ch1301.doc

Final exam May 12, 1999 Wednesday 1 pm (Please check accuracy)

ch1302.doc

ch1303.doc Examples of varying attractive forces:

intermolecular attractions

ch1303.doc

INTERMOLECULAR FORCES
· hydrogen bonding
· ion - ion interactions
· dipole dipole interactions
· London Forces

ch1303a.doc

HEAT CALCULATIONS
· heating matter: non-isothermal
· phase change of matter: isothermal

INTERMOLECULAR FORCES
· hydrogen bonding: strongest
· ion - ion interactions: ionic solids
· dipole dipole interactions: liquids
· London Forces: induced polarization

LIQUID STATE
· viscosity
· surface tension
· capillary action
· evaporation
· vapor pressure
· boiling points

SOLID STATE discussed next period

ch1304.doc

HYDROGEN BONDING (O, N. F)
H₂O_(l) ® H₂O_(g) requires 40.7 kJ/mole
main attractive force: hydrogen bonding
important: water, alcohols, DNA, amino acids,
H atom is on O, N, F
O¾ H is very polar
bonds with another O, N, or, F
O¾ H¼ ¼ O
hydrogen bond
intermolecular and intramolecular

ch1304.doc

INTERMOLECULAR FORCES

ION-ION attractions:

Ionic compounds with their melting pointsch1304.doc
Trend: Higher molecular weight, more polarization, the charge is more diffuse
hence easier to melt (overcome rigid attractive forces)

INTERMOLECULAR FORCES

DIPOLE-DIPOLE interactions
Polar Covalent molecules
H₂O, NH₃, HCl, Ethanol
Molecules with Permanent dipoles

ch1304.doc
INTERMOLECULAR FORCES

LONDON FORCES weak
· freezing and solidifying of inert gases and non-polar atoms
· dipole induced dipole
· 1/d⁷ dependence (force)
· polarized by nuclei
U µ a ²/d⁶ (bond energy) polarizability and distance

ch1305.doc

Contributions to overall energy

1. Educated or well-informed; schooled:

ch1305.doc
LIQUID STATE

Definition: poise (po¹z) n. A centimeter-gram-second unit of dynamic viscosity
equal to one dyne-second per square centimeter. Unit of Viscosity.

ch13006.doc

SURFACE TENSION:

draws the liquid together and forms the liquid-air interface

(Surface tension is) a Cohesive force that
Minimizes surface area
explains spherical droplets
bugs walk on water

ch13006.doc

CAPILLARY ACTION:

Adhesive force that causes liquids to rise against gravity up glass tubes
Roots
Meniscus: a) water; b) mercury (in figure)

ch1307.doc Heat

n. Physics. A form of energy associated with the motion of atoms or molecules
and capable of being transmitted through solid and fluid media
by conduction, through fluid media by convection, and through empty space by radiation

Energy (µn"…r-j¶) n. Physics. The capacity of a physical system to do work.

Calorie: heat required to raise 1.00 g H₂O 1.00 ^oC

Joule= 1/0.239 cal. ( 1 Newton meter)
Also 4.184 J =1.000 cal

Specific heat: Heat energy required to raise the temperature of 1.000 g of substance 1.000 ^oC

ch1307.doc

Q. How much heat is required to warm a cup of coffee at room temperature to 90 ^oC ?

A. Assume room temperature is 25 ^oC
Assume a cup holds 250 g of coffee.
Assume coffee is water.

Look up the specific heat of H₂O: 4.18 J/g

Energy = (90-25) x 250 g x 4.18 J/g ^oC = 67825 J

= 6.78 x 10⁺⁴ J

Q. How much heat must be removed to cool a cup of frozen coffee to -10 ^oC?

A. Look up specific heat of ice: 2.09 J/g ^oC
Other assumptions as just above.
Heat = (-10 - 0) x 250 g x 2.09 J/g ^oC = - 5225 J
(negative means heat evolved by system)
= -5.22 x 10³ J

ch1307.doc

The heat required to raise the temperature of exactly 1 mole of a substance 1.000 ^oC.

Q. How much heat is required to warm 3.00 mol water at room temperature to 90 ^oC?

A. 1st, look up or calculate molar heat capacity; 2nd, do the problem. We will actually calculate the molar heat capacity from specific heat
molar heat capacity= 4.18 J/g ^oC x 18.0 g/mol = 75.2 J/mol ^oC

Now we solve
Heat = 3.00 mol x 75.2 J/mol ^oC x 65 ^oC = 14664 J

Heat of Vaporization: Heat added to a liquid to vaporize 1.000 g at its boiling point.

Molar Heat of Vaporization: Heat added to a liquid to vaporize 1.000 mol
of liquid at its boiling point: D H_vap .

Q. How much heat is required to vaporize 250 g of water at its boiling point?

A. The process isisothermal at 100 ^oC: H₂O(l) ® H₂O(g)
? J = (250 g ) (2260 J/g) = 565 kJ

ch1307.doc

Heat of fusion: The heat required to melt 1.000 g of substance at its melting point.

Q. Calculate the heat that must be absorbed in melting 50.0 g ice at its normal melting point?

A. Look up the specific heat of fusion of ice: 334 J/g

50 g x 334 J/g = 16700 J or 16.7 kJ

Molar heat of fusion: The heat required to melt 1.000 mol solid at it's melting point.
?J/mol = 334 J/g x 18 g/mol = 6012 or 6.01 kJ

ch1308.doc

Combined example

A: How much heat is required to convert 10.0 g H₂O at 20 ^oC to water vapor at 120 ^oC.

Divide into 4 steps and add them up

step 1: Heat the water to 100 ^oC
10.0 g x (100 - 20 ^oC) x 4.18 J/g.^oC = 3.34 kJ

step 2: vaporize the water at 100 ^oC
10.0 g x 2.260 kJ/g = 22.6 kJ

step 3: heat the steam to 120 ^oC
10.0 g x (120 - 100 ^oC) x 2.03 J/g ^oC = 0. 407 kJ

step 4: combine individual heats
Total Energy= 3.34 kJ + 22.6 kJ + .407 kJ = 26.4 kJ heat

Q. Finger into the spout of boiling teakettle. Assume that the finger
condenses 0.5 g. steam. How much heat is absorbed by the finger?

The condensation of 1/2 g of water at its boiling point.
? J = 0.5 g x 2260 J/g = 11Z30 J or 1.1 kJ.

ch1309.doc

Summary of Our Understanding of Intermolecular Attractive Forces

Fill in the blanks with HIGH or LOW:

Practice a table like this for
a) water; b) motor oil; c) gasoline; d) ether; e) nail polish remover; f) turpentine; g) alcohol.

ch1310.doc
SOLIDS Important vocabulary

Solid
Liquid
gas
melting point
normal melting point
freezing
melting
boiling
condensing
endothermic process
exothermic process
temperature
phase (solid, liquid, gas)
Joule
gram
specific heat
heat of fusion
heat of solidification
heat of vaporization

ch1311.doc (water)

Q.How much heat is required to convert
25.0 g of ice at -10.0 ^oC to steam at 110.0 ^oC?

step 0: Phase change points and specific heats
Information: m. p. is 0.0 ⁰C, 334 J/g ; b. p. is 100.0 ^oC 2260 J/g.
Information: specific heats ice 2.09 J/g^oC; water 4.18 J/g ^oC; steam 2.03 J/g ^oC

step 1: Heat ice from -10.0 ^oC to 0.0 ^oC
25.0 g x (0.0 --10.0 ^oC) x 2.09 J/g ^oC = 522 J

step 2: Melt the ice at 0.0 ^oC
25.0 g x 334 J/g = 8350 J

step 3: Heat the water from 0.0 ^oC to 100.0 ^oC
25.0 g x (100.0 - 0.0 ^oC) x 4.18 J/g ^oC = 10450 J

step 4: Vaporize the water to steam at 100.0 ^oC
25.0 g x 2260 J/g = 56500 J

step 5: Heat the steam from 100.0 ^oC to 110.0 ^oC
25.0 g x (110.0 -100.0 ^oC) x 2.03 J/g ^oC = 507 J

step 6: combine individual steps
Total heat absorbed = 522 + 8350 + 10450 + 56500 + 507 = 76329 J or 76.3 kJ

ch1312.doc (benzene)

Q. How much heat is required to convert
25.0 g of frozen benzene at -10.0 ^oC to vapor at 110.0 ^oC?

m. p. is 5.5 ⁰C, 127 J/g ; b. p. is 80.1 ^oC 395 J/g.

specific heats frozen benzene 1.10 J/g^oC; liquid 1.74 J/g ^oC; vapor 1.04 J/g ^oC

25.0 g x (5.5 --10.0 ^oC) x 1.10 J/g ^oC = 426 J

25.0 g x 127 J/g = 3174 J

25.0 g x (80.1 - 5.5 ^oC) x 1.74 J/g ^oC = 3245 J step 4:Vaporize liquid benzene to vapor at 80.1 ^oC

25.0 g x 395 J/g = 9875 J

25.0 g x (110.0 -80.1 ^oC) x 1.04 J/g ^oC = 777 J

Total heat absorbed =

426 + 3174 + 3245 + 9875 + 777

= 17497 J or 17.5 kJ

benzene C₆H₆ vs H₂O

ch1313.doc

Q. Suppose 275 g boiling hot coffee is diluted with 475 g cream at 30 ^oC. What is the final temperature T of the coffee.?

A. Assume coffee = cream = water (probably a very good approximation)

The answer is reasonable:
T << 100 ^oC and T > 30 ^oC

ch1313.doc

Q. Suppose 175 g of liquid water at 0^oC is treated with 17.5 g superheated steam at 110 ^oC.
The final temperature is T?

A. Heat is conserved

cubic: a crystalline form that has three equal axes at right angles to each other; isometric.

tetragonal: a crystal system characterized by three axes at right angles of which only the two lateral axes are equal

orthorhombic: crystallization characterized by three unequal axes at right angles to each other

monoclinic: crystal system characterized by three unequal axes with one oblique (not 90^o) intersection

triclinic: crystal system having three unequal axes intersecting at oblique angles

rhombohedral: a parallelepiped (pairs of parallel planes) whose faces are rhombuses (equal sided parallelogram)

hexagonal: a crystal system characterized by three equal lateral axes intersecting at angles

of 60 degrees and a vertical axis of variable length at right angles

ch13defs.doc

cubic: a crystalline form that has three equal axes at right angles to each other; isometric.

a = b = c

a = b = g = 90^o

ch13defs.doc

tetragonal: a crystal system characterized by three axes at right angles of which only the two lateral axes are equal

a = b ¹ c

a = b = g = 90^o

ch13defs.doc

orthorhombic: crystallization characterized by three unequal axes at right angles to each other

a ¹ b ¹ c

a = b = g = 90^o

ch13defs.doc

monoclinic: crystal system characterized by three unequal axes with one oblique (not 90^o) intersection

a ¹ b ¹ c

a = g = 90^o, b ¹ 90^o

ch13defs.doc

triclinic: crystal system having three unequal axes intersecting at oblique angles

a ¹ b ¹ c

a ¹ b ¹ g ¹ 90^o

ch13defs.doc

rhombohedral: a parallelepiped (pairs of parallel planes) whose faces are rhombuses (equal sided parallelogram)

a = b = c

a = b = g ¹ 90^o

ch13defs.doc

hexagonal: a crystal system characterized by three equal lateral axes intersecting at angles

of 60 degrees and a vertical axis of variable length at right angles

a = b ¹ c

a = b = 90^o, g = 120^o

ch13face.doc

Counting atoms in cubic crystals (atoms/unit cell)

1 Atom/cell

8 corners only (corners are shared by 8 cells)

2 Atoms/cell

8 corners + 1 in the center (centers are not shared)

4 atoms/cell
8 corners + 6 faces (faces are shared by two cells)

ch1314.doc problem examples
Crystal density (g/cm³)

Q. Silver metal crystallizes as a solid whose lattice is face centered cubic with an edge 4.086 Å . Compute the density of silver.

108 g x 1 mole = 1.79 x 10^-22 g/atom
1 mol 6.02x10²³ atoms

volume of unit cell
V = (4.086 x 10^-8 cm)³
= 6.82 x 10^-23 cm³

Number of Ag atoms/unit cell
8 corners = 1 atom
6 faces = 3 atoms
Total = 4 atoms/unit cell

Total mass in unit cell
4 atoms/unit cell x 1.79 x 10^-22 g/atom = 7.16 x 10^-22 g/unit cell

Density
D = mass/unit volume
= 7.16 x 10^-22 g/unit cell = 10.5 g/cm³
6.82 x 10^-23 cm³