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(a)
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
(a)
Answer to Problem 11.56QE
The enthalpy of vaporization is expected to be greater for ethylene (C2H4).
Explanation of Solution
Both ethylene (C2H4) and methane (CH4) 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:
The liquid that is expected to have greater enthalpy of vaporization has to be given.
Concept Introduction:
Refer to part (a).
(b)
Answer to Problem 11.56QE
The enthalpy of vaporization is expected to be greater for chlorine (Cl2).
Explanation of Solution
Chlorine has London dispersion force and 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 ClF.
(c)
Interpretation:
The liquid that is expected to have greater enthalpy of vaporization has to be given.
Concept Introduction:
Refer to part (a).
(c)
Answer to Problem 11.56QE
The enthalpy of vaporization is expected to be greater for H2Te.
Explanation of Solution
Both H2S and H2Te have their intermolecular forces as dipole-dipole attractions. The enthalpy of vaporization of liquid depends on its boiling point. The boiling point of H2Te 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 H2Te would be greater than H2S.
(d)
Interpretation:
The liquid that is expected to have greater enthalpy of vaporization has to be given.
Concept Introduction:
Refer to part (a).
(d)
Answer to Problem 11.56QE
The enthalpy of vaporization is expected to be greater for NH3.
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:
The liquid that is expected to have greater enthalpy of vaporization has to be given.
Concept Introduction:
Refer to part (a).
(e)
Answer to Problem 11.56QE
The enthalpy of vaporization is expected to be greater for CHI3.
Explanation of Solution
Both methyl iodide (CH3I) and iodoform (CHI3) have London dispersion forces as their intermolecular attractions. The enthalpy of vaporization of liquid depends on its boiling point. The boiling point of iodoform (CHI3) 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 (CHI3) would be greater than methyl iodide (CH3I).
(f)
Interpretation:
The liquid that is expected to have greater enthalpy of vaporization has to be given.
Concept Introduction:
Refer to part (a).
(f)
Answer to Problem 11.56QE
The enthalpy of vaporization is expected to be greater for BBrI2.
Explanation of Solution
Both BBrI2 and BBr2I 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 BBrI2 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 BBrI2 would be greater than BBr2I.
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Chapter 11 Solutions
Chemistry: Principles and Practice
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