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

A â†’ B k1
B â†’ A k2
B+C â†’ 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.

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ajeigbeibraheem ajeigbeibraheem · Answer 1 · 2021-02-24T03:05:35+01:00

Answer:

Explanation:

From the given information:

A → B k₁

B → A k₂

B + C → D k₃

The rate law = [tex]\dfrac{d[D]}{dt}=k_3[B][C] --- (1)[/tex]

[tex]\dfrac{d[B]}{dt}=k[A] -k_2[B] -k_3[B][C][/tex]

Using steady-state approximation;

[tex]\dfrac{d[B]}{dt}=0[/tex]

[tex]k_1[A]-k_2[B]-k_3[B][C] = 0[/tex]

[tex][B] = \dfrac{k_1[A]}{k_2+k_3[C]}[/tex]

From equation (1), we have:

[tex]\mathbf{\dfrac{d[D]}{dt}= \dfrac{k_3k_1[A][C]}{k_2+k_3[C]}}[/tex]

when the pressure is high;

k₂ << k₃[C]

[tex]\dfrac{d[D]}{dt} = \dfrac{k_3k_1[A][C]}{k_3[C]}= k_1A \ \ \text{first order}[/tex]

k₂ >> k₃[C]

[tex]\dfrac{d[D]}{dt} = \dfrac{k_3k_1[A][C]}{k_2}= \dfrac{k_1k_3}{k_2}[A][C] \ \ \text{second order}[/tex]