Chapter 12, Problem 120QRT

(a)

Interpretation Introduction

Interpretation:

Reaction mechanism for the formation of $\text{NOBr}$ is given, for this reaction overall stoichiometric equation has to be given.

Explanation of Solution

Given reaction,

  $\begin{array}{l}\text{Step}\text{1:}{\text{NO}}_{\left(\text{g}\right)}\text{+}{\text{Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{NOBr}}_{\text{2}\left(\text{g}\right)}\\ \\ \text{Step}\text{2}\text{:}{\text{NOBr}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{NO}}_{\left(\text{g}\right)}\stackrel{}{\to }\text{2}{\text{NOBr}}_{\left(\text{g}\right)}\end{array}$

Write the stoichiometric equation by cancelling species that are present in both sides and add up the remaining species,

  $\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{NO}}_{\left(\text{g}\right)}\text{+}{\text{Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }\cancel{{\text{NOBr}}_{\text{2}\left(\text{g}\right)}}\\ \\ \cancel{{\text{NOBr}}_{\text{2}\left(\text{g}\right)}}\text{+}{\text{NO}}_{\left(\text{g}\right)}\stackrel{}{\to }\text{2}{\text{NOBr}}_{\left(\text{g}\right)}\end{array}}\\ \text{2}{\text{NO}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\to }\text{2}{\text{NOBr}}_{\left(\text{g}\right)}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Reaction mechanism for the formation of $\text{NOBr}$ is given, for this reaction the equilibrium constant in terms of the forward and reverse rate constant has to be given.

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,

  $\begin{array}{l}\text{Step}\text{1:}{\text{NO}}_{\left(\text{g}\right)}\text{+}{\text{Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{NOBr}}_{\text{2}\left(\text{g}\right)}\\ \\ \text{Step}\text{2}\text{:}{\text{NOBr}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{NO}}_{\left(\text{g}\right)}\stackrel{}{\to }\text{2}{\text{NOBr}}_{\left(\text{g}\right)}\end{array}$

The rate of forward reaction is equals to the rate of reverse reaction at equilibrium.

Write the equilibrium constant expression for the first step,

  $\begin{array}{l}\text{rate}\text{forward,}\text{step}\text{1}\text{=}\text{rate}\text{reverse,}\text{step}\text{1}\\ \\ {\text{k}}_{\text{1}}\left[\text{NO}\right]\left[{\text{Br}}_{\text{2}}\right]\text{=}{\text{k}}_{\text{-1}}\left[{\text{NOBr}}_{\text{2}}\right]\\ \\ \text{therefore,}{\text{K}}_{\text{C}}\text{,}\text{step}\text{1}\text{=}\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{-1}}}\text{=}\frac{\left[{\text{NOBr}}_{\text{2}}\right]}{\left[\text{NO}\right]\left[{\text{Br}}_{\text{2}}\right]}\end{array}$

Write the equilibrium constant expression for the second step,

  $\begin{array}{l}\text{rate}\text{forward,}\text{step}\text{2}\text{=}\text{rate}\text{reverse,}\text{step}\text{2}\\ \\ {\text{k}}_{\text{2}}\left[{\text{NOBr}}_{\text{2}}\right]\left[\text{NO}\right]\text{=}{\text{k}}_{\text{-2}}{\left[\text{NOBr}\right]}^{\text{2}}\\ \\ \text{therefore,}{\text{K}}_{\text{C}}\text{,}\text{step}\text{1}\text{=}\frac{{\text{k}}_{\text{2}}}{{\text{k}}_{\text{-2}}}\text{=}\frac{{\left[\text{NOBr}\right]}^{\text{2}}}{\left[{\text{NOBr}}_{\text{2}}\right]\left[\text{NO}\right]}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Reaction mechanism for the formation of $\text{NOBr}$ is given, for this reaction the overall stoichiometric equilibrium constant in terms of the forward and reverse rate constant has to be given.

Concept Introduction:

Refer part (b)

Explanation of Solution

Given reaction,

  $\begin{array}{l}\text{Step}\text{1:}{\text{NO}}_{\left(\text{g}\right)}\text{+}{\text{Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{NOBr}}_{\text{2}\left(\text{g}\right)}\\ \\ \text{Step}\text{2}\text{:}{\text{NOBr}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{NO}}_{\left(\text{g}\right)}\stackrel{}{\to }\text{2}{\text{NOBr}}_{\left(\text{g}\right)}\end{array}$

Write the stoichiometric equation, by cancelling species that are present in both sides and add up the remaining species,

  $\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{NO}}_{\left(\text{g}\right)}\text{+}{\text{Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }\cancel{{\text{NOBr}}_{\text{2}\left(\text{g}\right)}}\\ \\ \cancel{{\text{NOBr}}_{\text{2}\left(\text{g}\right)}}\text{+}{\text{NO}}_{\left(\text{g}\right)}\stackrel{}{\to }\text{2}{\text{NOBr}}_{\left(\text{g}\right)}\end{array}}\\ \text{2}{\text{NO}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\to }\text{2}{\text{NOBr}}_{\left(\text{g}\right)}\end{array}$

Write the overall stoichiometric equilibrium constant expression,

  $\begin{array}{l}{\text{K}}_{\text{C}\text{,}\text{overall}}\text{=}\frac{{\left[\text{NOBr}\right]}^{\text{2}}}{{\left[\text{NO}\right]}^{\text{2}}\left[{\text{Br}}_{\text{2}}\right]}\text{=}\left(\frac{\left[{\text{NOBr}}_{\text{2}}\right]}{\left[\text{BO}\right]\left[{\text{Br}}_{\text{2}}\right]}\right)\left(\frac{{\left[\text{NOBr}\right]}^{\text{2}}}{\left[\text{NO}\right]\left[{\text{NOBr}}_{\text{2}}\right]}\right)\\ \\ \text{=}\left(\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{-1}}}\right)\left(\frac{{\text{k}}_{\text{2}}}{{\text{k}}_{\text{-2}}}\right)\text{=}\frac{{\text{k}}_{\text{1}}{\text{k}}_{\text{2}}}{{\text{k}}_{\text{-1}}{\text{k}}_{\text{-2}}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Reaction mechanism for the formation of $\text{NOBr}$ is given, for this reaction the overall stoichiometric equilibrium constant in terms of the forward and reverse rate constant was found this has to be checked with the given statement.

Concept Introduction:

Refer part (b)

Explanation of Solution

Given reaction,

  $\begin{array}{l}\text{Step}\text{1:}{\text{NO}}_{\left(\text{g}\right)}\text{+}{\text{Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{NOBr}}_{\text{2}\left(\text{g}\right)}\\ \\ \text{Step}\text{2}\text{:}{\text{NOBr}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{NO}}_{\left(\text{g}\right)}\stackrel{}{\to }\text{2}{\text{NOBr}}_{\left(\text{g}\right)}\end{array}$

Write the stoichiometric equation, by cancelling species that are present in both sides and add up the remaining species,

  $\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{NO}}_{\left(\text{g}\right)}\text{+}{\text{Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }\cancel{{\text{NOBr}}_{\text{2}\left(\text{g}\right)}}\\ \\ \cancel{{\text{NOBr}}_{\text{2}\left(\text{g}\right)}}\text{+}{\text{NO}}_{\left(\text{g}\right)}\stackrel{}{\to }\text{2}{\text{NOBr}}_{\left(\text{g}\right)}\end{array}}\\ \text{2}{\text{NO}}_{\text{2}\left(\text{g}\right)}\text{+}{\text{Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\to }\text{2}{\text{NOBr}}_{\left(\text{g}\right)}\end{array}$

Write the overall stoichiometric equilibrium constant expression,

  $\begin{array}{l}{\text{K}}_{\text{C}\text{,}\text{overall}}\text{=}\frac{{\left[\text{NOBr}\right]}^{\text{2}}}{{\left[\text{NO}\right]}^{\text{2}}\left[{\text{Br}}_{\text{2}}\right]}\text{=}\left(\frac{\left[{\text{NOBr}}_{\text{2}}\right]}{\left[\text{BO}\right]\left[{\text{Br}}_{\text{2}}\right]}\right)\left(\frac{{\left[\text{NOBr}\right]}^{\text{2}}}{\left[\text{NO}\right]\left[{\text{NOBr}}_{\text{2}}\right]}\right)\\ \\ \text{=}\left(\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{-1}}}\right)\left(\frac{{\text{k}}_{\text{2}}}{{\text{k}}_{\text{-2}}}\right)\text{=}\frac{{\text{k}}_{\text{1}}{\text{k}}_{\text{2}}}{{\text{k}}_{\text{-1}}{\text{k}}_{\text{-2}}}\end{array}$

Therefore, the result obtained in part (c) confirms the statement that the equilibrium constant can be obtained by taking the product of the rate constants for all forward steps $\left({\text{here, k}}_{\text{1}}{\text{k}}_{\text{2}}\right)$ and dividing by the product of the rate constants for all reverse steps ${\text{(here, k}}_{\text{–1}}{\text{k}}_{\text{–2}}\text{)}$.