Chapter 17, Problem 17.40P

(a)

Interpretation Introduction

Interpretation:

The basis of the approximation that avoids the need to use quadratic formula to find an equilibrium concentration has to be explained.

Concept Introduction:

At equilibrium rate of forward reaction is equal to the rate of backward reaction. So concentration of reactant and products are constant at equilibrium.

Consider a reaction, a moles of A gives b moles of B as follows,

$\text{a A }\rightleftharpoons \text{ b B}$

At equilibrium, rate of formation of B will be equal to rate of decomposition of A. So, rate can be written as follows,

${\text{k}}_{\text{fwd}}{\text{[A]}}_{\text{eq}}^{\text{a}}{\text{ = k}}_{\text{rev}}{\text{[B]}}_{\text{eq}}^{\text{b}}$

Where,

    $\begin{array}{l}{\text{k}}_{\text{fwd}}\text{ = rate of forward reaction}\\ {\text{k}}_{\text{rev}}\text{ = rate of backward reaction}\\ {\text{[A]}}_{\text{eq}}^{}\text{= equilibrium concentration of A}\\ {\text{[B]}}_{\text{eq}}^{}\text{ = equilibrium concentration of B}\\ \text{ a = stoichiometric co-efficient of A}\\ \text{ b = stoichiometric co-efficient of A}\end{array}$

On rearranging the ratio of rate constant becomes equal to ratio of concentration which is equal to a constant called equilibrium constant K.

Equilibrium constant K can be written as follows,

$\text{K = }\frac{{\text{k}}_{\text{fwd}}}{{\text{k}}_{\text{rev}}}\text{ = }\frac{{\text{[B]}}_{\text{eq}}^{\text{b}}}{{\text{[A]}}_{\text{eq}}^{\text{a}}}$

So equilibrium constant K can be defined as the ratio of equilibrium concentration of product to reactant at a particular temperature.

(b)

Interpretation Introduction

Interpretation:

The situation when the approximation that avoids the need to use quadratic formula to find an equilibrium concentration cannot be used has to be explained.

Concept Introduction:

At equilibrium rate of forward reaction is equal to the rate of backward reaction. So concentration of reactant and products are constant at equilibrium.

Consider a reaction, a moles of A gives b moles of B as follows,

$\text{a A }\rightleftharpoons \text{ b B}$

At equilibrium, rate of formation of B will be equal to rate of decomposition of A. So, rate can be written as follows,

${\text{k}}_{\text{fwd}}{\text{[A]}}_{\text{eq}}^{\text{a}}{\text{ = k}}_{\text{rev}}{\text{[B]}}_{\text{eq}}^{\text{b}}$

Where,

    $\begin{array}{l}{\text{k}}_{\text{fwd}}\text{ = rate of forward reaction}\\ {\text{k}}_{\text{rev}}\text{ = rate of backward reaction}\\ {\text{[A]}}_{\text{eq}}^{}\text{= equilibrium concentration of A}\\ {\text{[B]}}_{\text{eq}}^{}\text{ = equilibrium concentration of B}\\ \text{ a = stoichiometric co-efficient of A}\\ \text{ b = stoichiometric co-efficient of A}\end{array}$

On rearranging the ratio of rate constant becomes equal to ratio of concentration which is equal to a constant called equilibrium constant K.

Equilibrium constant K can be written as follows,

$\text{K = }\frac{{\text{k}}_{\text{fwd}}}{{\text{k}}_{\text{rev}}}\text{ = }\frac{{\text{[B]}}_{\text{eq}}^{\text{b}}}{{\text{[A]}}_{\text{eq}}^{\text{a}}}$

So equilibrium constant K can be defined as the ratio of equilibrium concentration of product to reactant at a particular temperature.