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A) The Nature of the Reactants
B) The Concentration of the Reactants
C) Temperature
D) Presence of a Catalyst
E) Surface area in Heterogeneous Systems
In homogeneous reaction systems the rate is affected by:
a) The Nature of Reactants
Consider this reaction:
CaCO3(s) + HCl(aq) --> CO2(g) + H2O(l) +CaCl2(aq)
This reaction is quite fast. Why? HCl is completely
ionized. This means there are no HCl particles present in a
solution of HCl. Rather there are only H1+ ions and Cl1ions present. The H1+ ions are able to easily collide with
the CaCO3(s) particles and a successful reaction occurs.
This reaction however is quite slow.
CaCO3(s) + 2HC2H3O2(aq) -->CO2(g) + 2H2O(l) + Ca(C2H3O2)2
because HC2H3O2(aq) particles are only partially ionized.
The bonds within Acetic Acid are stronger so a
solution of this acid contains mostly HC2H3O2
particles and very few H1+ ions.
100
100
100
H-Cl
H1+ + Cl1100
3
3
H
|
H1+ + C2H3O21H-C-C=O
This is a stronger bond
| |
1+
than
the
bond
between
H
H O-H
and Cl1-
The Concentration of Reactants
If a reaction is to occur the reacting particles must
first collide. The more collisions the faster the
reaction proceeds.
Consider this reaction
H2(g) + I2(g) ------> 2HI(g)
It has been determined, experimentally, that if [H2]
doubles while [I2] remains constant the rate of
reaction doubles. This can be expressed as
rate a [H2]
Also if [I2] doubles while [H2] is constant, the
reaction rate (r) also doubles. So r a [I2]
So if r a [I2] and [H2],
Reaction Rate (r)= k [I2] [H2] where k is the
reaction rate constant which is specific to the
reaction and the conditions under which it is
carried out. This reaction rate expression
implies that the reaction in question must
proceed directly from reactants to products
with no intermediate steps. This kind of
reaction is called elementary. In general for an
elementary reaction
r = k[A]a[B]b
The Reaction Order is a + b
aA + bB -----> cC
Consider this elementary reaction (a
reaction which proceeds directly from
reactants to products with no
intermediate step)
2A + B ---> A2B the rate expression is
r = k[A]2[B]1
the reaction order is
2 + 1 = 3, or 2nd order for A, and 1st order for B,
Reaction Orders > 3 are usually not elementary (there
are intermediate steps). Now consider this reaction
4 HBr(g) + O2(g) ----> 2 H2O(g) + 2Br2(g)
if this is an elementary reaction then the rate
expression would be
r = k [HBr]4[O2]1, the reaction order is 4 + 1 = 5
However experimentation shows that for this
reaction if [HBr] doubles while [O2] is constant
the rate, r, doubles, so r a [HBr]; also if [O2] is
doubled and [HBr] is constant the rate, r, doubles
This means the rate expression is
r = k [HBr][O2] and the reaction order is 2.
Why are the rate expressions different?
This reaction must proceed in steps. The steps
1.
2
3
HBr + O2 ---> HOOBr
slow
HOOBr + HBr ---> 2 HOBr
fast
2HOBr + 2 HBr --> 2H2O + 2Br2 fast
adding steps 1+2+3 gives
4 HBr(g) + O2(g) ----> 2 H2O(g) + 2Br2(g)
A reaction can only go as fast as its slowest
step. The rate of step 1 is the same as the
overall rate.
Most reactions have reaction mechanisms
which limit overall rates. Data is collected in
an attempt to determine the rate limiting step.
Here is an example of how data is used….
For a reaction where A + B ----> C the
following data was collected.
Trial
[A] molL-1 [B] molL-1
#
1
2
3
4
0.5
0.5
1
2
0.3
0.6
0.6
1.2
Reaction Rate
molL-1s-1 of C
produced
0.027
0.216
0.216
1.728
This problem is solved in 2 steps. 1st to see how
[A] affects the rate consider 2 trials where [B] does
Trials 2 and 3.
not change.
Trial
[A] molL-1 [B] molL-1
#
1
2
3
4
0.5
0.5
1
2
0.3
0.6
0.6
1.2
Reaction Rate
molL-1s-1 of C
produced
0.027
0.216
0.216
1.728
When [A] doubles, the Rate is unchanged.
This means [A] has no effect on the rate.
To see how [B] affects the rate, two trials are
considered where [A] is unchanged.
Trials 1 and 2 fulfill this requirement.
Trial
[A] molL-1 [B] molL-1
#
1
2
3
4
0.5
0.5
1
2
0.3
0.6
0.6
1.2
Reaction Rate
molL-1s-1 of C
produced
0.027
0.216
0.216
1.728
When [B] doubles the Reaction Rate increases by a
factor of 8. What does this mean? Remember
This is the equation we are trying to solve
reaction rate = [B]x
Reaction Rate
molL-1s-1 of C
produced
0.027
0.216
0.216
1.728
Trial
[A] molL-1 [B] molL-1
#
1
2
3
4
0.5
0.5
1
2
0.3
0.6
0.6
1.2
x
[B] from trial 2
Reaction Rate from trial 2
=
Reaction Rate from trial 1 [B] from trial 1
8=
x
2
X=3
r
3
=k[B]
Now try this one yourself. Determine the rate
expression using this data
Trial
[A] molL-1 [B] molL-1
#
1
2
3
4
0.5
0.5
1
2
0.3
0.6
0.6
1.2
Reaction Rate
molL-1s-1 of C
produced
0.075
0.15
0.6
4.8
To see how the [A] affects the rate consider 2 trials
where [B] does not change.
Trials 2 and 3
Trial
[A] molL-1 [B] molL-1
#
1
2
3
4
0.5
0.5
1
2
0.3
0.6
0.6
1.2
Reaction Rate
molL-1s-1 of C
produced
0.075
0.15
0.6
4.8
x
[A] from trial 3
Reaction Rate from trial 3
=
Reaction Rate from trial 2 [A] from trial 2
4=
x
2
X=2
r
2
=k[A]
Trial
-1
-1
[A] molL
[B] molL
#
1
2
3
4
0.5
0.5
1
2
0.3
0.6
0.6
1.2
Reaction Rate
-1 -1
molL s of C
produced
0.075
0.15
0.6
4.8
To see how the [B] affects the rate
consider 2 trials where [A] does not
change.
Trials 1 and 2
Trial
[A] molL-1 [B] molL-1
#
1
2
3
4
0.5
0.5
1
2
0.3
0.6
0.6
1.2
Reaction Rate
molL-1s-1 of C
produced
0.075
0.15
0.6
4.8
x
[B] from trial 2
Reaction Rate from trial 2
=
Reaction Rate from trial 1 [B] from trial 1
2=
x
2
X=1
r
1
=k[B]
The overall rate expression is
2
1
=k[A] [B]
r
Let’s try one more
using this data
Trial
[A] molL-1 [B] molL-1
#
1
2
3
4
0.2
0.4
0.4
0.8
0.6
0.6
1.2
2.4
Reaction Rate
molL-1s-1 of C
produced
0.0432
0.0864
0.6912
11.0592
To see how the [A] affects the rate consider 2 trials
where [B] does not change. Trials 1and 2
x
[A] from trial 2
Reaction Rate from trial 2
=
Reaction Rate from trial 1 [A] from trial 1
2=
x
2
X=1
r =k[A]
Trial
[A] molL-1 [B] molL-1
#
1
2
3
4
0.2
0.4
0.4
0.8
0.6
0.6
1.2
2.4
Reaction Rate
molL-1s-1 of C
produced
0.0432
0.0864
0.6912
11.0592
To see how the [B] affects the rate consider 2 trials
where [A] does not change. Trials 2 and 3
x
[B] from trial 3
Reaction Rate from trial 3
=
Reaction Rate from trial 2 [B] from trial 2
8=
x
2
X=3
r
3
=k[B]
The overall rate expression is
r
3
=k[A][B]
As temperature increases Ek increases,
molecular collisions increase so the rate of
reaction, r, increases.
In general a 10 K increase in temperature
doubles r.
eg. If @ 290 K, r = 4.0 molL-1min-1 what is
the rate at 310 K?
16 molL-1min-1
If a successful reaction is going to
occur the reacting molecules must
reach the activated complex. To
do this they must come together
with sufficient velocity to
overcome electron-electron
repulsion. If the energy of all
reacting particles could be found
the following distribution curve
would be obtained.
Only particles to the right of Ea have sufficient
kinetic energy to collide and reach the activated
complex
E
a
T1
# of
particles
Ek
T2
Catalysts are substances which speed up chemical
reactions without themselves being consumed. This is
why a catalyst is written over the arrow between
reactants and products.
catalase
H2O2 ----------> H2O + 1/2O2
In general what a catalyst does is provide a new
reaction path containing a different, lower energy,
activated complex. Because less energy is needed to
reach the activated complex the reaction tends to go
faster. This can be illustrated by the following graph
Uncatalyzed reaction
catalyzed reaction
Potential
energy
Reaction Coordinate
If the reaction involves a heterogeneous
system (more than one phase) a chemical
reaction can only occur at the surface
between them.
A pile of starch burns very slowly, but a fine
particle spray of the same starch burns very
rapidly since the oxygen required for the
reaction can interact over a much larger
surface.
Limited
O2
Surface
Area
O2
O2
O2
Much
larger
O2
Surface
Area
O2
O2
O2
O2
O2
O2
O2