1. Consider the following molecules: H2S, PF3, CS2, IBr, and CBr4. Which molecules in this list are not polar? Justify your
reasoning.
We need to determine the polarity of the molecules using the geometries predicted by VSEPR Theory.
H2S Þ 2 bonding groups, 2 lone pairs on sulfur
bent molecular geometry Þ polar molecule
PF3 Þ 3 bonding groups, 1 lone pair on phosphorus
trigonal pyramidal molecular geometry Þ polar molecule
CS2 Þ 2 bonding groups, 0 lone pairs on carbon
linear molecular geometry Þ nonpolar molecule
IBr Þ diatomic molecule with 2 different atoms Þ polar molecule
CBr4 Þ 4 bonding groups, 0 lone pairs on carbon
tetrahedral molecular geometry Þ nonpolar molecule
Therefore, CS2 and CBr4 are the nonpolar molecules on this list.
2. Determine whether each of the following molecules is polar or nonpolar. Recall that the shape of a molecule is not
necessarily given by the Lewis structure. (Notice that lone pairs are omitted in the following structures.)
(a) (b)
polar polar
(c) (d) (e)
polar polar nonpolar
3. List the intermolecular attractive forces which must be overcome in order to convert each of the following substances
from a liquid to a gas.
(a) CO2
nonpolar molecule Þ dispersion forces
(b) CH3Br
polar molecule Þ dispersion forces and dipole-dipole forces
(c) Ne
noble gas atom Þ dispersion forces
(d) CH3NH2
polar molecule with a hydrogen atom covalently bonded to a nitrogen atom Þ dispersion forces
dipole-dipole forces
hydrogen “bonds”
4. Identify the types of intermolecular attractive forces that are present in each of the following substances and select the
substance in each pair that has: (a) the stronger overall intermolecular attractive forces and (b) the higher boiling point.
Justify your reasoning.
Remember that a molecular liquid with stronger intermolecular attractive forces will have a higher boiling point than a molecular liquid with weaker intermolecular attractive forces.
(i) HF or HCl
HF Þ dispersion forces, dipole-dipole forces, and hydrogen bonds
HCl Þ dispersion forces and dipole-dipole forces
Since hydrogen bonding is present between HF molecules and absent between HCl molecules, HF will have the stronger overall intermolecular attractive forces and the higher boiling point.
(ii) CH3CH3 or CH3CH2CH3
Both compounds have only dispersion forces present between their molecules. The molar mass of CH3CH2CH3 is greater than that of CH3CH3 (or CH3CH2CH3 has more electrons than does CH3CH3) so CH3CH2CH3 has stronger dispersion forces. Thus, CH3 CH2CH3 will have the stronger overall intermolecular attractive forces and the higher boiling point.
(iii) CH3CH3 or H2CO
CH3CH3 Þ nonpolar molecule Þ dispersion forces
H2CO Þ polar molecule Þ dispersion forces and dipole-dipole forces
Both compounds have essentially the same molar mass so their dispersion forces should be comparable in strength. H2CO will have the stronger overall intermolecular attractive forces and the higher boiling point due to the additional forces of attraction present between polar molecules.
5. Rationalize, in terms of the intermolecular attractive forces present, the observation that 1-propanol (CH3CH2CH2OH) has
a boiling point of 97.2EC, whereas a compound with the same molecular formula, ethyl methyl ether (CH3CH2OCH3),
boils at 7.4EC.
1-propanol is a polar molecule which can form hydrogen bonds, whereas ethyl methyl ether is a polar molecule which cannot form hydrogen bonds. Hence, 1-propanol has stronger intermolecular attractive forces which results in a higher boiling point.
6. List the substances CH3CH3, CH3OH, CH2Cl2, and CH4 in order of increasing DHvap. Explain your reasoning.
Remember the stronger the intermolecular attractive forces the greater the value of DHvap. We need to determine the intermolecular attractive forces present between molecules in each of these compounds as well as their relative strength.
CH3CH3 Þ dispersion forces
CH3OH Þ dispersion forces, dipole-dipole forces, and hydrogen bonds
CH2Cl2 Þ dispersion forces and dipole-dipole forces
CH4 Þ dispersion forces
The molar mass of CH4 is less than that of CH3CH3 (or CH4 has fewer electrons than CH3CH3) so CH4 has weaker dispersion forces than CH3CH3. Since CH3OH can form hydrogen bonds whereas CH2Cl2 cannot, CH3OH has stronger intermolecular attractive forces. We can list these compounds in order of the increasing strength of intermolecular attractive forces: CH4, CH3CH3, CH2Cl2, and CH3OH. Thus, we have the following list in order of increasing DHvap.
CH4 < CH3CH3 < CH2Cl2 < CH3OH
7. Which will evaporate more quickly: 25mL of propanal in a beaker or 25mL of 1-propanol in an identical beaker under
identical conditions? Is the vapor pressure of the two substances different? Explain your reasoning.
propanal 1-propanol
We need to determine which liquid has the weaker intermolecular attractive forces. A liquid with weaker intermolecular attractive forces will evaporate more quickly and will have a greater vapor pressure at a given temperature than a liquid with greater intermolecular attractive forces.
Propanal: Dispersion forces and dipole-dipole interactions
1-Propanol: Dispersion forces, dipole-dipole interactions, and hydrogen bonds
Since the intermolecular attractive forces are weaker between molecules of propanal than between molecules of 1-propanol, propanal will evaporate more quickly and will have a greater vapor pressure at a given temperature than 1-propanol.
8. How much heat (in kJ) is evolved in converting 1.00 mol of steam at 145.0°C to ice at -50.0°C? The heat capacity of
steam is 2.01 J/g·°C and of ice is 2.09 J/g·°C.
This process can be broken down into five steps.
1. Cool steam from 145.0°C to 100.0°C
2. Condense steam at 100.0°C
3. Cool water from 100.0°C to 0.0°C
4. Freeze water at 0.0°C
5. Cool ice from 0.0°C to – 50.0°C
The additional data required for this problem are given below.
molar mass of water = 18.05 g/mol cwater = 4.184 J/g·°C
DHcondense = – 40.7 kJ/mol DHfreeze = – 6.02 kJ/mol
1. q1 = mcsteamDT = (18.05 g)(2.01 J/g·°C)(100.0°C – 145.0°C) = (18.05 g)(2.01 J/g·°C)(-45.0°C)
q1 = -1.63 kJ
2. q2 = nDHcondense = (1.00mol)(-40.7 kJ/mol) = – 40.7 kJ
3. q3 = mcwaterDT = (18.05 g)(4.184 J/g·°C)( 0.0°C – 100.0°C) = (18.05 g)(4.184 J/g·°C)(-100.0°C)
q3 = -7.55 kJ
4. q2 = nDHfreeze = (1.00mol)(-6.02 kJ/mol) = – 6.02 kJ
5. q1 = mciceDT = (18.05 g)(2.09 J/g·°C)(-50.0°C – 0.0°C) = (18.05 g)(2.09 J/g·°C)(-50.0°C)
q1 = -1.89 kJ
qTotal = q1 + q2 + q3 + q4 + q5 = – 57.8 kJ
Thus 57.8 kJ of heat are evolved.
9. Textbook, problem 11.84.
(a) What is the normal boiling point for iodine? 184.4 °C
(b) What is the melting point for iodine at 1 atm?
This is the definition of the normal melting point which is 113.6 °C.
(c) What phase is present at room temperature and normal atmospheric pressure?
The solid phase is present under these conditions (20 – 25 °C and approximately 1 atm pressure).
(d) What phase is present at 186 °C and 1.0 atm?
The gas phase is present under these conditions.
