Unit 5
Phases of matter
Chapters 10, 11.4-11.6
Class starter
Answer the following questions using principles of molecular structure and intermolecular
forces.
CompoundEmpirical Formula
Solubility in Water
Boiling Point (oC)
1
C2H6O
Slightly soluble
-24
Soluble
78
2
C2H6O
Compounds 1 and 2 in the data table above have the same empirical formula, but they have
different physical properties.
a. The skeletal structure for one of the two compounds is shown below in Box X.
i. Complete the Lewis electron dot diagram of the molecule in Box X.
Include any
lone (nonbonding) pairs of electrons.
•
ii.
In Box Y above, draw the complete Lewis electron dot diagram for the other
compound, which is a structural isomer of the compound represented in Box X.
Include any lone (nonbonding) pairs of electrons.
b. On the basis of the complete Lewis electron dot diagrams you drew in part a and
the information in the data table above, identify which compound, 1 or 2, has the
structure represented in Box X. Justify your answer in terms of the intermolecular
forces present in each compound.
1 point is earned for
a correct
Lewis diagram
Compound 2 is in Box X. Compound 2 (X) would have intermolecular hydrogen
bonding. Compound 1 (Y) would have weaker dipole-dipole and London
dispersion forces (LDFs). Because compound 2 has stronger intermolecular
forces (IMFs) it has a higher boiling point. Also, compound 2 is capable of
forming more hydrogen bonds with H2O than compound 1 is, causing the
solubility difference noted in the table.
2 points are earned for identification of compound 2 and a
rationale that references the types of IMFs in each compound
while explaining relative boiling points and/or solubilities.
Unit 6.1
Changing phases, require or give
off energy
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 (continued)
 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. (higher temp. means
more kinetic energy)
(gases are more ideal at high temperatures/ low pressure
and weak attractions between molecules)
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
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 (101.3 kPa)
1 atmosphere = 1 atm = 760 mm Hg = 760 torr
Conversion practice:
Convert 0.876 atm to torr
Convert 786 mmHg to kPa
Measuring Pressure
The first device for
measuring atmospheric
pressure was developed by
Evangelista Torricelli
during the 17th century.
The device was called a
“barometer”
Baro = weight
Meter = measure
An Early Barometer
The normal pressure due to the atmosphere at sea
level can support a column of mercury that is 760
mm high.
Converting Celsius to Kelvin
Gas law problems involving temperature require
that the temperature be in KELVINS!
Kelvins = °C + 273
°C = Kelvins - 273
Standard Temperature and Pressure
“STP”
P = 1 atmosphere, 760 torr (sea level)
T = 0°C, 273 Kelvins
The molar volume of an ideal gas is 22.4
liters at STP
Gas laws
Pressure, temperature, and moles of
gas effect the volume of the gas
Boyle’s Law
Pressure is inversely proportional to volume
when temperature and moles of gas are held
constant.
P1V1 = P2V2
Charles’s Law
The volume of a gas is directly proportional to
temperature, and extrapolates to zero at zero Kelvin.
(Pressure and moles = constant)
V1 V2
=
T1 T2
Gay Lussac’s Law
The pressure and temperature of a gas are
directly related, provided that the volume
and moles remains constant.
P1 P2
=
T1 T2
The 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. (Remember temp. needs to be Kelvin)
Practice
• A sample of gas occupies a volume of 7.50 L at
0.988 atm and 28.0 ᴼC. Calculate the pressure
of gas if its volume is decreased to 4.89 L
while its temp. remains constant.
Practice
• If a sample of gas has a volume of 400 mL at
20 °C, what will it volume be at 30 °C if the
pressure remains constant?
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).
The more moles of gas there are, the higher the volume
if temp. and pressure are held constant
Unit 5.2
The ideal gas law
Class starter
1. Write a net ionic reaction and answer the question for each rxn.
a. Solid magnesium hydroxide is added to a solution of hydrobromic
acid
What volume, in mL, of 2.00 M hydrobromic acid is required to react
completely with 0.10 mol of solid magnesium hydroxide?
b. A copper wire is dipped into a solution of silver (I) nitrate?
Describe what is observed as the reaction proceeds.
2. At which of the following temperatures and pressures would a real
gas be most likely to deviate from ideal behavior?
Temperature(K)
Pressure (atm)
(a)
100
50
(b)
200
5
(c)
300
0.01
(d)
500
0.01
(e)
500
1
Class starter
1. Write a net ionic reaction and answer the question for each rxn.
a. Solid magnesium hydroxide is added to a solution of hydrobromic
acid. Mg(OH)2 + 2 H+ → Mg2+ + 2 H2O
What volume, in mL, of 2.00 M hydrobromic acid is required to react
completely with 0.10 mol of solid magnesium hydroxide? 100 mL
b. A copper wire is dipped into a solution of silver (I) nitrate?
Cu + 2 Ag+ → Cu2+ + 2 Ag
Describe what is observed as the reaction proceeds.
Silver metal will appear on the surface of the copper wire. OR The solution will turn
blue. OR
The copper wire will lose mass.
2. A
Ideal Gas Law
PV = nRT
P = pressure in atm
V = volume in liters
n = moles
R = proportionality constant
= 0.08206 L atm/ mol·Κ or 62.36 L torr/mol K
T = temperature in Kelvins
Holds closely at P < 1 atm
Units of P, V, n, and T must agree with R
Practice
• Calcium carbonate decomposes upon heating
to give calcium oxide and carbon dioxide. A
sample of calcium carbonate is decomposed
and the carbon dioxide is collected in a 250
mL flask. After the decomposition is
complete, the gas has a pressure of 1.3 atm at
a temp. of 31 C. How many moles of gas were
generated?
• Answer: 0.013 mol carbon dioxide
Density and the Ideal Gas Law
Combining the formula for density with the Ideal
Gas law, substituting and rearranging algebraically:
MP
D=
RT
M = Molar Mass
P = Pressure
R = Gas Constant
T = Temperature in Kelvins
Molar mass and ideal gas law
• Molar mass = g/mol
• Molar mass = dRT/P
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)
3 moles H2
3 liters H2
+
N2(g)

2NH3(g)
+ 1 mole N2

2 moles NH3
+ 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)
+
12 L H2
N2(g)
2NH3(g)

2 L NH3
3 L H2
=
8.0
L NH3
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)
50.0 g KClO3
1 mol KClO3
122.55 g KClO3
3 mol O2
22.4 L O2
2 mol KClO3
1 mol O2
= 13.7 L O2
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)
50.0 g KClO3
1 mol KClO3
122.55
122.55 gg KClO
KClO33
nRT
V=
P
=
33 mol
mol O
O22
2 mol KClO3
0.612
= “n”
mol O2
mol O2
L ⋅ atm
)(310 K)
mol ⋅ K
= 16.7 L
0.930 atm
(0.612 mol)(0.0821
Dalton’s Law of Partial Pressures
For a mixture of gases in a container,
PTotal = P1 + P2 + P3 + . . .
Gases act as if they are alone
This is particularly useful in calculating the
pressure of gases collected over water.
• A sample of potassium chlorate decomposes,
producing oxygen that is collected over water.
The volume of gas collected is 250 mL at 26 °C
and 765 torr total pressure.
• How many moles of oxygen gas are collected?
0.00992 mol
• How many grams of potassium chlorate were
decomposed?
0.811 g KClO3
• P (gas) = (mol gas/total mol) * P t
• If a 1 L container has a total pressure of 760
torr at 20 °C and there are 0.14 mol of
nitrogen and 0.86 mol of oxygen, how much
pressure is exerted by just the nitrogen?
(mol gas/total moles = mol fraction)
Kinetic Energy of Gas Particles
At the same conditions of temperature, all gases
have the same average kinetic energy.
1 2
KE = mv
2
At the same temperature, smaller molecules
move faster than larger molecules
(rms speed)
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.
Remember smaller molecules will diffuse and effuse faster because
they move at a higher velocity at the same temperature
Unit 5.3
Liquids and Solids
Class starter
1. At approximately what temperature will 40. g of argon gas at 2.0 atm occupy a
volume of 22.4 L?
(a) 1,200 K
(b) 600 K
(c) 550 K
(d) 270 K
(e) 140 K
Gas
Amount
Ar
0.35 mol
CH4
0.90 mol
N2
0.25 mol
2. Three gases in the amounts shown in the table above are added to a previously
evacuated rigid tank. If the total pressure in the tank is 3.0 atm at 25 oC, the partial
pressure of N2(g) in the tank is closest to
(a )0.75 atm
(b) 0.50 atm
(c) 0.33 atm
(d) 0.25 atm
(e) 0.17 atm
Some properties of liquids
• Surface Tension: The
resistance to an increase in
its surface area (polar
molecules, liquid metals).
• Capillary Action:
Spontaneous rising of a
liquid in a narrow tube.
Properties of liquids
Viscosity: Resistance to
flow
High viscosity is an indication
of strong Intermolecular
forces
Volatile liquids: evaporate
easily (weak
intermolecular forces)
Types of Solids
Crystalline Solids: highly
regular arrangement of
their components
Amorphous solids:
considerable disorder
in their structures
(glass).
Network atomic solids
some covalently bonded substances DO NOT form discrete molecules
(diamond, graphite, silicon oxides)
Diamond, a network of
covalently bonded
carbon atoms
Graphite, a network of
covalently bonded
carbon atoms
Phase Changes- require or release energy
• Heat of fusion: amount of energy it takes to melt
a solid
• Heat of vaporization: amount of energy it takes to
vaporize a liquid
• Heat of vaporization > heat of fusion
(break all bonds)
(break some bonds)
Heating curve
Energy
Equilibrium vapor pressure
• The pressure of the vapor present at equilibrium.
• Determined principally by the size of the
intermolecular forces in the liquid.
(strong forces have low vapor pressure)
• Increases significantly with temperature.
• Volatile liquids have high vapor pressures.
• A liquid will boil when its vapor pressure =
external pressure
• Strong intermolecular forces = low vapor pressure
= high boiling point
• Higher altitudes (low external pressure) = lower
boiling point
• Normal boiling point = boiling point at standard
pressure