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Chapter 12
Intermolecular Forces and the Physical Properties of Liquids and Solids
Intermolecular Forces and the Physical properties of Liquids and Solids
12
12.1 Intermolecular Forces
Dipole-Dipole Interactions
Hydrogen Bonding
Dispersion Forces
Ion-Dipole Interactions
12.2 Properties of Liquids
Surface Tension
Viscosity
Vapor Pressure
12.3 Crystal Structure
Unit Cells
Packing Spheres
Closest Packing
12.4 Types of Crystals
Ionic Crystals
Covalent Crystals
Molecular Crystals
Metallic Crystals
12.5 Amorphous Solids
12.6 Phase Changes
Liquid-Vapor Phase Transition
Solid-Liquid Phase Transition
Solid-Vapor Phase Transition
12.7 Phase Diagrams
Intermolecular Forces
Intermolecular forces are attractive forces that hold particles together in the condensed phases.
The magnitude (and type) of intermolecular forces is what determines whether the particles that make up a substance are a gas, liquid, or solid.
12.1
Gas
Liquid
Solid
ATTRACTIVE FORCES
electrostatic in nature
Intramolecular forces
bonding forces
These forces exist within each molecule.
They influence the chemical properties of the substance.
Intermolecular forces
nonbonding forces
These forces exist between molecules.
They influence the physical properties of the substance.
*
Intermolecular Forces
Intermolecular forces are the attractive forces holding particles together in the condensed (liquid and solid) phases of matter
Result from Coulombic attractions
Dependent on the magnitude of the charge
Dependent on distance between charges
Weaker than forces of ionic bonding
Involve partial charges
Intermolecular Forces
Intermolecular forces are attractive forces between molecules.
Intramolecular forces hold atoms together in a molecule.
Intermolecular vs Intramolecular
41 kJ to vaporize 1 mole of water (inter)
930 kJ to break all O-H bonds in 1 mole of water (intra)
Generally, intermolecular forces are much weaker than intramolecular forces.
“Measure” of intermolecular force
boiling point
melting point
DHvap
DHfus
DHsub
Types of Intermolecular Forces
van der Waals forces –between atoms and molecules of pure substances
Dipole-dipole interactions – attractive forces between polar molecules
Hydrogen bonding – attractive force in polar molecules containing a H atom bonded to a small, highly electronegative element (N, O and F)
(London) Dispersion forces – attractive forces arising from instantaneous dipoles and induced dipoles
Attractive forces that act between atoms or molecules in a pure substance are collectively called van der Waals forces.
Dipole-dipole interactions are attractive forces that act between polar molecules.
The magnitude of the attractive forces depends on the magnitude of the dipole.
Intermolecular Forces
Dipole - Dipole interactions – between polar molecules
Intermolecular Forces
Induced dipole – induced dipole interactions-
Attractive forces that arise as a result of temporary dipoles induced in atoms or molecules
ion-induced dipole interaction
dipole-induced dipole interaction
magnitude of the induced dipole moment, mind
the proportionality constant, a, is called the polarizability
Hydrogen bonding is a special type of dipole-dipole interaction.
Hydrogen bonding only occurs in molecules that contain H bonded to a small, highly electronegative atom such as N, O, or F.
Hydrogen Bonding
F
H
F
Hydrogen Bonding
Hydrogen Bond
The hydrogen bond is a special dipole-dipole interaction between the hydrogen atom in a polar N-H, O-H, or F-H bond and an electronegative O, N, or F atom.
A
H
…
B
A
H
…
A
or
A & B are N, O, or F
THE HYDROGEN BOND
a dipole-dipole intermolecular force
The elements which are so electronegative are N, O, and F.
A hydrogen bond may occur when an H atom in a molecule, bound to small highly electronegative atom with lone pairs of electrons, is attracted to the lone pairs in another molecule.
..
F
..
..
..
H
O
..
N
..
F
H
..
..
..
O
..
..
..
N
H
hydrogen bond
donor
hydrogen bond
acceptor
hydrogen bond
acceptor
hydrogen bond
donor
hydrogen bond
donor
hydrogen bond
acceptor
hydrogen bonds.
covalent bonds.
HF(l)
Dispersion forces or London dispersion forces result from the Coulombic attractions between instantaneous dipoles of non-polar molecules.
Dispersion Forces
Instantaneous and Induced Dipoles
Magnitude depends on the ability to be polarized which is greater for larger molecules.
Dispersion (or London) interactions
two identical atoms or nonpolar molecules
two nonidentical atoms or nonpolar molecules, A and B
instantaneous dipole moment.
Polarizability and Charged-Induced Dipole Forces
distortion of an electron cloud
Polarizability increases down a group
size increases and the larger electron clouds are further
from the nucleus
Polarizability decreases left to right across a period
increasing Zeff shrinks atomic size and holds the electrons
more tightly
Cations are less polarizable than their parent atom because they are smaller.
Anions are more polarizable than their parent atom because they are larger.
Dispersion forces increase with:
greater number of electrons
more diffuse electron cloud
molar mass
Molar mass and boiling point.
Molecular shape and boiling point.
more points for dispersion forces to act
fewer points for dispersion forces to act
Worked Example 12.1
Strategy Draw Lewis dot structures and apply VSEPR theory to determine whether each molecule is polar or nonpolar. Nonpolar molecules exhibit dispersion forces only. Polar molecules exhibit dipole-dipole interactions and dispersion forces. Polar molecules with N–H , F–H, or O–H bonds exhibit dipole-dipole interactions (including hydrogen bonding) and dispersion forces.
(a) (b) (c) (d)
What kind(s) of intermolecular forces exist in (a) CCl4(l), (b) CH3COOH(l), (c) CH3COCH3(l), and (d) H2S(l).
Worked Example 12.1 (cont.)
(a) (b) (c) (d)
Solution (a) CCl4 is nonpolar, so the only intermolecular forces are dispersion forces.
(b) CH3COOH is polar and contains an O–H bond, so it exhibits dipole-dipole interactions (including hydrogen bonding) and dispersion forces.
(c) CH3COCH3 is polar but does not contain N–H , F–H, or O–H bonds, so it exhibits dipole-dipole interactions and dispersion forces.
(d) H2S is polar but does not contain N–H , F–H, or O–H bonds, so it exhibits dipole-dipole interactions and dispersion forces.
Think About It Being able to draw correct Lewis structures is, once again, vitally important. Review, if you need to, the procedure for drawing them.
Ion-dipole interactions are Coulombic attractions between ions (either positive or negative) and polar molecules.
Ion-Dipole Interactions
Intermolecular Forces
What kind(s) of intermolecular forces exist in
CH2ClCH2COOH(l)
dispersion forces
dipole-dipole interactions
hydrogen bonding
Summary diagram for analyzing the intermolecular forces in a sample.
INTERACTING PARTICLES
(atoms, molecules, ions)
ions only
IONIC BONDING
ion + polar molecule
ION-DIPOLE FORCES
ions present
ions not present
polar molecules only
DIPOLE-DIPOLE
FORCES
HYDROGEN
BONDING
polar + nonpolar
molecules
DIPOLE-
INDUCED DIPOLE
FORCES
nonpolar
molecules only
DISPERSION
FORCES only
DISPERSION FORCES ALSO PRESENT
H bonded to
N, O, or F
Properties of Liquids
Surface tension is the amount of energy required to stretch or increase the surface of a liquid by a unit area.
The stronger the intermolecular forces, the higher the surface tension.
12.2
The molecular basis of surface tension.
hydrogen bonding
occurs in three
dimensions
hydrogen bonding
occurs across the surface
and below the surface
the net vector
for attractive
forces is downward
Surface Tension and Forces Between Particles
Substance
Formula
Surface Tension
(J/m2) at 200C
Major Force(s)
diethyl ether
ethanol
butanol
water
mercury
dipole-dipole; dispersion
H bonding
H bonding; dispersion
H bonding
metallic bonding
1.7x10-2
2.3x10-2
2.5x10-2
7.3x10-2
48x10-2
CH3CH2OCH2CH3
CH3CH2OH
CH3CH2CH2CH2OH
H2O
Hg
Effect of Surface Tension
Capillary action is the movement of a liquid up a narrow tube.
Two types of forces bring about capillary action:
cohesion is the attraction between like molecules
adhesion is the attraction between unlike molecules
Properties of Liquids
Adhesive forces are greater than cohesive forces
Cohesive forces are
greater than adhesive forces
Viscosity is a measure of a fluid’s resistance to flow.
The higher the viscosity the more slowly a liquid flows.
Liquids that have higher intermolecular forces have higher viscosities.
Properties of Liquids
Glycerol – high viscosity
due to
Three hydrogen bonding
sites
Molecular shape
Viscosity – a measure of a fluid’s resistance to flow
Units – N.s/m2
The higher the viscosity the greater the resistance to flow
Varies inversely with temperature
Stronger intermolecular forces produce higher viscocities
Vapor pressure is also dependent on intermolecular forces.
If a molecule at the surface of a liquid has enough kinetic energy, it can escape to the gas phase in a process called vaporization.
T1 < T2
Properties of Liquids
The number of molecules with enough kinetic energy to escape.
H2O(l) H2O(g)
Evaporation: H2O(l) → H2O(g)
Condensation: H2O(l) ← H2O(g)
The vapor pressure increases until the rate of evaporation equals the rate of condensation.
Properties of Liquids
When the forward process and reverse process are occurring at the same rate, the system is in dynamic equilibrium.
The vapor pressure increases until the rate of evaporation equals the rate of condensation.
Properties of Liquids
H2O(l) H2O(g)
The vapor pressure increases with temperature.
Properties of Liquids
The Clausius-Clapeyron equation relates the natural log of vapor pressure and the reciprocal of absolute temperature.
ln P = natural log of vapor pressure
ΔHvap = the molar heat of vaporization
R = the gas constant (8.314 J/K mol)
T = the kelvin temperature
C is an experimentally determined constant
Properties of Liquids
The Clausius-Clapeyron equation:
Plotting ln P versus 1/T is a line with a slope of ΔH/R.
ΔH is assumed to be independent of temperature.
Properties of Liquids
The Clausius-Clapeyron equation can be rearranged into a two point form:
Properties of Liquids
Worked Example 12.2
Strategy Given the vapor pressure at one temperature, P1, use the equation below to calculate the vapor pressure at a second temperature, P2.
Temperature must be expressed in kelvins, so T1 = 291.15 K and T2 = 305.15 K. Because the molar heat of vaporization is given in kJ/mol, we will have to convert it to J/mol for the units of R to cancel properly: ΔHvap = 2.6×104 J/mol. The inverse function of ln x is ex.
Diethyl ether is a volatile, highly flammable organic liquid that today is used mainly as a solvent. (It was used as an anesthetic during the nineteenth century and as a recreational intoxicant early in the twentieth century during prohibition, when ethanol was difficult to obtain.) The vapor pressure of diethyl ether is 401 mmHg at 18°C, and its molar heat of vaporization is 26 kJ/mol. Calculate its vapor pressure at 32°C.
ln =
P1
P2
ΔHvap
R
1
T2
1
T1
Worked Example 12.2 (cont.)
Solution
Think About It It is easy to switch P1 and P2 or T1 and T2 accidentally and get the wrong answer to a problem such as this. One way to help safeguard against this common error is to verify that the vapor pressure is higher at the higher temperature.
ln =
P1
P2
2.6×104 J/mol
8.314 J/K mol
1
305.15 K

1
291.15 K
= 0.4928
P1
P2
= e 0.4928 = 0.6109
P1
0.6109
= P2
P2 =
401 mmHg
0.6109
= 6.6×102 mmHg
The H-bonding ability of the water molecule.
hydrogen bond donor
hydrogen bond acceptor
The Unique Nature of Water
great solvent properties due to polarity and
hydrogen bonding ability
exceptional high specific heat capacity
high surface tension and capillarity
density differences of liquid and solid states
The hexagonal structure of ice.
The expansion and contraction of water.
The macroscopic properties of water and their atomic and molecular “roots”.
The major types of intermolecular forces in solutions.
LIKE DISSOLVES LIKE
Substances with similar types of intermolecular forces dissolve in each other.
When a solute dissolves in a solvent, solute-solute interactions and solvent-solvent interactions are being replaced with solute-solvent interactions. The forces must be comparable in strength in order to have a solution occur.
Hydration shells around an aqueous ion.
Like dissolves like: solubility of methanol in water.
water
methanol
A solution of methanol in water
(c) Ethanol - Diethyl ether can interact through a dipole and dispersion forces. Ethanol can provide both while water would like to H bond.
(b) Water - Hexane has no dipoles to interact with the -OH groups in ethylene glycol. Water can H bond to the ethylene glycol.
SAMPLE PROBLEM
Predicting Relative Solubilities of Substances
SOLUTION:
PROBLEM:
Predict which solvent will dissolve more of the given solute:
(a) Sodium chloride in methanol (CH3OH) or in propanol (CH3CH2CH2OH)
(b) Ethylene glycol (HOCH2CH2OH) in hexane (CH3CH2CH2CH2CH2CH3)
or in water.
(c) Diethyl ether (CH3CH2OCH2CH3) in water or in ethanol (CH3CH2OH)
PLAN:
Consider the intermolecular forces which can exist between solute molecules and consider whether the solvent can provide such interactions and thereby substitute.
(a) Methanol - NaCl is ionic and will form ion-dipoles with the -OH groups of both methanol and propanol. However, propanol is subject to the dispersion forces to a greater extent.
Correlation Between Boiling Point and Solubility in Water
Gas
Solubility (M)*
bp (K)
He
Ne
N2
CO
O2
NO
4.2 x 10-4
4.2
6.6 x 10-4
27.1
10.4 x 10-4
77.4
15.6 x 10-4
81.6
21.8 x 10-4
90.2
32.7 x 10-4
121.4
* At 273K and 1 atm
The physiological form of an amino acid.
C
H3N
R
C
O
O-
H
+
one of 20 different side chains
-carbon
A portion of a polypeptide chain.
The forces that maintain protein structure.
The structure and function of a soap.
The structure of lecithin.
lecithin
phospholipid found in all cell membranes
Intermolecular forces and membrane structure.
The mode of action of an antibiotic.
A portion of DNA polynucleotide chain.
The double helix of DNA.
1.08 nm
1.08 nm
The cyclic structure of glucose in aqueous solution.
The structure of cellulose.
Equilibrium in a saturated solution.
solute (undissolved) solute (dissolved)
Crystal Structure
A crystalline solid possess rigid and long-range order; its atoms, molecules, or ions occupy specific positions.
A unit cell is the basic repeating structural unit of a crystalline solid.
12.3
There are seven types of unit cells.
Crystal Structure
The coordination number is defined as the number of atoms surrounding an atom in a crystal lattice.
The value of the coordination number indicates how tightly the atoms are packed together.
The basic repeating unit in the array of atoms is called a simple cubic cell.
Crystal Structure
There are three types of cubic cells.
Crystal Structure
In a body-centered cubic cell (bcc) the spheres in each layer rest in the depressions between spheres in the previous layer.
The coordination number is 8.
Crystal Structure
In a face-centered cubic cell (fcc) the coordination number is 12.
Crystal Structure
Most of a cell’s atoms are shared by neighboring cells.
Crystal Structure
A corner atom is shared by eight unit cells.
An edge atom is shared by four unit cells.
A face-centered atom is shared by two unit cells.
A simple cubic cell has the equivalent of only one complete atom contained within the cell.
Crystal Structure
A body-centered cubic cell has two equivalent atoms:
A face-centered cubic cell contains four complete atoms:
Crystal Structure
Hexagonal close-packed (hcp) structure:
Crystal Structure
Close packing starts with a layer of atoms (A)
Atoms in the second layer (B) fit into the depressions of the first layer
Hexagonal close-packed structure.
Site directly over an atom in layer A
Cubic close-packed (ccp) structure:
Crystal Structure
Site directly over an atom in layer A
(hcp)
Site NOT directly over an atom in layer A (ccp)
Cubic close-packed structure
Closest packing:
Crystal Structure
Hexagonal close-packing (hcp)
Cubic close-packing (ccp) corresponds to a face-centered cubit cell.
Edge length (a) and radius (r) are related:
Crystal Structure
Simple cubic
Body-centered cubic
Face-centered cubic
Worked Example 12.3
Strategy Using the given density and the mass of gold continued within a face-centered cubic unit cell, determine the volumes of the unit cell. Then, use the volume to determine the value of a, and use the equation a = √8r to find r. Be sure to use consistent units for mass, length, and volume. The face-centered cubic unit cell contains a total of four atoms of gold [six faces, each shared by two unit cells, and eight corners, each shared by eight unit cells]. d = m/V and V = a3.
Gold crystallizes in a cubic close-packed structure (face-centered cubic unit cell) and has a density of 19.3 g/cm3. Calculate the atomic radius of an Au atom in angstroms ( ).
Solution First, we determine the mass of gold (in grams) contained within a unit cell:
m = × × = 1.31×10-21 g/unit cell
4 atoms
unit cell
1 mol
6.022×1023 atoms
197.0 g Au
1 mol Au
Worked Example 12.3 (cont.)
Solution
Then we calculate the volume of the unit cell in cm3:
V = = = 6.78×10-23 cm3
Using the calculated volume and the relationship V = a3 (rearranged to solve for a), we determine the length of a side of a unit cell:
a = = √6.78×10-23 cm3 = 4.08×10-8 cm
Using the relationship provided a = √8r (rearranged to solve for r), we determine the radius of a gold atom in centimeters.
r = = = 1.44×10-8 cm
Finally, we convert centimeters to angstroms:
1.44×10-8 cm × × = 1.44
m
d
1.31×10-21 g
19.3 g/cm3
3
a
√8
4.08×10-8 cm
√8
1×10-2 m
1 cm
1
1×10-10 m
Think About It Atomic radii tend to be on the order of 1 , so this answer is reasonable.
Ionic crystals are composed of charged ions that are held together by Coulombic attraction.
The unit cell of an ionic compound can be defined be either the positions of the anions or the positions of the cations.
Types of Crystals
12.4
Crystal structures of three ionic compounds:
Types of Crystals
CsCl
Simple cubic lattice
ZnS
Zincblende structure
(based on FCC)
CaF2
fluorite structure
(based on FCC)
Worked Example 12.4
Strategy Determine the contribution of each ion in the unit cell based on its position. Referring to the figure, the unit cell has four Zn2+ ions completely contained within the unit cell, and S2- ions at each of the eight corners and at each of the six faces. Interior ions (those completely contained within the unit cell) contribute one, those at the corners each contribute one-eighth, and those on the faces contribute one-half.
How many of each ion are contained within a unit cell of ZnS
Solution The ZnS unit cell contains four Zn2+ ions (interior) and four S2- ions [8 × (corners) and 6 × (faces)]
1
8
1
2
Think About It Make sure that the ratio of cations to anions that you determine for a unit cell matches the ratio expressed in the compound’s empirical formula.
Worked Example 12.5
Strategy Use the number of Na+ and Cl- ions in a unit cell (four of each) to determine the mass of a unit cell. Calculate volume using the edge length given in the problem statement. Density is mass divided by volume (d = m/V). Be careful to use units consistently.
The masses of Na+ and Cl- ions are 22.99 amu and 35.45 amu, respectively. The conversion factor from amu to grams is
so the masses of the Na+ and Cl- ions are 3.818×10-23 g and 5.887×10-23 g, respectively. The unit cell length is
564 pm × × = 5.64×10-8 cm
The edge length of the NaCl unit cell is 564 pm. Determine the density of NaCl in g/cm3.
1 g
6.022×1023 amu
1×10-12 m
1 pm
1 cm
1×10-2 m
Worked Example 12.5 (cont.)
Solution The mass of the unit cell is 3.882×10-22 g (4 × 3.818×10-23 g + 4 × 5.887×10-23 g). The volume of a unit cell is 1.794×10-22 cm3 [(5.64×10-8 cm)3]. Therefore, the density is given by
d = = 2.16 g/cm3
3.882×10-22 g
1.794×10-22 cm3
Think About It If you were to hold a cubic centimeter (1 cm3) of salt in your hand, how heavy would you expect it to be Common errors in this type of problem include errors of unit conversion–especially with regard to length and volume. Such errors can lead to results that are off by many orders of magnitude. Often you can use common sense to gauge whether or not a calculated answer is reasonable. For instance, simply getting the centimeter-meter conversion upside down would result in a calculated density of 2.16×1012 g/cm3! You know that a cubic centimeter of salt doesn’t have a mass that large. (That’s billions of kilograms!) If the magnitude of a result is not reasonable, go back and check your work.
In covalent crystals, atoms are held together in an extensive three-dimensional network entirely by covalent bonds.
Types of Crystals
In molecular crystals, the lattice points are occupied by molecules; the attractive forces between them are van der Waals forces and/or hydrogen bonding.
Types of Crystals
Worked Example 12.6
Strategy A face-centered metallic crystal contains four atoms per unit cell [8 × (corners) and 6 × (faces)]. Use the number of atoms per cell and the atomic mass to determine the mass of a unit cell. Calculate volume using the edge length given in the problem statement. Density is then mass divided by volume (d = m/V). Be sure to make all necessary unit conversions.
The mass of an Ir atom is 192.2 amu. The conversion factor from amu to grams is
so the mass of an Ir atom is 3.192×10-22 g. The unit cell length is
383 pm × × = 3.83×10-8 cm
The metal iridium (Ir) crystallizes with a face-centered cubic unit cell. Given that the length of the edge of a unit cell is 383 pm, determine the density of iridium in g/cm3.
1 g
6.022×1023 amu
1×10-12 m
1 pm
1 cm
1×10-2 m
1
8
1
2
Worked Example 12.6 (cont.)
Solution The mass of the unit cell is 1.277×10-21 g (4 × 3.192×10-22 g). The volume of a unit cell is 5.618×10-23 cm3 [(3.83×10-8 cm)3]. Therefore, the density is given by
d = = 22.7 g/cm3
1.277×10-21 g
5.62×10-23 cm3
Think About It Metals typically have high densities, so common sense can help you decide whether or not your calculated answer is reasonable.
In metallic crystals, every lattice point is occupied by an atom of the same metal.
Electrons are delocalized over
the entire crystal.
Delocalized electrons make metals
good conductors.
Large cohesive force resulting from
delocalization makes metals strong.
Types of Crystals
Types of Crystals
Amorphous Solids
Amorphous solids lack a regular three-dimensional arrangement of atoms.
Glass is an amorphous solid.
Glass is a fusion product.
SiO2 is the chief component.
Na2O and B2O3 are typically fused with molten SiO2 and allowed to cool without crystallizing.
12.5
Amorphous Solids
Amorphous Solids
Crystalline quartz
Noncrystalline (amorphous) quartz glass
Phase Changes
A phase is a homogeneous part of a system that is separated from the rest of the system by a well defined boundary.
When a substance goes from one phase to another phase, it has undergone a phase change.
Example Phase Change
Freezing of water H2O(l) → H2O(s)
Evaporation (or vaporization) of water H2O(l) → H2O(g)
Melting (fusion) of ice H2O(s) → H2O(l)
Condensation of water vapor H2O(g) → H2O(l)
Sublimation of dry ice CO2(s) → CO2(g)
Deposition of iodine I2(g) → I2(s)
12.6
The six possible phase changes
Phase Changes
The boiling point of a substance is defined as the temperature at which its vapor pressure equals the external atmospheric pressure.
The molar heat of vaporization (ΔHvap) is the amount of heat required to vaporize a mole of substance at its boiling point.
Phase Changes
The transformation of a liquid to a solid is called freezing.
The reverse process is called melting, or fusion.
The melting point (freezing point) of a solid (or liquid) is the temperature at which the solid and liquid phases coexist in equilibrium.
ice water
H2O(s) H2O(l)
In dynamic equilibrium, the forward and reverse process are occurring at the same rate.
Phase Changes
The molar heat of fusion (ΔHfus) is the energy required to melt 1 mol of a solid.
Phase Changes
Heating curves:
Phase Changes
Solid
Boiling point
Vapor
Liquid
Solid and liquid in equilibrium
Liquid and vapor in equilibrium
Time
Melting point
Sublimation is the process by which molecules go directly from the solid phase to the vapor phase.
Deposition is reverse process of sublimation.
The molar enthalpy of sublimation (ΔHsub) of a substance is the energy required to sublime 1 mole of a solid.
Phase Changes
ΔHsub = ΔHfus + ΔHvap
Solid I2 in equilibrium with its vapor
Worked Example 12.7
(a) Calculate the amount of heat deposited on the skin of a person burned by 1.00 g of liquid water at 100.0°C and (b) the amount of heat deposited by 1.00 g of steam at 100.0°C. (c) Calculate the amount of energy necessary to warm 100.0 g of water from 0.0°C to body temperature and (d) the amount of heat required to melt 100.0 g of ice 0.0°C and then warm it to body temperature. (Assume that body temperature is 37.0°C.)
Worked Example 12.7 (cont.)
Strategy For the purpose of following the sign conventions, we can designate the water as the system and the body as the surroundings. (a) Heat is transferred from hot water to the skin in a single step: a temperature change. (b) The transfer of heat from steam to the skin takes place in two steps: a phase change and a temperature change. (c) Cold water is warmed to body temperature in a single step: a temperature change. (d) The melting of ice and the subsequent warming of the resulting liquid water takes place in two steps: a phase change and a temperature change. In each case, the heat transferred during a temperature change depends on the mass of the water, the specific heat of the water, and the change in temperature. For the phase changes, the heat transferred depends on the amount of water (in moles) and the molar heat of vaporization (ΔHvap) or molar heat of fusion (ΔHfus). In each case, the total energy transferred or required is the sum of the energy changes for the individual steps.
The specific heat is 4.184 J/g °C for water and 1.99 J/g °C for steam. ΔHvap is 40.79 kJ/mol and ΔHfus is 6.01 kJ/mol. Note: The ΔHvap of water is the amount of heat required to vaporize a mole of water. However, we want to know how much heat is deposited when water vapor condenses, so we use 40.79 kJ/mol.
Worked Example 12.7 (cont.)
Solution (a) ΔT = 37.0°C – 100.0°C = –63.0°C
q = msΔT = 1.00 g × ×–63.0°C
Thus, 1.00 g of water at 100.0°C deposits 0.264 kJ of heat on the skin. (The negative sign indicates that heat is given off by the system and absorbed by the surroundings.)
(b)
q1 = nΔHvap = 0.0555 mol ×
q2 = msΔT = 1.00 g × ×–63.0°C
The overall energy deposited on the skin by 1.00 g of steam is the sum of q1 and q2:
–2.26 kJ + (–0.264 kJ) = –2.53 kJ
4.184 J
g °C
= –2.64×102 J = –0.264 kJ
1.00 g
18.02 g/mol
= 0.0555 mol water
40.79 kJ
mol
= –2.26 kJ
4.184 J
g °C
= –2.64×102 J = –0.264 kJ
Worked Example 12.7 (cont.)
Solution (c) ΔT = 37.0°C – 0.0°C = 37.0°C
q = msΔT = 1.00 g × ×37.0°C
The energy required to warm 100.0 g of water from 0.0°C to 37.0°C is 15.5 kJ.
(d)
q1 = nΔHfus = 5.55 mol ×
q2 = msΔT = 100.0 g × ×37.0°C
The energy rquired to melt 100.0 g of ice at 0.0°C and warm it to 37.0°C is the sum of q1 and q2:
33.4 kJ + 15.5 kJ = 48.9 kJ
4.184 J
g °C
= 1.55×104 J = 15.5 kJ
100.0 g
18.02 g/mol
= 5.55 mol water
6.01 kJ
mol
= 33.4 kJ
4.184 J
g °C
= 1.55×104 J = 15.5 kJ
Think About It In problems that include phase changes, the q values corresponding to the phase-change steps will be the largest contributions to the total. If you find that this is not the case in your solution, check to see if you have made the common error of neglecting to convert the q values corresponding to temperature changes from J to kJ.
Phase Diagrams
A phase diagram summarizes the conditions at which a substance exists as a solid, liquid, or gas.
The triple point is the
only combination of
pressure and temperature
where three phases of a
substance exist in
equilibrium.
12.6
triple point
The phase diagram of water:
Phase Diagrams
Worked Example 12.8
Strategy Each point on the phase diagram corresponds to a pressure-temperature combination. The normal boiling and melting points are the temperatures at which the substance undergoes phase changes. These points fall on the phase boundary lines. The triple point is where the three phase boundaries meet.
Using the following phase diagram, (a) determine the normal boiling point and the normal melting point of the substance, (b) determine the physical state of the substance at 2 atm and 110°C, and (c) determine the pressure and temperature that correspond to the triple point of the substance.
Worked Example 12.8 (cont.)
Solution By drawing lines corresponding to a given pressure and/or temperature, we can determine the temperature at which a phase change occurs, or the physical state of the substance under specified conditions.
(a)
The normal boiling and melting points are ~140°C and ~205°C, respectively.
Worked Example 12.8 (cont.)
Solution
(b)
At 2 atm and 110°C the substance is a solid.
Worked Example 12.8 (cont.)
Solution
(c)
The triple point occurs at ~0.8 atm and ~115°C.
Think About It The triple point of this substance occurs at a pressure below atmospheric pressure. Therefore, it will melt rather than sublime when it is heated under ordinary conditions.
Key Concepts
12
Intermolecular Forces
Dipole-Dipole Interactions
Hydrogen Bonding
Dispersion Forces
Ion-Dipole Interactions
Properties of Liquids
Surface Tension
Viscosity
Vapor Pressure
Crystal Structure
Unit Cells
Packing Spheres
Closest Packing
Types of Crystals
Ionic Crystals
Covalent Crystals
Molecular Crystals
Metallic Crystals
Amorphous Solids
Phase Changes
Liquid-Vapor Phase Transition
Solid-Liquid Phase Transition
Solid-Vapor Phase Transition
Phase Diagrams

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