Chemical Kinetics: The Rate of Reaction and Equilibria
undefined
1
Wanyera C
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Wanyera C
A
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Wanyera C
3
Some reactions are
too
fast
and instantaneous e.g.
neutralization of acid and bases/alkalis in aqueous
solution and double decomposition/precipitation.
Other reactions are
explosive
and very
risky
to
carry out safely e.g. reaction of potassium with
water and sodium with dilute acids.
The study of the rate of chemical reaction is
useful in knowing the
factors
that influence the
reaction so that
efficiency
and
profitability
is
maximized
in industries.
Wanyera C
4
Theories of rates of reaction.
The rate of a chemical reaction is defined as the rate of
change of concentration/amount of reactants in unit time.
It is also the rate of formation of given concentration of
products in unit time. i.e.
Rate of reaction =
Change in concentration/amount of reactants
Time taken for the change to occur
Rate of reaction =
Change in concentration
of products formed
Time taken for the products to form
For the above, therefore the rate of a chemical reaction is
rate of decreasing reactants to form an increasing product.
The SI unit of
time
is
second
(s) but minutes and hours are
also used.
Wanyera C
5
(
a
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Wanyera C
6
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Wanyera C
7
(vi)dissolving substances in a solvent ,make the
solvent a
medium
for the reaction to take place.
The solute particle distance is reduced as the
particle ions are free to move in the solvent
medium.
(vii)successful collisions take place if the particles
colliding have the required
energy
and right
orientation
which increases their
vibration
and
intensity
of successful / fruitful/ effective
collisions to form products.
Wanyera C
8
(b)The Activation Energy(Ea) theory
The
Enthalpy of activation(
∆
H
a
) /Activation
Energy(Ea)
is the minimum amount of energy
which the reactants must overcome before they
react.
Activation Energy(Ea) is usually required
/needed in
-
bond breaking
-
weakening of bonds holding the reacting
particles together.
-
Reorientation of reacting particles, ions ,atoms
and molecules.
Wanyera C
9
Bond breaking/weakening is an
endothermic process that require an energy
input.
The higher the bond energy the slower the
reaction to
start of
.
Activation energy does not influence
whether a reaction is exothermic or
endothermic.
Wanyera C
10
Energy
kJ
Reaction path/coordinate/path
Energy level diagram showing the activation
energy for exothermic processes /reactions.
Activated
complex
A-A
B-B
A A
B
B
A-
B
A-
B
∆
H
r
Wanyera C
11
The
activated complex
is a mixture of many
intermediate
possible products which may
not
exist
under normal physical conditions ,but can theoretically
exist.
Exothermic reaction proceed without further heating
/external energy because it generates its own energy/heat
to overcome activation energy.
Endothermic reaction cannot proceed without further
heating /external energy because it does not generates its
own energy/heat to
overcome
activation energy.
It generally therefore require
continuous
supply of
more energy/heat to
sustain
it to completion.
Wanyera C
12
3
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Wanyera C
13
The SI unit of time is the
second
.
Minutes and hours may be used where necessary
/specified.
The
stop watch/clock
is the standard
apparatus
used to
measure time.
Candidate/student should learn to:
(i)press
start
button concurrently with s
tarting
off
a reaction by using one hand for each.
Here, enough practice is very necessary.
Candidate/student should remember to
test
the
stop watch
before
using it.
It can
fail
to start
on/off
Wanyera C
14
(ii)Press
stop
button when reaction is
over
.
(iii)
Record
without
a
decimal point
the
time
immediately
and not try memorize
them.
e.g. “10 , 20, 23”…
not
“10.00” or “10.0”
(iv)
Avoid
accidental
pressing of reset
button before recording the time.
It can be very
frustrating
repeating a
whole procedure
Wanyera C
15
(v)
Ignore
time
beyond
the second (or minutes
when specified).
Hours may be beyond examination time limits at
this level.
(vi)Press
reset
button to
begin
another timing
from
O:OO:
OO seconds.
Rates of reaction practical test various principles
of Chemistry and patience is important.
Examining bodies/council
reward
this with
generous marks
Wanyera C
16
2
:
2
3
:
5
4
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Wanyera C
17
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Wanyera C
18
4.Factors influencing rate of reaction
The following
factors alter / influence /affect/
determine the rate of a chemical reaction:
(i)
concentration
(ii)Pressure
(iii)
Temperature
(iv)
Surface
area/particle size
(v)
Catalyst
Wanyera C
19
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r
o
d
u
c
t
s
.
P
r
a
c
t
i
c
a
l
l
y
a
n
i
n
c
r
e
a
s
e
i
n
c
o
n
c
e
n
t
r
a
t
i
o
n
r
e
d
u
c
e
s
t
h
e
t
i
m
e
t
a
k
e
n
f
o
r
t
h
e
r
e
a
c
t
i
o
n
t
o
t
a
k
e
p
l
a
c
e
.
Wanyera C
20
Practical determination of effect of
concentration
on reaction rate
Method 1
Reaction of sodium thisulphate with dilute
hydrochloric acid
Procedure:
Measure 20cm3 of 0.05M sodium thisulphate into a
50cm3 glass beaker.
Place the beaker on a white piece of filter paper with
ink
mark
‘X’ on it.
Measure 20cm3 of 0.1M hydrochloric acid solution
using a 50cm3 measuring cylinder.
Wanyera C
21
Put the acid into the beaker containing sodium
thiosulphate.
Immediately start off the stop watch/clock.
Determine the time taken for the ink
mark
‘X’ to
become invisible /obscured when viewed from
above.
Repeat the procedure by measuring different
volumes of the acid and adding the volumes of the
distilled water to complete table 1.
Wanyera C
22
Sample results
:
Table 1.
Wanyera C
23
F
o
r
m
o
s
t
e
x
a
m
i
n
i
n
g
b
o
d
i
e
s
/
c
o
u
n
c
i
l
s
/
b
o
a
r
d
s
t
h
e
a
b
o
v
e
r
e
s
u
l
t
s
s
c
o
r
e
f
o
r
:
(a)
c
o
m
p
l
e
t
e
t
a
b
l
e
a
s
e
v
i
d
e
n
c
e
f
o
r
a
l
l
t
h
e
p
r
a
c
t
i
c
a
l
w
o
r
k
d
o
n
e
a
n
d
c
o
m
p
l
e
t
e
d
.
(
b
)
(
i
)
C
o
n
s
i
s
t
e
n
t
u
s
e
o
f
a
d
e
c
i
m
a
l
p
o
i
n
t
o
n
t
i
m
e
a
s
e
v
i
d
e
n
c
e
o
f
u
n
d
e
r
s
t
a
n
d
i
n
g
/
k
n
o
w
l
e
d
g
e
o
f
t
h
e
d
e
g
r
e
e
o
f
a
c
c
u
r
a
c
y
o
f
s
t
o
p
w
a
t
c
h
e
s
/
c
l
o
c
k
.
(
i
i
)
C
o
n
s
i
s
t
e
n
t
u
s
e
o
f
a
m
i
n
i
m
u
m
o
f
f
o
u
r
d
e
c
i
m
a
l
p
o
i
n
t
s
o
n
i
n
v
e
r
s
e
/
r
e
c
i
p
r
o
c
a
l
o
f
t
i
m
e
a
s
e
v
i
d
e
n
c
e
o
f
u
n
d
e
r
s
t
a
n
d
i
n
g
/
k
n
o
w
l
e
d
g
e
o
f
t
h
e
d
e
g
r
e
e
o
f
a
c
c
u
r
a
c
y
o
f
s
c
i
e
n
t
i
f
i
c
c
a
l
c
u
l
a
t
o
r
.
(
c
)
a
c
c
u
r
a
c
y
a
g
a
i
n
s
t
a
s
c
h
o
o
l
v
a
l
u
e
b
a
s
e
d
o
n
c
a
n
d
i
d
a
t
e
’
s
t
e
a
c
h
e
r
s
-
r
e
s
u
l
t
s
s
u
b
m
i
t
t
e
d
.
(
d
)
c
o
r
r
e
c
t
t
r
e
n
d
(
t
i
m
e
i
n
c
r
e
a
s
e
a
s
m
o
r
e
w
a
t
e
r
i
s
a
d
d
e
d
/
a
c
i
d
i
s
d
i
l
u
t
e
d
)
i
n
c
o
n
f
o
r
m
i
t
y
w
i
t
h
e
x
p
e
c
t
e
d
t
h
e
o
r
e
t
i
c
a
l
r
e
s
u
l
t
s
.
Wanyera C
24
Sample questions
1. On separate graph papers plot a graph of:
(i)volume of acid used(x-axis) against time.
Label this graph I
(ii) volume of acid used(x-axis) against 1/t.
Label this graph II
2. Explain the shape of graph I
Diluting/adding water causes a
decrease
in
concentration.
Decrease in concentration
reduces
the rate of
reaction by increasing the time taken for reacting
particle to collide to form products.
Wanyera C
25
T
i
m
e
(
s
e
c
o
n
d
s
)
V
o
l
u
m
e
o
f
a
c
i
d
(
c
m
3
)
S
k
e
t
c
h
s
a
m
p
l
e
G
r
a
p
h
I
Wanyera C
26
1
/
t
S
e
c
-
1
x
1
0
-
2
V
o
l
u
m
e
o
f
a
c
i
d
(
c
m
3
)
Sketch sample Graph II
Wanyera C
27
3
.
F
r
o
m
g
r
a
p
h
I
I
,
d
e
t
e
r
m
i
n
e
t
h
e
t
i
m
e
t
a
k
e
n
f
o
r
t
h
e
c
r
o
s
s
t
o
b
e
o
b
s
c
u
r
e
d
/
i
n
v
i
s
i
b
l
e
w
h
e
n
t
h
e
v
o
l
u
m
e
o
f
t
h
e
a
c
i
d
i
s
:
(
i
)
1
3
c
m
3
F
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
1
/
t
a
t
1
3
c
m
3
o
n
t
h
e
g
r
a
p
h
=
>
2
.
7
5
x
1
0
-
2
t
=
1
/
2
.
7
5
x
1
0
-
2
=
3
6
.
3
6
3
6
s
e
c
o
n
d
s
(
i
i
)
1
5
c
m
3
F
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
1
/
t
a
t
1
5
c
m
3
o
n
t
h
e
g
r
a
p
h
=
>
3
.
3
5
x
1
0
-
2
t
=
1
/
3
.
3
5
x
1
0
-
2
=
2
9
.
8
5
0
7
s
e
c
o
n
d
s
Wanyera C
28
(
i
i
i
)
1
7
c
m
3
F
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
1
/
t
a
t
1
7
c
m
3
o
n
t
h
e
g
r
a
p
h
=
>
4
.
0
x
1
0
-
2
t
=
1
/
4
.
0
x
1
0
-
2
=
2
5
.
0
s
e
c
o
n
d
s
(
i
v
)
1
9
c
m
3
F
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
1
/
t
a
t
1
9
c
m
3
o
n
t
h
e
g
r
a
p
h
=
>
4
.
6
5
x
1
0
-
2
t
=
1
/
4
.
6
5
x
1
0
-
2
=
2
1
.
5
0
5
4
s
e
c
o
n
d
s
4
.
F
r
o
m
g
r
a
p
h
I
I
,
d
e
t
e
r
m
i
n
e
t
h
e
v
o
l
u
m
e
o
f
t
h
e
a
c
i
d
u
s
e
d
i
f
t
h
e
t
i
m
e
t
a
k
e
n
f
o
r
t
h
e
c
r
o
s
s
t
o
b
e
o
b
s
c
u
r
e
d
/
i
n
v
i
s
i
b
l
e
i
s
:
Wanyera C
29
(
i
)
2
5
s
e
c
o
n
d
s
1
/
t
=
>
1
/
2
5
=
4
.
0
x
1
0
-
2
R
e
a
d
i
n
g
f
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
;
4
.
0
x
1
0
-
2
c
o
r
r
e
s
p
o
n
d
t
o
1
7
.
0
c
m
3
(
i
i
)
3
0
s
e
c
o
n
d
s
1
/
t
=
>
1
/
3
0
=
3
.
3
3
x
1
0
-
2
R
e
a
d
i
n
g
f
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
;
3
.
3
3
x
1
0
-
2
c
o
r
r
e
s
p
o
n
d
t
o
1
4
.
7
c
m
3
(
i
i
i
)
4
0
s
e
c
o
n
d
s
1
/
t
=
>
1
/
4
0
=
2
.
5
x
1
0
-
2
R
e
a
d
i
n
g
f
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
;
2
.
5
x
1
0
-
2
c
o
r
r
e
s
p
o
n
d
t
o
1
2
.
3
c
m
3
Wanyera C
30
4
.
W
r
i
t
e
t
h
e
e
q
u
a
t
i
o
n
f
o
r
t
h
e
r
e
a
c
t
i
o
n
t
a
k
i
n
g
p
l
a
c
e
N
a
2
S
2
O
3
(
a
q
)
+
2
H
C
l
(
a
q
)
-
>
2
N
a
C
l
(
a
q
)
+
S
O
2
(
g
)
+
S
(
s
)
+
H
2
O
(
l
)
I
o
n
i
c
a
l
l
y
:
S
2
O
3
2
-
(
a
q
)
+
2
H
+
(
a
q
)
-
>
S
O
2
(
g
)
+
S
(
s
)
+
H
2
O
(
l
)
5
.
N
a
m
e
t
h
e
y
e
l
l
o
w
p
r
e
c
i
p
i
t
a
t
e
C
o
l
l
o
i
d
a
l
s
u
l
p
h
u
r
Wanyera C
31
Method 2
Reaction of Magnesium with dilute hydrochloric
acid
Procedure
Scub 10centimeter length of magnesium ribbon
with sand paper/steel wool.
Measure 40cm3 of 0.5M dilute hydrochloric acid
into a flask .
Fill a graduated gas jar with water and invert it
into a trough.
Stopper the flask and set up the apparatus to
collect the gas produced as in the set up below:
Wanyera C
32
Magnesium ribbon/shavings
Hydrochloric acid
Graduated
gas jar
Carefully remove the stopper, put the magnesium ribbon into the
flask . cork tightly.
Add the acid into the flask.
Connect the delivery tube into the gas jar.
Immediately start off the stop watch and determine the volume of
the gas produced after every 30 seconds to complete table II below.
water
Dropping funnel
Wanyera C
33
Sample practice questions
1.Plot a graph of volume of gas produced (y-axis) against
time
Sample results: Table II
Wanyera C
34
●
Curve I
V
o
l
u
m
e
o
f
g
a
s
T
i
m
e
i
n
m
i
n
u
t
e
s
Curve II
●
Wanyera C
35
2.Explain the shape of the graph.
The rate of reaction is faster when the concentration of
the acid is high .
As time goes on, the
concentration
of the acid
decreases
and therefore
less
gas is produced.
When
all
the acid has reacted, no more gas is produced
after 210 seconds and the graph flattens.
3.Calculate the rate of reaction at 120 seconds
From a tangent
at 120 seconds rate of reaction =
Change in volume of gas
Change in time
From the tangent
at 120seconds
V
2
- V
1
=
96-84
=
12
=
0.2cm3sec
-1
T
2
- T
1
150-90 60
Wanyera C
36
4. Write an ionic equation for the reaction taking place.
Mg
2+
(s) + 2H
+
(aq) -> Mg
2+
(aq) + H
2
(g)
5. On the same axis sketch then explain the curve that would be
obtained if:
(i) 0.1 M hydrochloric acid is used
–
Label this curve I
(ii)1.0 M hydrochloric acid is used
–
Label this curve II
Observation:
Curve I is to the right
Curve II is to the left
Explanation
A decrease in concentration shift the rate of reaction graph to the
right as more time is taken for completion of the reaction.
An increase in concentration shift the rate of reaction graph to the
left as less time is taken for completion of the reaction.
Both graphs
flatten
after some time indicating the
completion
of
the reaction.
Wanyera C
37
b)
Influence of
pressure
on rate of reaction
Pressure affects only gaseous reactants.
An increase in pressure reduces the volume(Boyles law)
in which the particles are contained.
Decrease in volume of the container bring the reacting
particles closer to each other which increases their
chances of effective/successful/fruitful collision to form
products.
An increase in pressure therefore increases the rate of
reaction by reducing the time for reacting particles of
gases to react.
At
industrial level,
the following are some reactions that
are affected by pressure:
Wanyera C
38
(
a
)
H
a
b
e
r
p
r
o
c
e
s
s
f
o
r
m
a
n
u
f
a
c
t
u
r
e
o
f
a
m
m
o
n
i
a
N
2
(
g
)
+
3
H
2
(
g
)
-
>
2
N
H
3
(
g
)
(
b
)
C
o
n
t
a
c
t
p
r
o
c
e
s
s
f
o
r
m
a
n
u
f
a
c
t
u
r
e
o
f
s
u
l
p
h
u
r
i
c
(
V
I
)
a
c
i
d
2
S
O
2
(
g
)
+
O
2
(
g
)
-
>
2
S
O
3
(
g
)
(
c
)
O
s
t
w
a
l
d
s
p
r
o
c
e
s
s
f
o
r
t
h
e
m
a
n
u
f
a
c
t
u
r
e
o
f
n
i
t
r
i
c
(
V
)
a
c
i
d
4
N
H
3
(
g
)
+
5
O
2
(
g
)
-
>
4
N
O
(
g
)
+
6
H
2
O
(
l
)
T
h
e
i
n
f
l
u
e
n
c
e
o
f
p
r
e
s
s
u
r
e
o
n
r
e
a
c
t
i
o
n
r
a
t
e
i
s
n
o
t
f
e
l
t
i
n
s
o
l
i
d
s
a
n
d
l
i
q
u
i
d
s
.
T
h
i
s
i
s
b
e
c
a
u
s
e
t
h
e
s
o
l
i
d
a
n
d
l
i
q
u
i
d
p
a
r
t
i
c
l
e
s
h
a
v
e
f
i
x
e
d
p
o
s
i
t
i
o
n
s
i
n
t
h
e
i
r
s
t
r
o
n
g
b
o
n
d
s
a
n
d
t
h
e
r
e
f
o
r
e
n
o
d
e
g
r
e
e
o
f
f
r
e
e
d
o
m
(
K
i
n
e
t
i
c
T
h
e
o
r
y
o
f
m
a
t
t
e
r
)
Wanyera C
39
c
)
I
n
f
l
u
e
n
c
e
o
f
t
e
m
p
e
r
a
t
u
r
e
o
n
r
a
t
e
o
f
r
e
a
c
t
i
o
n
A
n
i
n
c
r
e
a
s
e
i
n
t
e
m
p
e
r
a
t
u
r
e
i
n
c
r
e
a
s
e
s
t
h
e
k
i
n
e
t
i
c
e
n
e
r
g
y
o
f
t
h
e
r
e
a
c
t
i
n
g
p
a
r
t
i
c
l
e
s
b
y
i
n
c
r
e
a
s
i
n
g
t
h
e
i
r
c
o
l
l
i
s
i
o
n
f
r
e
q
u
e
n
c
y
.
I
n
c
r
e
a
s
e
i
n
t
e
m
p
e
r
a
t
u
r
e
i
n
c
r
e
a
s
e
s
t
h
e
p
a
r
t
i
c
l
e
s
w
h
i
c
h
c
a
n
o
v
e
r
c
o
m
e
t
h
e
a
c
t
i
v
a
t
i
o
n
e
n
e
r
g
y
(
E
a
)
.
A
1
0
o
C
r
i
s
e
i
n
t
e
m
p
e
r
a
t
u
r
e
d
o
u
b
l
e
s
t
h
e
r
a
t
e
o
f
r
e
a
c
t
i
o
n
b
y
r
e
d
u
c
i
n
g
t
h
e
t
i
m
e
t
a
k
e
n
f
o
r
t
h
e
r
e
a
c
t
i
o
n
t
o
c
o
m
p
l
e
t
e
b
y
a
h
a
l
f
.
P
r
a
c
t
i
c
a
l
d
e
t
e
r
m
i
n
a
t
i
o
n
o
f
e
f
f
e
c
t
o
f
T
e
m
p
e
r
a
t
u
r
e
o
n
r
e
a
c
t
i
o
n
r
a
t
e
.
M
e
t
h
o
d
1
Wanyera C
40
Reaction of sodium thisulphate with dilute hydrochloric
acid
Procedure:
Measure 20cm3 of 0.05M sodium thisulphate into a
50cm3 glass beaker.
Place the beaker on a white piece of filter paper with ink
mark
‘
X
’
on it.
Determine and record its temperature as room
temperature in table 2 below.
Measure 20cm3 of 0.1M hydrochloric acid solution
using a 50cm3 measuring cylinder.
Put the acid into the beaker containing sodium
thisulphate.
Wanyera C
41
Immediately start off the stop watch/clock.
Determine the time taken for the ink
mark
‘
X
’
to
become invisible /obscured when viewed from
above.
Measure another 20cm3 separate portion of the
thisulphate into a beaker, heat the solution to
30
o
C.
Add the acid into the beaker and repeat the
procedure above.
Complete table 2 below using different
temperatures of the thiosulphate.
Wanyera C
42
Sample results
:
Table 2.
Wanyera C
43
S
a
m
p
l
e
p
r
a
c
t
i
c
e
q
u
e
s
t
i
o
n
s
1
.
P
l
o
t
a
g
r
a
p
h
o
f
t
e
m
p
e
r
a
t
u
r
e
(
x
-
a
x
i
s
)
a
g
a
i
n
s
t
1
/
t
2
(
a
)
F
r
o
m
y
o
u
r
g
r
a
p
h
d
e
t
e
r
m
i
n
e
t
h
e
t
e
m
p
e
r
a
t
u
r
e
a
t
w
h
i
c
h
:
(
i
)
1
/
t
i
s
;
I
.
0
.
0
3
R
e
a
d
i
n
g
d
i
r
e
c
t
l
y
f
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
=
3
2
.
2
5
o
C
I
I
.
0
.
0
7
R
e
a
d
i
n
g
d
i
r
e
c
t
l
y
f
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
=
4
8
.
0
o
C
(
i
i
)
t
i
s
;
I
.
3
0
s
e
c
o
n
d
s
3
0
s
e
c
o
n
d
s
=
>
1
/
t
=
1
/
3
0
=
0
.
0
3
3
R
e
a
d
i
n
g
d
i
r
e
c
t
l
y
f
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
0
.
0
3
3
=
>
3
3
.
5
o
C
I
I
.
4
5
s
e
c
o
n
d
s
4
5
s
e
c
o
n
d
s
=
>
1
/
t
=
1
/
4
5
=
0
.
0
2
2
R
e
a
d
i
n
g
d
i
r
e
c
t
l
y
f
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
0
.
0
2
2
=
>
2
9
.
0
o
C
Wanyera C
44
I
I
I
.
2
5
s
e
c
o
n
d
s
2
5
s
e
c
o
n
d
s
=
>
1
/
t
=
1
/
2
5
=
0
.
0
4
R
e
a
d
i
n
g
d
i
r
e
c
t
l
y
f
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
0
.
0
4
=
>
3
6
.
0
o
C
(
b
)
F
r
o
m
y
o
u
r
g
r
a
p
h
d
e
t
e
r
m
i
n
e
t
h
e
t
i
m
e
t
a
k
e
n
f
o
r
t
h
e
c
r
o
s
s
t
o
b
e
c
o
m
e
i
n
v
i
s
i
b
l
e
a
t
:
(i)
5
7
.
5
o
C
R
e
a
d
i
n
g
d
i
r
e
c
t
l
y
f
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
a
t
5
7
.
5
o
C
=
0
.
0
9
4
=
>
1
/
t
=
0
.
0
9
4
t
=
1
/
0
.
0
9
4
=
>
1
0
.
6
3
8
3
s
e
c
o
n
d
s
(
i
i
)
4
5
o
C
R
e
a
d
i
n
g
d
i
r
e
c
t
l
y
f
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
a
t
4
5
o
C
=
0
.
0
6
2
=
>
1
/
t
=
0
.
0
6
2
t
=
1
/
0
.
0
9
4
=
>
1
6
.
1
2
9
0
s
e
c
o
n
d
s
(
i
i
i
)
3
5
o
C
R
e
a
d
i
n
g
d
i
r
e
c
t
l
y
f
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
a
t
3
5
o
C
=
0
.
0
4
7
=
>
1
/
t
=
0
.
0
4
7
t
=
1
/
0
.
0
4
7
=
>
2
1
.
2
7
6
6
s
e
c
o
n
d
s
Wanyera C
45
Method 2
Reaction of Magnesium with dilute hydrochloric acid
Procedure
Scub 5centimeter length of magnesium ribbon with sand
paper/steel wool.
Cut the piece into five equal one centimeter smaller pieces.
Measure 20cm3 of 1.0M dilute hydrochloric acid into a glass
beaker.
Put one piece of the magnesium ribbon into the acid, swirl.
Immediately start off the stop watch/clock.
Determine the time taken for the effervescence/fizzing/bubbling to
stop when viewed from above.
Record the time in table 2 at room temperature.
Wanyera C
46
Measure another 20cm3 portions of 1.0M dilute
hydrochloric acid into a clean beaker.
Heat separately one portion to 30
o
C, 40
o
C , 50
o
C
and 60
o
C and adding 1cm length of the ribbon
and determine the time taken for effervescence
/fizzing /bubbling to stop when viewed from
above .
Record each time to complete table 2 below using
different temperatures of the acid.
Wanyera C
47
S
a
m
p
l
e
r
e
s
u
l
t
s
:
T
a
b
l
e
1
.
S
a
m
p
l
e
p
r
a
c
t
i
c
e
q
u
e
s
t
i
o
n
s
•
P
l
o
t
a
g
r
a
p
h
o
f
t
e
m
p
e
r
a
t
u
r
e
(
x
-
a
x
i
s
)
a
g
a
i
n
s
t
1
/
t
Wanyera C
48
1/t
Temperature(
o
C)
Wanyera C
49
2
.
(
a
)
C
a
l
c
u
l
a
t
e
t
h
e
n
u
m
b
e
r
o
f
m
o
l
e
s
o
f
m
a
g
n
e
s
i
u
m
u
s
e
d
g
i
v
e
n
t
h
a
t
1
c
m
o
f
m
a
g
n
e
s
i
u
m
h
a
s
a
m
a
s
s
o
f
1
g
.
(
M
g
=
2
4
.
0
)
M
o
l
e
s
=
M
a
s
s
o
f
m
a
g
n
e
s
i
u
m
=
>
1
.
0
M
o
l
a
r
m
a
s
s
o
f
M
g
2
4
=
4
.
1
6
7
x
1
0
-
2
m
o
l
e
s
(
b
)
C
a
l
c
u
l
a
t
e
t
h
e
n
u
m
b
e
r
o
f
m
o
l
e
s
o
f
h
y
d
r
o
c
h
l
o
r
i
c
a
c
i
d
u
s
e
d
M
o
l
e
s
o
f
a
c
i
d
=
m
o
l
a
r
i
t
y
x
v
o
l
u
m
e
o
f
a
c
i
d
1
0
0
0
=
>
1
.
0
x
2
0
=
2
.
0
x
1
0
-
2
m
o
l
e
s
1
0
0
0
Wanyera C
50
(
c
)
C
a
l
c
u
l
a
t
e
t
h
e
m
a
s
s
o
f
m
a
g
n
e
s
i
u
m
t
h
a
t
r
e
m
a
i
n
u
n
r
e
a
c
t
e
d
M
o
l
e
r
a
t
i
o
M
g
:
H
C
l
=
1
:
2
M
o
l
e
s
M
g
=
½
m
o
l
e
s
H
C
l
=
>
½
x
2
.
0
x
1
0
-
2
m
o
l
e
s
=
1
.
0
x
1
0
-
2
m
o
l
e
s
M
a
s
s
o
f
r
e
a
c
t
e
d
M
g
=
m
o
l
e
s
x
m
o
l
a
r
m
a
s
s
=
>
1
.
0
x
1
0
-
2
m
o
l
e
s
x
2
4
=
0
.
2
4
g
M
a
s
s
o
f
u
n
r
e
a
c
t
e
d
M
g
=
O
r
i
g
i
n
a
l
t
o
t
a
l
m
a
s
s
-
M
a
s
s
o
f
r
e
a
c
t
e
d
M
g
=
>
1
.
0
g
–
0
.
2
4
=
0
.
7
6
g
Wanyera C
51
(
c
)
C
a
l
c
u
l
a
t
e
t
h
e
t
o
t
a
l
v
o
l
u
m
e
o
f
h
y
d
r
o
g
e
n
g
a
s
p
r
o
d
u
c
e
d
d
u
r
i
n
g
t
h
e
a
b
o
v
e
r
e
a
c
t
i
o
n
s
.
M
o
l
e
r
a
t
i
o
M
g
:
H
2
=
1
:
1
M
o
l
e
s
o
f
M
g
t
h
a
t
r
e
a
c
t
e
d
p
e
r
e
x
p
e
r
i
m
e
n
t
=
m
o
l
e
s
H
2
=
1
.
0
x
1
0
-
2
m
o
l
e
s
V
o
l
u
m
e
o
f
H
y
d
r
o
g
e
n
a
t
s
.
t
.
p
p
r
o
d
u
c
e
d
p
e
r
e
x
p
e
r
i
m
e
n
t
=
m
o
l
e
s
x
2
4
d
m
3
=
>
1
.
0
x
1
0
-
2
m
o
l
e
s
x
2
4
d
m
3
=
0
.
2
4
d
m
3
V
o
l
u
m
e
o
f
H
y
d
r
o
g
e
n
a
t
s
.
t
.
p
p
r
o
d
u
c
e
d
i
n
5
e
x
p
e
r
i
m
e
n
t
s
=
0
.
2
4
d
m
3
x
5
=
1
.
2
d
m
3
Wanyera C
52
3
.
(
a
)
A
t
w
h
a
t
t
e
m
p
e
r
a
t
u
r
e
w
a
s
t
h
e
t
i
m
e
t
a
k
e
n
f
o
r
m
a
g
n
e
s
i
u
m
t
o
r
e
a
c
t
e
q
u
a
l
t
o
:
(
i
)
7
0
s
e
c
o
n
d
s
7
0
s
e
c
o
n
d
s
=
>
1
/
t
=
1
/
7
0
=
0
.
0
1
4
2
9
R
e
a
d
i
n
g
d
i
r
e
c
t
l
y
f
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
0
.
0
1
4
2
9
=
>
2
8
.
0
o
C
(
i
i
)
4
0
s
e
c
o
n
d
s
4
0
s
e
c
o
n
d
s
=
>
1
/
t
=
1
/
4
0
=
0
.
0
2
5
R
e
a
d
i
n
g
d
i
r
e
c
t
l
y
f
r
o
m
a
c
o
r
r
e
c
t
l
y
p
l
o
t
t
e
d
g
r
a
p
h
0
.
0
2
5
=
>
3
2
.
0
o
C
Wanyera C
53
(b)What time would magnesium take to react
react at:
(i) 55.0
o
C
Reading directly from a correctly plotted graph at 55.0
o
C
1/t =
8.0 x 10
-2
=> t = 1/8.0 x 10
-2
=
12.5 seconds
(ii) 47.0
o
C
Reading directly from a correctly plotted graph at 47.0
o
C
1/t =
6.0 x 10
-2
=> t = 1/6.0 x 10
-2
=
16.6667 seconds
(iii) 33.0
o
C
Reading directly from a correctly plotted graph at 33.0
o
C
1/t =
2.7 x 10
-2
=> t = 1/2.7 x 10
-2
=
37.037 seconds
4. Explain the shape of the graph.
Increase in temperature increases the rate of reaction as particles
gain kinetic energy
increasing their frequency and intensity of
collision to form products.
Wanyera C
54
d
)
I
n
f
l
u
e
n
c
e
o
f
s
u
r
f
a
c
e
a
r
e
a
o
n
r
a
t
e
o
f
r
e
a
c
t
i
o
n
S
u
r
f
a
c
e
a
r
e
a
i
s
t
h
e
a
r
e
a
o
f
c
o
n
t
a
c
t
.
A
n
i
n
c
r
e
a
s
e
i
n
s
u
r
f
a
c
e
a
r
e
a
i
s
a
d
e
c
r
e
a
s
e
i
n
p
a
r
t
i
c
l
e
s
i
z
e
.
P
r
a
c
t
i
c
a
l
l
y
a
n
i
n
c
r
e
a
s
e
i
n
s
u
r
f
a
c
e
a
r
e
a
i
n
v
o
l
v
e
s
c
h
o
p
p
i
n
g
/
c
u
t
t
i
n
g
s
o
l
i
d
l
u
m
p
s
i
n
t
o
s
m
a
l
l
e
r
p
i
e
c
e
s
/
c
h
i
p
s
t
h
e
n
c
r
u
s
h
i
n
g
t
h
e
c
h
i
p
s
i
n
t
o
p
o
w
d
e
r
.
C
h
i
p
s
t
h
u
s
h
a
v
e
a
h
i
g
h
e
r
s
u
r
f
a
c
e
a
r
e
a
t
h
a
n
s
o
l
i
d
l
u
m
p
s
b
u
t
p
o
w
d
e
r
h
a
s
a
h
i
g
h
e
s
t
s
u
r
f
a
c
e
a
r
e
a
.
A
n
i
n
c
r
e
a
s
e
i
n
s
u
r
f
a
c
e
a
r
e
a
o
f
s
o
l
i
d
s
i
n
c
r
e
a
s
e
s
t
h
e
a
r
e
a
o
f
c
o
n
t
a
c
t
w
i
t
h
a
l
i
q
u
i
d
s
o
l
u
t
i
o
n
i
n
c
r
e
a
s
i
n
g
t
h
e
c
h
a
n
c
e
s
o
f
s
u
c
c
e
s
s
f
u
l
/
e
f
f
e
c
t
i
v
e
/
f
r
u
i
t
f
u
l
c
o
l
l
i
s
i
o
n
t
o
f
o
r
m
p
r
o
d
u
c
t
s
.
T
h
e
i
n
f
l
u
e
n
c
e
o
f
s
u
r
f
a
c
e
a
r
e
a
o
n
r
a
t
e
o
f
r
e
a
c
t
i
o
n
i
s
m
a
i
n
l
y
i
n
h
e
t
e
r
o
g
e
n
e
o
u
s
r
e
a
c
t
i
o
n
s
.
Wanyera C
55
Reaction of chalk/calcium carbonate on dilute
hydrochloric acid
Procedure
Measure 20cm3 of 1.0 M hydrochloric acid into
three separate conical flasks labeled C
1
C
2
and C
3
Using a watch glass weigh three separate 2.5g a
piece of white chalk.
Place the conical flask C
1
on an electronic
balance.
Reset the balance scale to 0.0.
Wanyera C
56
Put one weighed sample of the chalk into the acid
in the conical flask.
Determine the scale reading and record it at time
=0.0.
Simultaneously start of the stop watch.
Determine and record the scale reading after
every 30 seconds to complete Table I .
Repeat all the above procedure separately with C
2
and C
3
to complete Table II and Table III by
cutting the chalk into small pieces/chips for C
2
and crushing the chalk to powder for C
3
Wanyera C
57
S
a
m
p
l
e
r
e
s
u
l
t
s
:
T
a
b
l
e
1
I
.
S
a
m
p
l
e
r
e
s
u
l
t
s
:
T
a
b
l
e
I
I
I
.
S
a
m
p
l
e
r
e
s
u
l
t
s
:
T
a
b
l
e
1
.
Wanyera C
58
S
a
m
p
l
e
q
u
e
s
t
i
o
n
s
:
1
.
C
a
l
c
u
l
a
t
e
t
h
e
l
o
s
s
i
n
m
a
s
s
m
a
d
e
a
t
t
h
e
e
n
d
o
f
e
a
c
h
t
i
m
e
f
r
o
m
t
h
e
o
r
i
g
i
n
a
l
t
o
c
o
m
p
l
e
t
e
t
a
b
l
e
I
,
I
I
a
n
d
I
I
I
2
.
O
n
t
h
e
s
a
m
e
a
x
e
s
p
l
o
t
a
g
r
a
p
h
o
f
t
o
t
a
l
l
o
s
s
i
n
m
a
s
s
a
g
a
i
n
s
t
t
i
m
e
(
x
-
a
x
e
s
)
a
n
d
l
a
b
e
l
t
h
e
m
c
u
r
v
e
I
,
I
I
,
a
n
d
I
I
I
f
r
o
m
T
a
b
l
e
I
,
I
I
,
a
n
d
I
I
I
.
3
.
E
x
p
l
a
i
n
w
h
y
t
h
e
r
e
i
s
a
l
o
s
s
i
n
m
a
s
s
i
n
a
l
l
e
x
p
e
r
i
m
e
n
t
s
.
C
a
l
c
i
u
m
c
a
r
b
o
n
a
t
e
r
e
a
c
t
w
i
t
h
t
h
e
a
c
i
d
t
o
f
o
r
m
c
a
r
b
o
n
(
I
V
)
o
x
i
d
e
g
a
s
t
h
a
t
e
s
c
a
p
e
t
o
t
h
e
a
t
m
o
s
p
h
e
r
e
.
4
.
W
r
i
t
e
a
n
i
o
n
i
c
e
q
u
a
t
i
o
n
f
o
r
t
h
e
r
e
a
c
t
i
o
n
t
h
a
t
t
a
k
e
p
l
a
c
e
.
C
a
C
O
3
(
s
)
+
2
H
+
(
a
q
)
-
>
C
a
2
+
(
a
q
)
+
H
2
O
(
l
)
+
C
O
2
(
g
)
Wanyera C
59
5.Sulphuric(VI)acid cannot be used in the
above reaction. On the same axes sketch the
curve which would be obtained if the reaction
was attempted by reacting a piece of a lump of
chalk with 0.5M sulphuric(VI)acid. Label it
curve IV. Explain the shape of curve IV.
Calcium carbonate would react with dilute
0.5M sulphuric(VI)acid to form
insoluble
calcium
sulphate(VI) that
coat /cover
unreacted Calcium
carbonate stopping the reaction from reaching
completion.
Wanyera C
60
6
.
C
a
l
c
u
l
a
t
e
t
h
e
v
o
l
u
m
e
o
f
c
a
r
b
o
n
(
I
V
)
o
x
i
d
e
e
v
o
l
v
e
d
(
m
o
l
a
r
g
a
s
v
o
l
u
m
e
a
t
r
o
o
m
t
e
m
p
e
r
a
t
u
r
e
=
2
4
d
m
3
,
C
=
1
2
.
0
,
O
=
1
6
.
O
C
a
=
4
0
.
0
)
M
e
t
h
o
d
I
M
o
l
e
r
a
t
i
o
C
a
C
O
3
(
s
)
:
C
O
2
(
g
)
=
1
:
1
M
o
l
e
s
C
a
C
O
3
(
s
)
u
s
e
d
=
M
a
s
s
C
a
C
O
3
(
s
)
=
0
.
0
2
5
m
o
l
e
s
M
o
l
a
r
m
a
s
s
C
a
C
O
3
(
s
)
M
o
l
e
s
C
O
2
(
g
)
=
0
.
0
2
5
m
o
l
e
s
V
o
l
u
m
e
o
f
C
O
2
(
g
)
=
m
o
l
e
s
x
m
o
l
a
r
g
a
s
v
o
l
u
m
e
=
>
0
.
0
2
5
m
o
l
e
s
x
2
4
d
m
3
=
0
.
6
0
0
d
m
3
/
6
0
0
c
m
3
M
e
t
h
o
d
I
I
M
o
l
a
r
m
a
s
s
o
f
C
a
C
O
3
(
s
)
=
1
0
0
g
=
>
2
4
d
m
3
o
f
C
O
2
(
g
)
M
a
s
s
o
f
C
a
C
O
3
(
s
)
=
2
.
5
g
=
>
2
.
5
x
2
4
=
0
.
6
0
0
d
m
3
1
0
0
Wanyera C
61
7
.
F
r
o
m
c
u
r
v
e
I
,
d
e
t
e
r
m
i
n
e
t
h
e
r
a
t
e
o
f
r
e
a
c
t
i
o
n
(
l
o
s
s
i
n
m
a
s
s
p
e
r
s
e
c
o
n
d
)
a
t
t
i
m
e
1
8
0
s
e
c
o
n
d
s
o
n
t
h
e
c
u
r
v
e
.
F
r
o
m
t
a
n
g
e
n
t
a
t
1
8
0
s
e
c
o
n
d
s
o
n
c
u
r
v
e
I
R
a
t
e
=
M
2
-
M
1
=
>
2
.
0
8
–
1
.
3
7
5
=
0
.
6
2
5
=
T
2
-
T
1
2
2
2
-
1
3
2
9
0
=
0
.
0
0
6
9
4
4
g
s
e
c
-
1
8
.
W
h
a
t
i
s
t
h
e
e
f
f
e
c
t
o
f
p
a
r
t
i
c
l
e
s
i
z
e
o
n
t
h
e
r
a
t
e
o
f
r
e
a
c
t
i
o
n
?
A
l
a
r
g
e
r
s
u
r
f
a
c
e
a
r
e
a
i
s
a
r
e
d
u
c
t
i
o
n
i
n
p
a
r
t
i
c
l
e
s
i
z
e
w
h
i
c
h
i
n
c
r
e
a
s
e
s
t
h
e
a
r
e
a
o
f
c
o
n
t
a
c
t
b
e
t
w
e
e
n
r
e
a
c
t
i
n
g
p
a
r
t
i
c
l
e
s
i
n
c
r
e
a
s
i
n
g
t
h
e
i
r
c
o
l
l
i
s
i
o
n
f
r
e
q
u
e
n
c
y
.
Wanyera C
62
Theoretical examples
1. Excess marble chips were put in a beaker
containing 100cm3 of 0.2M hydrochloric acid.
The beaker was then placed on a balance and total
loss in mass recorded after every two minutes as in the
table below.
Wanyera C
63
(
a
)
W
h
y
w
a
s
t
h
e
r
e
a
l
o
s
s
i
n
m
a
s
s
?
C
a
r
b
o
n
(
I
V
)
o
x
i
d
e
g
a
s
w
a
s
p
r
o
d
u
c
e
d
t
h
a
t
e
s
c
a
p
e
t
o
t
h
e
s
u
r
r
o
u
n
d
i
n
g
(
b
)
C
a
l
c
u
l
a
t
e
t
h
e
a
v
e
r
a
g
e
r
a
t
e
o
f
l
o
s
s
i
n
m
a
s
s
b
e
t
w
e
e
n
:
(
i
)
0
t
o
2
m
i
n
u
t
e
s
A
v
e
r
a
g
e
r
a
t
e
=
M
2
-
M
1
=
>
1
.
8
0
–
0
.
0
=
1
.
8
=
T
2
-
T
1
2
.
0
–
0
.
0
2
=
9
.
0
0
g
m
i
n
-
1
(
i
)
6
t
o
8
m
i
n
u
t
e
s
A
v
e
r
a
g
e
r
a
t
e
=
M
2
-
M
1
=
>
3
.
2
0
–
2
.
9
5
=
0
.
2
5
T
2
-
T
1
8
.
0
–
6
.
0
2
=
0
.
1
2
5
g
m
i
n
-
1
Wanyera C
64
(
i
i
i
)
E
x
p
l
a
i
n
t
h
e
d
i
f
f
e
r
e
n
c
e
b
e
t
w
e
e
n
t
h
e
a
v
e
r
a
g
e
r
a
t
e
s
o
f
r
e
a
c
t
i
o
n
i
n
(
i
)
a
n
d
(
i
i
)
a
b
o
v
e
.
B
e
t
w
e
e
n
0
a
n
d
2
m
i
n
u
t
e
s
,
t
h
e
c
o
n
c
e
n
t
r
a
t
i
o
n
o
f
m
a
r
b
l
e
c
h
i
p
s
a
n
d
h
y
d
r
o
c
h
l
o
r
i
c
a
c
i
d
i
s
h
i
g
h
t
h
e
r
e
f
o
r
e
t
h
e
r
e
i
s
a
h
i
g
h
e
r
c
o
l
l
i
s
i
o
n
f
r
e
q
u
e
n
c
y
b
e
t
w
e
e
n
t
h
e
r
e
a
c
t
i
n
g
p
a
r
t
i
c
l
e
s
l
e
a
d
i
n
g
t
o
h
i
g
h
s
u
c
c
e
s
s
f
u
l
r
a
t
e
o
f
f
o
r
m
a
t
i
o
n
o
f
p
r
o
d
u
c
t
s
.
B
e
t
w
e
e
n
6
a
n
d
8
m
i
n
u
t
e
s
,
t
h
e
c
o
n
c
e
n
t
r
a
t
i
o
n
o
f
m
a
r
b
l
e
c
h
i
p
s
a
n
d
h
y
d
r
o
c
h
l
o
r
i
c
a
c
i
d
i
s
l
o
w
t
h
e
r
e
f
o
r
e
t
h
e
r
e
i
s
l
o
w
c
o
l
l
i
s
i
o
n
f
r
e
q
u
e
n
c
y
b
e
t
w
e
e
n
t
h
e
r
e
a
c
t
i
n
g
p
a
r
t
i
c
l
e
s
l
e
a
d
i
n
g
t
o
l
e
s
s
s
u
c
c
e
s
s
f
u
l
r
a
t
e
o
f
f
o
r
m
a
t
i
o
n
o
f
p
r
o
d
u
c
t
s
.
Wanyera C
65
(c)Write the equation for the reaction that takes place.
CaCO
3
(s) + 2HCl (aq) -> CaCO
3
(aq) + H
2
O(l) + CO
2
(g)
(d)State and explain three ways in which the rate of reaction
could be increased.
(i)
Heating the acid
- increasing the temperature of the
reacting particles increases their kinetic energy and thus collision
frequency.
(
ii
)
Increasing the concentration
of the acid
-increasing in
concentration reduces the distances between the reacting particles
increasing their chances of effective/fruitful/successful collision to
form products faster.
(
iii
)
Crushing
the marble chips to powder
-this reduces the
particle size/increase surface area increasing the area of contact
between reacting particles.
Wanyera C
66
(e)If the solution in the beaker was evaporated to dryness
then left overnight in the open, explain what would happen.
It becomes wet because calcium (II) chloride absorbs water
from the atmosphere and form solution/is deliquescent.
(f)When sodium sulphate (VI) was added to a portion of the
contents in the beaker after the reaction , a white precipitate
was formed .
(
i
)
Name the white precipitate.
Calcium(II)sulphate(VI)
(
ii
)
Write an ionic equation for the formation of the
white precipitate
Ca
2+
(aq) + SO
4
2-
(aq)->CaSO
4
(s)
(
iii
)
State one use of the white precipitate
-Making plaster for building
-Manufacture of plaster of Paris
-Making sulphuric(VI)acid
Wanyera C
67
(g)
(
i
)
Plot a graph of total loss in mass(y-axes)
against time
(
i
i
)
F
r
o
m
t
h
e
g
r
a
p
h
,
d
e
t
e
r
m
i
n
e
t
h
e
r
a
t
e
o
f
r
e
a
c
t
i
o
n
a
t
t
i
m
e
2
m
i
n
u
t
e
s
.
From a tangent/slope at 2 minutes;
Rate of reaction =
Average rate =
M
2
-M
1
T
2
- T
1
=>
2.25
–
1.30
=
0.95
=
0.3958g min
-1
3.20
–
0.8
2.4
(
iii
)Sketch on the same axes the graph that would
be obtained if 0.02M hydrochloric acid was used.
Label it curve II
Wanyera C
68
e)
Influence of
catalyst
on rate of reaction
Catalyst is a substance that alter the rate /speed of a
chemical reaction but remain chemically unchanged at the
end of a reaction.
Biological catalysts are called
enzymes
.
A catalyst does not alter the amount of products formed
but itself may be altered
physically
e.g. from solid to
powder to fine powder.
Most industrial catalysts are
transition metals
or their
compounds.
Catalyst works by lowering the Enthalpy of
activation(∆H
a
)/activation energy (Ea) of the reactants
.
The catalyst lowers the Enthalpy of activation
(∆H
a
)/activation energy (Ea) by:
Wanyera C
69
(i) forming short lived intermediate compounds
called
activated complex
that break up to form
the final product/s
(ii) being absorbed by the reactants thus providing
the surface area on which reaction occurs.
A catalyst has no effect on the enthalpy of
reaction ∆H
r
but only lowers the Enthalpy of
activation(∆H
a
)/activation energy (Ea)
It thus do not affect/influence whether the reaction
is exothermic or endothermic as shown in the
energy level diagrams below.
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70
B A
B A
Energy
kJ
Reaction path/coordinate/path
A
-A
B
-B
A
-
B
A-B
Energy level diagram showing the activation energy for exothermic processes /reactions.
Activated complex
Ea Catalysed
Activated complex
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71
A
A
B
B
Energy
kJ
Reaction path/coordinate/path
Ea
A
-A
B
-B
A
-
B
A-B
∆
Hr
Energy level diagram showing the activation energy for
endothermic processes /reactions.
Activated complex
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72
The following are some catalysed reaction processes.
(a)The contact process
Vanadium(V) Oxide(V
2
O
5
) or platinum(Pt) catalyses the
oxidation of sulphur(IV)oxide during the manufacture of
sulphuric(VI) acid from contact process.
SO
2
(g)
+
O
2
(g) ----
V
2
O
5
--> SO
3
(g)
To
reduce
industrial cost of manufacture of sulphuric (VI) acid
from contact process Vanadium(V) Oxide(V
2
O
5
) is used because it
is
cheaper
though it is
easily poisoned
by impurities.
(b)Ostwalds process
Platinum promoted with Rhodium catalyses the oxidation of
ammonia to nitrogen(II)oxide and water during the manufacture of
nitric(V)acid.
4NH
3
(g)
+
5O
2
(g) ----
Pt/Rh
--> 4NO (g) + 6H
2
O(l)
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73
(c)Haber process
Platinum or iron catalyses the combination of
nitrogen and hydrogen to form ammonia gas
N
2
(g) + 3H
2
(g) ---
Pt or Fe
---> 2NH
3
(g)
(d)Hydrogenation/Hardening of oil to fat
Nickel (Ni)
catalyses the hydrogenation of unsaturated
compound containing - C=C- or
–
C=C- to saturated
compounds without double or triple bond
This process is used is used in hardening oil to fat.
(e)Decomposition of hydrogen peroxide
Manganese(IV)oxide
speeds up the rate of
decomposition of hydrogen peroxide to water and oxygen
gas.
This is a common laboratory preparation of Oxygen.
2H
2
O
2
(g) ----
MnO
2
--> O
2
(g) + 2H
2
O(l)
Wanyera C
74
(f)Reaction of metals with dilute sulphuric (VI) acid
Copper(II)sulphate(VI)
speeds up the rate of
production of hydrogen gas from the reaction of Zinc and
dilute sulphuric(VI)acid. This process/reaction is used in
the school laboratory preparation of Hydrogen.
H
2
SO
4
(aq) + Zn(s) ----CuSO
4
--> ZnSO
4
(aq) + H
2
(g)
(g) Substitution reactions
When placed in bright
sunlight
or
U.V /ultraviolet
light , a mixture of a halogen and an alkane undergo
substitution reactions
explosively
to form
halogenoalkanes. When paced in
diffused
sunlight the
reaction is very
slow
.
e.g.
CH
4
(g)
+ Cl
2
(g) ---
u.v. light
--> CH
3
Cl(g) + HCl(g)
Wanyera C
75
(h)Photosynthesis
Plants convert carbon(IV)oxide gas from the
atmosphere and water from the soil to form glucose and
oxygen as a byproduct using sunlight / ultraviolet light.
6CO
2
(g) + 6H
2
O(l) -
u.v. light
--> C
6
H
12
O
6
(g) + O
2
(g)
(i)Photography
Photographic film contains silver bromide emulsion
which decomposes to silver and bromine on exposure to
sunlight
.
2AgBr(s) ---
u.v/sun light
--> 2Ag(s) + Br
2
(l)
When developed, the silver deposits give the picture
of the object whose photograph was taken depending on
intensity
of light. A picture photographed in
diffused
light is therefore
blurred
.
Wanyera C
76
Practical determination of effect of catalyst on
decomposition of hydrogen peroxide
Measure 5cm3 of 20 volume hydrogen peroxide and then
dilute to make 40cm3 in a measuring cylinder by adding
distilled water.
Divide it into two equal portions.
(
i
)Transfer one 20cm3volume hydrogen peroxide into a
conical/round bottomed/flat bottomed flask.
Cork and swirl for 2 minutes.
Remove the cork.
Test the gas produced using a glowing splint.
Clean the conical/round bottomed/flat bottomed flask.
Wanyera C
77
(
ii
)Put 2.0g of Manganese (IV) oxide into the clean
conical/round bottomed/flat bottomed flask.
Stopper the flask.
Transfer the second portion of the 20cm3volume
hydrogen peroxide into a conical/round bottomed/flat
bottomed flask through the dropping/thistle funnel.
Connect the delivery tube to a calibrated/graduated gas
jar as in the set up below.
Start off the stop watch and determine the volume of gas
in the calibrated/graduated gas jar after every 30 seconds
to complete Table 1.
Wanyera C
78
Manganese (IV) Oxide
Hydrogen peroxide
Oxygen gas
Graduated/calibrated
gas jar
Set up of apparatus to determine the effect of catalyst on rate of
reaction
Wanyera C
79
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80
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81
Catalysed reaction
Uncatalysed reaction
A graph of volume of gas produced against time(x-axes)
Wanyera C
82
b) On the same axes, plot a sketch graph for the
uncatalysed reaction.
(c) Explain the changes in mass of
manganese(IV)oxide before and after the reaction.
The mass of MnO
2
before and after the reaction is
the
same
but a
more fine
powder after the
experiment.
A catalyst therefore remains unchanged
chemically but
may
physically change.
L
e
g
a
l
c
a
u
t
i
o
n
!
!
!
D
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o
t
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c
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u
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a
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.
L
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a
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a
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r
c
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s
t
!
!
!
Wanyera C
83
Wanyera C
84
B.
EQUILIBRIA
(CHEMICAL CYBERNETICS)
Equilibrium is a state of balance.
Chemical equilibrium is state of balance between the
reactants and products.
As reactants form products, some products form back the
reactants.
Reactions in which the reactants form products to
completion
are said to be
irreversible
i.e.
A
+
B
->
C
+
D
Reactions in which the reactants form products and the
products can
reform
the reactants are said to be
reversible
.
A
+
B
C
+
D
Wanyera C
85
Reversible reactions may be:
(
a
)Reversible physical changes
(
b
)Reversible chemical changes
(
c
)Dynamic equilibrium
(
a
)
Reversible physical changes
Reversible physical change is one which involves:
(i) change of
state/phase
from solid, liquid, gas or
aqueous solutions.
States of matter are interconvertible and a reaction
involving a change from one state/phase can be reversed
back to the original.
(ii)
colour
changes. Some substances/compounds
change their colours without change in chemical
substance.
Wanyera C
86
Examples of reversible physical changes
(i)
colour change on heating and cooling:
I.
Zinc(II)Oxide changes from white when
cool/cold to yellow when hot/heated and back.
ZnO(
s
)
ZnO(
s
)
(white when cold)
(yellow when hot)
II.
Lead(II)Oxide changes from yellow when
cold/cool to brown when hot/heated and back.
PbO(
s
)
PbO(
s
)
(brown when hot)
(yellow when cold)
Wanyera C
87
(ii)Sublimation
I.
Iodine
sublimes
from a grey crystalline solid on
heating to purple vapour. Purple vapour undergoes
deposition
back to the grey crystalline solid.
I
2
(
s
)
I
2
(
g
)
(grey crystalline solid
(purple vapour
undergo sublimation)
undergo deposition)
II.
Carbon (IV)oxide gas undergoes
deposition
from a
colourless gas to a white solid at very high pressures in a
cylinder. It
sublimes
back to the colourless gas if pressure
is reduced
CO
2
(
s
)
CO
2
(
g
)
(white powdery solid
(colourless/odourless gas
undergo sublimation)
undergo deposition)
Wanyera C
88
(iii)Melting/ freezing and boiling/condensation
Ice on heating undergo
melting
to form a liquid/water.
Liquid/water on further heating
boil
/vaporizes to form gas/water
vapour.
Gas/water vapour on cooling,
condenses
/liquidifies to
water/liquid.
On further cooling, liquid water
freezes
to ice/solid.
(
iv
)
Dissolving/ crystallization/distillation
S
olid crystals of soluble substances (solutes) dissolve in water
/solvents to form a uniform mixture of the solute and
solvent/solution.
On crystallization /distillation /evaporation the solvent evaporate
leaving a solute back. e.g.
NaCl(
s
) + aq
NaCl(
aq
)
Wanyera C
89
(b)Reversible chemical changes
These are reactions that involve a chemical change of
the reactants which can be reversed back by recombining
the new substance formed/products.
Examples of Reversible chemical changes
(
i
)
Heating Hydrated/adding water to anhydrous salts.
When
hydrated
salts are heated they
lose
some/all
their
water
of crystallization and become
anhydrous
.
Heating an unknown substance /compound that forms a
colourless liquid droplets
on the
cooler
parts of a dry
test/boiling tube is in fact a
confirmation
inference that the
substance /compound being heated is
hydrated
.
Wanyera C
90
When anhydrous salts are added (back) some water they
form hydrated compound/salts.
Heating Copper(II)sulphate(VI)pentahydrate and
cobalt(II)chloride hexahydrate
(i)Heat about 5.0g of Copper(II)sulphate(VI)
pentahydrate in a clean dry test tube until there is no
further colour change on a
small
Bunsen flame.
Observe any changes on the side of the test/boiling
tube.
Allow the boiling tube to cool.
Add about 10 drops of distilled water.
Observe any changes.
Wanyera C
91
(ii)Dip a filter paper in a solution of
cobalt(II)chloride hexahydrate.
Pass one end of the filter paper to a
small
Bunsen flame repeatedly.
Take care for it not to catch fire
Observe any changes on the filter paper.
Dip the paper in a beaker containing distilled
water.
Observe any changes.
Wanyera C
92
Sample observations
Wanyera C
93
Wanyera C
94
When
blue
Copper(II)sulphate (VI) pentahydrate is
heated, it loses the five molecules of water of
crystallization to form
white
anhydrous Copper
(II) sulphate (VI).
Water of crystallization form and condenses as
colourless droplets on the cooler parts of a dry
boiling/test tube.
This is a chemical change that produces a new
substance.
On adding drops of water to an anhydrous
white
copper(II)sulphate(VI) the hydrated
blue
compound is formed back.
Wanyera C
95
The change from
hydrated
to
anhydrous
and
back
is
therefore
reversible chemical change.
Both anhydrous white copper(II)sulphate(VI) and
blue
cobalt(II)chloride hexahydrate are therefore used to test
for the presence of water when they turn to blue and
pink respectively.
CuSO
4
(
s
) + 5H
2
O(
l
)
CuSO
4
.
5H
2
O(
s
/
aq
)
(white/anhydrous)
(
blue
/hydrated)
CoCl
2
(
s
) + 6H
2
O(
l
)
CoCl
2
.
6H
2
O(
s
/
aq
)
(
blue
/anhydrous)
(
pink
/hydrated)
Wanyera C
96
(ii)Chemical sublimation
Some compounds sublime from solid to gas by
dissociating into new different compounds. e.g.
Heating ammonium chloride
(
i
)Dip a glass rod containing concentrated
hydrochloric acid.
Bring it near the mouth of a bottle containing
concentrated ammonia solution.
Explain the observations made.
When a glass rod containing hydrogen chloride gas is
placed near ammonia gas, they react to form ammonium
chloride solid that appear as
white fumes
.
Wanyera C
97
This experiment is used interchangeably to test for the
presence of hydrogen chloride gas (and hence Cl
-
ions)
and ammonia gas (and hence NH
4
+
ions)
(
ii
)Put 2.0 g of ammonium chloride in a
long
dry boiling
tube.
Place wet / moist /damp blue and red litmus papers
separately on the sides of the mouth of the boiling tube.
Heat the boiling tube gently then strongly.
Explain the observations made.
When ammonium chloride is heated it dissociates into
ammonia and hydrogen chloride gases.
Wanyera C
98
Since ammonia is less dense, it diffuses faster to
turn both litmus papers blue before hydrogen
chloride turn red because it is denser.
The heating and cooling of ammonium chloride is
therefore a
reversible chemical change
.
NH
3
(g) + HCl(g)
NH
4
Cl(s)
(Turns moist
(Turns moist
(forms white fumes)
litmus paper
blue
) litmus paper
red
)
Wanyera C
99
(c)Dynamic equilibria
For reversible reactions in a closed system:
(i) at the beginning;
-the reactants are decreasing in
concentration with time
-the products are increasing in concentration
with time
(ii) after some time a point is reached when as
the reactants are forming products the products
are forming reactants.
This is called equilibrium.
Wanyera C
100
R
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a
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q
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Wanyera C
101
For a system in equilibrium:
(i) a reaction from left to right (reactants to
products) is called forward reaction.
(ii) a reaction from right to left (products to
reactants) is called backward reaction.
(iii)a reaction in which the rate of forward reaction
is equal to the rate of backward reaction is called a
dynamic equilibrium.
A dynamic equilibrium
is therefore a balance of the rate
of formation of products and reactants.
This balance continues until the reactants or products are
disturbed/changed/ altered.
Wanyera C
102
The influence of different factors on a dynamic equilibrium
was first investigated from 1850-1936 by the French Chemist
Louis Henry Le Chatellier.
His findings were called Le Chatelliers Principle which states
that:
“
if a stress is applied to a system in dynamic equilibrium,
the system readjust/shift/move/behave so as to
remove/reduce/counteract the stress
”
Le Chatelliers Principle is applied in determining the
effect/influence of several factors on systems in dynamic
equilibrium.
The following are the main factors that influence/alter/affect
systems in dynamic equilibrium:
(a)Concentration
(b)Temperature
(c)Pressure
(d)Catalyst
Wanyera C
103
(a)Influence of concentration on dynamic
equilibrium
An increase/decrease in concentration of
reactants/products at equilibrium is a stress.
From Le Chatelliers principle the system readjust so as
to remove/add the excess/reduced concentration.
Examples of influence of concentration on dynamic
equilibrium
(i)Chromate(VI)/CrO
4
2-
ions in solution are
yellow
.
Dichromate(VI)/Cr
2
O
7
2-
ions in solution are
orange
.
The two solutions exist in equilibrium as in the
equation:
2H
+
(aq) + 2CrO
4
2-
(aq)
Cr
2
O
7
2-
(aq) + H
2
O(l)
(
Yellow
)
(
Orange
)
Wanyera C
104
I
. If an acid/H
+
(aq) is added to the equilibrium mixture a
stress is created on the reactant side where there is
already H
+
ions.
The equilibrium shift forward to the right to
remove/reduce the
excess
H
+
ions added.
Solution mixture becomes More
Cr
2
O
7
2-
ions formed in
the solution mixture make it to be more
orange
in colour.
II
. If a base/OH
-
(aq) is added to the equilibrium mixture
a stress is created on the reactant side on the H
+
ions. H
+
ions react with OH
-
(aq) to form water.
H
+
(aq) +OH
-
(aq) -> H
2
O(l)
Wanyera C
105
The equilibrium shift backward to the left to add/replace
the H
+
ions that have reacted with the OH
-
(aq) ions .
More of the CrO
4
2-
ions formed in the solution mixture
makes it to be more
yellow
in colour.
2OH
-
(aq) + 2Cr
2
O
7
2-
(aq)
CrO
4
2-
(aq) + H
2
O(l)
(
Orange)
(
Yellow
)
I
. If an acid/ H
+
(aq) is added to the equilibrium mixture a
stress is created on the reactant side on the OH
-
(aq). H
+
ions react
with OH
-
(aq) to form water.
H
+
(aq) +OH
-
(aq) -> H
2
O(l)
The equilibrium shift backward to the left to add/replace the
2OH
-
(aq) that have reacted with the H
+
(aq) ions .
More Cr
2
O
7
2-
(aq)ions formed in the solution mixture makes it
to be more
Orange
in colour.
Wanyera C
106
II
. If a base /OH
-
(aq) is added to the equilibrium
mixture a stress is created on the reactant side
where there is already OH
-
(aq) ions.
The equilibrium shift forward to the right to
remove/reduce the
excess
OH
-
(aq) ions added.
More
of the Cr
2
O
7
2-
ions are formed in the
solution mixture making it to be more
Orange
in
colour.
Wanyera C
107
Practical determination of the influence of alkali/acid on
Cr
2
O
7
2-
/CrO
4
2-
equilibrium mixture
Measure about 2 cm3 of Potassium dichromate (VI)
solution into a test tube.
Note that the solution mixture is orange.
Add three drops of 2M sulphuric(VI) acid. Shake the
mixture carefully.
Note that the solution mixture is remains orange.
Add about six drops of 2M sodium hydroxide solution.
Shake carefully.
Note that the solution mixture is turns yellow.
Wanyera C
108
Cr
2
O
7
2-
(aq)
CrO
4
2-
(aq)
Explanation
The above observations can be explained from the fact
that both the dichromate(VI)and chromate(VI) exist in
equilibrium.
Dichromate(VI) ions are stable in acidic solutions while
chromate(VI)ions are stable in basic solutions.
An equilibrium exist thus:
When an
acid
is added, the equilibrium shift
forward
to
the right and the mixture become more
orange
as more
Cr
2
O
7
2-
ions exist.
When a
base
is added, the equilibrium shift
backward
to the left and the mixture become more
yellow
as more
CrO
4
2-
ions exist.
Wanyera C
109
(ii)
Practical determination of the influence of
alkali/acid on bromine water in an equilibrium mixture
Measure 2cm3 of bromine water into a boiling tube.
Note its colour.
Bromine water is yellow
Add three drops of 2M sulphuric(VI)acid. Note any
colour change
Colour becomes more yellow
Add seven drops of 2M sodium hydroxide solution.
Note any colour change.
Solution mixture becomes colourless/Bromine
water is decolourized.
Wanyera C
110
Explanation
When added distilled water,an equilibrium exist between
bromine liquid (Br
2
(aq)) and the bromide ion(Br
-
), hydrobromite
ion(OBr
-
) and hydrogen ion(H
+
) as in the equation:
H
2
O(l) + Br
2
(aq) OBr
-
(aq) + H
+
(aq) + Br
-
(aq)
If an acid (H
+
)ions is added to the equilibrium mixture, it
increases the concentration of the ions on the product side which
shift backwards to the left to remove the excess H
+
ions on the
product side making the colour of the solution mixture more yellow.
If a base/alkali OH
-
is added to the equilibrium mixture, it reacts
with H
+
ions on the product side to form water.
H
+
(aq)+ OH
-
(aq) -> H
2
O(l)
This decreases the concentration of the H
+
ions on the product
side which shift the equilibrium forward to the right to replace H
+
ions making the solution mixture colourless/less yellow (Bromine
water is decolorized)
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111
(iii)
Practical determination of the influence of alkali/acid
on common acid-base indicators.
Place 2cm3 of phenolphthalein ,methyl orange and litmus
solutions each in three separate test tubes.
To each test tube add two drops of water. Record your
observations in Table 1 below.
To the same test tubes, add three drops of 2M
sulphuric(VI)acid. Record your observations in Table 1
below.
To the same test tubes, add seven drops of 2M sodium
hydroxide solution. Record your observations in Table 1
below.
To the same test tubes, repeat adding four drops of 2M
sulphuric(VI)acid.
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112
Explanation
Wanyera C
113
An indicator is a substance which shows whether
another
substance is an
acid
,
base
or
neutral
.
Most indicators can be regarded as very weak acids that
are partially dissociated into ions.
An equilibrium exist between the undissociated
molecules and the dissociated anions.
Both the molecules and anions are
coloured
. i.e.
HIn(aq)
H
+
(aq) + In
-
(aq)
(undissociated indicator
(dissociated indicator
molecule(
coloured
)) molecule(
coloured
))
Wanyera C
114
When an acid H
+
is added to an indicator, the H
+
ions
increase and equilibrium shift backward to remove excess
H
+
ions and therefore the
colour
of the undissociated
(
HIn
) molecule
shows/appears.
When a base/alkali OH
-
is added to the indicator, the
OH
-
reacts with H
+
ions from the dissociated indicator to
form water.
H
+
(aq)
+
OH
-
(aq)
-> H
2
O(l)
(from indicator) (from alkali/base)
The equilibrium shift forward to the right to replace the
H
+
ion and therefore the colour of dissociated (
In
-
)
molecule
shows
/
appears.
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115
Examples illustrating the changes in acid.base equilibrium
(i)Phenolphthalein indicator exist as:
HPh
H
+
(aq)
+
Ph
-
(aq)
(colourless molecule)
(Pink anion)
On adding an acid ,equilibrium shift backward to the left to
remove excess H
+
ions and solution mixture is
colourless.
When a base/alkali OH
-
is added to the indicator, the OH
-
reacts with H
+
ions from the dissociated indicator to form
water.
H
+
(aq)
+
OH
-
(aq)
-> H
2
O(l)
(from indicator) (from alkali/base)
The equilibrium shift forward to the right to replace the
removed /reduced H
+
ions. The
pink
colour of dissociated
(
Ph
-
) molecule
shows/appears.
Wanyera C
116
(ii)Methyl Orange indicator exists as:
HMe
H
+
(aq)
+
Me
-
(aq)
(Red molecule)
(Yellow/Orange anion)
On adding an acid ,equilibrium shift backward to the
left to remove excess H
+
ions and the solution mixture is
therefore
red.
When a base/alkali OH
-
is added to the indicator, the
OH
-
reacts with H
+
ions from the dissociated indicator to
form water.
H
+
(aq)
+
OH
-
(aq)
-> H
2
O(l)
(from indicator) (from alkali/base)
The equilibrium shift forward to the right to replace
the removed/reduced H
+
ions. The
Orange
colour of
dissociated (
Me
-
) molecule
shows/appears.
Wanyera C
117
(b)Influence of Pressure on dynamic
equilibrium
Pressure affects gaseous reactants/products.
Increase in
pressure
shift/
favours
the equilibrium
towards the side with
less volume/molecules
.
Decrease in pressure shift the equilibrium towards
the side with more volume/molecules.
More yield of products is obtained if high
pressures produce
less molecules
/volume of
products are formed.
Chemical kinetics explores the rate at which chemical reactions occur and the factors influencing them. This tutorial delves into the concepts of reaction rates, equilibrium, collision theory, and the role of concentration in determining reaction rates. By understanding these principles, industries can enhance efficiency and profitability in their processes.
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2012 The rate of reaction and Equilibria Comprehensive tutorial notes POWERPOINT VERSION Wanyera C 1
A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed. The rate of a chemical reaction is also the time taken for a given mass/amount of reactant to be consumed /used up. Some reactions are too slow to be determined. e.g rusting ,decomposition of hydrogen peroxide and weathering. Wanyera C 2
Some reactions are too fast and instantaneous e.g. neutralization of acid and bases/alkalis in aqueous solution and double decomposition/precipitation. Other reactions are explosive and very risky to carry out safely e.g. reaction of potassium with water and sodium with dilute acids. The study of the rate of chemical reaction is useful in knowing the factors that influence the reaction so that efficiency and profitability is maximized in industries. Wanyera C 3
Theories of rates of reaction. The rate of a chemical reaction is defined as the rate of change of concentration/amount of reactants in unit time. It is also the rate of formation of given concentration of products in unit time. i.e. Rate of reaction = Change in concentration/amount of reactants Time taken for the change to occur Rate of reaction = Change in concentration of products formed Time taken for the products to form For the above, therefore the rate of a chemical reaction is rate of decreasing reactants to form an increasing product. The SI unit of time is second(s) but minutes and hours are also used. Wanyera C 4
(a)The collision theory The collision theory is an application of the Kinetic Theory of matter which assumes matter is made up of small/tiny/minute particles like ions atoms and molecules. Thecollision theory proposes that (i)for a reaction to occur, reacting particles must collide. (ii)not all collisions between reacting particles are successful in a reaction. Collisions that initiate a chemical reaction are called successful / fruitful/ effective collisions Wanyera C 5
(iii)the speed at which particles collide is called collision frequency. The higher the collision frequency the higher the chances of successful / fruitful/ effective collisions to form products. (iv)the higher the chances of successful collisions, the faster the reaction. (v)the average distance between solid particles from one another is too big for them to meet and collide successfully. Wanyera C 6
(vi)dissolving substances in a solvent ,make the solvent a medium for the reaction to take place. The solute particle distance is reduced as the particle ions are free to move in the solvent medium. (vii)successful collisions take place if the particles colliding have the required energy and right orientation which increases their vibration and intensity of successful / fruitful/ effective collisions to form products. Wanyera C 7
(b)The Activation Energy(Ea) theory The Enthalpy of activation( Ha) /Activation Energy(Ea) is the minimum amount of energy which the reactants must overcome before they react. Activation Energy(Ea) is usually required /needed in - bond breaking - weakening of bonds holding the reacting particles together. - Reorientation of reacting particles, ions ,atoms and molecules. Wanyera C 8
Bond breaking/weakening is an endothermic process that require an energy input. The higher the bond energy the slower the reaction to start of. Activation energy does not influence whether a reaction is exothermic or endothermic. Wanyera C 9
Energy level diagram showing the activation energy for exothermic processes /reactions. Activated complex A A B B Energy kJ A-A B-B Hr A-B A-B Reaction path/coordinate/path Wanyera C 10
The activated complex is a mixture of many intermediate possible products which may not exist under normal physical conditions ,but can theoretically exist. Exothermic reaction proceed without further heating /external energy because it generates its own energy/heat to overcome activation energy. Endothermic reaction cannot proceed without further heating /external energy because it does not generates its own energy/heat to overcome activation energy. It generally therefore require continuous supply of more energy/heat to sustain it to completion. Wanyera C 11
3. Measuring the rate of a chemical reaction. The rate of a chemical reaction can be measured as: (i)Volume of a gas in unit time; - if reaction is producing a gas as one of the products. - if reaction is using a gas as one reactants (ii)Change in mass of reactants/products for solid products/reactants in unit time. (iii)formation of a given mass of precipitate in unit time (iv)a certain mass of reactants to completely form products/diminish. Wanyera C 12
The SI unit of time is the second. Minutes and hours may be used where necessary /specified. The stop watch/clock is the standard apparatus used to measure time. Candidate/student should learn to: (i)press start button concurrently with starting off a reaction by using one hand for each. Here, enough practice is very necessary. Candidate/student should remember to test the stop watch before using it. It can fail to start on/off Wanyera C 13
(ii)Press stop button when reaction is over. (iii)Record without a decimal point the time immediately and not try memorize them. e.g. 10 , 20, 23 not 10.00 or 10.0 (iv)Avoid accidental pressing of reset button before recording the time. It can be very frustrating repeating a whole procedure Wanyera C 14
(v)Ignore time beyond the second (or minutes when specified). Hours may be beyond examination time limits at this level. (vi)Press reset button to begin another timing from O:OO:OO seconds. Rates of reaction practical test various principles of Chemistry and patience is important. Examining bodies/council reward this with generous marks Wanyera C 15
Common School laboratory stop watch /clock Start/stop button Reset/zero button 2 : 23 :54 IGNORE this 1/100 seconds reading Number of minutes Number of seconds Stop watch reading = (2 x 60) + 23 = 143 seconds Wanyera C 16
Reactants may be homogenous or heterogenous. -Homogenous reactions involve reactants in the samephase/state e.g. solid-solid,gas-gas,liquid- liquid. e.g. Reaction of hydrogen gas and chlorine gas to form hydrogen chloride gas -Heterogenous reactions involve reactants in the differentphase/state e.g. solid-liquid, gas- liquid,solid-gas. e.g. Reaction of Magnesium solid and hydrochloric acid to form hydrogen gas and magnesium chloride Wanyera C 17
4.Factors influencing rate of reaction The following factors alter / influence /affect/ determine the rate of a chemical reaction: (i)concentration (ii)Pressure (iii)Temperature (iv)Surface area/particle size (v)Catalyst Wanyera C 18
a)Influence of concentration on rate of reaction The higher the concentration ,the higher the rate of a chemical reaction. An increase in concentration of the reactants reduces the distance between the reacting particles increasing their collision frequency to form products. Practically an increase in concentration reduces the time taken for the reaction to take place. Wanyera C 19
Practical determination of effect of concentration on reaction rate Method 1 Reaction of sodium thisulphate with dilute hydrochloric acid Procedure: Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark X on it. Measure 20cm3 of 0.1M hydrochloric acid solution using a 50cm3 measuring cylinder. Wanyera C 20
Put the acid into the beaker containing sodium thiosulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark X to become invisible /obscured when viewed from above. Repeat the procedure by measuring different volumes of the acid and adding the volumes of the distilled water to complete table 1. Wanyera C 21
Sample results:Table 1. Volume of acid(cm3) Volume of water(cm3) Volume of sodium thiosulphate (cm3) Time taken for mark X to be invisible /obscured (seconds) 20.0 Reciprocal of time 1 t 20.0 0.0 20.0 5.0 x 10-2 18.0 2.0 20.0 23.0 4.35 x 10-2 16.0 4.0 20.0 27.0 3.7 x 10-2 14.0 6.0 20.0 32.0 3.13 x 10-2 12.0 8.0 20.0 42.0 2.38 x 10-2 10.0 10.0 20.0 56.0 1.78 x 10-2 Wanyera C 22
For most examining bodies/councils/boards the above results score for: (a) complete table as evidence for all the practical work done and completed. (b) (i)Consistent use of adecimal point on time as evidence of understanding/knowledge of the degree of accuracy of stop watches/clock. (ii)Consistent use of a minimum of fourdecimal points on inverse/reciprocal of time as evidence of understanding/knowledge of the degree of accuracy of scientific calculator. (c) accuracy against a school value based on candidate s teachers-results submitted. (d) correct trend (time increase as more water is added/acid is diluted) in conformity with expected theoretical results. Wanyera C 23
Sample questions 1. On separate graph papers plot a graph of: (i)volume of acid used(x-axis) against time. Label this graph I (ii) volume of acid used(x-axis) against 1/t. Label this graph II 2. Explain the shape of graph I Diluting/adding water causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. Wanyera C 24
Sketch sample Graph I Time (seconds) Volume of acid (cm3) Wanyera C 25
Sketch sample Graph II 1/t Sec-1 x 10-2 Volume of acid (cm3) Wanyera C 26
3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds Wanyera C 27
(iii) 17cm3 From a correctly plotted graph 1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: Wanyera C 28
(i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Readingfrom a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Readingfrom a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Readingfrom a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 Wanyera C 29
4. Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Wanyera C 30
Method 2 Reaction of Magnesium with dilute hydrochloric acid Procedure Scub 10centimeter length of magnesium ribbon with sand paper/steel wool. Measure 40cm3 of 0.5M dilute hydrochloric acid into a flask . Fill a graduated gas jar with water and invert it into a trough. Stopper the flask and set up the apparatus to collect the gas produced as in the set up below: Wanyera C 31
Dropping funnel Hydrochloric acid Graduated gas jar water Magnesium ribbon/shavings Carefully remove the stopper, put the magnesium ribbon into the flask . cork tightly. Add the acid into the flask. Connect the delivery tube into the gas jar. Immediately start off the stop watch and determine the volume of the gas produced after every 30 seconds to complete table II below. Wanyera C 32
Sample results: Table II 0 30 60 90 120 150 180 210 240 Time (seconds) Volume of gas produced (cm3) 0.0 20.0 40.0 60.0 80.0 90.0 95.0 96.0 96.0 Sample practice questions 1.Plot a graph of volume of gas produced (y-axis) against time Wanyera C 33
Curve II Curve I Volume of gas Time in minutes Wanyera C 34
2.Explain the shape of the graph. The rate of reaction is faster when the concentration of the acid is high . As time goes on, the concentration of the acid decreases and therefore less gas is produced. When all the acid has reacted, no more gas is produced after 210 seconds and the graph flattens. 3.Calculate the rate of reaction at 120 seconds From a tangent at 120 seconds rate of reaction = Change in volume of gas Change in time From the tangent at 120seconds V2 - V1= 96-84 = 12 = 0.2cm3sec-1 T2 - T1 150-90 60 Wanyera C 35
4. Write an ionic equation for the reaction taking place. Mg2+(s) + 2H+(aq) -> Mg2+(aq) + H2 (g) 5. On the same axis sketch then explain the curve that would be obtained if: (i) 0.1 M hydrochloric acid is used Label this curve I (ii)1.0 M hydrochloric acid is used Label this curve II Observation: Curve I is to the right Curve II is to the left Explanation A decrease in concentration shift the rate of reaction graph to the right as more time is taken for completion of the reaction. An increase in concentration shift the rate of reaction graph to the left as less time is taken for completion of the reaction. Both graphs flatten after some time indicating the completion of the reaction. Wanyera C 36
b)Influence of pressure on rate of reaction Pressure affects only gaseous reactants. An increase in pressure reduces the volume(Boyles law) in which the particles are contained. Decrease in volume of the container bring the reacting particles closer to each other which increases their chances of effective/successful/fruitful collision to form products. An increase in pressure therefore increases the rate of reaction by reducing the time for reacting particles of gases to react. At industrial level, the following are some reactions that are affected by pressure: Wanyera C 37
(a)Haber process for manufacture of ammonia N2(g) + 3H2(g) -> 2NH3(g) (b)Contact process for manufacture of sulphuric(VI)acid 2SO2(g) + O2(g) -> 2SO3(g) (c)Ostwalds process for the manufacture of nitric(V)acid 4NH3(g) + 5O2(g) -> 4NO (g) + 6H2O (l) The influence of pressure on reaction rate is not felt in solids and liquids. This is because the solid and liquid particles have fixed positions in their strong bonds and therefore no degree of freedom (Kinetic Theory of matter) Wanyera C 38
c)Influence of temperature on rate of reaction An increase in temperature increases the kinetic energy of the reacting particles by increasing their collision frequency. Increase in temperature increases the particles which can overcome the activation energy (Ea). A 10oC rise in temperature doubles the rate of reaction by reducing the time taken for the reaction to complete by a half. Practical determination of effect of Temperature on reaction rate. Method 1 Wanyera C 39
Reaction of sodium thisulphate with dilute hydrochloric acid Procedure: Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark X on it. Determine and record its temperature as room temperature in table 2 below. Measure 20cm3 of 0.1M hydrochloric acid solution using a 50cm3 measuring cylinder. Put the acid into the beaker containing sodium thisulphate. Wanyera C 40
Immediately start off the stop watch/clock. Determine the time taken for the ink mark X to become invisible /obscured when viewed from above. Measure another 20cm3 separate portion of the thisulphate into a beaker, heat the solution to 30oC. Add the acid into the beaker and repeat the procedure above. Complete table 2 below using different temperatures of the thiosulphate. Wanyera C 41
Sample results:Table 2. 30 40 50 60 Temperature of Na2S2O3 Room temperature 50.0 40.0 20.0 15.0 10.0 Time taken for mark X to be obscured /invisible (seconds) Reciprocal of time(1/t) 0.02 0.025 0.05 0.0667 0.1 Wanyera C 42
Sample practice questions 1. Plot a graph of temperature(x-axis) against 1/t 2(a)From your graph determine the temperature at which: (i)1/t is ; I. 0.03 Reading directly from a correctly plotted graph = 32.25 oC II. 0.07 Reading directly from a correctly plotted graph = 48.0 oC (ii) t is; I. 30 seconds 30 seconds => 1/t =1/30 =0.033 Reading directly from a correctly plotted graph 0.033 => 33.5 oC II. 45 seconds 45 seconds => 1/t =1/45 =0.022 Reading directly from a correctly plotted graph 0.022 => 29.0 oC Wanyera C 43
III. 25 seconds Reading directly from a correctly plotted graph 0.04 => 36.0 oC (b) From your graph determine the time taken for the cross to become invisible at: (i) 57.5 oC Reading directly from a correctly plotted graph at 57.5 oC= 0.094 =>1/t = 0.094 t= 1/0.094 => 10.6383 seconds (ii) 45 oC Reading directly from a correctly plotted graph at 45 oC = 0.062 =>1/t = 0.062 t= 1/0.094 => 16.1290 seconds (iii) 35 oC Reading directly from a correctly plotted graph at 35 oC = 0.047 =>1/t = 0.047 t= 1/0.047 => 21.2766 seconds 25 seconds => 1/t =1/25 =0.04 Wanyera C 44
Method 2 Reaction of Magnesium with dilute hydrochloric acid Procedure Scub 5centimeter length of magnesium ribbon with sand paper/steel wool. Cut the piece into five equal one centimeter smaller pieces. Measure 20cm3 of 1.0M dilute hydrochloric acid into a glass beaker. Put one piece of the magnesium ribbon into the acid, swirl. Immediately start off the stop watch/clock. Determine the time taken for the effervescence/fizzing/bubbling to stop when viewed from above. Record the time in table 2 at room temperature. Wanyera C 45
Measure another 20cm3 portions of 1.0M dilute hydrochloric acid into a clean beaker. Heat separately one portion to 30oC, 40oC , 50oC and 60oC and adding 1cm length of the ribbon and determine the time taken for effervescence /fizzing /bubbling to stop when viewed from above . Record each time to complete table 2 below using different temperatures of the acid. Wanyera C 46
Sample results:Table 1. 30 40 50 60 Temperature of acid(oC) Room temperature 80.0 50.0 21.0 13.5 10.0 Time taken effervescence to stop (seconds) Reciprocal of time(1/t) 0.0125 0.02 0.0476 0.0741 0.1 Sample practice questions Plot a graph of temperature(x-axis) against 1/t Wanyera C 47
1/t Temperature(oC) Wanyera C 48
2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 Molar mass of Mg = 4.167 x 10 -2 moles (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 24 Wanyera C 49
(c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = moles HCl => x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g 0.24 = 0.76 g Wanyera C 50
