Consider the mechanism. step 1: a+b↽−−⇀ca+b↽−−⇀c equilibrium step 2: c+a⟶dc+a⟶d slow overall: 2a+b⟶d2a+b⟶d determine the rate law for the overall reaction, where the overall rate constant is represented as

rate = k[A][B] where k = k₂K

Explanation:

Your mechanism is a slow step with a prior equilibrium:

(The arrow in Step 1 should be equilibrium arrows).

1. Write the rate equations:

2. Derive the rate law

Assume k₋₁ ≫ k₂.  

Then, in effect, we have an equilibrium that is only slightly disturbed by C slowly reacting to form D.  

In an equilibrium, the forward and reverse rates are equal:

k₁[A][B] = k₋₁[C]

[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)

rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]

The rate law is  

rate = k[A][B] where k = k₂K

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa

idk what is it? is this a question that you know the answer or u dont