Chapter 12, Problem 86QRT

(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.

    $\begin{array}{l}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\underset{}{\rightleftharpoons }{\text{H}}_{\text{(aq)}}^{\text{+}}\text{+}{\text{OH}}_{\text{(aq)}}^{\text{-}}{\text{K}}_{\text{c}}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-14}}\\ {\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\underset{}{\rightleftharpoons }{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\text{+}{\text{H}}_{\text{(aq)}}^{\text{+}}{\text{K}}_{\text{c}}\text{=}\text{1}\text{.8}\text{×}{\text{10}}^{\text{-5}}\\ {\text{N}}_{\text{2}}{}_{\text{(g)}}{\text{+3H}}_{\text{2}}{}_{\text{(g)}}\underset{}{\rightleftharpoons }{\text{2NH}}_{\text{3}}{}_{\text{(g)}}^{}{\text{K}}_{\text{c}}\text{=}\text{3}\text{.5}\text{×}{\text{10}}^{\text{8}}\end{array}$

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,

  ${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\underset{}{\rightleftharpoons }{\text{H}}_{\text{(aq)}}^{\text{+}}\text{+}{\text{OH}}_{\text{(aq)}}^{\text{-}}{\text{K}}_{\text{c}}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-14}}$

The equilibrium constant expressions for above equation is,

    ${\text{K}}_{\text{c}}{\text{=[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\text{=1}\text{.0}\text{×}{\text{10}}^{\text{-14}}$

ICE table for the above equation is,

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

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

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{=}{\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \text{=}\text{(1}\text{.0+x)(1}\text{.0+x)=1}\text{.0}\text{×}{\text{10}}^{\text{-14}}\end{array}$.

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

  ${\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\underset{}{\rightleftharpoons }{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\text{+}{\text{H}}_{\text{(aq)}}^{\text{+}}{\text{K}}_{\text{c}}\text{=}\text{1}\text{.8}\text{×}{\text{10}}^{\text{-5}}$

The equilibrium constant expressions for above equation is,

    ${\text{K}}_{\text{c}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}}{{\text{[CH}}_{\text{3}}\text{COOH]}}\text{=1}\text{.8}\text{×}{\text{10}}^{\text{-5}}$

ICE table for the above equation is,

    $\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\underset{}{\rightleftharpoons }{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\text{+}{\text{H}}_{\text{(aq)}}^{\text{+}}\\ \text{Initial}\text{1}\text{.0}\text{1}\text{.0}\text{1}\text{.0}\\ \text{Change}\text{-x}\text{+x}\text{+x}\\ \text{Equilibrium}\text{1}\text{.0}\text{-}\text{x}\text{1}\text{.0}\text{+x}\text{1}\text{.0+x}\end{array}$

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

  ${\text{K}}_{\text{c}}\text{=}\frac{(1.0+x)(1.0+x)}{(1.0-x)}\text{=1}\text{.8}\text{×}{\text{10}}^{\text{-5}}$

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

  ${\text{N}}_{\text{2}}{}_{\text{(g)}}{\text{+3H}}_{\text{2(g)}}\underset{}{\rightleftharpoons }\text{2N}{\text{H}}_{\text{3(g)}}^{}{\text{K}}_{\text{c}}\text{=}\text{3}\text{.5}\text{×}{\text{10}}^{\text{8}}$

The equilibrium constant expressions for above equation is,

    ${\text{K}}_{\text{c}}\text{=}\frac{{{\text{[NH}}_3\text{]}}^2}{{\text{[N}}_2{\text{][H}}_2{\text{]}}^3}\text{=1}\text{.8}\text{×}{\text{10}}^{\text{-5}}$

ICE table for the above equation is,

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

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

  ${\text{K}}_{\text{c}}\text{=}\frac{{\text{(1}\text{.0+2x)}}^{\text{2}}}{\text{(1}\text{.0-x)(1}{\text{.0-3x)}}^{\text{3}}}\text{=3}\text{.5}\text{×}{\text{10}}^{\text{8}}$

(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,

  $\begin{array}{l}{\text{(1) H}}_{\text{2}}{\text{O}}_{\text{(l)}}\underset{}{\rightleftharpoons }{\text{H}}_{\text{(aq)}}^{\text{+}}\text{+}{\text{OH}}_{\text{(aq)}}^{\text{-}}\\ {\text{K}}_{\text{c}}\text{=}{\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \text{=}\text{(1}\text{.0+x)(1}\text{.0+x)=1}\text{.0}\text{×}{\text{10}}^{\text{-14}}\end{array}$

From the equilibrium constant expression, it yields quadratic equation.

  $\begin{array}{l}\text{(2)}{\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\underset{}{\rightleftharpoons }{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\text{+}{\text{H}}_{\text{(aq)}}^{\text{+}}\\ {\text{K}}_{\text{c}}\text{=}\frac{(1.0+x)(1.0+x)}{(1.0-x)}\text{=1}\text{.8}\text{×}{\text{10}}^{\text{-5}}\end{array}$

From the equilibrium constant expression, it yields quadratic equation.

  $\begin{array}{l}\text{(3)}{\text{N}}_{\text{2}}{}_{\text{(g)}}{\text{+3H}}_{\text{2(g)}}\underset{}{\rightleftharpoons }\text{2N}{\text{H}}_{\text{3(g)}}^{}\\ {\text{K}}_{\text{c}}\text{=}\frac{{\text{(1}\text{.0+2x)}}^{\text{2}}}{\text{(1}\text{.0-x)(1}{\text{.0-3x)}}^{\text{3}}}\text{=3}\text{.5}\text{×}{\text{10}}^{\text{8}}\end{array}$

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 3 reaction is,

  ${\text{N}}_{\text{2}}{}_{\text{(g)}}{\text{+3H}}_{\text{2(g)}}\underset{}{\rightleftharpoons }\text{2N}{\text{H}}_{\text{3(g)}}^{}{\text{K}}_{\text{c}}\text{=}\text{3}\text{.5}\text{×}{\text{10}}^{\text{8}}$

The equilibrium constant expressions for above equation is,

    ${\text{K}}_{\text{c}}\text{=}\frac{{{\text{[NH}}_3\text{]}}^2}{{\text{[N}}_2{\text{][H}}_2{\text{]}}^3}\text{=1}\text{.8}\text{×}{\text{10}}^{\text{-5}}$

ICE table for the above equation is,

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

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

  ${\text{K}}_{\text{c}}\text{=}\frac{{\text{(1}\text{.0+2x)}}^{\text{2}}}{\text{(1}\text{.0-x)(1}{\text{.0-3x)}}^{\text{3}}}\text{=3}\text{.5}\text{×}{\text{10}}^{\text{8}}$

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

Let, ${\text{H}}_{\text{2}}$ is a limiting reactant so the changes in stoichiometry limiting reactant taken as fallows,

    $\begin{array}{l}{\text{N}}_{\text{2}}{}_{\text{(g)}}\text{+}{\text{3H}}_{\text{2(g)}}\underset{}{\rightleftharpoons }\text{2N}{\text{H}}_{\text{3(g)}}^{}\\ \text{Initial}\text{1}\text{.0}\text{1}\text{.0}\text{1}\text{.0}\\ \text{limiting reacgent }\text{-0}\text{.33}\text{-1}\text{.0}\text{+0}\text{.67}\\ \text{Change}\text{-x}\text{-3x}\text{+2x}\\ \text{limiting reacgent }\text{0}\text{.7}\text{0}\text{.0}\text{1}\text{.7}\\ \text{Reverse}\text{reaction}\text{+x}\text{+3x}\text{-2x}\\ \text{Equilibrium}\text{0}\text{.7}\text{-}\text{x}\text{+3x}\text{1}\text{.7-2x}\end{array}$

The valve of x is calculated as,

    $\begin{array}{l}{\text{K}}_{\text{c}}\text{=}\frac{{\text{(1}\text{.7+2x)}}^{\text{2}}}{\text{(0}{\text{.7-x)(3x)}}^{\text{3}}}=\frac{{\text{(1}\text{.7)}}^{\text{2}}}{\text{(0}{\text{.7)(3x)}}^{\text{3}}}\text{=}\text{4}\text{×}{\text{10}}^{\text{8}}\\ \text{=}\text{0}\text{.0008}\text{mol/L}\end{array}$

The calculated value is, $\text{0}\text{.0008}\text{mol/L <<1}\text{.7 or 0}\text{.7}$ and it is negligible.