Kinetics (Pseudo-Order)

Reaction KineticsPseudo-order Demystified

Pseudo-order

1. Definition of pseudo-order2. Application in continuous expts3. Half-life for pseudo 1st order reactions

Demystified

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

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

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

Pseudo-orderWhen 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 2nd order

For expt where [B] >> [A], [B] = constantRate = k’[A]1 reaction appears to be 1st order

pseudo (‘fake’) 1st order

Definition of

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

Consider reaction where A and B reacts to form P:

A + B → P

General rate equation:

Rate = k[A]m[B]n

Continuous ExptsApplication in

Assume we carried out a continuous expt and monitored [A] vs t:

Continuous ExptsApplication in

Observation: half-life is constantConclusion: 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 ExptsApplication in

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 ExptsApplication in

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 ExptsApplication in

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 = 1m = 1Reaction is 1st order w.r.t. A

Rate = k[A]m[B]n

Continuous ExptsApplication in

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 ExptsApplication in

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 = 0From the graph,m + n = 1m = 1Reaction is 1st order w.r.t. A

Continuous ExptsApplication in

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 ExptsApplication in

For a 1st order reaction:- Half-life is constant- Value of half-life is independent on conc. of reactants

t1/2 =

pseudo 1st order exptsHalf-life for

ln2k

pseudo 1st order exptsHalf-life for

For a pseudo 1st order reaction:- Half-life is constant- Value of half-life is dependent on conc. of reactants

t1/2 =ln2k’

pseudo 1st order exptsHalf-life for

Example:

A + B → P

Rate = k[A]1[B]1

pseudo 1st order exptsHalf-life for

Expt 1 (t1/2):

[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

t1/2 = = =ln2k’

ln2k[B]

ln2k[0.100]

pseudo 1st order exptsHalf-life for

Expt 2 (t1/2’)

[A] = 0.0100 mol dm–3 [B] = 0.050 mol dm–3

t1/2’ = = = = 2 x t1/2ln2k’

ln2k[B]

ln2k[0.050]