Chapter 11, Problem 52SCQ

(a)

Interpretation Introduction

Interpretation:

The reason for ${\text{H}}_{\text{2}}\text{S}$ being a gas at normal pressure and water being a liquid has to be determined.

Concept Introduction:

• Partial pressure: The pressure of each gas in a mixture of gases is the partial pressure.
• Dalton’s law of partial pressure: The total pressure of gases in a mixture of gases is the sum of the pressure of each gas in the mixture
• Mole fraction: Quantity which defines the number of moles of a substance in a mixture divided by the total number of moles of all substances present.

  ${\text{x}}_{\text{a}}\text{=}\frac{{\text{n}}_{\text{a}}}{{\text{n}}_{\text{total}}}$

• $\begin{array}{l}{\text{p}}_{\text{a}}\text{=}{\text{x}}_{\text{a}}\text{×}{\text{P}}_{\text{total}}\\ \end{array}$

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

Explanation of Solution

The major reason behind this observation is hydrogen bonding The electronegativity difference between hydrogen and oxygen is very large. So the intermolecular force of attraction is high. But the electronegativity difference between hydrogen and sulphur is comparatively low.

Comparing ${\text{H}}_{\text{2}}\text{S}$ and water , energy required to overcome the forces of attraction is less for ${\text{H}}_{\text{2}}\text{S}$ than water. This energy can be easily taken from room temperature.

Thus ${\text{H}}_{\text{2}}\text{S}$ exist as a gas and water exist as a liquid at room temperature.

(b)

Interpretation Introduction

Interpretation:

The partial pressure of ${\text{H}}_{\text{2}}\text{S}$ has to determined.

Concept Introduction:

• Partial pressure: The pressure of each gas in a mixture of gases is the partial pressure.
• Dalton’s law of partial pressure: The total pressure of gases in a mixture of gases is the sum of the pressure of each gas in the mixture
• Mole fraction: Quantity which defines the number of moles of a substance in a mixture divided by the total number of moles of all substances present.

  ${\text{x}}_{\text{a}}\text{=}\frac{{\text{n}}_{\text{a}}}{{\text{n}}_{\text{total}}}$

• $\begin{array}{l}{\text{p}}_{\text{a}}\text{=}{\text{x}}_{\text{a}}\text{×}{\text{P}}_{\text{total}}\\ \end{array}$

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

Explanation of Solution

The partial pressure of ${\text{H}}_{\text{2}}\text{S}$ at given conditions.

Given:

$\begin{array}{c}\text{Concentration}\text{of}{\text{H}}_{\text{2}}\text{S}\text{=}\text{80}\text{ppm}\\ \text{V}\text{=}\text{1}\text{L}\\ {\text{P}}_{\text{Total}}\text{=}\text{725}\text{mm}\text{Hg}\end{array}$

$\text{M}\text{=}\frac{\text{ppm}}{\text{Molecular}\text{mass×1000}}$

Molarity of ${\text{H}}_{\text{2}}\text{S}$$\begin{array}{l}\text{=}\frac{\text{80}}{\text{34}\text{.08×1000}}\\ \text{=}\text{2}{\text{.347×10}}^{\text{-3}}\text{mol}/\text{L}\end{array}$

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{=}\text{Molarity×Volume}\\ \text{Number}\text{of}\text{moles}\text{of}{\text{H}}_{\text{2}}\text{S}\text{=}\text{2}{\text{.347×10}}^{\text{-3}}\text{×1}\\ \text{=}\text{2}{\text{.347×10}}^{\text{-3}}\text{moles}\end{array}$

Using the given data we can find out the total number of moles in the mixture.

$\text{PV}\text{=}\text{nRT}$

$\begin{array}{l}\text{n}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{=}\frac{\text{0}\text{.9539×1}}{\text{0}\text{.0831×295}}\\ \text{=}\text{0}\text{.039166}\text{moles}\end{array}$

$\begin{array}{l}\text{Mole}\text{fraction}\text{of}{\text{H}}_{\text{2}}\text{S}\text{=}\frac{\text{2}{\text{.347×10}}^{\text{-3}}\text{moles}}{\text{0}\text{.039166moles}}\\ \text{=}\text{0}\text{.0599}\end{array}$

$\begin{array}{c}{\text{p}}_{\text{a}}\text{=}{\text{x}}_{\text{a}}\text{×}{\text{P}}_{\text{total}}\\ {\text{p}}_{{\text{H}}_{\text{2}}\text{S}}\text{=}\text{0}\text{.05992×0}\text{.9539}\\ \text{=}\text{0}\text{.0571}\text{atm}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The amount of oxygen required to convert the given amount of ${\text{H}}_{\text{2}}\text{S}$ to ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ has to be given.

Concept Introduction:

• Partial pressure: The pressure of each gas in a mixture of gases is the partial pressure.
• Dalton’s law of partial pressure: The total pressure of gases in a mixture of gases is the sum of the pressure of each gas in the mixture
• Mole fraction: Quantity which defines the number of moles of a substance in a mixture divided by the total number of moles of all substances present.

  ${\text{x}}_{\text{a}}\text{=}\frac{{\text{n}}_{\text{a}}}{{\text{n}}_{\text{total}}}$

• $\begin{array}{l}{\text{p}}_{\text{a}}\text{=}{\text{x}}_{\text{a}}\text{×}{\text{P}}_{\text{total}}\\ \end{array}$

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

Explanation of Solution

Given:

${\text{H}}_{\text{2}}{\text{S+2O}}_{\text{2}}\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$

$\begin{array}{l}\text{Volume}\text{of}{\text{H}}_{\text{2}}\text{S}\text{=}\text{5}\text{.2}\text{L}\\ \text{P}\text{=}\text{130}\text{mm}\text{Hg}\text{=}\text{0}\text{.1710}\text{atm}\\ \text{T}\text{=}\text{25°C}\text{=298}\text{K}\end{array}$

Using ideal gas equation we can calculate the number of moles of ${\text{H}}_{\text{2}}\text{S}$

$\begin{array}{l}\text{n}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{=}\frac{\text{0}\text{.1710}\text{×5}\text{.2}}{\text{0}\text{.0829×298}}\\ \text{=}\text{0}\text{.036}\text{moles}\end{array}$

From the balanced equation it is clear that if one mole of ${\text{H}}_{\text{2}}\text{S}$ react with 2 moles of ${\text{O}}_{\text{2}}$ , we will get one mole of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$

Number of moles of ${\text{O}}_{\text{2}}$ required to complete the reaction will be $\text{0}\text{.072}\text{moles}{\text{O}}_{\text{2}}\text{(0}\text{.036×2)}$

Using the ideal gas equation we can calculate the volume of oxygen needed to complete the reaction.

$\text{PV}\text{=}\text{nRT}$

$\begin{array}{l}\text{V}\text{=}\frac{\text{nRT}}{\text{P}}\\ \text{=}\frac{\text{0}\text{.072}\text{×0}\text{.0829×298}}{\text{0}\text{.1710}}\\ \text{=}\text{10}\text{.4}\text{L}\end{array}$