1. Topic 16 Kinetics Rate expressions Reaction mechanism
Activation energy
A + B → C + D

2. 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

3. 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

4. What happens to the rate in the reaction A + B  C + …
Reaction rate
Order
Double the concentration
Keep constant
No change
20 = 1
Zero order
X 2
21 = 2
First order
X 4
22 = 4
Second order

5. Experiment [A] mol*dm-3 [B] [C] Initial rate
Reactants A, B and C in four experiments with altering concentrations: A + B + C => ……
Experiment
[A]
mol*dm-3
[B]
[C]
Initial rate
mol*dm-3*s-1 1
0.400
1.600
0.0600
4.86 2
0.800
9.72 3 4
0.1800
87.5

6. Compare Experiment 1 and 2:
mol*dm-3
[B]
[C]
Initial rate
mol*dm-3*s-1 1
0.400
1.600
0.0600
4.86 2
0.800
9.72
Initial rate: [A]: 2x
[B] and [C] : constant
2X rate => [A]1
The reaction is first order in [A]

7. Compare Experiment 1 and 3:
mol*dm-3
[B]
[C]
Initial rate
mol*dm-3*s-1 1
0.400
1.600
0.0600
4.86 3
0.800
Initial rate: [A]: constant [B]: ½
[C] : constant
same rate => [B]0
The reaction is zero order in [B]

8. Compare Experiment 2 and 4:
mol*dm-3
[B]
[C]
Initial rate
mol*dm-3*s-1 2
0.800
1.600
0.0600
9.72 4
0.1800
87.5
Initial rate: [A]: constant [B]: constant
[C] : 3X
32 = 9X=> [C]2
The reaction is second order in [C]

9. 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

10. Exercises
1-2 on page 120

11. 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.

12. 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

14. 16.2 Reaction mechanism Types of reactions: Molecularity
A  Products Unimolecular
A + B  Products Bimolecular
In a Bimolecular reaction the reactants collide and form an activated complex

15. 2 molecules
Nucleophilic Substitution bimolecular, SN2- topic 10

16. If we study the reaction: 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:

18. 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
Intermediate

19. Activated complex = Transition state, T

20. Exercises
1 and 2 on page 122

21. Temperature  Average speed
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

22. 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

23. The equation can also be given in a logarithmic form:

24. Exersize: Consider the following graph of ln k against (temperature in Kelvin)
for the second order decomposition of N2O into N2 and O
N2O → N2 + 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 dm3 mol–1 s–1 at 750 °C.
A sample of N2O of concentration mol dm–3 is allowed to decompose.
Calculate the rate when 10 % of the N2O has reacted.

25. 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=
Therefore:
Ea=gradient*R=-30000*8.31= 2,49*105
Answer: Ea= 2,49*105

26. Solution:
(c) The rate expression for this reaction is rate = k [N2O]2 and the rate constant is dm3 mol–1 s–1 at 750 °C.
A sample of N2O of concentration mol dm–3 is allowed to decompose.
Calculate the rate when 10 % of the N2O has reacted.
10 % has reacted → 90 % left → *0.9= mol/dm3
Rate= * [0.180]2 = 7.91*10-3