Liquids and Solids
Chapter 10
Solids and Liquids
• the “condensed” states of matter. Forces
of attraction (“cohesive” forces) are
stronger than the repulsive forces that
make the Ideal Gas Law.
Cohesive Forces
• def forces between particles in a lattice or
a liquid. There are five types of cohesive
forces.
1. Ionic – two ions connected together,
one atom gives up electrons while the
other atom gains the electrons
2. Covalent – two atoms sharing electons
Covalent Forces
3. hydrogen bonding
4. dipole-dipole
5. Van der Waals (London dispersion forces)
• Ionic and covalent forces are intramolecular
forces (inside the molecules)
• Hydrogen, dipole-dipole and van der Waals
forces are intermolecular forces (between
molecules)
Relative Magnitudes of Forces
• The types of bonding forces vary in their
strength as measured by average bond
energy.
Strongest
Ionic bonds
Covalent bonds (400 kcal/mol)
Hydrogen bonding (12-16
kcal/mol )
Dipole-dipole interactions (20.5 kcal/mol)
Weakest
London forces (less than 1
kcal/mol)
Hydrogen bonding
• Two consequences of hydrogen bonding
– Gives substances high boiling points
– Substances with hydrogen bonding tend to be
more viscous
• Both are explained by increased
attractions making them more difficult to
separate
Dipole-Dipole
• result when the partial positive (d+) and partial
negative (d-) charges of neighboring POLAR
covalent molecules attract (roughly 1% as
strong as a “typical” covalent bond)
• Arrange themselves so the negative and positive
ends attract one another
• Eventually arrange in order to find the best
compromise between attraction and repulsion
London Dispersion (van der Waals)
• Small electrostatic forces are caused
by the movement of electrons in the
covalent bonds of molecules
• London forces increase with the size
of the molecules and with surface
area.
London Dispersion (van der Waals)
• As one approaches another, the electrons
of one or both are temporarily displaced
• This movement causes small, temporary dipoles
to be set up which attract one another
– Large molecules with big surface area have more
electrons and more dispersion forces and greater
attraction and, therefore, higher boiling points
The stronger the cohesive forces
• THE HIGHER THE MELTING POINT
• THE HIGHER THE BOILING POINT
• This is because you gotta add more
energy to overcome the stronger cohesive
forces
Solids
• Ionic structures
– Lattice structures – a three dimensional
system of points; held together by strong
electrostatic interactions
– Strong bonds give them high melting points
– Can only conduct electricity when molten or in
solution
Solids
• Molecular structures
– Held in lattice structures held together by
weak dispersion forces
– Fairly low melting points
– A nonconductor
– Example - iodine
Liquids
• Substances that have insufficient energy
to overcome intermolecular forces but
enough energy to break away from the
ordered state of solids
Liquids
• Capillary action – spontaneous rising of a
liquid in a narrow tube
– Cohesive forces – between the molecules
– Adhesive forces – between molecules and
container
VAPOR PRESSURE AND PHASE
DIAGRAMS
• 2 important vapor pressure factors
1. the lower the cohesive forces in the
condensed state, the higher the vapor
pressure
2. the higher the system
temperature, the higher the vapor
pressure
Phase Diagrams
• A way of combining most of this chapter
(info on condensed states) into one
diagram: a graph of Pressure versus
Temperature under all conditions, that
then shows physical state at that point.
Phase Diagrams
• Represents phases as a function of
temperature and pressure.
• Critical temperature: temperature above
which the vapor can not be liquefied.
Phase Diagrams
Colligative Properties: General
Solutions Properties
Chapter 11
Colligative Properties
• Those properties that depend only on the
number of solute particles present, not
their nature
Colligative Properties
Non-electrolyte solutions
Electrolyte solutions
Covalent solute
Ionic solute
No dissociation
Dissocation
Freezing
point
depression
Van’t Hoff factor
Osmotic
Pressure
Boiling
point
elevation
Vapor
pressure
Raoult’s law
Nonvolatile
solute
Freezing
point
depression
2 nonvolatile
components
Deviations
Osmotic
pressure
Boiling
point
elevation
More Solution Concentration
Units
• Since volumes of solutions are temperature
dependent, molarity of solutions changes
slightly with temp. Chemists have three other
solutions concentration units that are NOT temp
dependent, since they are related only to MASS
and NOT VOLUME. They are mass percent,
mole fraction, and molality.
Mass Percent and Mole
Fraction
• Mass Percent Equation
– Mass percent = (mass of solute/mass of
solution) x 100
• Mole Fraction Equation
– Mole fraction of A = cA = nA/(nA+nB)
Molality
• Molality = m = moles solute/kilograms
solvent
• Ex You add 5.84 g formaldehyde,
H2CO, to 100.0 g H2O. The final
volume is 104.0 ml. Find molarity,
molality, mass %,and c H2CO?
Solubility Trends
• The solubility of MOST solids increases with
temperature.
• The rate at which solids dissolve increases with
increasing surface area of the solid.
• The solubility of gases decreases with increases
in temperature.
• The solubility of gases increases with the
pressure above the solution.
Therefore…
• Solids tend to dissolve best when:
– Heated
– Stirred
– Ground into small particles
• Gases tend to dissolve best when:
– The solution is cold
– Pressure is high
Pressure/Temperature
Effects on Solubility of
Solutes
• Pressure has does not effect how well a solid or liquid
dissolve into a solution.
• Pressure INCREASES gas solubility. There are more
collision on a given surface of the solvent with increased
gas pressure. The pressure –concentration relationship
is linear (at LOW gas concentrations) and only applies to
gases that do not chemically react with the solvent to
make new products.
Pressure/Temperature
Effects on Solubility of
Solutes
• This is related mathematically by HENRY’S
LAW (this relates gas pressure and solubility).
• c = KP
– c = molar concentration (mol/L)
– K = a constant that depends on temperature (mol/L
atm)
– P = pressure in atm (atmospheres)
• Ex The solubility of O2 is 2.2 X 10-4 M
at 0 oC and 0.10 atm. What is the
solubility if you increase pressure to
0.35 atm?
General Temperature Effects
• As a GENERAL RULE, increased
temperature…
• INCREASES solid/liquid solubility into
solvents
• DECREASES gas solubility because this
increases vapor pressure
Pressure
• Is caused by the collisions of molecules
with the walls of a container
• is equal to force/unit area
• SI units = Newton/meter2 = 1 Pascal (Pa)
• 1 atmosphere = 101,325 Pa
• 1 atmosphere = 1 atm = 760 mm Hg = 760
torr
Pressure - examples
• Convert 5.6 atm to torr.
• Convert 920mmHg to atm.
• Convert 99.8kPa to torr.
Vapor Pressure Lowering
•
Two ways to explain why the vapor
pressure is lowered when a solute is added
1. The process of solvent molecules escaping is
partially blocked by the interactions with solute
particles at the surface, causing vapor pressure
to fall
2. When a solute is added to a pure solvent, the
entropy (disorder) dramatically increases. This
is the preferred situation. Prior to the solute
being added, the solvent could only increase the
disorder by evaporating. Adding a solute
created disorder and there is less incentive for
the solvent to evaporate.
Raoult’s Law (Non-volatile
Solute)
• explains nonvolatile solute effects on
vapor pressure.
• Non-volatile solutes do not have their own
vapor pressure
Raoult’s Law (Non-volatile
Solute)
• Vapor pressure will be lower than that of
the pure solvent
• Vp = cPo
– Vp = vapor pressure
– c = mole fraction of the solvent
– Po = vapor pressure of the pure solvent
Raoult’s Law
• Ex What is the vapor pressure of a
glycerin (nonvolatile), C3H8O3, solution
made by adding 164 grams of it to 338 ml
of H2O at 39.8 °C? The PH2O at that temp
is 54.74 torr and the density is 0.992 g/mL
Raoult’s Law
• Raoult’s Law also holds for electrolytes
(soluble salts), but you actually have more
solute present because of ionization.
Raoult’s Law
• Ex What would be the vapor pressure
effect of adding 105.3 grams of NaCl to
the water in the previous example?
Raoult’s Law
• ex At 29.6 oC, PoH2O = 31.1 torr. You make
a solution using 86.7 grams of unknown
“M” in 350.0 grams of H2O, and determine
the resulting vapor pressure to be 28.6
torr. What is the molar mass of M?
Raoult’s Law for volatile
components
• If a solute that has its OWN vapor pressure (is volatile),
you must do a separate Raoult’s calculation on each
volatile component.
Ptotal = cAPAo + cBPBo
• Ex The Po of hexane is 573 torr at 60.0°C (C6H14). The
Po of benzene (C6H6) is 391 torr. What would be the
vapor pressure if you mixed 58.9 grams hexane with
44.0 grams benzene?
Boiling Point Elevation
• Boiling point of a solution is reached when the
external atmospheric pressure is equal to its
vapor pressure allowing a change from liquid to
gas
• Since the addition of a solute lowers the vapor
pressure of a solution it must raise the boiling
point
Boiling Point Elevation
DTb = Kbm
DTb = change in boiling point of the solvent
Kb = molal boiling point constant for the
solvent
m = molality (mol/kg) of the solute
Examples
• Ex What is the boiling point of a solution
made by adding 31.65 g of NaCl to 218.7
grams of H2O at 34.0°C? Kb H2O 0.51 oC
kg/mol.
Freezing Point Depression
• Freezing points of a solvent are lowered by the addition
of solutes
• Can be explained in two ways
– Solute particles get in the way of the solid structure being
formed. If the solidification process is hindered, more heat must
be removed from the solution and the freezing point is lowered.
– The solution has more disorder than the pure solvent so it needs
more energy removed from it in order to create the more ordered
solid
Freezing Point Depression
• Change in freezing point for non-electrolyte
solutions can be summarized as
m
DTf = Kf
DTf = change in freezing point of the solvent
Kf = molal freezing point constant for the solvent
m = molality of the solute
Examples
• Ex How many grams of glycerin, C3H8O3,
must be added to 350.0 grams of water in
order to lower the freezing point to –
3.84oC? Kf H2O = 1.86 oC kg/mol.
Osmotic Pressure
• Osmotic pressure () is the pressure needed to
stop osmosis of a solvent across a
semipermeable membrane. It can also be used
to determine molar mass.
•  = MRT
– M = molarity
– R = 0.08206 L atm/K mol (constant)
– T = temperature in Kelvin
Osmotic Pressure
• ex The osmotic pressure of a 26.5 mg/L
solution of aspartame is 1.70 torr at 30 oC.
What is the molar mass (formula weight)
of aspartame?
Electrolyte Solutions and the
van’t Hoff Factor
• Van’t Hoff factor (i) corrects for the
dissociation of ions in solution
– Added to the formulas previously used
m
DT = iK m
DTb = iKb
f
f
 = iMRT
– Given as
• i = actual number of particles (ions or molecules) in
solution after dissociation
Gases
Chapter Five
General Properties and Kinetic Theory
• Gases are made up of particles that have
(relatively) large amounts of energy
• A gas has no definite shape or volume and
will expand to fill as much space as
possible
• As a result of the large amount of empty
space in a volume of gas, gases are easily
compressed
The Nature of Gases
• Gases expand to fill their containers
• Gases are fluid – they flow
• Gases have low density
– 1/1000 the density of the equivalent liquid
or solid
• Gases are compressible
• Gases effuse and diffuse
Standard Temperature and Pressure
• P = 1 atmosphere, 760 torr
• T = 0°C, 273 Kelvins
• The molar volume of an ideal gas is 22.42
liters at STP
Ideal Gases
• Ideal gases are imaginary gases that
perfectly fit all of the assumptions of the
kinetic molecular theory.
• Gases consist of tiny particles that are far
apart relative to their size.
• Collisions between gas particles and
between particles and the walls of the
container are elastic collisions
• No kinetic energy is lost in elastic
collisions
Ideal Gases
• Gas particles are in constant, rapid
motion. They therefore possess kinetic
energy, the energy of motion.
• There are no forces of attraction
between gas particles
• The average kinetic energy of gas
particles depends on temperature, not
on the identity of the particle.
Kinetic Molecular Theory
• Particles of matter are ALWAYS in motion
• Volume of individual particles is  zero.
• Collisions of particles with container walls
cause pressure exerted by gas.
• Particles exert no forces on each other.
• Average kinetic energy  Kelvin
temperature of a gas.
Boyle’s Law
• Pressure is inversely proportional to
volume when temperature is held
constant.
• As pressure increases, volume
decreases
P1V1 = P2V2
Boyle’s Law - Example
• In an automobile engine the gaseous fuel-air
mixture enters the cylinder and is compressed
by a moving piston before it is ignited. In a
certain engine the initial cylinder volume is
0.725L. After the piston moves up, the volume is
0.075L. The fuel-air mixture initially has a
pressure of 1.00atm. Calculate the pressure of
the compressed fuel-air mixture, assuming that
both the temperature and the amount of gas
remain constant.
Charles’s Law
• The volume of a gas is directly proportional
to temperature, and extrapolates to zero at
zero Kelvin.
(P = constant)
• As temperature increases, volume increases
V1 = V2
T1 T2
Charles’s Law - example
• In former times, gas volume was used as
a way to measure temperature using
devices called gas thermometers.
Consider a gas that has a volume of
0.675L at 35oC and 1atm pressure. What
is the temperature (in units of oC) of a
room where this gas has a volume of
0.535L at 1atm of pressure.
Avogadro’s Law
• For a gas at constant temperature and pressure,
the volume is directly proportional to the number
of moles of gas (at low pressures).
• As the number of moles increases, volume
increases
V1 = V2
n1 n2
n = number of moles of gas
Avogadro’s Law - example
• Suppose we have a 12.2L sample
containing 0.50mol of oxygen gas at a
pressure of 1atm and a temperature of
25oC. If all of this oxygen gas is converted
to ozone, O3, at the same temperature and
pressure, what will be the volume of the
ozone formed?
Gay Lussac’s Law
• The pressure and temperature of a gas
are directly related, provided that the
volume remains constant.
• As temperature increases, pressure
increases
P1 = P2
T1 T2
Gay Lussac’s Law Example
• A gas at 25oC in a closed container has its
pressure raised from 150. atm to 160. atm.
What is the final temperature of the gas?
Combined Gas Law
• The combined gas law expresses the
relationship between pressure, volume
and temperature of a fixed amount of
gas.
P1V1 = P2V2
T1
T2
• Boyle’s law, Gay-Lussac’s law, and
Charles’ law are all derived from this by
holding a variable constant.
Combined Gas Law
• A sample of diborane gas, B2H6, a
substance that bursts into flames when
exposed to air, has a pressure of 0.454atm
at a temperature of -15oC and a volume of
3.48L. If conditions are changed so that
the temperature is 36oC and the pressure
is 0.616atm, what will the new volume be?
Ideal Gas Law
PV = nRT
•
•
•
•
P= pressure in atm
V = volume in liters
n = moles
R = proportionality constant
– = 0.08206 L atm/ mol·K
• T = temperature in Kelvins
• Holds closely at P < 1 atm and T > 273 K
Gas Laws Example
• Anesthetic gas is normally given to a
patient when the room temperature is
20oC and the patient’s body temperature
is 37oC. What would this temperature
change do to 1.60L of gas if the pressure
remains constant?
Gas Laws Example 2
• What volume is occupied by 0.250mol of
carbon dioxide gas at 25oC and 371torr?
Gas Density
• Density = mass
= molar mass
volume
molar volume
… so at STP …
Density = molar mass
22.4 L
Gas Stoichiometry #1
• If reactants and products are at the same
conditions of temperature and pressure, then
mole ratios of gases are also volume ratios.
3 H2(g)
+ N2(g)

2NH3(g)
3 moles H2 + 1 mole N2 
2 moles NH3
3 liters H2 + 1 liter N2

2 liters NH3
Gas Stoichiometry #2
• How many liters of ammonia can be
produced when 12 liters of hydrogen react
with an excess of nitrogen?
3 H2(g)
+ N2(g)

2NH3(g)
Gas Stoichiometry #3
• How many liters of oxygen gas, at STP, can be
collected from the complete decomposition of
50.0 grams of potassium chlorate?
2 KClO3(s)  2 KCl(s) + 3 O2(g)
Gas Stoichiometry #4
• How many liters of oxygen gas, at 37.0C and
0.930 atmospheres, can be collected from the
complete decomposition of 50.0 grams of
potassium chlorate?
2 KClO3(s)  2 KCl(s) + 3 O2(g)
Mole Fractions
• Partial Pressures are determined by the
mole of each gas present.
• Find the number of moles of each gas
present – total moles creates total
pressure
• Pgas = Xgas*Ptotal
• Xgas = mole gas / total moles
Dalton’s Law of Partial Pressure
• For a mixture of gases in a container,
PTotal = P1 + P2 + P3 + . . .
• This is particularly useful in calculating the
pressure of gases collected over water.
• Can also get the following equation
Ptotal = ntotal RT
V
Real Gases (van der Waal’s equation)
• Must correct ideal gas behavior when
at high pressure (smaller volume) and
low temperature (attractive forces
become important).
[Pobs + a(n/V)2] x (V-nb) = nRT
corrected pressure
corrected volume
Kinetic Energy of Gas Particles
• At the same conditions of temperature,
all gases have the same average kinetic
energy.
KE = (1/2)mv2
The Meaning of Temperature
(KE)avg = (3/2)RT
• Kelvin temperature is an index of the
random motions of gas particles (higher T
means greater motion.)
Root Mean Square Velocity
• The average velocity (speed) of a gas
particle
• R = 8.3145J/K.mol or 8.3145 kg.m2/s2
• M = molar mass in kg/mol
Root Mean Square Velocity
• Example: High above the stratosphere, the
temperature of the atmosphere may reach
1000oC. However, spacecraft and
astronauts do not burn up because the
concentration of molecules is very low and
the impacts transfer very little energy.
What is the rms velocity of nitrogen
molecules at that temperature?
Diffusion
• Diffusion: describes
the mixing of gases.
The rate of diffusion is
the rate of gas mixing.
Effusion
• Effusion: describes the passage of gas
into an evacuated chamber.
Graham’s Law of Effusion