Question

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 ????.

Answer 1

The rate law for the overall reaction is:
`\[ \text{Rate}_{\text{overall}} = k_{\text{overall}}[A]^2[B] \]`

How to determine rate law?

The given reaction mechanism involves two steps:

`\[ \text{Step 1: } A + B \underset{k_1}{\overset{k_(-1)}{\rightleftharpoons}} C \]`

`\[ \text{Step 2: } C + A \xrightarrow{k_2} D \]`

The overall reaction is the sum of these steps:

`\[ 2A + B \xrightarrow{k_{\text{overall}}} D \]`

To determine the rate law for the overall reaction, consider the rate-determining step, which is the slowest step in the mechanism. In this case, it's Step 2.

The rate law for Step 2 is given by:

`\[ \text{Rate}_{\text{Step 2}} = k_2[C][A] \]`

At equilibrium for Step 1:

`\[ K_{\text{eq}} = ([C])/([A][B]) \]`

Now, rearrange the equation to solve for [C]:

`\[ [C] = K_{\text{eq}}[A][B] \]`

Now substitute this expression for [C] into the rate law for Step 2:

`\[ \text{Rate}_{\text{Step 2}} = k_2K_{\text{eq}}[A][B][A] \]`

Combine like terms:

`\[ \text{Rate}_{\text{Step 2}} = k_{\text{overall}}[A]^2[B] \]`

Therefore, the rate law for the overall reaction is:

`\[ \text{Rate}_{\text{overall}} = k_{\text{overall}}[A]^2[B] \]`

The overall rate constant
`\(k_{\text{overall}}\)` incorporates the rate constants for both steps.

Answer 2

rate = k[A][B] where k = k₂K

Step-by-step explanation:

Your mechanism is a slow step with a prior equilibrium:

`\begin{array}{rrcl}\text{Step 1}:& \text{A + B} & \xrightarrow [k_(-1)]{k_(1)} & \text{C}\\\text{Step 2}: & \text{C + A} & \xrightarrow [ ]{k_(2)} & \text{D}\\\text{Overall}: & \text{2A + B} & \longrightarrow \, & \text{D}\\\end{array}`

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

1. Write the rate equations:

`-\frac{\text{d[A]}}{\text{d}t} = -\frac{\text{d[B]}}{\text{d}t} = -k_(1)[\text{A}][\text{B}] + k_(1)[\text{C}]\\\\\frac{\text{d[C]}}{\text{d}t} = k_(1)[\text{A}][\text{B}] - k_(2)[\text{C}]\\\\\frac{\text{d[D]}}{\text{d}t} = k_(2)[\text{C}]`

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