Chapter 5, Problem 160CP

Interpretation Introduction

Interpretation: The mole fraction of $\text{He}$ in the original mixture needs to be determined.

Concept Introduction: The mole fraction of $\text{He}$ in the original mixture can be calculated using partial and total densities.

The partial density of a compound is determined by the formula:

  $\frac{1\text{ mol of compound}\times \text{Molar mass of compound in g/mol}}{\text{Volume at STP (22}\text{.4 L)}}\text{ = Partial demsity of compound in g/L}$

The formula relating partial density and total density is:

  $\sum \text{partial density}\times \text{mole fraction = total density}$

Answer to Problem 160CP

The mole fraction of $\text{He}$ in the original mixture is $0.112$ .

Explanation of Solution

Given:

Equimolar mixture of gases, ${\text{SO}}_{\text{2}}{\text{ and O}}_{\text{2}}$ along with some $\text{He}$ .

The density of mixture at STP is $\text{1}\text{.924 g/L}$ .

The partial density of each gas, ${\text{SO}}_{\text{2}}{\text{, O}}_{\text{2}}\text{ and He}$ is calculated as:

Molar mass of ${\text{SO}}_{\text{2}}{\text{, O}}_{\text{2}}\text{ and He}$ are $\text{64}\text{.07 g/mol, 32}\text{.00 g/mol and 4}\text{.003 g/mol}$ respectively.

  $\begin{array}{l}\frac{{\text{1 mol SO}}_2\left(\frac{\text{64}{\text{.07 g SO}}_2}{{\text{1 mol SO}}_2}\right)}{\text{22}\text{.4 L}}\text{ = 2}{\text{.86 g/L SO}}_2\\ \frac{{\text{1 mol O}}_2\left(\frac{\text{32}{\text{.00 g O}}_2}{{\text{1 mol O}}_2}\right)}{\text{22}\text{.4 L}}\text{ = 1}{\text{.43 g/L O}}_2\\ \frac{\text{1 mol He}\left(\frac{\text{4}\text{.003 g He}}{\text{1 mol He}}\right)}{\text{22}\text{.4 L}}\text{ = 0}\text{.179 g/L He}\end{array}$

Let $\text{x}$ be the mole fraction of ${\text{SO}}_{\text{2}}$ . Since, ${\text{SO}}_{\text{2}}{\text{ and O}}_{\text{2}}$ are in equimolar so, mole fraction of ${\text{O}}_{\text{2}}$ is also $\text{x}$ . For mixture of gases, the sum of mole fraction is 1, thus, the mole fraction of $\text{He}$ is:

  $\begin{array}{l}{\text{x + x + x}}_{\text{He}}\text{ = 1}\\ {\text{x}}_{\text{He}}\text{ = }1\text{ - 2x}\end{array}$

Now, using the relation,

  $\sum \text{partial density}\times \text{mole fraction = total density}$

Substituting the values:

  $\begin{array}{l}\left[\text{x(2}{\text{.86 SO}}_{\text{2}}\text{) + x(1}{\text{.43 O}}_{\text{2}}\text{) + (1 - 2x)(0}\text{.179 He)}\right]\text{ = 1}\text{.924}\\ \text{2}\text{.86x + 1}\text{.43x + 0}\text{.179 - 0}\text{.358x = 1}\text{.924}\\ \text{3}\text{.932x = 1}\text{.924 - 0}\text{.179}\\ \text{x = }\frac{\text{1}\text{.745}}{\text{3}\text{.932}}\\ \text{x = 0}\text{.444}\end{array}$

Substituting the value of $\text{x}$ in mole fraction of $\text{He}$ as:

  ${\text{x}}_{\text{He}}\text{ = }1\text{ - 2x}$

  $\begin{array}{l}{\text{x}}_{\text{He}}\text{ = }1\text{ - 2}\left(\text{0}\text{.444}\right)\\ {\text{x}}_{\text{He}}\text{ = }1\text{ - 0}\text{.888}\\ {\text{x}}_{\text{He}}\text{ = }0.112\end{array}$

Hence, the mole fraction of $\text{He}$ in the original mixture is $0.112$ .

Interpretation Introduction

Interpretation: The density of gas mixture after the completion of reaction needs to be determined.

Concept Introduction: The mole fraction of $\text{He}$ in the original mixture can be calculated using partial and total densities.

The partial density of a compound is determined by the formula:

  $\frac{1\text{ mol of compound}\times \text{Molar mass of compound in g/mol}}{\text{Volume at STP (22}\text{.4 L)}}\text{ = Partial demsity of compound in g/L}$

The formula relating partial density and total density is:

  $\sum \text{partial density}\times \text{mole fraction = total density}$

Answer to Problem 160CP

The density of gas mixture after the completion of reaction is $\text{2}\text{.47 g/L}$ .

Explanation of Solution

Given:

The ${\text{SO}}_2$ and ${\text{O}}_2$ reacts completely to form ${\text{SO}}_3$ .

The balanced reaction for the formation of ${\text{SO}}_3$ from ${\text{SO}}_2$ and ${\text{O}}_2$ is:

  ${\text{SO}}_2\text{ + }\frac{1}{2}{\text{O}}_2\to {\text{ SO}}_3$

From the balanced reaction, it can be observed that 1 mole of ${\text{SO}}_2$ combines with $\frac{1}{2}$ mole of ${\text{O}}_2$ to form 1 mole of ${\text{SO}}_3$ and the moles of $\text{He}$ remains unchanged.

Now, determining the limiting reagent:

Since, from part (a) the initial moles of the reactants are $\text{0}\text{.444 mol}$ so, the moles of ${\text{O}}_2$ required for $\text{0}\text{.444 mol}$ of ${\text{SO}}_2$ is:

  $\text{0}{\text{.444 mol of SO}}_2\times \frac{1{\text{ mol of O}}_2}{2{\text{ mol of SO}}_2}\text{ = 0}{\text{.222 mol of O}}_2$

Thus, ${\text{SO}}_2$ is the limiting reactant. Hence, the final moles on the basis of balanced reaction is:

  $\begin{array}{l}{\text{SO}}_2\text{ = 0 mol}\\ {\text{O}}_2\text{ = 0}{\text{.444 mol of SO}}_2\times \frac{1{\text{ mol of O}}_2}{2{\text{ mol of SO}}_2}\text{ = 0}{\text{.222 mol of O}}_2\\ \text{He = 0}\text{.112 mol}\\ {\text{SO}}_3\text{ = 0}{\text{.444 mol of SO}}_2\times \frac{1{\text{ mol of SO}}_3}{{\text{1 mol of SO}}_2}\text{ = 0}{\text{.444 mol of O}}_3\end{array}$

The density of the gas mixture is calculated using formula:

  $\sum \text{partial density}\times \text{mole fraction = total density}$ - (1)

The partial density of ${\text{SO}}_3$ is:

Molar mass of ${\text{SO}}_3$ is $\text{80}\text{.07 g/mol}$ .

Substituting the values:

  $\frac{1\text{ mol }\left(\frac{\text{80}{\text{.07 g SO}}_3}{{\text{1 mol SO}}_3}\right)}{\text{22}\text{.4 L}}\text{= 3}{\text{.57 g/L SO}}_3$

Calculating the mole fraction of each component as:

The sum of total final moles is:

  $\begin{array}{l}\text{total final moles = (0}\text{.444 + 0}\text{.112 + 0}\text{.222) mol}\\ \text{total final moles = 0}\text{.778 mol}\end{array}$

Now, the mole fraction of each component is calculated using formula:

  $\text{mole fraction = }\frac{\text{mole of component}}{\text{total number of moles in the mixture}}$

So:

  $\begin{array}{l}{\text{X}}_{{\text{O}}_2}\text{ = }\frac{0.222\text{ mol}}{0.778\text{ mol}}\\ {\text{X}}_{{\text{O}}_2}\text{ = 0}\text{.285}\end{array}$

  $\begin{array}{l}{\text{X}}_{\text{He}}\text{ = }\frac{0.112\text{ mol}}{0.778\text{ mol}}\\ {\text{X}}_{\text{He}}\text{ = 0}\text{.144}\end{array}$

  $\begin{array}{l}{\text{X}}_{{\text{SO}}_2}\text{ = }\frac{0.444\text{ mol}}{0.778\text{ mol}}\\ {\text{X}}_{{\text{SO}}_2}\text{ = 0}\text{.571}\end{array}$

Substituting the values in equation (1):

  $\begin{array}{l}\text{Total density = }\left[(0.285)(1.43{\text{ g/L O}}_2)+(0.144)(0.179\text{ g/L He)+(0}\text{.571)(3}{\text{.57 g/L SO}}_3)\right]\\ \text{Total density = }0.40755\text{ g/L }+\text{ 0}\text{.025776 g/L + 2}\text{.03847 g/L }\\ \text{Total density = 2}\text{.47 g/L}\end{array}$

Hence, the density of gas mixture after the completion of reaction is $\text{2}\text{.47 g/L}$ .