Chapter 11, Problem 11.56QE

(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 11.56QE

The enthalpy of vaporization is expected to be greater for ethylene $\left({\text{C}}_2{\text{H}}_4\right).$

Explanation of Solution

Both ethylene $\left({\text{C}}_2{\text{H}}_4\right)$ 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 ethylene 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 ethylene would 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 11.56QE

The enthalpy of vaporization is expected to be greater for chlorine $\left({\text{Cl}}_2\right)\text{.}$

Explanation of Solution

Chlorine has London dispersion force and $\text{ClF}$ has their intermolecular forces as dipole-dipole attractions.  The enthalpy of vaporization of liquid depends on its boiling point.  The boiling point of chlorine 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 iodine would be greater than $\text{ClF}\text{.}$

(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 11.56QE

The enthalpy of vaporization is expected to be greater for ${\text{H}}_{\text{2}}\text{Te}\text{.}$

Explanation of Solution

Both ${\text{H}}_{\text{2}}\text{S}$ and ${\text{H}}_{\text{2}}\text{Te}$ have their intermolecular forces as dipole-dipole attractions.  The enthalpy of vaporization of liquid depends on its boiling point.  The boiling point of ${\text{H}}_{\text{2}}\text{Te}$ 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 ${\text{H}}_{\text{2}}\text{Te}$ would be greater than ${\text{H}}_{\text{2}}\text{S}\text{.}$

(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 11.56QE

The enthalpy of vaporization is expected to be greater for ${\text{NH}}_{\text{3}}.$

Explanation of Solution

The intermolecular force present in ammonia is hydrogen bonding and the intermolecular force present in phosphine is dipole-dipole attraction/London dispersion force.  The enthalpy of vaporization of liquid depends on its boiling point.  The boiling point of ammonia 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 ammonia would greater than phosphine.

(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 11.56QE

The enthalpy of vaporization is expected to be greater for ${\text{CHI}}_{\text{3}}.$

Explanation of Solution

Both methyl iodide $\left({\text{CH}}_{\text{3}}\text{I}\right)$ and iodoform $\left({\text{CHI}}_{\text{3}}\right)$ have London dispersion forces as their intermolecular attractions.  The enthalpy of vaporization of liquid depends on its boiling point.  The boiling point of iodoform $\left({\text{CHI}}_{\text{3}}\right)$ 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 iodoform $\left({\text{CHI}}_{\text{3}}\right)$ would be greater than methyl iodide $\left({\text{CH}}_{\text{3}}\text{I}\right).$

(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 11.56QE

The enthalpy of vaporization is expected to be greater for ${\text{BBrI}}_{\text{2}}\text{.}$

Explanation of Solution

Both ${\text{BBrI}}_{\text{2}}$ and ${\text{BBr}}_{\text{2}}\text{I}$ 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 ${\text{BBrI}}_{\text{2}}$ 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 ${\text{BBrI}}_{\text{2}}$ would be greater than ${\text{BBr}}_{\text{2}}\text{I}\text{.}$