Transcript Camp 1 - Quynh Nguyen

Mark S. Cracolice
Edward I. Peters
www.cengage.com/chemistry/cracolice
Chapter 15
Gases, Liquids, and Solids
Mark S. Cracolice • The University of Montana
Dalton’s Law of Partial Pressures
Dalton’s Law of Partial Pressures
The partial pressure of a component in a gas mixture
is the pressure that component would exert if it alone
occupied the same volume at the same temperature.
The total pressure exerted by a mixture of gases is the
sum of the partial pressures.
Ptotal
= p1
+ p2
+ p3
+ p4
+----
Dalton’s Law of Partial Pressures
Example:
The partial pressures in a mixture of gases are:
Helium, 344 torr
Neon, 298 torr
Argon, 109 torr
What is the total pressure?
Solution:
P = p1 + p2 + p3 = pHe + pNe + pAr
= (344 + 298 + 109) torr = 751 torr
Laboratory Collection of Oxygen
• A mixture of oxygen and water vapor are collected at 220C, at
which water vapor pressure is 19.8 torr. If the total pressure of
the mixture is 755 torr, what is the oxygen pressure?
Laboratory Collection of Oxygen
Ptotal = Patm when the water levels are equalized.
Poxygen = Ptotal – Pwater vapor = 755 torr – 19.8 torr = 735 torr
Properties of Liquids
Comparison of Properties of Liquids
with Properties of Gases
1. Gases may be compressed; liquids cannot.
2. Gases expand to fill their containers; liquids do not.
3. Gases have low densities; liquids have relatively high
densities.
4. Gases may be mixed in a fixed volume; liquids cannot.
Properties of Liquids
Gas particles are far apart.
Forces between the particles are negligible.
In a liquid, the particles are very close to one another.
The intermolecular attractions in a liquid are strong.
Properties of Liquids: Vapor Pressure
The partial pressure exerted by a vapor in equilibrium
with its liquid phase is the vapor pressure.
The stronger the intermolecular forces, the lower the
vapor pressure.
The vapor pressure is higher at higher temperature.
Properties of Liquids: Vapor Pressure
Properties of Liquids: Heat of Vaporization
The energy required to vaporize one gram of
substance is called heat of vaporization (the unit is
kJ/g)
The energy required to vaporized one mole is called
molar heat of vaporization (the unit is kJ/mol).
The stronger the intermolecular attraction, the higher
the molar heat of vaporization.
.
Properties of Liquids: Heat of Vaporization
Properties of Liquids: Boiling Point
Boiling point is the temperature at which the vapor
pressure of a liquid is equal to the external pressure
above its surface.
The temperature at which the vapor pressure is equal
to one atmosphere is called the normal boiling point.
Properties of Liquids: Boiling Point
Properties of Liquids: Boiling Point
In order for a stable bubble to form in a boiling liquid,
the vapor pressure within the bubble must be high
enough to push back the surrounding liquid and the
atmosphere above the liquid
The boiling point of a substance increases with the
surrounding pressure.
Properties of Liquids
Properties of Liquids: Viscosity
Viscosity
A measure of the resistance of a liquid to flow.
The stronger the intermolecular attractive forces, the
greater the resistance to flow, so the higher the
viscosity.
Properties of Liquids: Viscosity
A measure of the resistance of a liquid to flow.
water<syrup<honey
Properties of Liquids: Surface tension
Surface Tension
The tendency for a liquid to form the
minimum possible surface area.
The stronger the intermolecular attractive forces, the
stronger each particle is attracted in all directions by
the particles around it, so the higher the surface
tension.
Properties of Liquids: Surface Tension
Molecules at the surface possess a higher potential energy
than those within the body of the liquid. To minimize this
energy, a liquid tends to have minimum surface.
Three Kind of Intermolecular Forces
Dipole forces
Induced dipole forces
Hydrogen bonds
Of the three kinds of intermolecular attractions,
hydrogen bonds are the strongest. Dipole forces are
next, and dispersion forces are the weakest.
Types of Intermolecular Forces
Dipole Forces
Polar molecules have permanent dipole moments.
The dipole force is the force of attraction between the
positive pole of one molecule and the negative pole
of another molecule.
Types of Intermolecular Forces
Types of Intermolecular Forces
Types of Intermolecular Forces:
Induced Dipole Forces
Induced Dipole: The presence of an external
electric field may cause an induced, or temporarily,
polarization of the molecule.
Molecules in which a dipole can be induced in this
way are said to be polarizable.
Attractions between induced dipoles are also called
dispersion forces.
Types of Intermolecular Forces:
Induced Dipole Forces
Attractive forces between instantaneous electric dipoles
on neigbboring molecules
Types of Intermolecular Forces
Induced Dipole Forces
Large molecules are more easily polarized than
small molecules. Induced dipole forces increase
with molar mass
Br2 molecules have 70 electrons.
Bromine (left) is a liquid.
Iodine, I2 , molecules have 106 e
Iodine (right) is a solid.
.
Types of Intermolecular Forces
Hydrogen bonds
The covalent bond between hydrogen atom and atom of
nitrogen, oxygen, or fluorine is strongly polar.
Types of Intermolecular Forces:
Hydrogen bonds
Hydrogen bonds: The attractive forces between
molecules that have a hydrogen atom covalently
bonded to nitrogen, oxygen, or fluorine atom.
Hydrogen atom lies between two small strongly
electronegative atoms with lone pairs of electrons.
A hydrogen bond is an intermolecular force, an
attraction between hydrogen atom of one molecule
with the negative pole of another molecule.
Types of Intermolecular Forces:
Hydrogen bonds between H2O molecules
A hydrogen bond is denoted by dots to distinguish it
from a covalent bond.
The Solid State
Solid substances can be classified based on
how the particles are arranged in the solid:
Crystalline Solid
Particles are arranged in a geometric pattern that
repeats itself over and over in three dimensions.
Amorphous Solid
No long-range ordering of the particles in the solid.
The Solid State
The Solid State
Crystalline Solid
and
amorphous solid
Types of Crystalline Solids
Crystalline solids can be divided into four classes
on the basis of the type of forces that
hold particles together in the crystal lattice:
Ionic crystals
Molecular crystals
Covalent network solids
Metallic crystals
Types of Crystalline Solids: Ionic Crystals
Ionic Crystals
Ionic crystals have ions at the points of the
lattice that describes the structure of the solid.
Examples are sodium chloride, calcium
carbonate.
Ionic Crystals: NaCl
Ionic Crystals: CaCO3
Molecular Crystals
Molecular Crystals
A molecular solid has covalently bonded
molecules at each of its lattice points.
Examples are sugar, I2, S8
Molecular Crystal of Sulfur
Molecular Crystal of S8
and
liquid form of sulfur
Covalent Network Solids
Covalent Network Solids
The atoms bond to each other with strong
directional covalent bonds that lead to giant
molecules, or networks, of atoms.
Examples are diamond, graphite, quartz.
Diamond
Covalent Network Solids: Quartz
Covalent Network Solids: Quartz
Types of Crystalline Solids: Metallic Crystals
Metallic Crystals
A crystal of positive ions through which
valence electrons move freely.
Examples: Al, Cu, Fe
Metallic Crystal
Metal ions in an electron sea
Metallic Crystal: Aluminum Metal
Properties of Crystalline Solids
Energy and Change of State
Energy and Change of State: Vaporization
To boil a liquid (vaporization), heat must be added.
For a pure substance at its boiling point, the amount of
heat added to vaporize the liquid is proportional to
the mass of the liquid:
q = ∆Hvap x m
∆Hvap is the enthalpy of vaporization or the heat of
vaporization.
Energy and Change of State
Energy and Change of State
Example:
How much energy is required to vaporize 19.6 g of water at
100°C?
Solution:
Solve with algebra.
GIVEN: 19.6 g; ∆Hvap = 2.26 kJ/g (Table 15.4)
WANTED: Energy (assume kJ)
2.26kJ
q = m  H vap = 19.6g 
= 44.3kJ
g
Energy and Change of State
To melt a solid (fusion), heat must be added.
For a pure substance at its melting point, the amount of
heat added to melt the solid is proportional to the
mass of the solid:
q = ∆Hfus x m
∆Hfus is the enthalpy of fusion or the heat of fusion.
Energy and Change of State
Example:
How much energy is required to melt 19.6 g of ice at 0°C?
Solution:
Solve with algebra.
GIVEN: 19.6 g; ∆Hfus = 333 J/g (Table 15.4)
WANTED: Energy (assume kJ)
q = m  H fus
333J
3
= 19.6g 
= 6.53 10 J
g
Energy and ∆T: Specific Heat
Experiments indicate that the heat flow, q, in heating or
cooling a substance is proportional to the mass, m, and
its temperature change, ∆T
q = m x c x ∆T
C is the specific heat of the substance
Energy and ∆T: Specific Heat
q
c º
m ´ ∆T
Specific Heat, c
The heat flow required to change the temperature
of one gram of a substance by one degree Celsius.
A substance with a low specific heat gains little energy in
warming through a temperature change, as compared with a
substance with a higher specific heat.
Energy and ∆T: Specific Heat
Energy and ∆T: Specific Heat
Example:
How much energy is required to change the temperature of
25.0 g of water from 10.0°C to 50.0°C?
Solution:
Solve with algebra.
GIVEN: 25.0 g; c = 4.18 J/g • °C (Table 15.5)
WANTED: Energy (assume J)
4.18 J
25.0 g ´
´ (50.0 – 10.0)°C = 4.18 ´ 10 3 J
g × °C
Changes in Temperature & State
If you steadily apply heat to a pure substance
in the solid state, five things will happen:
1. The solid will warm to its melting point temperature.
2. The solid will change to liquid at the melting point
temperature.
3. The liquid will warm to its boiling point temperature.
4. The liquid will change to gas at the boiling point temperature.
5. The gas will become hotter.
Changes in Temperature & State
Changes in Temperature & State
Example:
Calculate the energy that must be added to 19.6 g of ice at
–12°C to change it to steam at 115°C. Answer in kJ.
Solution:
Step 1 is to sketch a graph of temperature vs. heat flow.
Changes in Temperature & State
Step 2 is to calculate the heat flow for each step.
A to B: Solid at –12°C to solid at 0°C
GIVEN: 19.6 g; c = 2.06 J/g • °C (Table 15.5)
WANTED: Energy in kJ
2.06J
1 kJ
Q(A  B)  19.6g 
 [0 — (—12)]C 
= 0.48kJ
g  C
1000J
Changes in Temperature & State
B to C: Solid at 0°C to liquid at 0°C
GIVEN: 19.6 g; ∆Hfus = 333 J/g (Table 15.4)
WANTED: Energy in kJ
333J
1 kJ
Q(B  C)  19.6g 

= 6.53kJ
g
1000J
Changes in Temperature & State
C to D: Liquid at 0°C to liquid at 100°C
GIVEN: 16.9 g; c = 4.18 J/g • °C (Table 15.5)
WANTED: Energy in kJ
4.18J
1 kJ
Q(C  D)  16.9 g 
 (100— 0)C 
= 8.19kJ
g  C
1000J
Changes in Temperature & State
D to E: Liquid at 100°C to gas at 100°C
GIVEN: 19.6 g; ∆Hvap = 2.26 kJ/g (Table 15.4)
WANTED: Energy in kJ
2.26kJ
Q(D  E)  19.6g 
= 44.3 kJ
g
Changes in Temperature & State
E to F: Gas at 100°C to gas at 115°C
GIVEN: 19.6 g; c = 2.00 J/g • °C (Table 15.5)
WANTED: Energy in kJ
2.00J
1 kJ
Q(E  F)  19.6g 
 (115- 100)C 
= 0.59kJ
g  C
1000J
Changes in Temperature & State
Step 3 is to add the heat flows.
qtotal = qA to B + qB to C + qC to D + qD to E + qE to F =
0.48 kJ + 6.53 kJ + 8.19 kJ + 44.3 kJ + 0.59 kJ =
60.1 kJ