Topic 16 Kinetics

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Topic 16 Kinetics
• Rate expressions
• Reaction mechanism
• Activation energy
A+B→C+D
16.1 Rate expression
• Change in concentration usually affects the
rate of reaction
• The change in rate isn’t the same for all
reactants (A and B)
• Must be determined by experiment. (Change
the concentrations of one reactant and hold
the others constant)
A+B→C+D
rate =k*[A]a*[B]b
If a reaction involves the reactants A, B etc =>
The Rate expression
Rate of reaction = - d[A]/dt = k[A]a[B]b
k = rate constant
• Order of reaction: “a” in substance A and “b”
in substance B
• Overall order of reaction: a+b
What happens to the rate in the reaction
A+BC+…
[A]
[B]
Reaction
rate
Double the
concentration
Keep constant
20 = 1
Double the
concentration
Keep constant
No
change
X2
Double the
concentration
Keep constant
X4
22 = 4
Order
21 = 2
Zero
order
First
order
Second
order
Reactants A, B and C in four experiments with altering
concentrations:
A + B + C => ……
Experiment
[B]
mol*dm-3
1.600
[C]
mol*dm-3
0.0600
mol*dm-3*s-1
1
[A]
mol*dm-3
0.400
Initial rate
2
0.800
1.600
0.0600
9.72
3
0.400
0.800
0.0600
4.86
4
0.800
1.600
0.1800
87.5
4.86
Compare Experiment 1 and 2:
Experiment
[B]
mol*dm-3
1.600
[C]
mol*dm-3
0.0600
mol*dm-3*s-1
1
[A]
mol*dm-3
0.400
2
0.800
1.600
0.0600
9.72
Initial rate: [A]: 2x
[B] and [C] : constant
2X rate => [A]1
The reaction is first order in [A]
Initial rate
4.86
Compare Experiment 1 and 3:
Experiment
[B]
mol*dm-3
1.600
[C]
mol*dm-3
0.0600
mol*dm-3*s-1
1
[A]
mol*dm-3
0.400
3
0.400
0.800
0.0600
4.86
Initial rate: [A]: constant [B]: ½
[C] : constant
same rate => [B]0
The reaction is zero order in [B]
Initial rate
4.86
Compare Experiment 2 and 4:
Experiment
[B]
mol*dm-3
1.600
[C]
mol*dm-3
0.0600
mol*dm-3*s-1
2
[A]
mol*dm-3
0.800
4
0.800
1.600
0.1800
87.5
Initial rate: [A]: constant [B]: constant
[C] : 3X
32 = 9X=> [C]2
The reaction is second order in [C]
Initial rate
9.72
Conclusion
• Rate = k*[A]1’*[B]0*[C]2 = k*[A]1*[C]2
• Overall order 1+2 = 3
• k can be calculated using the data from one of
the experiments above
Exercises
• 1-2 on page 120
The order can also be found in a graph where initial
concentration is set against initial rate.
The gradient of the graph => rate of the reaction.
First order reactions
They show an exponential decrease:
the time to half the concentration is equal to go from ½
to 1/4
Half life, t½ = 0.693/k
16.2 Reaction mechanism
Types of reactions:
• A  Products
• A + B  Products
Molecularity
Unimolecular
Bimolecular
• In a Bimolecular reaction the reactants collide
and form an activated complex
2 molecules
Nucleophilic Substitution bimolecular, SN2- topic 10
If we study the reaction:
H
CH3COCH3 + I2
CH3COCH2I + HI
+
• It could be a bimolecular process with a rate
expression rate = k*[CH3COCH3] *[I2]
• The rate is independent of [I2], but first-order in
propanone and acid
=> rate = k*[CH3COCH3]*[H+]
• The reaction must proceed through a series of
steps, a mechanism must be found:
+
H
H
O
O
C
+
F a s t e q u ilib riu m
C
C H3
H3C
C H3
H3C
P ro p a n o n e
H
H
O
O
+
C
C
C H3
H3C
C H3
H2C
+
H+
S lo w R a te D e te rm in in g
S te p
H
O
+
O
I2
C
H2C
C H3
C
+
C H3
H2C
I
HI
Fast
• CH3C(OH+)CH3 is known as a intermediate, not
an activated complex, though it occur at an
energy minimum. In the mechanism there will
be several activated complexes
H
O
+
C
H3C
CH3
Intermediate
Activated complex = Transition state, T
O
C
O
+
H3C
H
3
C
CH3
C
H2C
C
C H
3
I
H
O
C
O
CH3
H2C
H
C H3
Exercises
• 1 and 2 on page 122
16.3 Activation energy
Recall: Maxwell-Boltzmann energy distribution curve.
Temperature  Average speed
Higher temperature
=>More particles with higher speed
=> Greater proportion of particles with energy enough
to react
The Arrhenius equation
• The rate constant, k, can be given if collision rate
and orientation is given
• Ea = activation energy
• T = temperature, K
• R = Gas constant
The equation can also be given in a logarithmic form:
Exersize: Consider the following graph of ln k against (temperature in Kelvin)
for the second order decomposition of N2O into N2 and O
N2 O → N 2 + O
(a) State how the rate constant, k varies with temperature, T
(b) Determine the activation energy, Ea, for this reaction.
(c) The rate expression for this reaction is rate = k [N2O]2 and the rate constant is
0.244 dm3 mol–1 s–1 at 750 °C.
A sample of N2O of concentration 0.200 mol dm–3 is allowed to decompose.
Calculate the rate when 10 % of the N2O has reacted.
Solution:
(a) State how the rate constant, k varies with temperature, T
The Arrhenius equation (it’s in the Data booklet): k=Ae(-Ea/RT)
Logaritming the equation on both sides:
lnk=lnA –Ea/RT
In the graph we see that the gradient is negative
Answer: when k increases T decreases
(b) Determine the activation energy, Ea, for this reaction
The gradient (-Ea/R) can be calculated from the graph:
DY/DX= -3/0,1*10-3= -3*104= -30000
Therefore:
Ea=gradient*R=-30000*8.31= 2,49*105
Answer: Ea= 2,49*105
Solution:
(c) The rate expression for this reaction is rate = k [N2O]2 and the rate constant is
0.244 dm3 mol–1 s–1 at 750 °C.
A sample of N2O of concentration 0.200 mol dm–3 is allowed to decompose.
Calculate the rate when 10 % of the N2O has reacted.
10 % has reacted → 90 % left → 0.200*0.9= 0.180 mol/dm3
Rate= 0.244 * [0.180]2 = 7.91*10-3

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