Chapter 12, Problem 60QRT

(a)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $\text{NO}$ with ${\text{N}}_{\text{2}}\text{and}{\text{O}}_{\text{2}}$ is given, for the given conditions whether the system is at equilibrium or not has to be checked.

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 reaction,

  $\begin{array}{l}{\text{N}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{O}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }\text{2}{\text{NO}}_{\left(\text{g}\right)}{\text{K}}_{\text{C}}\text{=}\text{1}\text{.7}\text{×}{\text{10}}^{\text{-3}}\text{at}\text{2300}\text{K}\\ \\ \left[{\text{N}}_{\text{2}}\right]\text{=}\left[{\text{O}}_{\text{2}}\right]\text{=}\text{0}\text{.25}\text{M}\text{and}\left[\text{NO}\right]\text{=}\text{0}\text{.0042}\text{M}\end{array}$

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

  $\begin{array}{l}{\text{Q}}_{\text{C}}\text{=}\frac{{\left(\text{Conc}\text{NO}\right)}^{\text{2}}}{\left(\text{Conc}{\text{N}}_{\text{2}}\right)\left(\text{Conc}{\text{O}}_{\text{2}}\right)}\\ \\ \text{=}\frac{{\left(\text{0}\text{.0042}\right)}^{\text{2}}}{\left(\text{0}\text{.25}\right)\left(\text{0}\text{.25}\right)}\text{=}\text{2}\text{.8}\text{×}{\text{10}}^{\text{-4}}\end{array}$

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

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

Therefore, the given reaction is not at equilibrium for the given conditions.

(b)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $\text{NO}$ with ${\text{N}}_{\text{2}}\text{and}{\text{O}}_{\text{2}}$ is given, for the given conditions the direction of the reaction has to be given.

Concept Introduction:

Refer part (a)

Explanation of Solution

Given reaction,

  $\begin{array}{l}{\text{N}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{O}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }\text{2}{\text{NO}}_{\left(\text{g}\right)}{\text{K}}_{\text{C}}\text{=}\text{1}\text{.7}\text{×}{\text{10}}^{\text{-3}}\text{at}\text{2300}\text{K}\\ \\ \left[{\text{N}}_{\text{2}}\right]\text{=}\left[{\text{O}}_{\text{2}}\right]\text{=}\text{0}\text{.25}\text{M}\text{and}\left[\text{NO}\right]\text{=}\text{0}\text{.0042}\text{M}\end{array}$

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

  $\begin{array}{l}{\text{Q}}_{\text{C}}\text{=}\frac{{\left(\text{Conc}\text{NO}\right)}^{\text{2}}}{\left(\text{Conc}{\text{N}}_{\text{2}}\right)\left(\text{Conc}{\text{O}}_{\text{2}}\right)}\\ \\ \text{=}\frac{{\left(\text{0}\text{.0042}\right)}^{\text{2}}}{\left(\text{0}\text{.25}\right)\left(\text{0}\text{.25}\right)}\text{=}\text{2}\text{.8}\text{×}{\text{10}}^{\text{-4}}\end{array}$

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

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

The given reaction is not at equilibrium for the given conditions.

And also ${\text{Q}}_{\text{C}}<{\text{K}}_{\text{C}}$ so the given reaction must proceeds toward products to reach equilibrium condition.

(c)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $\text{NO}$ with ${\text{N}}_{\text{2}}\text{and}{\text{O}}_{\text{2}}$ is given, concentrations of all three substances has to be calculated.

Concept Introduction:

Refer part (a)

Explanation of Solution

Given reaction,

  $\begin{array}{l}{\text{N}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{O}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }\text{2}{\text{NO}}_{\left(\text{g}\right)}{\text{K}}_{\text{C}}\text{=}\text{1}\text{.7}\text{×}{\text{10}}^{\text{-3}}\text{at}\text{2300}\text{K}\\ \\ \left[{\text{N}}_{\text{2}}\right]\text{=}\left[{\text{O}}_{\text{2}}\right]\text{=}\text{0}\text{.25}\text{M}\text{and}\left[\text{NO}\right]\text{=}\text{0}\text{.0042}\text{M}\end{array}$

Construct ICE table,

  $\begin{array}{l}{\text{N}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{O}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }\text{2}{\text{NO}}_{\left(\text{g}\right)}\\ \text{Initial}\text{conc}\text{.}\left(\text{M}\right)\text{0}\text{.25}\text{0}\text{.25}\text{0}\text{.0042}\\ \text{Change}\text{in}\text{conc}\text{.}\left(\text{M}\right)\text{-}\text{x}\text{-}\text{x}\text{+}\text{2}\text{x}\\ \text{Equilibrium}\text{conc}\left(\text{M}\right)\text{0}\text{.25}\text{-}\text{x}\text{0}\text{.25}\text{-}\text{x}\text{0}\text{.0042}\text{+}\text{2}\text{x}\end{array}$

At equilibrium,

  $\begin{array}{l}{\text{K}}_{\text{C}}\text{=}\frac{{\left[\text{NO}\right]}^{\text{2}}}{\left[{\text{N}}_{\text{2}}\right]\left[{\text{O}}_{\text{2}}\right]}\\ \\ \text{=}\frac{{\left(\text{0}\text{.0042}\text{+}\text{2x}\right)}^{\text{2}}}{\left(\text{0}\text{.25}\text{-}\text{x}\right)\left(\text{0}\text{.25}\text{-}\text{x}\right)}\text{=}\text{1}\text{.7}\text{×}{\text{10}}^{\text{-3}}\end{array}$

Solve for x,

  $\begin{array}{l}\text{take}\text{square}\text{root}\text{of}\text{each}\text{side,}\\ \\ \frac{\left(\text{0}\text{.0042}\text{+}\text{2x}\right)}{\left(\text{0}\text{.25}\text{-}\text{x}\right)}\text{=}\text{0}\text{.041}\\ \\ \text{0}\text{.0042}\text{+}\text{2x}\text{=}\text{0}\text{.010}\text{-}\text{0}\text{.041}\text{x}\\ \\ \left(\text{2}\text{+}\text{0}\text{.041}\right)\text{x}\text{=}\text{0}\text{.010}\text{-}\text{0}\text{.0042}\text{=}\text{0}\text{.006}\\ \\ \text{therefore,}\text{x}\text{=}\text{3}\text{×}{\text{10}}^{\text{-3}}\text{M}\end{array}$

Calculate the concentrations,

  $\begin{array}{l}\left[{\text{N}}_{\text{2}}\right]\text{=}\left[{\text{O}}_{\text{2}}\right]\text{=}\text{0}\text{.025}\text{-}\text{x}\text{=}\text{0}\text{.025}\text{-}\text{3}\text{×}{\text{10}}^{\text{-3}}\text{M}\\ \\ \text{=}\text{0}\text{.25}\text{M}\\ \\ \left[\text{NO}\right]\text{=}\text{0}\text{.0042}\text{+}\text{2}\text{x}\text{=}\text{0}\text{.0042}\text{M}\text{+}\text{2}\left(\text{3}\text{×}{\text{10}}^{\text{-3}}\text{M}\right)\\ \\ \text{=}\text{0}\text{.0042}\text{M}\text{+}\text{0}\text{.006}\text{=}\text{0}\text{.010}\text{M}\end{array}$

Therefore, the concentration of ${\text{N}}_{\text{2}}\text{and}{\text{O}}_{\text{2}}$ is $\text{0}\text{.25}\text{M}$ and the concentration of $\text{NO}$ is $\text{0}\text{.010}\text{M}$.