Chapter 24, Problem 24.19P

a)

Interpretation Introduction

Interpretation:

The enthalpy of vaporization of Octane has to be calculated.

Concept Introduction:

The Trouton’s rule for non-bonding Hydrogen solutes states that the heat of vaporization is directly proportional to the boiling temperature.

The enthalpy of vaporization can be calculated as,

${\text{ΔH}}_{\text{vap}}^{\text{°}}\cong \left(\text{88}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\right){\text{•T}}_{\text{bp}}$

To calculate the enthalpy of vaporization of Octane

Answer to Problem 24.19P

The enthalpy of vaporization of Octane is $\text{35}\text{.12}\text{kJ}{\text{mol}}^{\text{-1}}$ .

Explanation of Solution

Given,

Boiling point of Octane = $\text{126°C}$

Boiling point = $\text{126°C+273}\text{.15}$

Boiling point = $\text{339}\text{.15}\text{K}$

The enthalpy of vaporization of Octane is calculated as,

${\text{ΔH}}_{\text{vap}}^{\text{°}}\cong \left(\text{88}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\right){\text{•T}}_{\text{bp}}$

$\begin{array}{l}{\text{ΔH}}_{\text{vap}}^{\text{°}}=\left(\text{88}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\right)\text{•}\left(\text{399}\text{.15}\text{K}\right)\\ \\ {\text{ΔH}}_{\text{vap}}^{\text{°}}\text{=35125}\text{.2}\text{J}{\text{mol}}^{\text{-1}}\\ \\ {\text{ΔH}}_{\text{vap}}^{\text{°}}\text{=35}\text{.12}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

The enthalpy of vaporization of Octane = $\text{35}\text{.12}\text{kJ}{\text{mol}}^{\text{-1}}$

b)

Interpretation Introduction

Interpretation:

The vapour pressure of Octane has to be calculated.

Concept Introduction:

The vapour pressure of compound can be related to heat of vaporization ${\text{ΔH}}_{\text{vap}}$ , through Clausius-Clapeyron equation that is given as,

$\text{ln}\left(\frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\right)\text{=}\frac{{\text{-ΔH}}_{\text{vap}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{2}}}\right]$

Here, $\left(\frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\right)$= vapour pressure of the compounds

${\text{-ΔH}}_{\text{vap}}$= Heat of vaporization

${\text{T}}_{\text{1}}\text{and}{\text{T}}_{\text{2}}$ = Temperatures (in K)

To calculate the vapour pressure of Octane

Answer to Problem 24.19P

The vapour pressure of Octane is $\text{0}\text{.18}\text{bar}$ .

Explanation of Solution

Given,

${\text{T}}_{\text{2}}\text{=70°C}$

${\text{T}}_{\text{1}}\text{=126°C+273}\text{.15=399}\text{.15}\text{K}$

${\text{T}}_{\text{2}}\text{=70°C+273}\text{.15=343}\text{.15}\text{K}$

$\begin{array}{l}\text{ln}\text{P=}\left[\frac{\text{35125}\text{.2}\text{J}{\text{mol}}^{\text{-1}}}{\text{8}\text{.314}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}}\right]\left[\frac{\text{1}}{\text{343}\text{.15}}\text{-}\frac{\text{1}}{\text{399}\text{.15}}\right]\\ \\ \text{ln P=}\left[\frac{\text{35125}\text{.2}\text{J}{\text{mol}}^{\text{-1}}}{\text{8}\text{.314}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}}\right]\left[\frac{\text{399}\text{.15-343}\text{.15}}{\left(\text{343}\text{.15}\right)\left(\text{399}\text{.15}\right)}\right]\\ \\ \text{ln}\text{P=}\frac{\text{35125}\text{.2×56}}{\text{8}\text{.314×136968}\text{.32}}\\ \\ \text{ln}\text{P=}\frac{\text{1967011}\text{.2}}{\text{1138754}\text{.61}}\\ \\ {\text{P=e}}^{\text{-1}\text{.727}}\\ \\ \text{P=0}\text{.18}\text{bar}\end{array}$

The vapour pressure of Octane = $\text{0}\text{.18}\text{bar}$

c)

Interpretation Introduction

Interpretation:

The vapour pressure of Hexane has to be calculated.

Concept Introduction:

The vapour pressure of compound can be related to heat of vaporization ${\text{ΔH}}_{\text{vap}}$ , through Clausius-Clapeyron equation that is given as,

$\text{ln}\left(\frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\right)\text{=}\frac{{\text{-ΔH}}_{\text{vap}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{2}}}\right]$

Here, $\left(\frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\right)$= vapour pressure of the compounds

${\text{-ΔH}}_{\text{vap}}$= Heat of vaporization

${\text{T}}_{\text{1}}\text{and}{\text{T}}_{\text{2}}$ = Temperatures (in K)

To calculate the vapour pressure of Hexane

Answer to Problem 24.19P

The vapour pressure of Hexane is $\text{1}\text{.05}\text{bar}$ .

Explanation of Solution

$\begin{array}{l}\text{ln P=}\frac{{\text{-ΔH}}_{\text{vap}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{2}}}\right]\\ \\ \text{ln}\text{P=}\left[\frac{\text{-30197}\text{.2}\text{J}{\text{mol}}^{\text{-1}}}{\text{8}\text{.314}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}}\right]\left[\frac{\text{1}}{\text{343}\text{.15}}\text{-}\frac{\text{1}}{\text{342}\text{.15}}\right]\\ \\ \text{P=1}\text{.05bar}\\ \end{array}$

The vapour pressure of Hexane = $\text{1}\text{.05}\text{bar}$

d)

Interpretation Introduction

Interpretation:

The relationship between vapour pressure of solute and retention has to be given.

To give the relationship between vapour pressure of solute and retention

Explanation of Solution

The vapour pressure of solute and the retention are inversely proportional to each other.

If the vapour pressure would be lower, then the retention would be greater.

e)

Interpretation Introduction

Interpretation:

The reason has to be explained for the name of technique as “gas chromatography” if the analytes retained are partially vaporized.

To explain why the technique is called as “gas chromatography” if the analytes retained are partially vaporized

Explanation of Solution

When the analytes retained are partially vaporized, the technique is called as gas chromatography since the mobile phase would be gaseous in form.