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Phase Diagrams A phase diagram allows for the prediction of the state of matter at any given temperature & pressure. PowerPoint Presentation
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Phase Diagrams A phase diagram allows for the prediction of the state of matter at any given temperature & pressure. - PowerPoint PPT Presentation


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Phase Diagrams A phase diagram allows for the prediction of the state of matter at any given temperature & pressure. Key aspects: -critical point -normal boiling point -triple point. SPECIFIC HEAT re-visited. The quantity of heat required to raise the temperature of one

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slide2

Phase Diagrams

A phase diagram allows for the prediction of the state of matter at any given temperature & pressure.

Key aspects:

-critical point

-normal boiling point

-triple point

slide3

SPECIFIC HEAT re-visited

The quantity of heat required to raise the temperature of one

gram of a substance by one degree Celsius (or one Kelvin)

q = s x m xT

ENTHALPY OF A PHASE CHANGE

The heat energy required to undergo a change in phase

occurs at constant temperature and is associated with the average change in distance between molecules.

For water:

Hº fus = 335 J/g or 6.02 kJ/mol

Hºvap = 2260 J/g or 40.7 kJ/mol

slide4

HEATING - COOLING CURVE

Calculate the amount of energy required to convert

50.0 g of ice at 0.0 ºC to steam at 100.0ºC

Hvap

g

100 -

T (oC)

0 -

l

Hfus

s

Energy (J)

qtotal = q(s) + DHfus + q(l) + DHvap + q(g)

slide5

Water and the Changes of State

Q. How many kilojoules of energy are needed to change 15.0 g of ice at -5.00oC to steam at 125.0 oC?

The first step is to design a pathway:

q1 = msDT for ice from -5.0 to 0.0 oC, the specific heat of ice is 4.213 J/g oC

q2 = DHfus for ice to liquid at 0.0oC

q3 = msDT for liquid 0.0oC to 100.0 oC

q4 = DHvap for liquid to steam at 100.0oC

q5 = msDT for steam 100.0 to 125.0 oC; the specific heat of steam is 1.900 J/g oC

so qT = q1 + q2 + q3 + q4 + q5

The next step is to calculate each q:

q1= (15.0 g) (4.213 J/g oC) (0.0 - (-5.0) oC) = 316 J

q2 = (335 J / g) (15.0 g) = 5025 J

q3= (15.0 g) (4.184 J/g oC) (100.0 - (0.0) oC) = 6276 J

q4 = (2260 J / g) (15.0 g) = 33900 J

q5= (15.0 g) (1.900 J/g oC) (110 - 100 oC) = 285 J

qT = 316 J + 5025 J + 6276 J + 33900 J + 285 J = 45.8 kJ

slide6

INTERMOLECULAR FORCES

INTRAMOLECULAR > INTERMOLECULAR

(covalent, ionic) (van der Waals, etc)

“ between atoms” “between molecules”

TYPES

Neutral Molecules: 1. Dipole-dipole forces

2. London Dispersion

3. Hydrogen bonding

Ions: 1. Ion-dipole force

slide7

INTERMOLECULAR FORCES

STRENGTH:

BOILING POINTS AND MELTING POINTS ARE DEPENDENT ON STRENGTH OF INTERMOLECULAR FORCES.

STRONG FORCE  HIGH BOILING POINT

Type of interaction Approximate Energy

(kJ/mol)

Intermolecular

van der Waals 0.1 to 10

(London, dipole-dipole)

Hydrogen bonding 10 to 40

Chemical bonding

Ionic 100 to 1000

Covalent 100 to 1000

slide8

ION-DIPLE FORCES

- between ions and polar molecules

- strength is dependent on charge of the ions or polarity of the bonds

- usually involved with salts & H20

DIPOLE-DIPOLE FORCES

- between neutral polar molecules

- weaker force than ion-dipole

- positive dipole attracted to negative dipole

- molecules should be relatively close together

- strength is dependent on polarity of bonds.

LONDON DISPERSION FORCES

- all molecules and compounds

- involves instantaneous dipoles

- strength is dependent on Molar Mass (size)

- contributes more than dipole-dipole

- shape contributes to strength

slide9

HYDROGEN BONDING

AN INTERMOLECULAR ATTRACTION THAT EXISTS

BETWEEN A HYDROGEN ATOM IN A POLAR BOND

AND AN UNSHARED ELECTRON PAIR ON A NEARBY

ELECTRONEGATIVE SPECIES, USUALLY

O, F, and N

NOTE: (A special type of dipole-dipole interaction)

- stronger than dipole-dipole and London dispersion forces

- Accounts for water’s unusual properties

- high boiling point

- solid Less dense than liquid

- universal solvent

- high heat capacity

slide10

FLOWCHART OF INTERMOLECULAR FORCES

Interacting molecules or ions

Are polar Are ions Are polar

molecules involved? molecules

involved? and ionsboth

present?

Are hydrogen

atoms bonded to N,

O, or F atoms?

London forcesDipole-dipolehydrogen bondingIon-dipoleIonic

only (induced forcesforcesBonding

dipoles)

Examples: Examples: Examples Example: Examples:

Ar(l), I2(s) H2S, CH3Cl liquid and solid KBr in NaCl,

H2O, NH3, HF H2O NH4NO3

NO

NO

YES

NO

YES

Yes

YES

NO

Van der Waals forces

slide11

PROPERTIES OF LIQUIDS

VISCOSITY

- The resistance of a liquid to flow

- Depends on attractive forces between molecules

and structural features which cause greater

interaction (entanglement).

SURFACE TENSION

- The energy required to increase the surface area of a liquid by a unit amount (E/A)

- Due to interactions between molecules and the lack of interaction if there are no molecules to interact with.

slide12

VAPOR PRESSURE

The pressure exerted by a vapor in equilibrium with

its liquid or solid state.

1. Vapor pressure changes with intermolecular forces and temperature

2. Vapor pressure involves a dynamic equilibrium

liquidgas

3. Volatile vs nonvolatile

4. Clausius - Clapeyron equation

-relates vapor pressure and liquid temperature

slide13

CLAUSIUS - CLAPEYRON EQUATION

In general, the higher the temperature, the weaker the

intermolecular forces, and therefore the higher the vapor pressure. The non-linear relationship between vapor pressure and temperature is given by the Clausius - Clapeyron equation.

In P = (-Hvap/RT) + C : a straight line if lnP vs. 1/T

It describes the amount of energy required to vaporize 1 mole of

molecules in the liquid state

R = 8.31 J/mol K T = Kelvin P = vapor pressure

In P2 = -Hvap1 - 1

P1 R T2 T1

Q. The vapor pressure of ethanol at 34.9ºC is 100.0 mmHg and at 78.5ºC it’s 760.0 mmHg. What is the heat of vaporization of ethanol?

slide14

CYRSTALLINE SOLIDS

Type of solid lattice site Type of force properties of examples

particle type between particles solids

IONIC positive & electrostatic high M.P. NaCl

negative ions attraction nonvolatile Ca(NO3)2

hard & brittle

poor conductor

POLAR polar dipole-dipole & moderate M.P. Sucrose,

MOLECULAR molecules London Dispersion moderate C12H22O11

forces volatility Ice, H2O

NONPOLAR Nonpolar London Dispersion low M.P., Argon, Ar,

MOLECULAR molecules & forces volatile Dry Ice, CO2 atoms

MACRO- atoms covalent bonds extremely high Diamond, C

MOLECULAR between atoms M.P. nonvolatile Quartz, SiO2

Covalent- Arranged in Very Hard

Network Network Poor conductor

METALLIC metal atoms attraction between variable M.P. Cu, Fe

outer electrons low volatility Al, W

and positive good conductor

atomic centers

slide15

TYPE OF MELTING POINT HARDNESS ELECTRICAL

SOLID OF SOLID & BRITTLENESS CONDUCTIVITY

Molecular Low soft & brittle Nonconducting

Metallic Variable Variable hardness, conducting

malleable

Ionic High to very hard & brittle Nonconducting

high solid

(conducting

liquid)

Covalent Very high Very hard Usually

Network nonconducting

slide16

CRYSTALLINE SOLIDS

- Composed of crystal lattices

- the geometric arrangement of lattice points of a crystal consists of unit cells

- smallest unit from which atoms can be stacked in 3-D

- unit cells (there are different types; see transparencies)

- edge lengths and angles are used to describe the unit cell

3 types of unit cells

- primitive (simple)

- body center cubic

- face centered cubic

- metals and salts are usually cubic

Ni = FCC Na=BCC NaCl=FCC

In FCC corners and face are shared with other units

Question: Determine the net number of ions in LiF (FCC)

Li+ = 1/4 (Li per edge) (12 edges) = 3

1 (center) (1 center) = 1

F- = 1/8 (per corner) (8 corners) = 1

1/2 (face) (6 faces) = 2

slide17

MOLECULAR SOLIDS

- frozen noble gases, Ne

- needs large number of atoms surrounding center for maximum attraction (see transparency)

- close packing arrangement variations:

-- hexagonal close-packing structure (ABABABA…); 6-unit cell (hcp)

-- cubic close-packed structure (ccp); ABC ABC ABC ...

Similar to FCC

Coordination number = the number of nearest neighbors; 12 for close-pack structures

METALLIC SOLIDS

- sea of electrons; delocalized (bonding is non-directional)

-many metals are cubic or hexagonal close-packed crystals

COVALENT NETWORK

-directional covalent bonds

-diamond, Si, Ge, gray Sn are tetrahedral, sp3 hybridized, FCC

-graphite is hexagonal flat sheets; sp2 hybridized; electrical properties are due to the increased delocalization of the electrons.

slide18

X-RAY DIFFRACTION

- Determining crystal structure

x-ray diffraction used to obtain structure of

proteins (1962 Nobel Prize myoglobin) &

hemoglobin

- Due to order structure - crystals consists of repeating planes

- Planes act as reflecting surfaces

- X-ray reflecting off surface creates diffraction pattern

constructive interference gives more intense (higher

amplitude) weaves.

- only at certain angles will x-rays stay in-phase

- creates light and dark areas on photographs

- used to determine type of unit cell and size

- if molecular, then can determine position of each atom

nl = 2d sin q Bragg’s equation

d=distance between atomic planes q =angle of reflection