Chapter 12, Problem 63QRT

Interpretation Introduction

Interpretation:

Equilibrium reaction of $\text{CO}$ with $\text{C}\left(\text{graphite}\right)\text{and}{\text{CO}}_{\text{2}}$ is given, the observation has to be given when $\text{3}\text{.5}\text{mol}\text{CO}$ and $\text{3}\text{.5}\text{mol}{\text{CO}}_{\text{2}}$ are mixed in a $\text{1}\text{.5}\text{L}$ sealed graphite container.

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

Given information,

  $\begin{array}{l}\text{C}\left(\text{graphite}\right)\text{+}{\text{CO}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }\text{2}{\text{CO}}_{\left(\text{g}\right)}{\text{K}}_{\text{C}}\text{=}\text{3}\text{.7}\text{×}{\text{10}}^{\text{-23}}\text{at}{\text{25}}^{\text{o}}\text{C}\\ \\ \text{CO}\text{=}\text{3}\text{.5}\text{mol}\text{and}{\text{CO}}_{\text{2}}\text{=}\text{3}\text{.5}\text{mol}\\ \\ \text{Volume}\text{of}\text{container}\text{=}\text{1}\text{.5}\text{L}\end{array}$

Calculate the value of ${\text{Q}}_{\text{C}}$,

  $\begin{array}{l}\text{Conc}\text{.}\text{CO}\text{=}\text{Conc}\text{.}{\text{CO}}_{\text{2}}\text{=}\frac{\text{3}\text{.5}\text{mol}}{\text{1}\text{.5}\text{L}}\text{=}\text{2}\text{.3}\text{M}\\ \\ {\text{Q}}_{\text{C}}\text{=}\frac{{\left(\text{Conc}\text{.}\text{CO}\right)}^{\text{2}}}{\text{Conc}\text{.}{\text{CO}}_{\text{2}}}\text{=}\frac{{\left(\text{2}\text{.3}\right)}^{\text{2}}}{\text{2}\text{.3}}\text{=}\text{2}\text{.3}\end{array}$

Compare the value of ${\text{Q}}_{\text{C}}$ and ${\text{K}}_{\text{C}}$,

  ${\text{Q}}_{\text{C}}\text{>}{\text{K}}_{\text{C}}$

So the given reaction must proceeds toward reactants to reach equilibrium condition.  So first take the reaction all the way to reactants as much as stoichiometrically possible, after that take it back to equilibrium with a very small x.

Construct ICE table,

  $\begin{array}{l}{\text{C}}_{\left(\text{s}\right)}\text{+}{\text{CO}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }\text{2}{\text{CO}}_{\left(\text{g}\right)}\\ \text{Initial}\text{conc}\text{.}\left(\text{M}\right)\text{solid}\text{2}\text{.3}\text{2}\text{.3}\\ \text{change}\text{going}\text{to}\text{reactants}\left(\text{M}\right)\text{+}\text{some}\text{+2}\text{.3/2}\text{-}\text{2}\text{.3}\\ \text{final}\text{conc}\left(\text{M}\right)\text{solid}\text{+}\text{some}\text{3}\text{.45}\text{0}\\ \text{change}\text{as}\text{reaction}\text{equilibrates}\left(\text{M}\right)\text{-}\text{some}\text{-}\text{x}\text{+}\text{2x}\\ \text{Equilibrium}\text{conc}\text{.}\left(\text{M}\right)\text{solid}\text{3}\text{.45}\text{-}\text{x}\text{2x}\end{array}$

At equilibrium,

  $\begin{array}{l}{\text{K}}_{\text{C}}\text{=}\frac{{\left[\text{CO}\right]}^{\text{2}}}{\left[{\text{CO}}_{\text{2}}\right]}\\ \\ \text{=}\frac{{\left(\text{2x}\right)}^{\text{2}}}{\left(\text{3}\text{.45}\text{+}\text{x}\right)}\text{=}\text{3}\text{.7}\text{×}{\text{10}}^{\text{-23}}\end{array}$

Assume, $\text{3}\text{.45}\text{>>}\text{x}$, so $\text{3}\text{.45}\text{+}\text{x}\text{=}\text{3}\text{.45}$

Solve for x,

  $\begin{array}{l}\frac{{\left(\text{2x}\right)}^{\text{2}}}{\text{3}\text{.45}}\text{=}\text{3}\text{.7}\text{×}{\text{10}}^{\text{-23}}\\ \\ {\text{4x}}^{\text{2}}\text{=}\text{1}\text{.3}\text{×}{\text{10}}^{\text{-23}}\text{;}\text{therefore,}\text{x}\text{=}\text{5}\text{.6}\text{×}{\text{10}}^{\text{-12}}\text{M}\end{array}$

Calculate the concentration of the substances,

  $\begin{array}{l}\left[\text{CO}\right]\text{=}\text{2}\text{x}\text{=}\text{2}\text{×}\left(\text{5}\text{.6}\text{×}{\text{10}}^{\text{-12}}\text{M}\right)\text{=}\text{1}\text{.1}\text{×}{\text{10}}^{\text{-11}}\text{M}\\ \\ \text{moles}\text{of}{\text{CO}}_{\text{2}}\text{=}\text{3}\text{.45}\text{M}\text{×}\text{1}\text{.5}\text{L}\text{=}\text{5}\text{.2}\text{mol}\\ \\ \left[\text{CO}\right]\text{=}\text{1}\text{.1}\text{×}{\text{10}}^{\text{-11}}\text{M}\text{×}\text{1}\text{.5}\text{L}\text{=}\text{2}\text{.5}\text{×}{\text{10}}^{\text{-11}}\text{mol}{\text{CO}}_{\text{2}}\end{array}$

Almost all the $\text{CO}$ converts to ${\text{CO}}_{\text{2}}$, depositing some new solid graphite in the container, leaving $\text{5}\text{.2}\text{moles}$ of $\text{CO}$ and $\text{2}\text{.5}\text{×}{\text{10}}^{\text{-11}}\text{moles}\text{of}{\text{CO}}_{\text{2}}$.