Chapter 1, Problem 1.27P

(a)

Interpretation Introduction

Interpretation:

The expression for the equilibrium constant for the reaction, ${\text{3Ag}}^{+}{}_{\text{(aq)}}{\text{ + PO}}_4^{3-}{}_{\text{(aq)}}\stackrel{}{\rightleftharpoons }{\text{ Ag}}_{\text{3}}{\text{PO}}_{\text{4(s)}}$ has to be written.

Concept Introduction:

Equilibrium constant$\left({\text{K}}_{\text{c}}\right)$:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

    $\text{aA}\rightleftharpoons \text{bB}$

    $\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}{\left[\text{A}\right]}^{\text{a}}{\text{=k}}_{\text{r}}{\left[\text{B}\right]}^{\text{b}}\end{array}$

On rearranging,

    $\frac{{\left[\text{B}\right]}^{\text{b}}}{{\left[\text{A}\right]}^{\text{a}}}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}{\text{=K}}_{\text{c}}$

Where,

    ${\text{k}}_{\text{f}}$ is the rate constant of the forward reaction.

    ${\text{k}}_{\text{r}}$ is the rate constant of the reverse reaction.

    ${\text{K}}_{\text{c}}$ is the equilibrium constant.

Interpretation Introduction

(b)

Interpretation:

The expression for the equilibrium constant for the reaction, ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{}_{\text{(l)}}\text{ + }\frac{15}{2}{\text{O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\rightleftharpoons }{\text{ 3H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{ + 6CO}}_{\text{2(g)}}$ has to be written.

Concept Introduction:

Equilibrium constant$\left({\text{K}}_{\text{p}}\right)$:

In gas phase reactions, partial pressure is used to write equilibrium equation than molar concentration.  Equilibrium constant $\left({\text{K}}_{\text{p}}\right)$ is defined as ratio of partial pressure of products to partial pressure of reactants.  Each partial pressure term is raised to a power, which is same as the coefficients in the chemical reaction

Consider the reaction where A reacts to give B.

    $\text{aA}\rightleftharpoons \text{bB}$

    $\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}{\text{P}}_{\text{A}}{}^{\text{a}}{\text{=k}}_{\text{r}}{\text{P}}_{\text{B}}{}^{\text{a}}\end{array}$

On rearranging,

    $\frac{{\text{P}}_{\text{B}}{}^{\text{b}}}{{\text{P}}_{\text{A}}{}^{\text{a}}}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}{\text{=K}}_{\text{p}}$

Where,

    ${\text{k}}_{\text{f}}$ is the rate constant of the forward reaction.

    ${\text{k}}_{\text{r}}$ is the rate constant of the reverse reaction.

    ${\text{K}}_{\text{p}}$ is the equilibrium constant

The concentration of solids and pure liquids do not change, so their concentration terms are not included in the equilibrium constant expression.