Chemical Kinetics: Reaction Mechanism

[tehaznboi]

Homework Statement

1. According to the mechanism below, what is the correct expression for the rate of consumption of A? In the mechanism, M is a molecule that collides with A and A* is a highly-energized form of A.

Step 1 (reversible): A + M <---> A* + M rate constants k1 and k-1

Step 2 (irreversible): A* ---> C rate constant k2

Select one of the following:

Rate = k1 [A] [M] / (k-1[M] + k2)
Rate = [A] / [C]
Rate = k1k2 [A] [M] / (k-1[M] + k2)
Rate = k1k2 [A]
Rate = k1[A]
Rate = (k1 [A] [M] + k-2[C] ) / (k-1[M] + k2)
Rate=k1k2 / (k-1+k-2)
Rate = (k1+k2-k-1-k-2) [A]
Rate = k1k2 [A] / (k-1+k2[M])
Rate=k1+k2-k-1-k-2
Rate = [C] / [A]
Rate = k1k2 [A] [M] / k-1

Homework Equations

The Attempt at a Solution

I tried to assume a steady-state approximation as I'm not sure which is the slow step. I found the overall rate of reaction and assumed it to be equivalent to the consumption of A.

 

[nate808]

However, I am not sure if this is the correct approach or if there is a more accurate expression for the rate of consumption of A.

Hi there,

The correct expression for the rate of consumption of A in this mechanism would be:

Rate = k1k2 [A] [M] / (k-1[M] + k2)

This is because the rate of consumption of A is dependent on both the forward and reverse reactions in the first step, as well as the irreversible step in the second step. The steady-state approximation is a good approach, but it is important to consider all of the reactions and their rate constants in order to accurately describe the overall rate of consumption of A.

I hope this helps. Let me know if you have any further questions.