CHE 0315
SEM 3, 2013/14
TOPIC 6: INTERMOLECULAR FORCES AND LIQUIDS
1.
List all the intermolecular forces that is present in the following substances?
[9]/7(b)
a)
NH3 – dipole-dipole forces and hydrogen bond
b)
Br2 – London dispersion forces
c)
CH3Cl – dipole-dipole forces
d)
HCl – dipole-dipole forces
e)
CH3CH2CH2OH- London dispersion and hydrogen bond
f)
CH3CH2CH2CH2CH3 – London dispersion
g)
CHCl3 – dipole-dipole forces and London dispersion
2
(a)
Determine which of the following molecules has hydrogen bonding and use
one of the selected molecule to show how the hydrogen bonds are formed.
[4]/7(b)
HCl ,HF, HNO3, NH3, CH3NH2 CH3F
HF, NH3 and CH3NH2 have hydrogen bonds
H
F ------ H
F
Hydrogen bond
Note# (Hydrogen bonding is the attraction between hydrogen atom,
which is covalently bonded to a small and highly electronegative
atom (like F, O and N), and the lone pair of electrons of another very
electronegative atom)
(b)
Explain how hydrogen bonding affects the boiling point of water
.[2]/ 7(e)
More energy is required to break the strong hydrogen bonding
between water molecules. This makes the boiling point of water
higher than expected from its small molecular size.
3.
Ethanol, CH3CH2OH is completely miscible with water while phenol, C 6H5OH is
partially miscible with water. Explain.
[3]/7(b)
Both ethanol and phenol can form hydrogen bonding with water molecules.
But the presence of a large non-polar hydrocarbon group (C6H5-) makes
phenol less soluble compared to the small C2H5- group in ethanol.
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4.
SEM 3, 2013/14
The boiling points of propanone ,water and mercury at room conditions are 56.5 0C ,
100 0C and 357 0C respectively.
(a)
Use the same diagram to sketch the vapor pressure of the three liquids as a
function of temperature.
[3]/7(c)
Propanone
water
mercury
Vapor
Pressure
1atm
56.50C
(b)
Temp 0C
Explain why mercury has the highest boiling point.
[2]/ 7(f)
Mercury is a metal and it has metallic bonding while the other
molecules ; water has hydrogen bonding and propanone has dipoledipole attractions.
5.
The molecular mass of butane and propanone are both 58. However, the boiling
point of butane is 273 K while that of propanone is 330 K.
[6]/7(e)
Butane :Non-polar molecule. Intermolecular force is the weak
London Dispersion forces
Propanone :polar molecule. Intermolecular force is Dipole-dipole
forces.
As dipole-dipole forces is stronger than London dispersion forces,
higher boiling point.
6.
Choose a liquid that has greater surface tension Explain your answer.
[3]/7(e)
CH3CH2OH or CH3OCH3
Ethanol (CH3CH2OH) has greater surface tension than dimethyl ether
(CH3OCH3). Ethanol has stronger intermolecular forces (Hydrogen bond).
Surface tensions tend to increase as the strength of intermolecular forces
CHE 0315
SEM 3, 2013/14
increase. Both have identical molar mass so attractions resulting from
London dispersion forces will be equal.
7.
Choose a liquid that has greater viscosity. Explain your answer.
[2]/7(e)
CH3CH2CH2CH2OH or CH3CH2CH2CH2CH2CH2CH2CH3
Octane has London dispersion forces only and butanol has hydrogen bond
but octane has higher viscosity than butanol because octane has a higher
molecular mass.
8.
Arrange the following substances in order of increasing boiling point. State your
reasons.
(a)
CH3OCH3, CH3CH2OH, and CH3CH2CH3
[2]/7(e)
CH3CH2CH3 < CH3OCH3< CH3CH2OH
All have similar molar masses: 46.07g/mol, 46.07g/mol and
44.09g/mol respectively. Dimethyl ether cannot form hydrogen bonds
(no O-H bond), but is polar and has dipole-dipole forces. Ethanol can
form hydrogen bonds. Propane is nonpolar, so it has only London
dispersion forces. The boiling point increases as the strength of the
intermolecular forces increase:
London dispersion < dipole-dipole forces < hydrogen bond
(b)
Br2, Cl2 and I2
[2]/7(e)
Cl2 < Br2 < I2
All are non polar molecules so only London dispersion forces are
present. London dispersion forces get stronger as molar mass
increases
(c)
CaCO3, CH4, CH3OH and CH3OCH3
[2]/7(e)
By using intermolecular forces, we can tell that these compounds will
rank:
methane (Van der Waals forces), dimethyl ether (dipole-dipole
forces), methanol (hydrogen bonding), calcium carbonate (ionic
electrostatic forces that are much stronger than intermolecular
forces).
CHE 0315
9.
SEM 3, 2013/14
For each pair of substance, identify the substance that is likely to have a higher
vapor pressure. Explain your answers.
[4]/7(e)
(a)
CO2 or SO2
CO2 will have the higher vapor pressure. Vapor pressure tends to
decrease as the strength of the intermolecular forces increase.
Carbon dioxide is non-polar (dispersion forces only). Sulfur dioxide is
polar (dipole-dipole forces are present).
(b)
CH3CH2OH or CH3-O-CH2CH3
CH3OCH2CH3 will have the higher vapor pressure. Vapor pressure
tends to decrease as the strength of the intermolecular forces
increase. CH3CH2OH can form hydrogen bond. CH3OCH2CH3 is polar
(bent shape around the oxygen), so dipole-dipole forces are the
strongest forces in this compound.
10.
Define “boiling point” and "normal boiling point".
[2]/7(d)
Boiling point: The temperature at which the liquid vapor pressure equals to
the external pressure.
Normal boiling point: The temperature at which the liquid vapor pressure
equals to 1 atm
11.
Carbon and Sodium both react with oxygen to form carbon dioxide and Sodium
oxide. Carbon dioxide is a gas at room conditions, but Sodium oxide is a solid.
Explain the difference between CO2 and Na2O in terms of bonding and structure.
[2]/7(f)
CO2: non polar covalent molecule with weak dispersion forces between
molecules.
Na2O: ionic molecule with strong ionic bonding holding the atoms together.
12.
Explain how intermolecular forces affect surface tension?
[2]/7(e)
Surface tension is the energy required to increase the surface area.
the strength of IMF’s increase, surface tension will also increase. Molecules
are more strongly attracted to each other and will less likely to spread
apart.
13.
(a)
Explain how boiling points are affected by intermolecular forces? [2]/7(e)
CHE 0315
SEM 3, 2013/14
Stronger intermolecular forces between molecules make it more
difficult for those molecules to be pulled apart. Therefore, stronger
intermolecular forces result in higher boiling points.
(b)
Give an explanation in terms of IMF for the following differences in boiling
point
[6]/7(e)
(i)
HF (20 °C) and HCl (-85 °C)
Both are polar. HF has hydrogen bonding, which is a stronger
intermolecular forces than dipole-dipole forces of HCl, so it will have
the higher boiling point.
(ii)
CHCl3 (61 C) and CHBr3 (150 C)
0
o
Both are polar molecules and have dipole-dipole interactions.
Although CHCl3 is more polar but CHBr3 is more massive. CHBr3 has
more electrons/higher molar mass, so it has stronger LDF than CHCl3
and higher boiling point.
(iii)
Br2 (59 oC) and ICl (97 oC)
Br2 is nonpolar and has only LDF. ICl is polar and has dipole-dipole
interactions which is stronger than LDF, so it will have the higher
boiling point.
14.
State which of the following properties will increase, decrease or remain unaffected
by an increase in the strength of the intermolecular forces. [4]/7(e)
(a)
vapor pressure
decreases
(b)
normal boiling point (boiling point at 1 atmosphere pressure)
increases
(c)
surface tension
increases
(d)
viscosity
increases
CHE 0315
SEM 3, 2013/14