[toffeefan]

In a catalytic cycle:

R₁ + C  R₁+C (fast)
R₁C + R₂  Product + C (slow)

The first step is in virtual equilibrium, and the second therefore slow (rate determining).

The question they have ask is, how does the form of the overall kinetics of the RDS compare with the SSA? I don't really know how to answer this question to be honest.

I have already work out the overall rate for both the SSA and RDS:

RDS 

r(overall) = K₁K₃ [R₁][C][R₂]
                K₂

SSA

r(overall) = K₁K₃ [R₁][C][R₂]
                K₂ + K₃[R₂]

I notice that when comparing both the RDS and SSA overall rate, there is an extra K₃[R₂] in the denominator of the SSA equation. What does that imply? Can i just state that since there is an extra K₃[R₂]term in the denominator, the overall rate will be slower compared to the RDS equation?

[Yggdrasil]

Sure, that's a good answer.  It's may also be a useful exercise to consider under what conditions the SSA and RDS approaches give similar answers, and under which conditions they differ.

[toffeefan]

Sure, that's a good answer.  It's may also be a useful exercise to consider under what conditions the SSA and RDS approaches give similar answers, and under which conditions they differ.

Hi thanks for replying. For your questions:

1) what conditions the SSA and RDS approaches give similar answers?

Can i say that: When SSA is assume, we treat all rates as equal, that is why we have the extra K3[R2] term in the denominator. To make the SSA answer the same as RDS, we first need to know that the last step of the catalytic cycle is a slow process. So that K3[R2] = 0, and the answer of the SSA will then be similar to the RDS.

2) under which conditions they differ.

As in the question, the overall rate for the SSA and RDS is different because when SSA is assume here, we do not know which is the RDS. So we've to include all rates of the catalytic cycle to come out with the overall rate for SSA. Should we know that the last step is a slow process, the overall rate equation will only consist of the slow process step.

[Yggdrasil]

Hi thanks for replying. For your questions:

1) what conditions the SSA and RDS approaches give similar answers?

Can i say that: When SSA is assume, we treat all rates as equal, that is why we have the extra K3[R2] term in the denominator. To make the SSA answer the same as RDS, we first need to know that the last step of the catalytic cycle is a slow process. So that K3[R2] = 0, and the answer of the SSA will then be similar to the RDS.

Great.  Another way of writing this condition is k₂ >> k₃[R₂].

2) under which conditions they differ.

As in the question, the overall rate for the SSA and RDS is different because when SSA is assume here, we do not know which is the RDS. So we've to include all rates of the catalytic cycle to come out with the overall rate for SSA. Should we know that the last step is a slow process, the overall rate equation will only consist of the slow process step.

Another great answer.  Good job.

[toffeefan]

Hi thanks for the help.