Consider the following mechanism.
Step1: A+B<---->c (equilibrium)
Step 2: C+A---->d (slow).
Overall: 2A+B---->D.
Determine the rate law for the overall reaction (where the overall rate constant is represented as k).
To determine the rate law for the overall reaction, we need to examine the slow step, which is step 2: C+A → D. The rate law for the slow step will determine the overall rate law of the reaction.
Since the stoichiometry of the slow step shows that one C and one A are involved in the formation of one D, the rate law for the slow step can be expressed as:
rate = k(slow) * [C] * [A]
Where k(slow) is the rate constant for the slow step, [C] represents the concentration of C, and [A] represents the concentration of A.
Now, let's go back to the equilibrium step (Step 1) where A + B ↔ C. From the given information, we know that this step is in equilibrium. Therefore, we can assume that the rate of the forward reaction (A + B → C) is equal to the rate of the reverse reaction (C → A + B), which can be expressed as:
rate_forward = k(forward) * [A] * [B]
rate_reverse = k(reverse) * [C]
Since the forward reaction and the reverse reaction share the same rate, we can equate the two expressions:
k(forward) * [A] * [B] = k(reverse) * [C]
Now, we can solve for [C] in terms of [A] and [B]:
[C] = (k(forward) / k(reverse)) * [A] * [B]
Substituting this expression for [C] into the rate law of the slow step, we get:
rate = k(slow) * (k(forward) / k(reverse)) * [A] * [B] * [A]
Simplifying this equation, we have:
rate = k(overall) * [A]^2 * [B]
Here, k(overall) represents the overall rate constant for the reaction, and [A] and [B] are the concentrations of A and B, respectively.
Thus, the rate law for the overall reaction is:
rate = k(overall) * [A]^2 * [B]