Question

How to set up the rate expressions for the following mechanism?

A → B k1
B → A k2
B+C → D k3

If the concentration of B is small compared with the concentrations of A, C, and D, the steady-state approximation may be used to derive the rate law. Derive the rate law, and show that this reaction may follow the first-order equation at high pressures and the second-order equation at low pressures.

Answer 1

Step-by-step explanation:

From the given information:

A → B k₁

B → A k₂

B + C → D k₃

The rate law =
`(d[D])/(dt)=k_3[B][C] --- (1)`

`(d[B])/(dt)=k[A] -k_2[B] -k_3[B][C]`

Using steady-state approximation;

`(d[B])/(dt)=0`

`k_1[A]-k_2[B]-k_3[B][C] = 0`

`[B] = (k_1[A])/(k_2+k_3[C])`

From equation (1), we have:

`\mathbf{(d[D])/(dt)= (k_3k_1[A][C])/(k_2+k_3[C])}`

when the pressure is high;

k₂ << k₃[C]

`(d[D])/(dt) = (k_3k_1[A][C])/(k_3[C])= k_1A \ \ \text{first order}`

k₂ >> k₃[C]

`(d[D])/(dt) = (k_3k_1[A][C])/(k_2)= (k_1k_3)/(k_2)[A][C] \ \ \text{second order}`