Chapter 14, Problem 14.113QP

(a)

Interpretation Introduction

Interpretation:

The mole fraction of the gases has to be calculated using the data.

Concept Introduction:

Mole fraction calculation:

$\text{Mole}\text{fraction}\text{of}\text{solute}\text{x}\text{in}\text{the}\text{mixture}\text{of}\text{x}\text{and}\text{y}\text{=}\frac{{\text{n}}_{\text{x}}}{{\text{n}}_{\text{x}}\text{+}{\text{n}}_{\text{y}}}$

Explanation of Solution

Given data:

$\text{Molar}\text{mass}\text{of}\text{the}\text{reacting}\text{mixture}\text{=}\text{70}\text{.6}\text{g}/\text{mol}$

The reaction is given below:

${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{g}\right)\underset{}{\rightleftharpoons }\text{2}{\text{NO}}_{\text{2}}\left(\text{g}\right)$

The sum of the mole fractions in a mixture must equal one.

$\begin{array}{l}{\text{X}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{+}{\text{X}}_{{\text{NO}}_{\text{2}}}\text{=}\text{1}\\ {\text{X}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{=}{\text{1-X}}_{{\text{NO}}_{\text{2}}}\end{array}$

It is known that the average molar mass is the sum of the products of the mole fraction of each gas and its molar mass.

$\begin{array}{c}\left({\text{X}}_{{\text{NO}}_{\text{2}}}\text{×}\text{46}\text{.01}\text{g}\right)\text{+}\left({\text{X}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{×}\text{92}\text{.02}\text{g}\right)\text{=}\text{70}\text{.6}\text{g}\\ \\ \left({\text{X}}_{{\text{NO}}_{\text{2}}}\text{×}\text{46}\text{.01}\text{g}\right)\text{+}\left(\left[{\text{1-X}}_{{\text{NO}}_{\text{2}}}\right]\text{×}\text{92}\text{.02}\text{g}\right)\text{=}\text{70}\text{.6}\text{g}\\ \\ \left({\text{X}}_{{\text{NO}}_{\text{2}}}\text{×}\text{46}\text{.01}\text{g}\right)\text{+}\text{92}\text{.02}\text{g-92}\text{.02}{\text{g×X}}_{{\text{NO}}_{\text{2}}}\text{=}\text{70}\text{.6}\text{g}\\ \\ \text{46}\text{.01}{\text{gX}}_{{\text{NO}}_{\text{2}}}\text{-92}\text{.02}{\text{g×X}}_{{\text{NO}}_{\text{2}}}\text{=}\text{70}\text{.6}\text{g-92}\text{.02}\text{g}\\ \\ \text{-46}\text{.01}{\text{X}}_{{\text{NO}}_{\text{2}}}\text{=}\text{-21}\text{.42}\\ \\ {\text{X}}_{{\text{NO}}_{\text{2}}}\text{=}\text{0}\text{.466}\\ \\ {\text{X}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{=}{\text{1-X}}_{{\text{NO}}_{\text{2}}}\text{=}\text{1-0}\text{.466}\text{=}\text{0}\text{.534}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{p}}$ for the reaction if the total pressure was $\text{1}\text{.2}\text{atm}$ has to be given.

Concept Introduction

• Equilibrium constant at constant pressure:$K_p$

It is used to express the relationship between product pressures and reactant pressures.

For a general reaction, $\text{aA+bB}\underset{}{\rightleftharpoons }\text{cC+dD}$

$\text{Equilibrium}\text{constant}{\text{K}}_p\text{=}\frac{P_C{}^{\text{c}}{P}_D{}^{\text{d}}}{P_A{}^{\text{a}}{P}_B{}^{\text{b}}}$

• $\begin{array}{l}{\text{K}}_{\text{p}}\text{=}{\text{K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{Δn}}\\ \text{Δn}\text{-}\text{change}\text{in}\text{number}\text{of}\text{moles}\\ \text{R}\text{-}\text{Gas}\text{constant}\\ \text{T-Temperature}\end{array}$

Explanation of Solution

Given data:

${\text{P}}_{\text{total}}\text{=}\text{1}\text{.2}\text{atm}$

The reaction is given below:

${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{g}\right)\underset{}{\rightleftharpoons }\text{2}{\text{NO}}_{\text{2}}\left(\text{g}\right)$

${\text{P}}_{\text{total}}\text{=}{\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{+}{\text{P}}_{{\text{NO}}_{\text{2}}}\text{=}\text{1}\text{.2}\text{atm}$

The partial pressure of a gas in a mixture is the product of the mole fraction of the gas and the total pressure.

$\begin{array}{l}{P}_{N{O}_2}=X_{N{O}_2}{P}_{total}\\ where{X}_{N{O}_2}isthemolefraction\\ P_{N_2{O}_4}=X_{N_2{O}_4}{P}_{total}\\ where{X}_{N_2{O}_4}isthemolefraction\end{array}$

${\text{P}}_{{\text{NO}}_{\text{2}}}\text{=}{\text{X}}_{{\text{NO}}_{\text{2}}}{\text{P}}_{\text{total}}\text{=}\left(\text{0}\text{.466}\right)\left(\text{1}\text{.2}\text{atm}\right)\text{=}\text{0}\text{.56}\text{atm}$

${\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{=}{\text{X}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}{\text{P}}_{\text{total}}\text{=}\left(\text{0}\text{.534}\right)\left(\text{1}\text{.2}\text{atm}\right)\text{=}\text{0}\text{.64}\text{atm}$

${\text{K}}_{\text{P}}\text{=}\frac{{\text{P}}^{\text{2}}{}_{{\text{NO}}_{\text{2}}}}{{\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}}\text{=}\frac{{\left(\text{0}\text{.56}\right)}^{\text{2}}}{\text{0}\text{.64}}\text{=}\text{0}\text{.49}$

(c)

Interpretation Introduction

Interpretation:

The mole fractions if the pressure were increased to $\text{4}\text{.0}\text{atm}$ by reducing the volume t the same temperature has to be determined.

Concept Introduction

The partial pressure of a gas in a mixture is the product of the mole fraction of the gas and the total pressure.

$\begin{array}{l}{\text{P}}_{\text{A}}\text{=}{\text{X}}_{\text{A}}{\text{P}}_{\text{total}}\\ \text{where}{\text{X}}_{\text{A}}\text{is}\text{the}\text{mole}\text{fraction}\text{of}\text{the}\text{component}\text{A}\end{array}$

Explanation of Solution

Given data:

${\text{P}}_{\text{total}}\text{=}\text{4}\text{.0}\text{atm}$

The reaction is given below:

${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{g}\right)\rightleftarrows \text{2}{\text{NO}}_{\text{2}}\left(\text{g}\right)$

The partial pressure of a gas in a mixture is the product of the mole fraction of the gas and the total pressure.

$\begin{array}{l}{P}_{N{O}_2}=X_{N{O}_2}{P}_{total}\\ where{X}_{N{O}_2}isthemolefraction\\ P_{N_2{O}_4}=X_{N_2{O}_4}{P}_{total}\\ where{X}_{N_2{O}_4}isthemolefraction\end{array}$

The sum of the mole fractions will be equal to one.

$\begin{array}{l}{\text{X}}_{{\text{NO}}_{\text{2}}}{\text{+X}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{=}\text{1}\\ {\text{X}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{=}{\text{1-X}}_{{\text{NO}}_{\text{2}}}\end{array}$

${\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{=}{\text{X}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}{\text{P}}_{\text{total}}\text{=}\left({\text{1-X}}_{{\text{NO}}_{\text{2}}}\right){\text{P}}_{\text{total}}$

Substituting the values in the equilibrium constant expression:

$\begin{array}{c}{\text{K}}_{\text{P}}\text{=}\frac{{\text{P}}^{\text{2}}{}_{{\text{NO}}_2}}{{\text{P}}_{N_{\text{2}}{O}_4}}\text{=}\frac{{\left({\text{X}}_{{\text{NO}}_2}{\text{P}}_{\text{T}}\right)}^{\text{2}}}{\left({\text{1-X}}_{{\text{NO}}_2}\right){\text{P}}_{\text{T}}}\\ \text{0}\text{.487=}\frac{{\left({\text{X}}_{{\text{NO}}_2}4.0\right)}^{\text{2}}}{\left({\text{1-X}}_{{\text{NO}}_2}\right)4.0}\\ \text{0}\text{.487}\text{=}\frac{{\text{16X}}^2{}_{{\text{NO}}_2}}{4.0-4.0{\text{X}}_{{\text{NO}}_2}}\\ 1.95-1.95{\text{X}}_{{\text{NO}}_2}={\text{16X}}^2{}_{{\text{NO}}_2}\\ 0={\text{16X}}^2{}_{{\text{NO}}_2}+1.95{\text{X}}_{{\text{NO}}_2}-1.95\end{array}$

The value of ${\text{X}}_{{\text{NO}}_2}$ will be obtained by solving the quadratic equation.

$\begin{array}{l}{\text{X}}_{{\text{NO}}_{\text{2}}}\text{=}\text{0}\text{.29}\\ \\ {\text{X}}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}\text{=}{\text{1-X}}_{{\text{NO}}_{\text{2}}}\text{=}\text{1-0}\text{.29}\text{=}\text{0}\text{.71}\end{array}$