Why are the Two SSA Equations the Same in a Catalytic Cycle?

[i_love_science]

The following is a basic surface-catalyzed isomerization reaction:
A -> P
This can be modeled by the following mechanism:
A + S -> A-S
A-S -> P + S

The text is trying to derive the rate law for the formation of P.
d[P]/dt = k[A-S]
It can also be written as
d[P]/dt = k'[theta_A], where theta_A is the fraction of sites to which a reactant is bound.

We are trying to find an expression for theta_A, which the text gives as
theta_A = [ A-S ] / [S]+[ A-S ]

Since theta_A contains intermediates, we need to get rid of both [S] and [A-S], by applying the steady-state approximation to both.

d[A-S]/dt = k₁[A] - (k₂ + k_-1)[A-S]
d[S]/dt = -k₁[A] + (k₂ + k_-1)[A-S]

This is the part I don't understand:
Because we have a catalytic cycle, the two halves of the cycle fit together perfectly, and the two SSA equations are the same.

Can anyone explain to me why that is the case? Thank you!

 

[BvU]

Hi,

which the text gives as
theta_A = [ A-S ] / +[ A-S ]

Are you certain this is what the text says ? [ A-S ] / [ S ] seems so much more sensible !?

what are k1, k2 and k-1 (or did you mean k_-1 ? Makes a difference ! )

the second d/dt : of what ?

 

[mfb]

The forum interprets [S] as command to strike-through the following text. To avoid that you can write [plain][S][/plain], it will be shown as [S]. Or simply add a space: [ S] is ignored by the forum. I fixed the initial post, it should look as intended now.

 

[i_love_science]

Hi,
Are you certain this is what the text says ? [ A-S ] / [ S ] seems so much more sensible !?

what are k1, k2 and k-1 (or did you mean k_-1 ? Makes a difference ! )

the second d/dt : of what ?

Yes, I meant k_-1, k₁, and k₂. It is d[S]/dt.

I made the edits, could anyone help me out with my question? Thanks!

 

[DrStupid]

The following is a basic surface-catalyzed isomerization reaction:
A -> P
This can be modeled by the following mechanism:
A + S -> A-S
A-S -> P + S

This is a Michaelis-Menten kinetics. It is typical for enzyme reactions but it also applies to many other catalytic reactions with a preceding equilibrium.

d[P]/dt = k'[theta_A], where theta_A is the fraction of sites to which a reactant is bound.

We are trying to find an expression for theta_A, which the text gives as
theta_A = [ A-S ] / [S]+[ A-S ]

I assume you actually mean theta_A = [ A-S ] / ([S]+[ A-S ])

d[A-S]/dt = k₁[A] - (k₂ + k_-1)[A-S]
d[S]/dt = -k₁[A] + (k₂ + k_-1)[A-S]
This is the part I don't understand:
Because we have a catalytic cycle, the two halves of the cycle fit together perfectly, and the two SSA equations are the same.

You are right. The two equations are redundant and give you [A-S] only. [S] (and therefore also theta_A) remains unknown.

S is obviously assumed to be present in large access. That justifies the steady state for [S] and turns the first elementary reaction A + S -> A-S into a pseudo-first-order reaction with [S] included into the rate constant k₁. But that would mean that [S] is eliminated from the rate equations while it is still present in theta_A. I have no idea how to solve this problem.

I would use the original second-order kinetics instead:

d[A-S]/dt = k’₁[A][S] - (k₂ + k_-1)[A-S] = 0

That results in

[A-S] = k’₁[A][S]/ (k₂ + k_-1)

and

theta_A = [ A-S ] / ([S]+[ A-S ]) = k’₁[A]/( k₂ + k_-1 + k’₁[A])

[S] is still hidden in k' but that's not a problem because it is a constant factor as long as it remains stationary. It is rather confusing that k' also includes [A]:

k’ = k·([S] + [A-S]) = k₂ [S](1+ k’₁[A]/(k₂ + k_-1))

In the equation d[P]/dt = k'[theta_A] it looks like a constant but it actually isn't.
Maybe the intermediate [A-S] is assumed to be very instable. That would result in

k’₁[A]/(k₂ + k_-1) << 1

and therefore in

k’ $\approx$ k₂ [S]

That would also imply

theta_A $\approx$ k’₁[A]/(k₂ + k_-1)

and therefore

d[P]/dt $\approx$ k₁k₂/(k₂ + k_-1)[A]

PS: I just realize that k’₁[A]/(k₂ + k_-1) << 1 may also result from very low concentrations of A (and not only from an instable complex [A-S]). However, I'm still not sure how to finish the derivation using the pseudo-first-order kinetics for A + S -> A-S.

 

[i_love_science]

The following is a basic surface-catalyzed isomerization reaction:
A -> P
This can be modeled by the following mechanism:
A + S -> A-S
A-S -> P + S

d[A-S]/dt = k1[A] - (k₂ + k_-1)[A-S]
d[S]/dt = -k₁[A] + (k₂ + k_-1)[A-S]

Because we have a catalytic cycle, the two halves of the cycle fit together perfectly, and the two SSA equations are the same.

You are right. The two equations are redundant and give you [A-S] only. [S] (and therefore also theta_A) remains unknown.

Why do d[A-S]/dt and d[S]/dt turn out to be the same in terms of what is going on in the reaction?

Is the rate d[P]/dt considering both reaction steps? Is [A-S] considered the product and [S] considered the reactant, or vice versa?

Thanks!

 

[DrStupid]

Why do d[A-S]/dt and d[S]/dt turn out to be the same in terms of what is going on in the reaction?

The reaction rates must be equal and opposite because every S reacts to an A-S and every A-S reacts back to an S. You don’t even need the reactions to see that. It is just stoichiometry. The sum of the free and occupied surface areas remains constant:

[S]₀ = [S] + [A-S] = const.

That results in the rates

d[S]₀/dt = d[S]/dt + d[A-S]/dt = 0

Is the rate d[P]/dt considering both reaction steps?

Yes, it does. The rate equation for P is based on the last reaction step only. However, this equation contains the concentration [A-S] of the intermediate which is formed in the first step. Therefore it depends on the first step too. There is no A-S -> P + S without A + S -> A-S.

Is [A-S] considered the product and [S] considered the reactant, or vice versa?

That depends on the reaction steps. A-S is the product of the reaction step A + S -> A-S and the reactant of the reaction steps A-S -> A + S and A-S -> P + S. For S it is the other way around.

 

[i_love_science]

Thank you! I understand now.