Chapter 15, Problem 122AP

In the gas phase, nitrogen dioxide is actually a mixture of nitrogen dioxide $\text{(N}{\text{O}}_{\text{2}}\text{)}$ and dinitrogen tetroxide $\text{(}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{)}$. If the density of such a mixture is 2.3 g/L at $\text{74°C}$ and 1.3 atm, calculate the partial pressures of the gases and $K_P$ for the dissociation of $\text{(}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{)}$.

Interpretation Introduction

Interpretation:

The partial pressure of ${\text{NO}}_2$ and ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ gases, and the value of $K_p$ for dissociation of ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ are to be calculated.

Concept introduction:

According to Dalton’s law, the total pressure of a mixture oftwo or more nonreactive gases is equal to the individual partial pressure of the gases present in the mixture. The pressure exerted by a gas in the mixture of gases is called its partial pressure.

Answer to Problem 122AP

Solution: $P_{{\text{NO}}_{\text{2}}}=1.2\text{atm}$, $P_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}=0.12\text{atm}$, and $K_p=12$.

Explanation of Solution

Given information: The density of mixture is $2.3\text{g}/\text{L}$.

The pressure of the gas is $1.3\text{atm}$.

The average mass of a gaseous mixture is calculated by the expression as follows:

$\overline{M}=\frac{d\text{R}T}{P}$

Here, $\overline{M}$ is the average mass of gases, $d$ is the density, $\text{R}$ is the gas constant, $T$ is the temperature, and $P$ is the pressure of the gas.

Substitute the values of $d,\text{R},T$, and $P$ in the equation given above,

$\begin{array}{c}\overline{M}=\frac{\left(2.3\text{g}/\cancel{\text{L}}\right)\left(0.0821\cancel{\text{L}}\text{.}\cancel{\text{atm}}/\cancel{\text{K}}\text{.mol}\right)\left(347\cancel{\text{K}}\right)}{1.3\cancel{\text{atm}}}\\ =\frac{65.524\text{g}/\text{mol}}{1.3}\\ =50.4\text{g}/\text{mol}\end{array}$

The sum of molar masses of each component is equal to the average molar mass with respective mole fractions.

The mole fraction of each component is calculated by the expression as follows:

$\overline{M}=X_{{\text{NO}}_{\text{2}}}{M}_{{\text{NO}}_{\text{2}}}+X_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}{M}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}$

Here, $X_{{\text{NO}}_{\text{2}}}$ is the mole fraction of ${\text{NO}}_2$, $M_{{\text{NO}}_{\text{2}}}$ is the molar mass of ${\text{NO}}_2$, $X_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}$ is the mole fraction of ${\text{N}}_2{\text{O}}_4$, and $M_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}$ is the molar mass of ${\text{N}}_2{\text{O}}_4$.

Substitute the values of $M_{{\text{NO}}_{\text{2}}},X_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}},M_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}$, and $\overline{M}$ in the equation given above,

$\begin{array}{c}{X}_{{\text{NO}}_{\text{2}}}\left(46.01\text{g}/\text{mol}\right)+\left(1-X_{{\text{NO}}_{\text{2}}}\right)\left(92.01\text{g}/\text{mol}\right)=50.4\text{g}/\text{mol}\\ X_{{\text{NO}}_{\text{2}}}\left(46.01\text{g}/\text{mol}\right)+92.01\text{g}/\text{mol}-X_{{\text{NO}}_2}\left(92.01\text{g}/\text{mol}\right)=50.4\text{g}/\text{mol}\\ X_{{\text{NO}}_{\text{2}}}\left(46.01\text{g}/\text{mol}-92.01\text{g}/\text{mol}\right)=50.4\text{g}/\text{mol}-92.01\text{g}/\text{mol}\end{array}$

On solving further, we get,

$\begin{array}{c}{X}_{{\text{NO}}_{\text{2}}}\left(-46\text{g}/\text{mol}\right)=-41.61\text{g}/\text{mol}\\ X_{{\text{NO}}_{\text{2}}}=\frac{-41.61\text{g}/\text{mol}}{-46\text{g}/\text{mol}}\\ =0.905\end{array}$

Partial pressure of a gas is the product of its mole fraction and the total pressure of the mixture of gases.

The partial pressure of $N{O}_2$ gas is calculated as follows:

$\begin{array}{c}{P}_{{\text{NO}}_{\text{2}}}=X_{{\text{NO}}_{\text{2}}}\left(\text{total pressure}\right)\\ =\left(0.905\right)\left(1.3\text{atm}\right)\\ =1.2\text{atm}\end{array}$

The partial pressure of $N_2{O}_4$ gas is calculated as follows:

$\begin{array}{c}{P}_{{\text{N}}_2{\text{O}}_4}=X_{{\text{N}}_2{\text{O}}_4}\left(\text{total pressure}\right)\\ =\left(1-0.905\right)\left(1.3\text{atm}\right)\\ =0.12\text{atm}\end{array}$

The equilibrium constant for dissociation of $N_2{O}_4$ is calculated as follows:

$K_p=\frac{P_{{}_{{\text{NO}}_{\text{2}}}}^2}{P_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}}$

Substitute the values of $P_{{\text{NO}}_{\text{2}}}$ and $P_{{\text{N}}_2{\text{O}}_4}$ in the equation given above,

$\begin{array}{c}{K}_p=\frac{{\left(1.2\right)}^2}{0.12}\\ =\frac{1.44}{0.12}\\ =12\end{array}$

Conclusion

The partial pressure of ${\text{NO}}_2$ and ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}$ gases is $1.2\text{atm}$ and $0.12\text{atm}$, respectively, and the value of the equilibrium constant is $12$.