Chapter 19, Problem 19C.1P

(a)

Interpretation Introduction

Interpretation:

The rate law for the reaction between A and B according to the given mechanism has to be written.

Explanation of Solution

  $\begin{array}{l}{A}_{\left(g\right)}\rightleftharpoons A_{\left(ads\right)}\\ \\ B_{\left(g\right)}\rightleftharpoons B_{\left(ads\right)}\\ \\ A_{\left(ads\right)}+B_{\left(ads\right)}\underrightarrow{slow}A{B}_{\left(ads\right)}\underrightarrow{fast}A{B}_{\left(g\right)}\end{array}$

The rate of the reaction or the rate law is given by

  Rate $=k{\theta }_A{\theta }_B=\frac{k{b}_A{P}_A{b}_B{P}_B}{{\left(1+b_A{P}_A+b_B{P}_B\right)}^2}$

Where,

$k=$ Rate constant

$P_A=$ Partial pressure of A

$P_B=$ partial pressure of B

$\begin{array}{l}{b}_A=\frac{k_{A,ad}}{k_{A,d}}\\ \\ b_B=\frac{k_{B,ad}}{k_{B,d}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The rate law for the case where the partial pressure of the reactants is low has to be found out.

Explanation of Solution

  $\begin{array}{l}{A}_{\left(g\right)}\rightleftharpoons A_{\left(ads\right)}\\ \\ B_{\left(g\right)}\rightleftharpoons B_{\left(ads\right)}\\ \\ A_{\left(ads\right)}+B_{\left(ads\right)}\underrightarrow{slow}A{B}_{\left(ads\right)}\underrightarrow{fast}A{B}_{\left(g\right)}\end{array}$

The rate of the reaction or the rate law is given by

  Rate $=k{\theta }_A{\theta }_B=\frac{k{b}_A{P}_A{b}_B{P}_B}{{\left(1+b_A{P}_A+b_B{P}_B\right)}^2}$

Consider the case where the partial pressure of the reactants that is here, A and B are low.

So, $b_A{P}_A\ll 1$ and $b_B{P}_B\ll 1$. Both the terms can be neglected in the denominator and the rate law becomes

  Rate $=k{\theta }_A{\theta }_B=k{b}_A{P}_A{b}_B{P}_B=k\text{'}{P}_A{P}_B$

Where, $k\text{'}=k{b}_A{b}_B$

This implies that first order in both the reactants. Hence, the overall order is 2.

(c)

Interpretation Introduction

Interpretation:

The given mechanism could ever account for zeroth order kinetics has to be determined.

Explanation of Solution

  $\begin{array}{l}{A}_{\left(g\right)}\rightleftharpoons A_{\left(ads\right)}\\ \\ B_{\left(g\right)}\rightleftharpoons B_{\left(ads\right)}\\ \\ A_{\left(ads\right)}+B_{\left(ads\right)}\underrightarrow{slow}A{B}_{\left(ads\right)}\underrightarrow{fast}A{B}_{\left(g\right)}\end{array}$

The rate of the reaction or the rate law is given by

  Rate $=k{\theta }_A{\theta }_B=\frac{k{b}_A{P}_A{b}_B{P}_B}{{\left(1+b_A{P}_A+b_B{P}_B\right)}^2}$

Consider the case of mixed pressure.

So, $b_A{P}_A\ll 1\ll b_B{P}_B$ or $b_B{P}_B\ll 1\ll b_A{P}_A$.

Consider $b_A{P}_A\ll 1\ll b_B{P}_B$. The term $b_A{P}_A$ can be neglected in the denominator and also 1 can be neglected in comparison to $b_B{P}_B$.  Then the rate law becomes

  Rate $=k{\theta }_A{\theta }_B=\frac{k{b}_A{P}_A}{b_B{P}_B}=\frac{k{P}_A}{P_B}$

This implies that first order in A and negative first order in B.  Hence, the overall order is zero.

Yes, this mechanism can account for zeroth order kinetics in the case of mixed pressure condition.