13 Properties of Liquids
Liquid water provides the basis for our bodies as well as
recreational sports like windsurfing.
Foundations of College Chemistry, 14th Ed.
Morris Hein and Susan Arena
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Chapter Outline
13.1 States of Matter: A Review
13.2 Properties of Liquids
A. Evaporation
B. Vapor Pressure
C. Surface Tension
13.3 Boiling Point and Melting Point
13.4 Changes of State
13.5 Intermolecular Forces
13.6 Hydrates
13.7 Water, a Unique Liquid
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States of Matter: A Review
Gases: contain particles that are far apart,
in random motion, and independent of one another.
Little contact
No attractive forces
Random motion
Solids: contain particles very close in space and
maintain a rigid shape. Significant attractive forces
exist between particles.
Close contact
Strong attractive forces
Rigid shape
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States of Matter: A Review
Liquids: intermediate between gases and solids.
Contain particles close to one another but have
fluidity (can assume the shape of a container).
Significant attractive forces exist between
particles in a liquid.
Close contact
Some attractive forces
Fluid shape
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Evaporation
Evaporation or Vaporization:
Escape of molecules from the liquid to the gas phase.
Liquid
Vapor
Molecules in the liquid state have
different kinetic energies (KEs).
Those with higher KEs can overcome
attractive forces between particles
and escape to the gas phase.
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Sublimation
Phase change from the solid to gas phase that
bypasses the liquid state.
Solid
Vapor
Examples
CO2 (s)
I2 (s)
CO2 (g)
I2 (g)
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Vapor Pressure
Molecules from the liquid phase can escape to
the vapor phase through evaporation.
Molecules in the gas phase can strike the surface
of the liquid and return to the liquid phase.
This process is called condensation.
In a closed container, an equilibrium develops
between molecules evaporating and condensing.
liquid
evaporation
condensation
vapor
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Vapor Pressure
Vapor pressure: pressure exerted by a vapor
in equilibrium with its liquid phase.
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Vapor Pressure
Vapor pressure: pressure exerted by a vapor
in equilibrium with its liquid phase.
Independent of the quantity of liquid or its surface area.
Increases with increasing temperature.
Depends on the strength of attraction between
molecules in the liquid state.
Volatile liquids: very weak attractive forces
between molecules. Evaporate very rapidly at ambient
temperature. Have high vapor pressures as a result.
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Vapor Pressure
Measuring Vapor Pressure of a Liquid
Measure using a barometer.
Vapor from the liquid exerts a
force on the Hg and pushes the
column downward.
The difference in height relative to
vacuum provides the vapor
pressure for the liquid.
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Surface Tension
Resistance of a liquid to an increase in surface area.
Molecules on a liquid surface are strongly attracted
by molecules within the liquid.
Surface tension increases with increasing attractive
interactions between molecules.
Mercury droplets form spheres
due to surface tension.
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Capillary Action
Spontaneous rise of a liquid in a narrow tube.
Cohesive forces exist between water molecules in a liquid.
Adhesive forces exist between water molecules
and the walls of the container.
When the cohesive forces between molecules are less
than the adhesive forces between liquid and container,
the liquid will move up the walls of the container.
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Capillary Action
Capillary Action in Action
Shape of the meniscus reflects the relative strength
of cohesive forces within the liquid and adhesive
forces between the liquid and the tube.
If convex:
adhesive forces < cohesive forces
If concave:
adhesive forces > cohesive forces
Hg
H2 O
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Boiling Point
Temperature at which the vapor pressure of a liquid
is equal to the external pressure above the liquid.
Where is the boiling point of a liquid higher,
at or above sea level?
At sea level. The atmospheric pressure is higher.
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Boiling Point
Normal boiling point:
temperature when the vapor pressure is 1 atm
Vapor Pressure Curve
Normal boiling points:
Water: 100 ºC
Ether: 35 ºC
Ethyl Alcohol: 78 ºC
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Boiling Point Practice
The vapor pressure curve for water is given below.
What is the boiling point of water at 300 mmHg?
a. 100 ºC
b. 86 ºC
c. 76 ºC
d. 30 ºC
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Freezing Point or Melting Point
Freezing/melting point: the temperature at which the
solid phase of a substance is in equilibrium
with its liquid phase.
solid
melting
freezing
liquid
While both phases are present,
the temperature remains constant.
The energy is used to change the solid to the liquid phase.
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Changes of State
Heat of fusion: energy required to change 1 g
of a solid at its melting point to a liquid.
Heat of
vaporization
Boiling
Melting
Heat of fusion
Heat of vaporization: energy required to change 1 g
of a liquid to vapor at its normal boiling point.
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Energy and Phase Changes
Heat of fusion: energy required to change 1 g
of a solid at its melting point to a liquid.
The heat of fusion for water is 335 J/g. Calculate
the amount of heat needed to melt 25.0 g of water.
Use the heat of fusion as a conversion factor!
25.0 g
×
335 J
1g
= 8380 J
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Energy and Phase Changes
Heat of vaporization: energy required to change 1 g
of a liquid to vapor at its normal boiling point.
The heat of vaporization for water is 2259 J/g.
Calculate the amount of heat needed to vaporize
25.0 g of water at 100 ºC.
Use the heat of vaporization as a conversion factor!
25.0 g
×
2259 J
1g
= 56,500 J
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Phase Change Practice
Calculate the energy needed to convert 25.0 g
of ice at 0 ºC to steam at 100 ºC?
Given: heat of fusion = 335 J/g
heat of vaporization = 2259 J/g
specific heat of liquid water = 4.184 J/gºC
Plan
The conversion of ice to steam is a three step process:
1. ice melts (total energy = mass x heat of fusion)
2. liquid water is warmed from 0° to 100°C (energy = mass x specific heat x ΔT)
3. water evaporates (energy = mass x heat of vaporization)
The overall energy required for the process
is the sum of the 3 steps.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Phase Change Practice
Calculate the energy needed to convert 25.0 g
of ice at 0 ºC to steam at 100 ºC?
Calculate
1. ice melts
2. warm water
25.0 g ×
335 J
1g
= 8375 J
25.0 g ×
4.184 J
1 g ºC
× 100 ºC
3. evaporate water
25.0 g ×
Total energy required:
2259 J
1g
= 10,460 J
= 56,475 J
E = Step 1 + Step 2 + Step 3 = 8375 J + 10,460 J + 56475 J = 75,300 J
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Phase Change Practice
How many joules of energy are needed to change
10.0 g of ice at 0.00 ºC to water at 20.0 ºC?
Given: heat of fusion = 335 J/g
specific heat of liquid water = 4.184 J/gºC
a. 4.19 x 103 J
b. 478 J
c. 3.35 x 103 J
d. 2.51 x 103 J
Plan
The conversion of ice to liquid water is a two step process.
1. ice melts (total energy = mass x heat of fusion)
2. liquid water is warmed from 0° to 20 °C (energy = mass x specific heat x ΔT)
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Phase Change Practice
How many joules of energy are needed to change
10.0 g of ice at 0.00 ºC to water at 20.0 ºC?
Calculate
1. ice melts
2. warm water
10.0 g
335 J
×
10.0 g
1g
×
4.184 J
1 g ºC
= 3350 J
× 20 ºC
= 837 J
Total energy required:
Energy = Step 1 + Step 2 = 3350 J + 837 J = 4.19 x 103 J
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Intermolecular Forces
Attractive forces between molecules.
These forces allow for formation of liquids and solids.
The degree of intermolecular forces correlates
with a compound’s physical properties.
Example:
The stronger the interaction between molecules in a liquid,
the higher the boiling point and the lower
the vapor pressure.
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Types of Intermolecular Forces
Dipole-Dipole Attractions
In covalent molecules, due to different atoms having
different electronegativities, molecules are polar.
When polar molecules are put together,
they will align to permit interaction between
oppositely polarized portions of the molecules.
The interaction of two polar
H2O molecules.
These interactions between dipoles in different molecules
are called dipole-dipole forces.
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The Hydrogen Bond
Water has very high melting and boiling points,
and heats of fusion and vaporization.
These anomalous properties are due to strong attraction
between water molecules due to hydrogen bonding,
a special type of dipole-dipole attraction.
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The Hydrogen Bond
Hydrogen bonds: one type of strong intermolecular
force/attraction between molecules.
Hydrogen bonds are much weaker than ionic or
covalent bonds which are intramolecular forces.
Hydrogen Bonding between H2O molecules.
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The Hydrogen Bond
To form hydrogen bonds, a compound must have
covalent bonds between hydrogen and F, O or N
(a very electronegative element).
Can hydrogen bond.
Cannot hydrogen bond.
(No H attached to oxygen).
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The Hydrogen Bond Practice
Which of the following molecules would be expected
to participate in hydrogen bonding?
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London Dispersion Forces
Molecules without dipoles can also interact
with one another.
These interactions between nonpolar molecules and
noble gases are called London dispersion forces.
London forces arise from uneven, instantaneous
charge distributions due to electron movement
in nonpolar molecules.
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London Dispersion Forces
This instantaneous dipole can then induce a dipole in a
neighboring nonpolar molecule, resulting in a small
attraction between particles.
London forces are very weak forces.
Generally become more important as the size of the
molecule increases. Larger sizes provide more
possible electrons to provide dipoles.
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London Dispersion Forces
Which of the following molecules would have the
largest London dispersion forces?
a. CH4
b. C4H10
c. C15H32
d. C8H18
The largest hydrocarbon
(i.e., having the largest molar mass)
will have the strongest London forces.
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Hydrates
Hydrates are solids that contain water molecules as
part of their crystalline structure.
The formula lists the anhydrous (without water)
formula of the compound. The number of waters present
per structural unit (water of hydration) are then given.
CaCl2∙2H2O
Hydrates are named by placing a prefix corresponding
to the number of water molecules, followed by hydrate
CaCl2∙2H2O
FeCl3∙6H2O
calcium chloride dihydrate
iron(III) chloride hexahydrate
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Hydrates
Hydrates will often decompose by
losing water upon heating.
CuSO4∙5 H2O (s)
250°C
CuSO4 (s) + 5 H2O (g)
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Hydrates Practice
Calculate the percent water in Epsom salts
MgSO4∙7H2O (s).
1. Calculate the molar mass of the compound.
Molar Mass MgSO4∙7H2O (s) = 120.4 + 7(18.02) = 246.5 g
2. Calculate the % water of the compound.
Mass water
x 100 =
% water =
Molar mass
126.1 g
246.5 g
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x 100 = 51.16 %
Hydrates Practice
What is the percent water in CuSO4∙5H2O (s)?
a. 56.46%
1. Calculate the molar mass of the compound.
b. 36.08%
c. 63.92%
Molar Mass CuSO4∙5H2O (s)
= 159.6 + 5(18.02) = 249.2 g
d. 61.57%
2. Calculate the % water of the compound.
Mass water
x 100 =
% water =
Molar mass
90.10 g
259.2 g
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x 100 = 36.08 %
Water: A Unique Liquid
Water covers 75% of the Earth’s surface;
97% of all water resides in the oceans.
Water constitutes 70% of a human body by mass.
Physical Properties of Water
Colorless, odorless, tasteless liquid.
More dense in liquid than solid phase (why ice floats).
High boiling point, high heat of fusion/vaporization
due to hydrogen bonding.
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Water: A Unique Liquid
Structure of Water Molecules
Two OH bonds are formed by the overlap of 1s orbitals
on H with orbitals on the O.
The molecular geometry of water is
bent, due to the two lone pairs on
oxygen.
Water has a permanent dipole due to
the molecules’ shape and the polar
O-H bonds.
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Water: A Unique Liquid
Sources of Water for Human Consumption
Climate change and increased demand for fresh water
make finding and sustaining sources of potable
water critical for future generations.
Strategies to Sustain Water Supplies
1. Reclamation of wastewater
Currently used in agriculture and industry
2. Desalination of seawater
Expensive, but useful for countries near the ocean.
3. Low temperature distillation
At low pressure, water’s boiling point is reduced.
Less energy is required to separate the salts by boiling.
4. Combustion of H2
H2 and O2 react very exothermically to produce water.
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Chemistry in Action: Osmosis
Osmosis: process by which water flows through a
membrane from a region of more pure water
to a region of less pure water.
Water flows into the raisin to dilute the sugar.
The size of the raisin expands.
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Chemistry in Action: Reverse Osmosis
Reverse Osmosis: process by which water flows through
a membrane from a region of less pure water to a region
of more pure water, due to the presence of an external
stimulus (typically pressure)
Process often used in water purification.
A semipermeable membrane is used and
only water can pass through.
By applying pressure, only water passes through
the membrane. The water is now pure!
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Learning Objectives
13.1 States of Matter: A Review
Review the states of matter and their properties.
13.2 Properties of Liquids
Explain why liquids tend to form drops and the process
of evaporation and its relationship to vapor pressure.
13.3 Boiling Point and Melting Point
Define boiling and melting points and determine
the boiling point of a liquid from a graph of
temperature and vapor pressure.
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Learning Objectives
13.4 Changes of State
Calculate the amount of energy involved
in a change of state.
13.5 Intermolecular Forces
Describe the three types of intermolecular forces
and explain their significance in liquids.
13.6 Hydrates
Explain what hydrates are, write formulas for hydrates
and calculate the percent water in a hydrate.
13.7 Water, A Unique Liquid
Describe the characteristics of water in terms of its
structure and list the sources of drinking water.
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