Kinetics (Pseudo-Order)

1. Reaction Kinetics
Pseudo-order Demystified

2. Pseudo-order
1.Definition of pseudo-order
2.Application in continuous expts
3.Half-life for pseudo 1
st
order reactions
Demystified

3. Pseudo-order
Consider reaction where A and B reacts to form P:
A + B → P
Assume we already know the rate eqn:
Rate = k[A]
1
[B]
1
Overall order = 2
Definition of

4. Pseudo-order
When an expt is carried out with:
[A] = 0.0100 mol dm
–3
[B] = 0.100 mol dm
–3
10 times more concentrated
If the reaction were to go to completion,
[B] remaining would be 0.099 ≈ 0.100 mol dm
–3

i.e. [B] remains constant throughout the expt.
Definition of

5. Pseudo-order
Rate = k[A]
1
[B]
1
For expt where [B] >> [A], [B] = constant
Rate = (k[B]
1
) [A]
1
Rate= k’[A]
1
where k’ = k[B]
1
Experimental results will show that the overall order is 1
instead of 2.
Definition of

6. Pseudo-order
When a reactant is made much more concentrated than the
other, it is made to have no effect on the rate of that
particular expt i.e. appears to be zero order.
Rate = k[A]
1
[B]
1
reaction is 2
nd
order
For expt where [B] >> [A],
[B] = constant
Rate = k’[A]
1
reaction appears to be 1st order
pseudo (‘fake’) 1
st
order
Definition of

7. Continuous Expts
The concept of pseudo-order is especially important in
continuous expts.
Recall:
Continuous expt  single run, monitor conc. of one rxt
with time.
Application in

8. Consider reaction where A and B reacts to form P:
A + B → P
General rate equation:
Rate = k[A]
m
[B]
n
Continuous Expts
Application in

9. Assume we carried out a continuous expt and monitored
[A] vs t:
Continuous Expts
Application in

10. Observation: half-life is constant
Conclusion: reaction is first order
Rate = k[A]
m
[B]
n
Question:
Is m=1 or n=1?
Is it first order w.r.t. A or B?
Continuous Expts
Application in

11. Is it first order w.r.t. to A or B?
Most students would answer ‘A’ as that is the case for
typical qns.
The correct answer is: the order observed in a continuous
expt is the overall order (m+n)
Then why is it that for typical qns, we can conclude the
reaction is first order w.r.t. A?
Continuous Expts
Application in

12. Then why is it that for typical qns, we can conclude the
reaction is first order w.r.t. A?
That is because for continuous expts to be meaningful, we
isolate one reactant by making the concentrations of the
other reactant(s) much higher i.e. the other reactant(s)
appear to be zero order.
Continuous Expts
Application in

13. Assume the expt was carried out with:
[A] = 0.0100 mol dm
–3
[B] = 0.100 mol dm
–3
Since [B] >> [A], [B] appears to be zero order (n=0)
From the graph,
m + n = 1
m = 1
Reaction is 1
st
order w.r.t. A
Rate = k[A]
m
[B]
n
Continuous Expts
Application in

14. Another possible scenario where pseudo-order kinetics is
observed is when one reactant is a catalyst.
Catalyst is chemically unchanged/ regenerated through the
expt; concentration of catalyst is constant i.e. appears to be
zero order
Continuous Expts
Application in

15. Assume the expt was carried out with:
[A] = 0.0100 mol dm
–3

[B] = 0.0100 mol dm
–3
(B is a catalyst)
B is a catalyst i.e. n = 0
From the graph,
m + n = 1
m = 1
Reaction is 1st order w.r.t. A
Continuous Expts
Application in

16. Rate = k[A]
m
[B]
n
To find m and n using continuous expt, we need to isolate
one reactant at a time by making the concentration of
the other reactant much higher.
To find m, we perform an expt where [B] >> [A]
To find n, we perform an expt where [A] >> [B]
Continuous Expts
Application in

17. For a 1
st
order reaction:
- Half-life is constant
- Value of half-life is independent on conc. of reactants
t
1/2
=
pseudo 1
st
order expts
Half-life for
ln2
k

18. pseudo 1
st
order expts
Half-life for
For a pseudo 1
st
order reaction:
- Half-life is constant
- Value of half-life is dependent on conc. of reactants
t
1/2
=
ln2
k’

19. pseudo 1
st
order expts
Half-life for
Example:
A + B → P
Rate = k[A]
1
[B]
1

20. pseudo 1
st
order expts
Half-life for
Expt 1 (t
1/2
):
[A] = 0.0100 mol dm
–3
[B] = 0.100 mol dm
–3
Rate = (k[B]
1
) [A]
1
Rate= k’[A]
1
where k’ = k[B]
1
pseudo first order
t
1/2
= = =
ln2
k’
ln2
k[B]
ln2
k[0.100]

21. pseudo 1
st
order expts
Half-life for
Expt 2 (t
1/2
’)
[A] = 0.0100 mol dm
–3
[B] = 0.050 mol dm
–3
t
1/2’ = = = = 2 x t
1/2
ln2
k’
ln2
k[B]
ln2
k[0.050]