spheres will travel. Record the distance in the datasheet.
3.
Set the four graduated cylinders containing a viscous liquid (ie - STP oil) each at
a different temperature. Each group is responsible to take measurements at two
temperatures.
Suggested temperatures for Group 1:
(1)
approximately 0oC
(2)
approximately 60 oC
Suggested temperatures for Group 2:
(1)
approximately room temperature
(2)
approximately 40oC
4.
Make sure the oil level in the graduated cylinders is below the water level in the
ice and water baths. Allow the oil to come to thermal equilibrium with the
bath temperature. It may take 20-30 minutes for the oil to reach the extreme
temperatures of the baths.
5.
Measure the actual temperatures of the ice and water baths with a thermometer.
Record the actual temperatures of the ice and water baths.
6
Drop a metal sphere in the center of the graduated cylinder ensuring that the
sphere does not touch the sides of the cylinder throughout its travel. Start the timer as the
sphere crosses the first mark and stop the timer as the sphere crosses the bottom mark.
Repeat this at least four times at each of the four different temperatures so that an
average time, therefore, an average velocity can be determined for each temperature.
4-4
Part B - Effect of Temperature on Rate Constant of Reaction
1.
Pipet:
10.00 mL of phenol solution,
10.00 mL of bromide/bromate solution
into an Erlenmeyer flask. Add 4 drops of methyl orange indicator.
2.
Pipet 5.00 mL 0.5 M H2SO4 into a second Erlenmeyer flask.
3.
Immerse the two Erlenmeyer flasks into a water bath at 75oC and allow
the contents to reach equilibrium temperature.
4.
Mix the contents of the two Erlenmeyer flasks and note the time in seconds
required for the red colour to disappear.
5.
Loosely stopper the flasks before putting them into the water baths.
6.
Repeat the above procedures at 65oC, 55oC, 45oC, 35oC, and 25oC.
4-5
DATA SHEET
Part A - Energy Barrier in Liquid Flow
Temperature, T
( oC )
(K)
1/T
time
Average
time
( K-1 )
( sec )
( sec )
Distance
Of falling
spheres
(m)
Average
velocity,
v
( m/s )
log v
#1
#2
#3
#4
#1
#2
#3
#4
#1
#2
#3
#4
#1
#2
#3
#4
4-6
DATA SHEET
Part B - Effect of Temperature on Rate Constant of Reaction
Temperature, T
( oC )
(K)
1/T
time, t
1/t
( K-1 )
(s)
( s-1 )
log (1/t)
4-7
TREATMENT OF DATA
Part A - Energy Barrier in Liquid Flow
1.
Tabulate data neatly in the data sheet.
2.
Calculate the velocity of fall in meters per second (m/s) of the metal sphere.
3.
Plot "log v versus
4.
From the graph, determine the slope of the line.
5.
Using eqn (4-7) determine the energy barrier to viscous flow, Evis.
[R = 8.314 J K-1mole-1]
1
" and draw the best straight line through the points.
T
Part B - Effect of Temperature on Rate Constant of Reaction
1.
Tabulate data neatly in the data sheet.
2.
Plot a graph of "log
1
1
versus " and draw the best straight line through the
t
T
points.
3.
From the graph, determine the slope of the line.
4.
Using eqn (4-12) determine the activation energy, Eact.
[R = 8.314 J K-1mole-1]
4-8