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STATES OF MATTER

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STATES OF

MATTER

Intermolecular Forces of Attraction

- All matter is composed of atoms that are in constant motion

- All phases of matter express the degree that they reflect the kinetic theory through their kinetic energy
- kinetic energy is measured by temperature
- phase changes involve changes in temperature due to the existence threshold temperature of each phase (i.e. ice naturally is found at cold not hot temperatures)

- While gases have a great deal of random motion, solids and liquids exist at lower temperatures, thus allowing other forces of attraction to act upon them
- these forces are the van der Waals forces

- Bonds are intramolecular forces of attraction
- Forces of attraction between molecules are called intermolecular forces of attraction
- intermolecular forces of attraction are commonly called van der Waals forces

- Solids and Liquids
- Physical properties of the condensed phases reflect the degree of intermolecular forces (i.e. boiling point)

- Exist between neutral polar molecules
- work best the closer the molecules are to each other
- the greater the polarity of the molecules, the greater the force of attraction

- Special case of dipole-dipole interaction specifically between H of one polar molecule with N, O or F and an unshared electron pair of another nearby small electronegative ion (usually N, O, or F on another molecule)
- VERY STRONG

- Induced dipoles
- not really dipoles on the AVERAGE, but instantaneously dipole conditions can exist thus allowing for pseudopolar regions to occur

- No matter how strong the van der Waal force of attraction is, it is still not stronger than attractions involving ions

- Attraction between ions and the partial charge on the end of a polar molecule
- ex. NaCl in water solution

INTRAMOLECULAR:

INTERMOLECULAR:

FORCES WITHIN A MOLECULE

FORCES

FORCES BETWEEN MOLECULES

STRONG INTRAMOLECULMR FORCES

O = O

VERY WEMK INTERMOLECULAR FORCES

MUCH WEAKER THAN EITHER

IONIC OR COVALENT BONDS

ELECTROSTATIC

DICTATE WHETHER MOLEULAR

SUBTANCE IS GAS, LIQUID OR SOLID

AT ROOM CONDITIONS

GAS: NEGLIGIBLE

LIQUID: WEAK TO MODERATE

SOLID: MODERATE TO STRONG

HELP DEFINE STANDARD STATE OF

A SUBSTANCE

INTERMOLECULAR

FORCES:

HELP DEFINE COMMON PHYSICAL

PROPERTIESOF A SUBSTANCE

ARE MUCH WEAKER THAN BONDS

(INTRA MOLECULAR FORCES)

BOILING/MELTING POINTS

SURFACE TENSION

CAPILLARY ACTION

VISCOSITY

IONIC BONDS:

~300 - 1000’S kJ/MOL

COVALENT BONDS:

~150 - ~800 kJ/MOL

INTERMOLCULAR FORCES:

~1 - 40 kJ/MOL

STRENGTH DIMINISHES WITH INCREASING DISTANCE

d+

d-

d+

d-

d-

d+

d-

d+

d-

d+

IONIC COMPOUNDS:

MOLECULAR

COMPOUNDS:

MOLECULAR

DIPOLE

“INDUCED” TEMPORARY DIPOLE

DISPERSION

FORCES

PRESENT IN ALL MOLECULAR SUBSTANCES!

INCREASES WITH MOLECULAR MASS

F2

Cl2

I2

-188 oC

58.8

184

Cl

Cl C

Cl

O

H H

O C O

Cl

F

d+

d+

DIPOLE-DIPOLE FORCES

PERMANENT DIPOLE

NATURAL ASSYMETRIC CHAGE DISTRIBUTION

POLAR

MOLECULE

OCCURS IF CENTERS OF CHARGES DO NOT COINCIDE

2d-

THE MORE POLAR THE MOLECULE,

THE STRONGER THEINTERMOLECULAR FORCE

CH3 - CH2 - CH3 CH3CN

44 = MASS = 41

0.1 = DIPOLE MOMENT = 3.9

231 = BOILING POINT (K) = 355

SPECIAL CASE OF DIPOLE FORCES

HYDROGEN

BOND

H ATOM ATTACHED TO F, O, N

LARGE ELECTRONEGATIVITY DIFFERENCE

SMALL SIZE ALLOWS H TO GET CLOSE

H2O

H2S

H2Se

H2Te

18

34

81

130

~ - 73

oC

- 60.7

-41.5

-2

100

THIS PROPERTY AFFECTS

LIFE AND MANY OTHER

PROPERTIES

DIFFUSION:

MOLECULES MOVING THRU MOLECULES

VISCOSITY:

RESISTANCE TO FLOW

SURFACE

TENSION:

GAS (VAPOR)

LIQUID

SYSTEM ENERGY

SYSTEM ENERGY

SOLID

CONDENSATION

VAPORIZATION

SUBLIMATION

DEPOSITION

MELTING

(FUSION)

FREEZING

MELTING POINT

DHf = HEAT REQUIRED TO MELT

SUBSTANCEAT ITS MELTING POINT

HEAT REQUIRED TO BREAK DOWN

INTERMOLECULAR FORCES

> 0 J/g OR J/mol

VAPOR

LIQUID +

VAPOR

LIQUID

ENTHALPY

TEMPERATURE

LIQUID +

SOLID

SOLID

FREEZING POINT

TIME

bar

Patm

/A = PRESSURE

F

Dh

14.7 lb/in2 = 1 ATM = 760

mm Hg

torr

GAS

P = Patm - Dh

1 atm

760 mm Hg

535 mm Hg x

760 mm Hg

1 atm

0.645 atm x

1 atm

760 torr

480 torr x

WHAT IS THE PRESSURE OF A GAS (ATM) IF IT SUPPORTS

A COLUMN OF MERCURY TO A HEIGHT OF 535 mm?

= 0.704 atm

IF THE ATMOSPHERIC PRESSURE DROPS TO 0.645 atm,

HOW HIGH IS THE COLUMN OF MERCURY SUPPORTED?

= 490 OR 4.9 x 102 mm Hg

IF THE BAROMETRIC PRESSURE IS 755 mm Hg AND A GAS

CREATES A Dh OF 275 mm IN A MANOMETER, WHAT IS

THE PRESSURE OF THE GAS?

P = Patm - Dh

= 755 - 275

= 480 mm Hg

= 480 torr

= 0.632 atm

P x V = k (n, T)

OR P a 1/V

BOYLE’S

LAW

VOLUME

PRESSURE

VOLUME

CONSTANT PRESSURE!!!

CHARLES’ LAW

V = k(n, P) x T

OR V a T

TEMPERATURE

n = 4

n = 2

VOLUME

TEMPERATURE

0 oC

ABSOLUTE OR KELVIN SCALE

K = oC + 273

-273.15 oC

n = 1.0

n = 0.5

USE IN ALL CALCULATIONS!!!

AVOGADRO: EQUAL VOLUMES OF GASES AT SAME

T & P CONTAIN EQUAL NUMBER OF MOLECULES!

V = k(n, T) x n

P x V = k (n, T)

V = k(n, P) x T

IDEAL GAS

LAW

PV = nRT

PV = nkT

R = 0.0821 L.atm.mol-1.K-1

1.5 atm x 1473 K

298 K

1 mol x 0.0821 L.atm.mol-1.K-1 x 273 K

1.0 atm

nRT

P

THE PRESSURE IN AN AEROSOL CAN AT 25 oC IS 1.5 atm.

A FIRE CAN REACH 1200 oC. WHAT IS THE PRESSURE OF

THE CAN AT THAT TEMPERATURE?

P V = n R T

INIT

FINAL

1.5

Vi

n

0.0821

25 + 273

X

Vi

n

0.0821

1200 + 273

Pf / Tf =

R=

Pi / Ti

Pi / Ti

STP

Pf =

7.4 atm

WHAT IS THE V OF 1.0 MOL GAS AT 1.0 atm AND 0 oC?

= 22.4 L

V =

IDEAL VS. REAL GAS

MOLECULES FAR APART

FAR APART

NO COLLISIONS

SOME COLLISIONS

LOW INTERACTIONS

NO INTERACTIONS

OCCUPY NO SPACE &

HAVE NO VOLUME

MATTER; MUST OCCUPY

SPACE & HAVE VOLUME

PV/RT = 1

PV/RT > 1

SINCE GASES ARE REAL:

- CAN NEVER ACHIEVE ABSOLUTE 0

- APPROACH IDEAL GAS AT HIGH VOLUMES

LOW P, HIGH T

KINETIC MOLECULAR THEORY

1. THE VOLUME OF GAS MOLECULES IS NEGLIBLE

COMPARED TO THE VOLUME OF THE CONTAINER

2. PARTICLES UNDERGO CONSTANT RANDOM MOTION

AND DO NOT INTERACT WITH ONE ANOTHER

3. AVERAGE KINETIC ENERGY OF THE PARTICLES IS

PROPORTIONAL TO ABSOLUTE TEMPERATURE

TEMPERATURE IS THE MEASURE OF THE AVG.

KINETIC ENERGY OF THE PARTICLES IN THE SYSTEM

E ~ RT

R = 8.314 J.mol-1.K-1

LIQUID VAPOR EQUILIBRIA

DYNAMIC EQUILIBRIUM!

EVAPORATION

VAPOR PRESSURE

ALL NON-GASEOUS

MATERIALS EXERT A

VAPOR PRESSURE.

FOR SOLIDS: VERY LOW

ASSUMED TO BE 0

760

VAPOR

PRESSURE

(mm Hg)

VOLATILE

NON-VOLATILE

DHVAP

AMOUNT OF HEAT

REQUIRED TO

VAPORIZE SOME

AMOUNT OF LIQUID

T, oC

NORMAL BOILING POINT

VS BOILING POINT

1.0

P,

a

t

m

T, oC

BOILING POINT

LIQUID

MELTING POINT

SOLID

VAPOR (GAS)

TRIPLE POINT

INTERMOLECULAR

>

<

CAPILLARY ACTION

ADHESIVE VS COHESIVE FORCES

ADHESIVE COHESIVE

IONIC

DISPERSION, DIPOLE

& H-BONDING

DISPERSION + DIPOLE

OR ANY OF THE PHYSICAL

PROPERTIES

DISPERSION ONLY

X

BOILING

POINT

DISPERSION: INCREASING MASS

DIPOLE-DIPOLE: INCREASING POLARITY

HYDROGEN BOND: INCREASING “NUMBER”

X

X

X

INTERMOLECULAR FORCES

WHAT FORCES ARE PRESENT IN:

A) NaBr

B) NF3

C) CH2OHCH2CH2OH

D) Ar

IONIC

DISPERSION + DIPOLE-DIPOLE

DISPERSION + DIPOLE + H-BOND

DISPERSION

WHICH OF THE FOLLOWING HAVE THE HIGHER BP?

A) C6H14 C10H22 C2H6

B) C6H14 C6H13OH C6H12(OH)2

C) CCl4 CCl3F CF4

D) HCl HF F2

UNIT CELLS

THE REPEATING PATTERN IN A

THREE DIMENSIONAL ARRAY

CRYSTAL LATTICE:

A A A A

A A A A A

A A A A

A A A A A

A A A A

A A A A A

NOTE: THIS ATOM

IS SHARED BY MORE

THAN ONE UNIT CELL

NOT 5 ATOMS PER CELL

IS 1 + 1/4(4) = 2 FULL ATOMS

CONSIDER FACES, EDGES,

CORNERS AND THOSE

TOTALLY WITHIN CELL

CRYSTALLINE SOLIDS

AMORPHOUS SOLIDS

WELL DEFINED POSTIONS

FOR EMCH ATOM

ORDERED REPETITION OF

PATTERN

LONG RANGE ORDER!

ILL DEFINED POSTIONS

ORDER EXTENDS OVER SHORT RANGE

LOCAL ORDER!

6 FACES

12 EDGES

EIGHT CORNERS

UNIT CELL STOICHIOMETRY

ATOM LOCATED ENTIRELY WITHIN CELL CONTRIBUTES

1 FULL ATOM TO CELL STOICHIOMETRY

2 CELLS

4 CELLS

8 CELLS

FACE ATOM CONTRIBUTES 1/2 x 6 = 3 ATOMS TO UNIT CELL

EDGE ATOM CONTRIBUTES 1/4 x 12 = 3 ATOMS TO UNIT CELL

CORNER ATOM CONTRIBUTES 1/8 x 8 = 1 ATOM TO UNIT CELL

WHAT ARE THESE UNIT CELLS?

c

g

b

a

b

a

SIMPLE CUBIC (SC)

a = b = c

a = b = c = 90o

SC = 1 ATOM/UNIT CELL

BODY CENTERED CUBIC (bcc)

CONTAINED WITHIN CELL

BCC = 2 ATOMS/UNIT CELL

FACE CENTERED CUBIC (fcc)

FCC = 4 ATOMS/UNIT CELL

52%

68%

74%

ATOMS/UNIT CELL x MASS ATOM

SIDE3

MASS OF SUBSTANCE

VOLUME OF SUBSTANCE

COORDINATION NUMBER (CN) OR GEOMETRY

4 PARTICLES CONNECTED TO CENTRAL ATOM

= TETRAHEDRON

CN = 4

CN = 8

CN = 6

CN = 12

PACKING EFFICIENCY: % OF UNIT CELL OCCUPIED BY

ATOMS, IONS OR MOLECULES

DENSITY IS DETERMINED BY HOW CLOSE (EFFICIENCY)

PARTICLES ARE PACKED INTO A UNIT CELL

DENSITY IS A MEASURE OF HOW CONCENTRATED IS THE

MASS OF A PURE SUBSTANCE

....OR HOW TIGHTLY PACKED

d =

METALLIC

IONIC

COVALENT

MOLECULAR

FOUR TYPES OF CRYSTALLINE SOLIDS:

COVALENT TYPE BOND IN METALS

METALLIC:

DELOCALIZED “SEA” OF ELECTRONS

ENERGY

BAND

2 MOLECULAR ORBITALS

2 ATOMIC ORBITALS

16

6 x 1023

4

8

ENERGY

EMPTY

ORBITALS

FERMI

LEVEL

BAND

GAP

FILLED

ORBITALS

NON-

CONDUCTOR

SEMI-

CONDUCTOR

METAL

EXTENDED SOLIDS

IONIC

CRYSTALS

HIGH MELTING & BOILING POINTS

IONIC COMPOUNDS

LARGE NETWORKS

COVALENT

SOLIDS

HIGH MELTING, HARD SOLIDS

DIAMOND, MOST SEMICONDUCTORS, SiO2

MOLECULAR

SOLIDS

INDIVIDUAL MOLECULES IN A LATTICE

LOW MELTING, SOFT

ICE, SUGAR, IODINE, SOLID HYDROGEN

- Airbags fill with N2 gas in an accident.
- Gas is generated by the decomposition of sodium azide, NaN3.
- 2 NaN3 ---> 2 Na + 3 N2

- There is a lot of “free” space in a gas.
- Gases can be expanded infinitely.
- Gases fill containers uniformly and completely.
- Gases diffuse and mix rapidly.

Gas properties can be modeled using math. Model depends on—

- V = volume of the gas (L)
- T = temperature (K)
- ALL temperatures in the entire chapter MUST be in Kelvin!!! No Exceptions!

- n = amount (moles)
- P = pressure (atmospheres)

Pressure of air is measured with a BAROMETER (developed by Torricelli in 1643)

Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). (Just like a straw in a soft drink)

P of Hg pushing down related to

- Hg density
- column height

Column height measures Pressure of atmosphere

- 1 standard atmosphere (atm) *
= 760 mm Hg (or torr) *

= 14.7 pounds/in2 (psi)

= 101.3 kPa (SI unit is PASCAL)

= about 34 feet of water!

* Memorize these!

A. What is 475 mm Hg expressed in atm?

1 atm

760 mm Hg

B. The pressure of a tire is measured as 29.4 psi.

What is this pressure in mm Hg?

760 mm Hg

14.7 psi

475 mm Hg x

= 0.625 atm

29.4 psi x

= 1.52 x 103 mm Hg

A. What is 2 atm expressed in torr?

B. The pressure of a tire is measured as 32.0 psi.

What is this pressure in kPa?

P α 1/V

This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down.

P1V1 = P2 V2

Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.

P proportional to 1/V

A bicycle pump is a good example of Boyle’s law.

As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.

If n and P are constant, then V α T

V and T are directly proportional.

V1 V2

=

T1 T2

- If one temperature goes up, the volume goes up!

Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist.

Charles’s original balloon

Modern long-distance balloon

If n and V are constant, then P α T

P and T are directly proportional.

P1 P2

=

T1 T2

- If one temperature goes up, the pressure goes up!

Joseph Louis Gay-Lussac (1778-1850)

P proportional to T

- The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION!
P1 V1 P2 V2

=

T1 T2

No, it’s not related to R2D2

If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law!

=

P1

V1

P2

Boyle’s Law

Charles’ Law

Gay-Lussac’s Law

V2

T1

T2

A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

Set up Data Table

P1 = 0.800 atm V1 = 180 mL T1 = 302 K

P2 = 3.20 atm V2= 90 mL T2 = ??

P1 = 0.800 atm V1 = 180 mL T1 = 302 K

P2 = 3.20 atm V2= 90 mL T2 = ??

P1 V1 P2 V2

= P1 V1T2 = P2 V2 T1

T1T2

T2 = P2 V2 T1

P1 V1

T2 = 3.20 atm x 90.0 mL x 302 K 0.800 atm x 180.0 mL

T2 = 604 K - 273 = 331 °C

= 604 K

A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?

OK, so it’s really not THIS kind of STP…

STP in chemistry stands for Standard Temperature and Pressure

Standard Pressure = 1 atm (or an equivalent)

Standard Temperature = 0° C (273 K)

STP allows us to compare amounts of gases between different pressures and temperatures

A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?

twice as many molecules

Equal volumes of gases at the same T and P have the same number of molecules.

V = n (RT/P) = kn

V and n are directly related.

The gases in this experiment are all measured at the same T and V.

P proportional to n

P V = n R T

Brings together gas properties.

Can be derived from experiment and theory.

BE SURE YOU KNOW THIS EQUATION!

P = Pressure

V = Volume

T = Temperature

N = number of moles

R is a constant, called the Ideal Gas Constant

Instead of learning a different value for R for all the possible unit combinations, we can just memorize one value and convert the units to match R.

R = 0.0821

L • atm

Mol • K

How much N2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to 745 mm Hg at 25 oC?

Solution

1. Get all data into proper units

V = 27,000 L

T = 25 oC + 273 = 298 K

P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm

And we always know R, 0.0821 L atm / mol K

How much N2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC?

Solution

2. Now plug in those values and solve for the unknown.

PV = nRT

RT RT

n = 1.1 x 103 mol (or about 30 kg of gas)

Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mm Hg) in the tank in the dentist office?

A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?

- Real molecules have volume.
The ideal gas consumes the entire amount of available volume. It does not account for the volume of the molecules themselves.

- There are intermolecular forces.
An ideal gas assumes there are no attractions between molecules. Attractions slow down the molecules and reduce the amount of collisions.

- Otherwise a gas could not condense to become a liquid.

The % of gases in air Partial pressure (STP)

78.08% N2593.4 mm Hg

20.95% O2159.2 mm Hg

0.94% Ar 7.1 mm Hg

0.03% CO2 0.2 mm Hg

PAIR = PN + PO + PAr + PCO = 760 mm Hg

2 2 2

Total Pressure760 mm Hg

- 2 H2O2 (l) ---> 2 H2O (g) + O2 (g)
- 0.32 atm 0.16 atm

What is the total pressure in the flask?

Ptotal in gas mixture = PA + PB + ...

Therefore,

Ptotal = PH2O + PO2 = 0.48 atm

Dalton’s Law: total P is sum of PARTIAL pressures.

John Dalton

1766-1844

When a scuba diver is several hundred feet under water, the high pressures cause N2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O2 in scuba tanks used for deep descents.

- Gases, since they mix with other gases readily, must be collected in an environment where mixing can not occur. The easiest way to do this is under water because water displaces the air. So when a gas is collected “over water”, that means the container is filled with water and the gas is bubbled through the water into the container. Thus, the pressure inside the container is from the gas AND the water vapor. This is where Dalton’s Law of Partial Pressures becomes useful.

A student collects some hydrogen gas over water at 20 degrees C and 768 torr. What is the pressure of the H2 gas?

768 torr – 17.5 torr = 750.5 torr

Low

density

High

density

22.4 L of ANY gas AT STP = 1 mole

2 H2O2 (l) ---> 2 H2O (g) + O2 (g)

Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP?

Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

2 H2O2 (l) ---> 2 H2O (g) + O2 (g)

Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP?

Solution

1.1 gH2O2 1 mol H2O2 1 mol O2 22.4 L O2

34 g H2O2 2 mol H2O2 1 mol O2

= 0.36 L O2 at STP

A. What is the volume at STP of 4.00 g of CH4?

B. How many grams of He are present in 8.0 L of gas at STP?

- 1. Do the problem like it was at STP. (V1)
- 2. Convert from STP (V1, P1, T1) to the stated conditions (P2, T2)

How many L of O2 are needed to react 28.0 g NH3 at24°C and 0.950 atm?

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

diffusion is the gradual mixing of molecules of different gases.

effusion is the movement of molecules through a small hole into an empty container.

Graham’s law governs effusion and diffusion of gas molecules.

Rate of effusion is inversely proportional to its molar mass.

Thomas Graham, 1805-1869. Professor in Glasgow and London.

Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is

- proportional to T
- inversely proportional to M.
Therefore, He effuses more rapidly than O2 at same T.

He

- HCl and NH3 diffuse from opposite ends of tube.
- Gases meet to form NH4Cl
- HCl heavier than NH3
- Therefore, NH4Cl forms closer to HCl end of tube.