Chapter 17, Problem 17.3IA

(a)

Interpretation Introduction

Interpretation:

The rate law of the base-catalysed reaction in which $\text{AH}$ goes to products according to the following scheme has to be deduced.

  $\begin{array}{l}\text{AH}\text{+}\text{B}\underset{{\text{k}}_{\text{a}}^{\text{'}}}{\overset{{\text{k}}_{\text{a}}}{\rightleftarrows }}{\text{BH}}^{\text{+}}\text{+}{\text{A}}^{\text{-}}\\ \\ {\text{A}}^{\text{-}}\text{+}\text{AH}\stackrel{{\text{k}}_{\text{b}}}{\to }\text{product}\text{(rate-determining)}\end{array}$

Explanation of Solution

Rate-determining step is the slow step of the reaction.  The rate law for the slow reaction is

  $\text{υ}\text{=}{\text{k}}_{\text{b}}{\text{[A}}^{\text{-}}\text{][AH]}$

${\text{A}}^{\text{-}}$ is the intermediate.  So,

  $\begin{array}{l}\frac{{\text{d[A}}^{\text{-}}\text{]}}{\text{dt}}\text{=}\text{0}\\ \\ {\text{k}}_{\text{a}}{\text{[AH][B]-k}}_{\text{a}}^{\text{'}}{\text{[BH}}^{\text{+}}{\text{][A}}^{\text{-}}{\text{]-k}}_{\text{b}}{\text{[A}}^{\text{-}}\text{][AH]}\text{=}\text{0}\\ \\ {\text{k}}_{\text{a}}^{\text{'}}{\text{[BH}}^{\text{+}}{\text{][A}}^{\text{-}}{\text{]+k}}_{\text{b}}{\text{[A}}^{\text{-}}\text{][AH]}\text{=}{\text{k}}_{\text{a}}\text{[AH][B]}\\ \\ {\text{[A}}^{\text{-}}\text{]}\left({\text{k}}_{\text{a}}^{\text{'}}{\text{[BH}}^{\text{+}}{\text{]+k}}_{\text{b}}\text{[AH]}\right)\text{=}{\text{k}}_{\text{a}}\text{[AH][B]}\\ \\ {\text{[A}}^{\text{-}}\text{]}\text{=}\frac{{\text{k}}_{\text{a}}\text{[AH][B]}}{{\text{k}}_{\text{a}}^{\text{'}}{\text{[BH}}^{\text{+}}{\text{]+k}}_{\text{b}}\text{[AH]}}\end{array}$

If the rate of conversion of intermediates to reactants is greater than rate of conversion of intermediate to products, then pre-equilibrium will be present.

  $\begin{array}{l}{\text{k}}_{\text{a}}^{\text{'}}{\text{[BH}}^{\text{+}}{\text{]>>k}}_{\text{b}}\text{[AH]}\\ \\ {\text{[A}}^{\text{-}}\text{]}\text{will}\text{be}\text{changed}\text{as}\frac{{\text{k}}_{\text{a}}\text{[AH][B]}}{{\text{k}}_{\text{a}}^{\text{'}}{\text{[BH}}^{\text{+}}\text{]}}\end{array}$

The rate law can be calculated as

  $\begin{array}{l}\text{υ}\text{=}{\text{k}}_{\text{b}}\frac{{\text{k}}_{\text{a}}\text{[AH][B]}}{{\text{k}}_{\text{a}}^{\text{'}}{\text{[BH}}^{\text{+}}\text{]}}\text{[AH]}\\ \\ \text{υ}\text{=}\frac{{\text{k}}_{\text{b}}{\text{k}}_{\text{a}}{\text{[AH]}}^{\text{2}}\text{[B]}}{{\text{k}}_{\text{a}}^{\text{'}}{\text{[BH}}^{\text{+}}\text{]}}\\ \\ \text{υ}\text{=}\frac{{\text{k}}_{\text{b}}\frac{{\text{k}}_{\text{a}}}{{\text{k}}_{\text{a}}^{\text{'}}}{\text{[AH]}}^{\text{2}}\text{[B]}}{{\text{[BH}}^{\text{+}}\text{]}}\end{array}$

The equilibrium constant for the fast reaction is

  $\text{K}\text{=}\frac{{\text{k}}_{\text{a}}}{{\text{k}}_{\text{a}}^{\text{'}}}$

The rate law of base-catalysed reaction is

  $\text{υ}\text{=}\frac{{\text{k}}_{\text{b}}{\text{K[AH]}}^{\text{2}}\text{[B]}}{{\text{[BH}}^{\text{+}}\text{]}}$

(b)

Interpretation Introduction

Interpretation:

The rate law of the acid-catalysed reaction in which $\text{HA}$ goes to products according to the following scheme has to be deduced.

  $\begin{array}{l}\text{HA}\text{+}{\text{H}}^{\text{+}}\underset{{\text{k}}_{\text{a}}^{\text{'}}}{\overset{{\text{k}}_{\text{a}}}{\rightleftarrows }}{\text{HAH}}^{\text{+}}\\ \\ {\text{HAH}}^{\text{+}}\text{+}\text{B}\stackrel{{\text{k}}_{\text{b}}}{\to }{\text{BH}}^{\text{+}}\text{+}\text{AH}\text{(rate-determining)}\end{array}$

Explanation of Solution

Rate-determining step is the slow step of the reaction.  The rate law for the slow reaction is

  $\text{υ}\text{=}{\text{k}}_{\text{b}}{\text{[HAH}}^{+}\text{][B]}$

${\text{HAH}}^{\text{+}}$ is the intermediate.  So,

  $\begin{array}{l}\frac{{\text{d[HAH}}^{\text{+}}\text{]}}{\text{dt}}\text{=}\text{0}\\ \\ {\text{k}}_{\text{a}}{\text{[HA][H}}^{\text{+}}{\text{]-k}}_{\text{a}}^{\text{'}}{\text{[HAH}}^{\text{+}}{\text{]-k}}_{\text{b}}{\text{[HAH}}^{\text{+}}\text{][B]}\text{=}\text{0}\\ \\ {\text{k}}_{\text{a}}^{\text{'}}{\text{[HAH}}^{\text{+}}\text{]}\text{+}{\text{k}}_{\text{b}}{\text{[HAH}}^{\text{+}}\text{][B]}\text{=}{\text{k}}_{\text{a}}{\text{[HA][H}}^{\text{+}}\text{]}\\ \\ {\text{[HAH}}^{\text{+}}\text{]}\left({\text{k}}_{\text{a}}^{\text{'}}\text{+}{\text{k}}_{\text{b}}\text{[B]}\right)\text{=}{\text{k}}_{\text{a}}{\text{[HA][H}}^{\text{+}}\text{]}\\ \\ {\text{[HAH}}^{\text{+}}\text{]}\text{=}\frac{{\text{k}}_{\text{a}}{\text{[HA][H}}^{\text{+}}\text{]}}{{\text{k}}_{\text{a}}^{\text{'}}\text{+}{\text{k}}_{\text{b}}\text{[B]}}\end{array}$

If the rate of conversion of intermediates to reactants is greater than rate of conversion of intermediate to products, then pre-equilibrium will be present.

  $\begin{array}{l}{\text{k}}_{\text{a}}^{\text{'}}{\text{>>k}}_{\text{b}}\text{[B]}\\ \\ {\text{[HAH}}^{\text{+}}\text{]}\text{will}\text{be}\text{changed}\text{as}\frac{{\text{k}}_{\text{a}}{\text{[HA][H}}^{\text{+}}\text{]}}{{\text{k}}_{\text{a}}^{\text{'}}}\end{array}$

The rate law can be calculated as

  $\begin{array}{l}\text{υ}\text{=}{\text{k}}_{\text{b}}\frac{{\text{k}}_{\text{a}}{\text{[HA][H}}^{+}\text{]}}{{\text{k}}_{\text{a}}^{\text{'}}}\text{[B]}\\ \\ \text{υ}\text{=}\frac{{\text{k}}_{\text{b}}{\text{k}}_{\text{a}}{\text{[HA][H}}^{+}\text{][B]}}{{\text{k}}_{\text{a}}^{\text{'}}}\end{array}$

The equilibrium constant for the fast reaction is

  $\text{K}\text{=}\frac{{\text{k}}_{\text{a}}}{{\text{k}}_{\text{a}}^{\text{'}}}$

The rate law of acid-catalysed reaction is

  $\text{υ}\text{=}{\text{k}}_{\text{b}}{\text{K[HA][H}}^{\text{+}}\text{][B]}$