Chapter 17.5, Problem 17.9AFP

(a)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of ${\text{[I}}_{\text{2}}]$ and $\text{[I]}$ at 600 K for the given halogen decomposition reaction has to be calculated.

Concept Introduction:

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

Given data:

The decomposition reaction of halogen is,

  ${\text{I}}_{\text{2(g)}}\text{}\stackrel{}{\rightleftharpoons }{\text{ 2I}}_{\text{(g)}}$

The equilibrium constant for the above reaction is ${\text{K}}_{\text{c}}\text{ = 2}\text{.94}\times {\text{10}}^{-10}$. For the above to reaction to occur $0.50\text{ mol}$ of ${\text{I}}_{\text{2}}$ was heated in a $\text{2}\text{.5 L}$ vessel at 600 K.

Determination of concentration at equilibrium:

From the balanced equation, it is known that one mole of ${\text{I}}_{\text{2}}$ decomposes to give two moles of iodine gas.

Calculate the initial concentration of ${\text{I}}_{\text{2}}$ using the given moles.

  $\begin{array}{l}{{\text{[I}}_{\text{2}}}_{\text{]}}\text{ = }\frac{{\text{moles of I}}_{\text{2}}}{\text{Volume}}\\ =\frac{\text{0}\text{.50 mol}}{\text{2}\text{.5 L}}\\ =0.20\text{}\text{}\text{ M}\end{array}$

Determine the concentration of all reactants and product using the below table.

$\begin{array}{l}\text{Conc}\text{. (M)}{\text{I}}_{\text{2(g)}}\text{}\stackrel{}{\rightleftharpoons }{\text{ 2I}}_{\text{(g)}}\\ \text{Initial}0.200\\ \text{Change}\text{-x}\text{+2x}\\ \overline{\text{Equilibrium}0.20\text{-x}2\text{x}}\end{array}$

The equilibrium expression for the given reaction is,

  ${\text{K}}_{\text{c}}\text{= }\frac{{\text{[I]}}^2}{{\text{[I}}_2\text{]}}$

Solve for x using the above expression and obtain the concentration of all reactants and products.

 $\begin{array}{l}{\text{ K}}_{\text{c}}\text{= }\frac{{\text{[I]}}^2}{{\text{[I}}_2\text{]}}\\ 2.94\times {10}^{-10}=\frac{{\text{[2x]}}^2}{\text{[0}\text{.20 - x]}}\\ \text{Assume x is negligible and 0}\text{.20 - x}\approx 0.20\\ 2.94\times {10}^{-10}=\frac{{\text{[2x]}}^2}{\text{[0}\text{.20]}}\\ {\text{ 4x}}^2\text{ = (2}\text{.94}\times {\text{10}}^{-10})(0.20)\\ \text{x}\text{=}\text{3}\text{.834}\times {10}^{-6}\\ \approx \text{3}\text{.8}\times {10}^{-6}\end{array}$

Check the assumption by calculating the % error.

  $\begin{array}{l}\frac{[\text{Change]}}{[\text{Initial]}}\times 100\\ \frac{3.8\times {10}^{-6}}{0.20}\times 100\text{ =0}\text{.0019\%}\end{array}$

The above value is smaller than 5%.  Thus the assumption is valid.

At equilibrium,

  $\begin{array}{l}{\text{[I]}}_{\text{eq}}\text{= 2x}\\ \text{=2(3}\text{.834}\times {\text{10}}^{-6})\\ =\text{7}\text{.668}\times {10}^{-6}\\ =7.7\times {10}^{-6}\text{ M}\\ {{\text{[I}}_{\text{2}}}_{\text{]}}=0.20\text{ - x}\\ \text{=0}\text{.20 - 3}\text{.834}\times {10}^{-6}\\ =\text{0}\text{.199996}\\ \approx \text{0}\text{.20 M}\end{array}$

Therefore, the equilibrium concentrations of ${\text{[I}}_{\text{2}}]$ and $\text{[I]}$ is,

$\begin{array}{l}{\text{[I]}}_{\text{eq}}\text{= }7.7\times {10}^{-6}\text{ M}\\ {{\text{[I}}_{\text{2}}}_{\text{]}}=\text{0}\text{.20 M}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of ${\text{[I}}_{\text{2}}]$ and $\text{[I]}$ at 2000 K for the given halogen decomposition reaction has to be calculated.

Concept Introduction:

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

Given data:

The decomposition reaction of halogen is,

  ${\text{I}}_{\text{2(g)}}\text{}\stackrel{}{\rightleftharpoons }{\text{ 2I}}_{\text{(g)}}$

The equilibrium constant for the above reaction is ${\text{K}}_{\text{c}}\text{ = }0.209$ .  For the above to reaction to occur $0.50\text{ mol}$ of ${\text{I}}_{\text{2}}$ was heated in a $\text{2}\text{.5 L}$ vessel at 2000 K.

Determination of concentration at equilibrium:

From the balanced equation, it is known that one mole of ${\text{I}}_{\text{2}}$ decomposes to give two moles of iodine gas.

Calculate the initial concentration of ${\text{I}}_{\text{2}}$ using the given moles.

  $\begin{array}{l}{{\text{[I}}_{\text{2}}}_{\text{]}}\text{ = }\frac{{\text{moles of I}}_{\text{2}}}{\text{Volume}}\\ =\frac{\text{0}\text{.50 mol}}{\text{2}\text{.5 L}}\\ =0.20\text{}\text{}\text{ M}\end{array}$

Determine the concentration of all reactants and product using the below table.

$\begin{array}{l}\text{Conc}\text{. (M)}{\text{I}}_{\text{2(g)}}\text{}\stackrel{}{\rightleftharpoons }{\text{ 2I}}_{\text{(g)}}\\ \text{Initial}0.200\\ \text{Change}\text{-x}\text{+2x}\\ \overline{\text{Equilibrium}0.20\text{-x}2\text{x}}\end{array}$

The equilibrium expression for the given reaction is,

  ${\text{K}}_{\text{c}}\text{= }\frac{{\text{[I]}}^2}{{\text{[I}}_2\text{]}}$

Solve for x using the above expression and obtain the concentration of all reactants and products.

$\begin{array}{l}{\text{ K}}_{\text{c}}\text{= }\frac{{\text{[I]}}^2}{{\text{[I}}_2\text{]}}\\ 0.209=\frac{{\text{[2x]}}^2}{\text{[0}\text{.20 - x]}}\\ \text{Assume x is negligible and 0}\text{.20 - x}\approx 0.20\\ 0.209=\frac{{\text{[2x]}}^2}{\text{[0}\text{.20]}}\\ {\text{ 4x}}^2\text{ = (0}\text{.209})(0.20)\\ \text{x}\text{=}0.102225\\ \approx 0.102\end{array}$

Check the assumption by calculating the % error.

  $\begin{array}{l}\frac{[\text{Change]}}{[\text{Initial]}}\times 100\\ \frac{0.102}{0.20}\times 100\text{ =51\%}\end{array}$

The above value is larger than 5% so the assumption is not valid.  Thus, solve using quadratic equation.

  $\begin{array}{l}\frac{{[\text{2x]}}^2}{[\text{0}\text{.20-x]}}=0.209\\ {\text{4x}}^2\text{ + 0}\text{.209x - 0}\text{.0418}\text{= 0}\\ \text{x}\text{=}\frac{-0.209\pm \sqrt{{(0.209)}^2-4(4)(-0.0418)}}{2(4)}\\ =\text{0}\text{.0793857 (or) }-\text{0}\text{.1316}\end{array}$

Choose the positive root value, $\text{x = 0}\text{.0793857}$

At equilibrium,

  $\begin{array}{l}{\text{[I]}}_{\text{eq}}\text{= 2x}\\ \text{=2(0}\text{.0793857)}\\ =\text{0}\text{.1587714}\\ \approx 0.16\text{ M}\\ {{\text{[I}}_{\text{2}}}_{\text{]}}=0.20\text{ - x}\\ \text{=0}\text{.20 - 0}\text{.0793857}\\ =\text{0}\text{.1206143}\\ \approx \text{0}\text{.12 M}\end{array}$

Therefore, the equilibrium concentrations of ${\text{[I}}_{\text{2}}]$ and $\text{[I]}$ is,

  $\begin{array}{l}{\text{[I]}}_{\text{eq}}\text{= }0.16\text{ M}\\ {{\text{[I}}_{\text{2}}}_{\text{]}}=\text{0}\text{.12 M}\end{array}$