Chapter 11, Problem 11.139QP

(a)

Interpretation Introduction

Interpretation:

Consider following compounds,

${\text{CH}}_{\text{3}}\text{CHO}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ and ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ given statements has to be explained.

Concept introduction:

Intermolecular forces are Van der Waals forces.  They are weak and have three types viz., London dispersion forces, dipole-dipole forces and hydrogen bonding. Hydrogen bonding is relatively the strongest one.

• Intermolecular forces are the forces acting between molecules whereas Intramolecular forces are the forces that operate within a molecule.
• Hydrogen bonding is a special type of Dipole-dipole forces but stronger than the former.
• London dispersion forces exist in non-polar covalent compounds whereas dipole-dipole forces exist in polar covalent compounds but both are weak.
• Larger the molecular size, stronger the London dispersion force.
• Arrangement of major types of intermolecular forces in increasing order of strength:

$\text{London}\text{dispersion}\text{forces}\text{<}\text{Dipole-dipole}\text{forces}\text{<}\text{Hydrogen}\text{bonding}$

There exist no deviations in this arrangement.

Explanation of Solution

Intermolecular force in each compound

${\text{CH}}_{\text{3}}\text{CHO}$ is found to be polar molecule, therefore it has London and dipole-dipole forces present.

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ is nonpolar molecule, so it has London force only.

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ is found to be polar molecule with London forces, dipole-dipole forces and hydrogen bonding.

(b)

Interpretation Introduction

Interpretation:

Consider following compounds,

${\text{CH}}_{\text{3}}\text{CHO}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ and ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ given statements has to be explained.

Concept introduction:

Intermolecular forces are Van der Waals forces.  They are weak and have three types viz., London dispersion forces, dipole-dipole forces and hydrogen bonding. Hydrogen bonding is relatively the strongest one.

• Intermolecular forces are the forces acting between molecules whereas Intramolecular forces are the forces that operate within a molecule.
• Hydrogen bonding is a special type of Dipole-dipole forces but stronger than the former.
• London dispersion forces exist in non-polar covalent compounds whereas dipole-dipole forces exist in polar covalent compounds but both are weak.
• Larger the molecular size, stronger the London dispersion force.
• Arrangement of major types of intermolecular forces in increasing order of strength:

$\text{London}\text{dispersion}\text{forces}\text{<}\text{Dipole-dipole}\text{forces}\text{<}\text{Hydrogen}\text{bonding}$

There exist no deviations in this arrangement.

Explanation of Solution

To determine: heat of vaporization of Methanol

Given

Heat of vaporization of these three compounds (not in particular order) is,

$\text{25}\text{.8kJ/mol,}\text{38}\text{.6kJ/mol and 19}\text{.0kJ/mol}$

The heat of vaporization of the compound increases with increasing intermolecular forces. So, the compound ${\text{CH}}_{\text{3}}\text{CHO}$ has stronger intermolecular force than ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ and weaker intermolecular forces than ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$.

Hence, the heat of vaporization of ${\text{CH}}_{\text{3}}\text{CHO}$ is $\text{25}\text{.8kJ/mol}\text{.}$

(c)

Interpretation Introduction

Interpretation:

Consider following compounds,

${\text{CH}}_{\text{3}}\text{CHO}$, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{3}}$ and ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ given statements has to be explained.

Concept introduction:

Intermolecular forces are Van der Waals forces.  They are weak and have three types viz., London dispersion forces, dipole-dipole forces and hydrogen bonding. Hydrogen bonding is relatively the strongest one.

• Intermolecular forces are the forces acting between molecules whereas Intramolecular forces are the forces that operate within a molecule.
• Hydrogen bonding is a special type of Dipole-dipole forces but stronger than the former.
• London dispersion forces exist in non-polar covalent compounds whereas dipole-dipole forces exist in polar covalent compounds but both are weak.
• Larger the molecular size, stronger the London dispersion force.
• Arrangement of major types of intermolecular forces in increasing order of strength:

$\text{London}\text{dispersion}\text{forces}\text{<}\text{Dipole-dipole}\text{forces}\text{<}\text{Hydrogen}\text{bonding}$

There exist no deviations in this arrangement.

Explanation of Solution

To determine: vapour pressure of ${\text{CH}}_{\text{3}}\text{CHO}$ at $\text{15°C}$

Given

Normal boiling point of ${\text{CH}}_{\text{3}}\text{CHO}$ is $\text{21°C}$

By using Clapeyron equation vapour pressure has to be calculated.

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

$\text{ln}\frac{{\text{P}}_{\text{2}}}{\text{760 mmHg}}\text{=}\frac{\text{25}{\text{.8×10}}^{\text{3}}\text{J/mol}}{\text{8}\text{.3145J/K•mol}}\text{(}\frac{\text{1}}{\text{294K}}\text{-}\frac{\text{1}}{\text{288K}}\text{) = -0}\text{.2198}$

Taking antilog of both sides,

$\text{ln}\frac{{\text{P}}_{\text{2}}}{\text{760 mmHg}}{\text{= e}}^{\text{-0}\text{.2198}}\text{= 0}\text{.8027}$

${\text{P}}_{\text{2}}\text{ = 0}\text{.8027×760mmHg = 610}{\text{.1×10}}^{\text{2}}\text{mmHg}$