Chapter 3, Problem 3.7E

(a)

Interpretation Introduction

Interpretation:

Balanced equation for decomposition of ${\text{N}}_2{\text{O}}_4$ has to be determined.

Concept Introduction:

Dolton’s law of partial pressure gives relation between total pressure of mixture of gases and partial pressure of individual gases. The expression of relation can be represented as follows:

  $P=P_{\text{A}}+P_{\text{B}}+\cdot \cdot \cdot$

Here,

A and B are individual gases.

$P_{\text{A}}$ is partial pressure of gas A.

$P_{\text{B}}$ is partial pressure of gas B.

$P$ is total pressure of mixture of gases.

Ideal gas law can be represented as equation for state for any imaginary gas. Expression for ideal gas equation is as follows:

  $PV=nRT$

Here,

$P$ is pressure of the gas.

$V$ is volume of gas.

$n$ denotes moles of gas.

$R$ is gas constant.

$T$ is temperature of gas.

Explanation of Solution

Decomposition of ${\text{N}}_2{\text{O}}_4$ gas produces ${\text{NO}}_{\text{2}}$ gas. This reaction can be represented as follows:

  ${\text{N}}_2{\text{O}}_4\to {\text{NO}}_{\text{2}}$        (1)

To balance the reaction (1), add coefficient 2 for ${\text{NO}}_{\text{2}}$ as follows:

  ${\text{N}}_2{\text{O}}_4\to 2{\text{NO}}_{\text{2}}$

Hence balance equation for decomposition of ${\text{N}}_2{\text{O}}_4$ gas is as follows:

  ${\text{N}}_2{\text{O}}_4\to 2{\text{NO}}_{\text{2}}$

(b)

Interpretation Introduction

Interpretation:

Pressure of ${\text{N}}_2{\text{O}}_4$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3.7E

Pressure of ${\text{N}}_2{\text{O}}_4$ at $25°\text{C}$ is $2.33\text{ atm}$.

Explanation of Solution

Number of moles of ${\text{N}}_2{\text{O}}_4$ can be calculated as follows:

  $\begin{array}{c}\text{Mole}=\frac{\text{Given mass}}{\text{Molar mass}}\\ =\frac{43.78\text{ g}}{92.011\text{ g/mol}}\\ =0.4758\text{ mol}\end{array}$

The expression to calculate pressure of gas is as follows:

  $P=\frac{nRT}{V}$        (2)

The conversion factor to convert $25\text{°C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{temperature}\text{in}\text{Kelvin}=\text{temperature}\text{in}\text{celsius}+273.15\\ =25\text{°C}+273.15\\ =298.15\text{K}\end{array}$

Substitute $0.4758\text{ mol}$ for $n$, $5.00\text{ L}$ for $V$, $8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ and $298.15\text{ K}$ for $T$ in equation (2).

  $\begin{array}{c}P=\frac{\left(0.4758\text{ mol}\right)\left(8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(298.15\text{ K}\right)}{5.00\text{ L}}\\ =2.33\text{ atm}\end{array}$

Hence pressure of ${\text{N}}_2{\text{O}}_4$ at $25°\text{C}$ is $2.33\text{ atm}$.

(c)

Interpretation Introduction

Interpretation:

Pressure of gas if whole ${\text{N}}_2{\text{O}}_4$ is converted into ${\text{NO}}_2$ gas has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3.7E

Pressure of ${\text{NO}}_2$ at $25°\text{C}$ is $\text{4}\text{.65 atm}$.

Explanation of Solution

Balance equation for decomposition of ${\text{N}}_2{\text{O}}_4$ gas is as follows:

  ${\text{N}}_2{\text{O}}_4\to 2{\text{NO}}_{\text{2}}$

Since $1\text{ mol}$ of ${\text{N}}_2{\text{O}}_4$ gives $2\text{ mol}$ of ${\text{NO}}_2$. At constant volume and temperature, pressure of gas is directly proportional to number of moles thus pressure of gas if whole ${\text{N}}_2{\text{O}}_4$ is converted into ${\text{NO}}_2$ gas is twice of pressure if it was ${\text{N}}_2{\text{O}}_4$.

Hence pressure of ${\text{NO}}_2$ is $\text{4}\text{.65 atm}$.

(b)

Interpretation Introduction

Interpretation:

Mole fractions of ${\text{N}}_2{\text{O}}_4$ and ${\text{NO}}_2$ after reaction have to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3.7E

Mole fraction for ${\text{N}}_2{\text{O}}_4$ is $0.574$ and same for ${\text{NO}}_2$ is $0.426$.

Explanation of Solution

Balanced equation for decomposition of ${\text{N}}_2{\text{O}}_4$ gas is as follows:

  ${\text{N}}_2{\text{O}}_4\to 2{\text{NO}}_{\text{2}}$

Initial partial pressure of ${\text{N}}_2{\text{O}}_4$ at $25°\text{C}$ is $2.33\text{ atm}$. Consider ICE table below for reaction.

$\begin{array}{cccc}& {\text{N}}_2{\text{O}}_4& \to & 2{\text{NO}}_{\text{2}}\\ \text{Initial}\left(\text{atm}\right)& 2.33\text{ atm}& & -\\ \text{Change}\left(\text{atm}\right)& -x& & +2x\\ \text{Equilibrium}\left(\text{atm}\right)& \left(2.33-x\right)\text{ atm}& & 2x\text{ atm}\end{array}$

Total pressure at equilibrium $\left(\left(2.33+x\right)\text{ atm}\right)$ is $2.96\text{ atm}$, thus value of x can be calculated as follows:

  $\begin{array}{c}\text{x}=\left(2.96-2.33\right)\text{ atm}\\ \text{=0}\text{.63 atm}\end{array}$

Thus at equilibrium, pressure of ${\text{N}}_2{\text{O}}_4$ is $1.70\text{ atm}$ and pressure of ${\text{NO}}_{\text{2}}$ is $1.26\text{ atm}$.

Expression for mole fraction is follows:

  $\text{Mole fraction}=\frac{n_{\text{A}}}{n_{\text{A}}+n_{\text{B}}}$        (3)

Here,

$n_{\text{A}}$ are moles of gas A.

$n_{\text{B}}$ are moles of gas B.

Since pressure is directly proportional to moles at constant temperature and volume thus equation (3) for ${\text{N}}_2{\text{O}}_4$ gas can be deduced as follows:

  $\text{Mole fraction}=\frac{P_{{\text{N}}_2{\text{O}}_4}}{P_{{\text{N}}_2{\text{O}}_4}+P_{{\text{NO}}_2}}$        (4)

Substitute $1.70\text{ atm}$ for $P_{{\text{N}}_2{\text{O}}_4}$ and $1.26\text{ atm}$ for $P_{{\text{NO}}_2}$ in equation (4).

  $\begin{array}{c}\text{Mole fraction}=\frac{1.70\text{ atm}}{1.70\text{ atm}+1.26\text{ atm}}\\ =0.574\end{array}$

Equation (1) for ${\text{NO}}_2$ gas can be deduced as follows:

  $\text{Mole fraction}=\frac{P_{{\text{NO}}_2}}{P_{{\text{N}}_2{\text{O}}_4}+P_{{\text{NO}}_2}}$        (5)

Substitute $1.70\text{ atm}$ for $P_{{\text{N}}_2{\text{O}}_4}$ and $1.26\text{ atm}$ for $P_{{\text{NO}}_2}$ in equation (5).

  $\begin{array}{c}\text{Mole fraction}=\frac{1.26\text{ atm}}{1.70\text{ atm}+1.26\text{ atm}}\\ =0.426\end{array}$

Hence mole fraction for ${\text{N}}_2{\text{O}}_4$ is $0.574$ and same for ${\text{NO}}_2$ is $0.426$.