Chapter 12, Problem 87QRT

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant expressions in terms of the unknown variable x for each given reactions has to be written by using the reaction table (ICE table) approach.

    $\begin{array}{l}{\text{2O}}_{\text{3}}{}_{\text{(g)}}\underset{}{\rightleftharpoons }{\text{3O}}_{\text{2(g)}}^{}{\text{K}}_{\text{c}}\text{=}\text{7}\text{×}{\text{10}}^{\text{56}}\\ {\text{2NO}}_{\text{2(g)}}\underset{}{\rightleftharpoons }{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{{}^{}}_{\text{(g)}}{\text{K}}_{\text{c}}\text{=}\text{1}\text{.7×}{\text{10}}^{\text{2}}\\ {\text{HCOO}}_{\text{(aq)}}^{\text{-}}\text{+}{\text{H}}_{\text{(aq)}}^{\text{+}}\underset{}{\rightleftharpoons }{\text{HCOOH}}_{\text{(aq)}}{\text{K}}_{\text{c}}\text{=}\text{5}\text{.6}\text{×}{\text{10}}^{\text{3}}\\ {\text{Ag}}_{\text{(aq)}}^{\text{+}}{\text{+I}}_{\text{(aq)}}^{\text{-}}\underset{}{\rightleftharpoons }{\text{AgI}}_{\text{(s)}}{\text{K}}_{\text{c}}\text{=}\text{6}\text{.7}\text{×}{\text{10}}^{\text{15}}\end{array}$

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.

Explanation of Solution

The equilibrium constant expressions in terms of the unknown variable x for 1 reaction is,

  ${\text{2O}}_{\text{3}}{}_{\text{(g)}}\underset{}{\rightleftharpoons }{\text{3O}}_{\text{2(g)}}^{}{\text{K}}_{\text{c}}\text{=}\text{7}\text{×}{\text{10}}^{\text{56}}$

The equilibrium constant expressions for above equation is,

    ${\text{K}}_{\text{c}}\text{=}\frac{{{\text{[O}}_{\text{2}}\text{]}}^{\text{3}}}{{{\text{[O}}_{\text{3}}\text{]}}^{\text{2}}}$

ICE table for the above equation is,

    $\begin{array}{l}{\text{2O}}_{\text{3}}{}_{\text{(g)}}\underset{}{\rightleftharpoons }{\text{3O}}_{\text{2(g)}}^{}\\ \text{Initial}\text{1}\text{.0}\text{1}\text{.0}\\ \text{Change}\text{-x}\text{+}\frac{\text{3}}{\text{2}}\text{x}\\ \text{Equilibrium}\text{1}\text{.0-x}\text{1}\text{.0}\text{+}\frac{\text{3}}{\text{2}}\text{x}\end{array}$

The equilibrium constant expressions in terms of the unknown variable x for 1 reaction is,

  ${\text{K}}_{\text{c}}\text{=}\frac{{\left(\text{1}\text{.0+}\frac{\text{3}}{\text{2}}\text{x}\right)}^3}{{\left(\text{1}\text{.0-x}\right)}^{\text{2}}}{\text{=7×10}}^{\text{56}}$

The equilibrium constant expressions in terms of the unknown variable x for 2 reaction is,

  ${\text{2NO}}_{\text{2}}{}_{\text{(g)}}\underset{}{\rightleftharpoons }{\text{N}}_{\text{2}}{\text{O}}_{\text{4(g)}}^{}{\text{K}}_{\text{c}}\text{=}\text{1}\text{.7}\text{×}{\text{10}}^{\text{2}}$

The equilibrium constant expressions for above equation is,

    ${\text{K}}_{\text{c}}\text{=}\frac{{\text{[N}}_{\text{2}}{\text{O}}_{\text{4}}\text{]}}{{{\text{[NO}}_{\text{2}}\text{]}}^{\text{2}}}$

ICE table for the above equation is,

    $\begin{array}{l}{\text{2NO}}_{\text{2}}{}_{\text{(g)}}\underset{}{\rightleftharpoons }{\text{N}}_{\text{2}}{\text{O}}_{\text{4(g)}}^{}\\ \text{Initial}\text{1}\text{.0}\text{1}\text{.0}\\ \text{Change}\text{-x}\text{+}\frac{\text{3}}{\text{2}}\text{x}\\ \text{Equilibrium}\text{1}\text{.0-x}\text{1}\text{.0}\text{+}\frac{1}{\text{2}}\text{x}\end{array}$

The equilibrium constant expressions in terms of the unknown variable x for 2 reaction is,

  ${\text{K}}_{\text{c}}\text{=}\frac{{\left(\text{1}\text{.0+}\frac{1}{\text{2}}\text{x}\right)}^{}}{{\left(\text{1}\text{.0-x}\right)}^{\text{2}}}\text{=1}{\text{.7×10}}^2$

The equilibrium constant expressions in terms of the unknown variable x for 3 reaction is,

  ${\text{HCOOH}}_{\text{(aq)}}\underset{}{\rightleftharpoons }{\text{HCOO}}^{\text{-}}{}_{\text{(aq)}}\text{+}{\text{H}}_{\text{(aq)}}^{\text{+}}{\text{K}}_{\text{c}}\text{=}\text{5}\text{.6×}{\text{10}}^{\text{3}}$

The equilibrium constant expressions for above equation is,

    ${\text{K}}_{\text{c}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][HCOO}}^{\text{-}}\text{]}}{\text{[HOOH]}}\text{=5}\text{.6×}{\text{10}}^3$

ICE table for the above equation is,

    $\begin{array}{l}{\text{HCOOH}}_{\text{(aq)}}\underset{}{\rightleftharpoons }{\text{HCOO}}^{\text{-}}{}_{\text{(aq)}}\text{+}{\text{H}}_{\text{(aq)}}^{\text{+}}\\ \text{Initial}\text{1}\text{.0}\text{1}\text{.0}\text{1}\text{.0}\\ \text{Change}\text{-x}\text{+x}\text{+x}\\ \text{Equilibrium}\text{1}\text{.0}\text{-}\text{x}\text{1}\text{.0}\text{+x}\text{1}\text{.0+x}\end{array}$

The equilibrium constant expressions in terms of the unknown variable x for given reaction is,

  ${\text{K}}_{\text{c}}\text{=}\frac{\text{(1}\text{.0+x)(1}\text{.0+x)}}{\text{(1}\text{.0-x)}}\text{= 5}\text{.6}\text{×}{\text{10}}^{\text{8}}$

The equilibrium constant expressions in terms of the unknown variable x for 4 reaction is,

  ${\text{Ag}}_{\text{(aq)}}^{\text{+}}{\text{+I}}_{\text{(aq)}}^{\text{-}}\underset{}{\rightleftharpoons }{\text{AgI}}_{\text{(s)}}{\text{K}}_{\text{c}}\text{=}\text{6}\text{.7}\text{×}{\text{10}}^{\text{15}}$

The equilibrium constant expressions for above equation is,

    ${\text{K}}_{\text{c}}\text{=}\frac{\text{1}}{{\text{[Ag}}^{\text{+}}{\text{][I}}^{\text{-}}\text{]}}\text{=}\text{6}\text{.7×}{\text{10}}^{\text{15}}$

ICE table for the above equation is,

    $\begin{array}{l}{\text{Ag}}_{\text{(aq)}}^{\text{+}}\text{+}{\text{I}}_{\text{(aq)}}^{\text{-}}\underset{}{\rightleftharpoons }{\text{AgI}}_{\text{(s)}}\\ \text{Initial}\text{1}\text{.0}\text{1}\text{.0}\text{-}\\ \text{Change}\text{-x}\text{-x}\text{-}\\ \text{Equilibrium}\text{1}\text{.0}\text{-}\text{x}\text{1}\text{.0}\text{-x}\text{-}\end{array}$

The equilibrium constant expressions in terms of the unknown variable x for given reaction is,

  ${\text{K}}_{\text{c}}\text{=}\frac{\text{1}}{\text{(1}\text{.0-x)(1}\text{.0-x)}}\text{=}\text{6}\text{.7×}{\text{10}}^{\text{15}}$

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant expressions in terms of the unknown variable x for each given reactions has to be written, which of these expressions yield quadratic equations has to be given.

Concept Introduction:

Refer part (a).

Explanation of Solution

The given reactions and it’s the equilibrium constant expressions in terms of the unknown variable x are,

  $\begin{array}{l}{\text{(1) 2O}}_{\text{3}}{}_{\text{(g)}}\underset{}{\rightleftharpoons }{\text{3O}}_{\text{2(g)}}^{}\\ {\text{K}}_{\text{c}}\text{=}\frac{{\left(\text{1}\text{.0+}\frac{\text{3}}{\text{2}}\text{x}\right)}^3}{{\left(\text{1}\text{.0-x}\right)}^{\text{2}}}{\text{=7×10}}^{\text{56}}\end{array}$

From the equilibrium constant expression, it does not yield quadratic equation.

  $\begin{array}{l}\text{(2)}{\text{2NO}}_{\text{2}}{}_{\text{(g)}}\underset{}{\rightleftharpoons }{\text{N}}_{\text{2}}{\text{O}}_{\text{4(g)}}^{}\\ {\text{K}}_{\text{c}}\text{=}\frac{{\left(\text{1}\text{.0+}\frac{1}{\text{2}}\text{x}\right)}^3}{{\left(\text{1}\text{.0-x}\right)}^{\text{2}}}\text{=1}{\text{.7×10}}^2\end{array}$

From the equilibrium constant expression, it yields quadratic equation.

  $\begin{array}{l}\text{(3)}{\text{HCOOH}}_{\text{(aq)}}\underset{}{\rightleftharpoons }{\text{HCOO}}^{\text{-}}{}_{\text{(aq)}}\text{+}{\text{H}}_{\text{(aq)}}^{\text{+}}\\ {\text{K}}_{\text{c}}\text{=}\frac{\text{(1}\text{.0+x)(1}\text{.0+x)}}{\text{(1}\text{.0-x)}}\text{= 5}\text{.6}\text{×}{\text{10}}^{\text{8}}\end{array}$

From the equilibrium constant expression, it yields quadratic equation.

  $\begin{array}{l}\text{(4)}{\text{Ag}}_{\text{(aq)}}^{\text{+}}{\text{+I}}_{\text{(aq)}}^{\text{-}}\underset{}{\rightleftharpoons }{\text{AgI}}_{\text{(s)}}\\ {\text{K}}_{\text{c}}\text{=}\frac{\text{1}}{\text{(1}\text{.0-x)(1}\text{.0-x)}}\text{=}\text{6}\text{.7×}{\text{10}}^{\text{15}}\end{array}$

From the equilibrium constant expression, it yields quadratic equation.

(c)

Interpretation Introduction

Interpretation:

The equilibrium constant expression in terms of the unknown variable x for given reaction has to be written and solving of x has to be explained.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The equilibrium constant expressions in terms of the unknown variable x for 1 reaction is,

  ${\text{2O}}_{\text{3}}{}_{\text{(g)}}\underset{}{\rightleftharpoons }{\text{3O}}_{\text{2(g)}}^{}{\text{K}}_{\text{c}}\text{=}\text{7}\text{×}{\text{10}}^{\text{56}}$

The equilibrium constant expressions for above equation is,

    ${\text{K}}_{\text{c}}\text{=}\frac{{{\text{[O}}_{\text{2}}\text{]}}^{\text{3}}}{{{\text{[O}}_{\text{3}}\text{]}}^{\text{2}}}$

ICE table for the above equation is,

    $\begin{array}{l}{\text{2O}}_{\text{3}}{}_{\text{(g)}}\underset{}{\rightleftharpoons }{\text{3O}}_{\text{2(g)}}^{}\\ \text{Initial}\text{1}\text{.0}\text{1}\text{.0}\\ \text{Change}\text{-x}\text{+}\frac{\text{3}}{\text{2}}\text{x}\\ \text{Equilibrium}\text{1}\text{.0-x}\text{1}\text{.0}\text{+}\frac{\text{3}}{\text{2}}\text{x}\end{array}$

The equilibrium constant expressions in terms of the unknown variable x for 1 reaction is,

  ${\text{K}}_{\text{c}}\text{=}\frac{{\left(\text{1}\text{.0+}\frac{\text{3}}{\text{2}}\text{x}\right)}^3}{{\left(\text{1}\text{.0-x}\right)}^{\text{2}}}{\text{=7×10}}^{\text{56}}$

The above expression is not a quadratic equation so it is solved as shown below,

The changes in stoichiometry of limiting reactant, the reaction is left favors so the x is calculated as fallows,

    $\begin{array}{l}{\text{2O}}_{\text{3}}{}_{\text{(g)}}\underset{}{\rightleftharpoons }{\text{3O}}_{\text{2(g)}}^{}\\ \text{Initial}\text{1}\text{.0}\text{1}\text{.0}\\ \text{final}\text{0}\text{2}\text{.5}\\ \text{Change}\text{-x}\text{-}\frac{\text{3}}{\text{2}}\text{x}\\ \text{Equilibrium}\text{1}\text{.0-x}\text{2}\text{.5-}\frac{\text{3}}{\text{2}}\text{x}\end{array}$

The valve of x is calculated as,

    $\begin{array}{l}{\text{K}}_{\text{c}}\text{=}\frac{{\left(\text{2}\text{.5-}\frac{\text{3}}{\text{2}}\text{x}\right)}^{\text{3}}}{{\text{x}}^{\text{2}}}\text{=}\frac{{\text{(2}\text{.5)}}^{\text{3}}}{{\text{x}}^{\text{2}}}\text{=}\text{7×}{\text{10}}^{\text{56}}\\ \text{=}{\text{1×10}}^{\text{-28}}\text{mol/L}\end{array}$

The calculated vale is, ${\text{1×10}}^{\text{-28}}\text{mol/L<<<}\text{2}\text{.5}$ and it is so negligible.