Chapter 11, Problem 11.55QE

(a)

Interpretation Introduction

Interpretation:

The liquid that is expected to have greater enthalpy of vaporization has to be given.

Concept Introduction:

Intermolecular forces:  Intermolecular forces are electrostatic in nature and include van der Waals forces and hydrogen bonds.  Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions.  The three major types of intermolecular interactions are,

• Dipole-dipole interactions
• London dispersion forces
• Hydrogen bonds

Answer to Problem 11.55QE

The enthalpy of vaporization is expected to be greater for propane $\left({\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\right).$

Explanation of Solution

Both propane $\left({\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\right)$ and methane $\left({\text{CH}}_{\text{4}}\right)$ have their intermolecular forces as London dispersion force.  The enthalpy of vaporization of liquid depends on its boiling point.  The boiling point of propane is said to be greater because of its increase in molar mass.  Stronger the intermolecular force, higher the boiling point, greater will be the enthalpy of vaporization.  Therefore, the enthalpy of vaporization of propane is expected to be greater than methane.

(b)

Interpretation Introduction

Interpretation:

The liquid that is expected to have greater enthalpy of vaporization has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 11.55QE

The enthalpy of vaporization is expected to be greater for iodine $\left({\text{I}}_{\text{2}}\right)\text{.}$

Explanation of Solution

Iodine has London dispersion force and $\text{ICl}$ has their intermolecular forces as dipole-dipole attractions.  The enthalpy of vaporization of liquid depends on its boiling point.  The boiling point of iodine is said to be greater because of its increase in molar mass.  Stronger the intermolecular force, higher the boiling point, greater will be the enthalpy of vaporization.  Therefore, the enthalpy of vaporization of iodine would be greater than $\text{ICl}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The liquid that is expected to have greater enthalpy of vaporization has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 11.55QE

The enthalpy of vaporization is expected to be greater for disulphide dichloride $\left({\text{S}}_{\text{2}}{\text{Cl}}_{\text{2}}\right)\text{.}$

Explanation of Solution

Both disulphide difluoride $\left({\text{S}}_{\text{2}}{\text{F}}_{\text{2}}\right)$ and disulphide dichloride $\left({\text{S}}_{\text{2}}{\text{Cl}}_{\text{2}}\right)$ have their intermolecular forces as London dispersion force.  The enthalpy of vaporization of liquid depends on its boiling point.  The boiling point of disulphide dichloride $\left({\text{S}}_{\text{2}}{\text{Cl}}_{\text{2}}\right)$ is said to be greater because of its increase in molar mass.  Stronger the intermolecular force, higher the boiling point, greater will be the enthalpy of vaporization.  Therefore, the enthalpy of vaporization of disulphide dichloride $\left({\text{S}}_{\text{2}}{\text{Cl}}_{\text{2}}\right)$ would be greater than disulphide difluoride $\left({\text{S}}_{\text{2}}{\text{F}}_{\text{2}}\right).$

(d)

Interpretation Introduction

Interpretation:

The liquid that is expected to have greater enthalpy of vaporization has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 11.55QE

The enthalpy of vaporization is expected to be greater for water $\left({\text{H}}_{\text{2}}\text{O}\right)\text{.}$

Explanation of Solution

The intermolecular force present in water is hydrogen bonding and the intermolecular force present in ${\text{H}}_{\text{2}}\text{Se}$ is dipole-dipole attraction.  The enthalpy of vaporization of liquid depends on its boiling point.  The boiling point of water is said to be greater.  Stronger the intermolecular force, higher the boiling point, greater will be the enthalpy of vaporization.  Therefore, the enthalpy of vaporization of water would be greater than ${\text{H}}_{\text{2}}\text{Se}\text{.}$

(e)

Interpretation Introduction

Interpretation:

The liquid that is expected to have greater enthalpy of vaporization has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 11.55QE

The enthalpy of vaporization is expected to be greater for dichloromethane $\left({\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}\right)\text{.}$

Explanation of Solution

Both dichloromethane $\left({\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}\right)$ and chloromethane $\left({\text{CH}}_{\text{3}}\text{Cl}\right)$ have dipole-dipole attractions and London dispersion forces as their intermolecular attractions.  The enthalpy of vaporization of liquid depends on its boiling point.  The boiling point of dichloromethane $\left({\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}\right)$ is said to be greater because of its increase in molar mass.  Stronger the intermolecular force, higher the boiling point, greater will be the enthalpy of vaporization.  Therefore, the enthalpy of vaporization of dichloromethane $\left({\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}\right)$ would be greater than chloromethane $\left({\text{CH}}_{\text{3}}\text{Cl}\right).$

(f)

Interpretation Introduction

Interpretation:

The liquid that is expected to have greater enthalpy of vaporization has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 11.55QE

The enthalpy of vaporization is expected to be greater for $\text{NOCl}\text{.}$

Explanation of Solution

Both $\text{NOF}$ and $\text{NOCl}$ have dipole –dipole attractions and London dispersion forces as their intermolecular attractions.  The enthalpy of vaporization of liquid depends on its boiling point.  The boiling point of $\text{NOCl}$ is said to be greater because of its increase in molar mass.  Stronger the intermolecular force, higher the boiling point, greater will be the enthalpy of vaporization.  Therefore, the enthalpy of vaporization of $\text{NOCl}$ would be greater than $\text{NOF}\text{.}$