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### Intermolecular Forces of Attraction

MATTER

Kinetic Molecular Theory

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

Kinetic Theory Facts

- 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

Definitions liquids exist at lower temperatures, thus allowing other forces of attraction to act upon them

- 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

The Condensed Phases liquids exist at lower temperatures, thus allowing other forces of attraction to act upon them

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

Dipole-dipole forces liquids exist at lower temperatures, thus allowing other forces of attraction to act upon them

- 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

H bonding liquids exist at lower temperatures, thus allowing other forces of attraction to act upon them

- 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

London dispersion forces liquids exist at lower temperatures, thus allowing other forces of attraction to act upon them

- 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

Ion-dipole forces 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: is, it is still not stronger than attractions involving ions

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 is, it is still not stronger than attractions involving ions

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 is, it is still not stronger than attractions involving ions+

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 is, it is still not stronger than attractions involving ions

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 is, it is still not stronger than attractions involving ions

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: is, it is still not stronger than attractions involving ions

MOLECULES MOVING THRU MOLECULES

VISCOSITY:

RESISTANCE TO FLOW

SURFACE

TENSION:

GAS (VAPOR) is, it is still not stronger than attractions involving ions

LIQUID

SYSTEM ENERGY

SYSTEM ENERGY

SOLID

CONDENSATION

VAPORIZATION

SUBLIMATION

DEPOSITION

MELTING

(FUSION)

FREEZING

MELTING POINT is, it is still not stronger than attractions involving ions

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 is, it is still not stronger than attractions involving ions

Patm

/A = PRESSURE

F

Dh

14.7 lb/in2 = 1 ATM = 760

mm Hg

torr

GAS

P = Patm - Dh

1 atm is, it is still not stronger than attractions involving ions

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) is, it is still not stronger than attractions involving ions

OR P a 1/V

BOYLE’S

LAW

VOLUME

PRESSURE

VOLUME is, it is still not stronger than attractions involving ions

CONSTANT PRESSURE!!!

CHARLES’ LAW

V = k(n, P) x T

OR V a T

TEMPERATURE

n = 4 is, it is still not stronger than attractions involving ions

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: is, it is still not stronger than attractions involving ions 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 is, it is still not stronger than attractions involving ions

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 is, it is still not stronger than attractions involving ions 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 is, it is still not stronger than attractions involving ions

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 is, it is still not stronger than attractions involving ions

DYNAMIC EQUILIBRIUM!

EVAPORATION

VAPOR PRESSURE

ALL NON-GASEOUS

MATERIALS EXERT A

VAPOR PRESSURE.

FOR SOLIDS: VERY LOW

ASSUMED TO BE 0

760 is, it is still not stronger than attractions involving ions

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 is, it is still not stronger than attractions involving ions

P,

a

t

m

T, oC

BOILING POINT

LIQUID

MELTING POINT

SOLID

VAPOR (GAS)

TRIPLE POINT

INTERMOLECULAR is, it is still not stronger than attractions involving ions

>

<

CAPILLARY ACTION

ADHESIVE VS COHESIVE FORCES

ADHESIVE COHESIVE

IONIC is, it is still not stronger than attractions involving ions

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: is, it is still not stronger than attractions involving ions

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 is, it is still not stronger than attractions involving ions

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 is, it is still not stronger than attractions involving ions

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 is, it is still not stronger than attractions involving ions

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 is, it is still not stronger than attractions involving ions

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% is, it is still not stronger than attractions involving ions

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 is, it is still not stronger than attractions involving ions

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 is, it is still not stronger than attractions involving ions

EMPTY

ORBITALS

FERMI

LEVEL

BAND

GAP

FILLED

ORBITALS

NON-

CONDUCTOR

SEMI-

CONDUCTOR

METAL

EXTENDED SOLIDS is, it is still not stronger than attractions involving ions

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

GASES is, it is still not stronger than attractions involving ions

Importance of Gases is, it is still not stronger than attractions involving ions

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

THREE STATES OF MATTER is, it is still not stronger than attractions involving ions

General Properties of Gases is, it is still not stronger than attractions involving ions

- 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.

Properties of Gases is, it is still not stronger than attractions involving ions

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 is, it is still not stronger than attractions involving ions

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

Pressure is, it is still not stronger than attractions involving ions

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!

Pressure is, it is still not stronger than attractions involving ionsConversions

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

Pressure Conversions is, it is still not stronger than attractions involving ions

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?

Boyle is, it is still not stronger than attractions involving ions’s Law

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.

Boyle is, it is still not stronger than attractions involving ions’s Law and Kinetic Molecular Theory

P proportional to 1/V

Boyle is, it is still not stronger than attractions involving ions’s Law

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.

Charles is, it is still not stronger than attractions involving ions’s Law

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 is, it is still not stronger than attractions involving ions’s original balloon

Modern long-distance balloon

Charles is, it is still not stronger than attractions involving ions’s Law

Gay-Lussac is, it is still not stronger than attractions involving ions’s Law

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)

Gas Pressure, Temperature, and Kinetic Molecular Theory is, it is still not stronger than attractions involving ions

P proportional to T

Combined Gas Law is, it is still not stronger than attractions involving ions

- 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

Combined Gas Law is, it is still not stronger than attractions involving ions

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

Combined Gas Law Problem is, it is still not stronger than attractions involving ions

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 = ??

Calculation is, it is still not stronger than attractions involving ions

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

Learning Check is, it is still not stronger than attractions involving ions

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?

One More Practice Problem is, it is still not stronger than attractions involving ions

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?

And now, we pause for this commercial message from STP is, it is still not stronger than attractions involving ions

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

Try This One is, it is still not stronger than attractions involving ions

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 is, it is still not stronger than attractions involving ions

Avogadro’s HypothesisEqual 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.

Avogadro is, it is still not stronger than attractions involving ions’s Hypothesis and Kinetic Molecular Theory

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

P proportional to n

IDEAL GAS LAW is, it is still not stronger than attractions involving ions

P V = n R T

Brings together gas properties.

Can be derived from experiment and theory.

BE SURE YOU KNOW THIS EQUATION!

Using PV = is, it is still not stronger than attractions involving ionsnRT

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

Using PV = is, it is still not stronger than attractions involving ionsnRT

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 is, it is still not stronger than attractions involving ions

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 N is, it is still not stronger than attractions involving ions2 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)

Learning Check is, it is still not stronger than attractions involving ions

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?

Learning Check is, it is still not stronger than attractions involving ions

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?

Deviations from is, it is still not stronger than attractions involving ionsIdeal Gas Law

- 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.

Gases in the Air is, it is still not stronger than attractions involving ions

The % of gases in air Partial pressure (STP)

78.08% N2 593.4 mm Hg

20.95% O2 159.2 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 Pressure 760 mm Hg

Dalton is, it is still not stronger than attractions involving ions’s Law of Partial Pressures

- 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.

Health Note is, it is still not stronger than attractions involving ions

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.

Collecting a gas is, it is still not stronger than attractions involving ions“over water”

- 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.

Table of Vapor Pressures for Water is, it is still not stronger than attractions involving ions

Solve This! is, it is still not stronger than attractions involving ions

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 is, it is still not stronger than attractions involving ions

density

High

density

GAS DENSITY22.4 L of ANY gas AT STP = 1 mole

Gases and Stoichiometry is, it is still not stronger than attractions involving ions

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.

Gases and Stoichiometry is, it is still not stronger than attractions involving ions

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

Gas Stoichiometry: Practice is, it is still not stronger than attractions involving ions!

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?

What if it is, it is still not stronger than attractions involving ions’s NOT 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)

Try this one! is, it is still not stronger than attractions involving ions

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.

GAS DIFFUSION AND EFFUSIONGAS DIFFUSION AND EFFUSION gases.

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.

GAS DIFFUSION AND EFFUSION gases.

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

Gas Diffusion gases.relation of mass to rate of diffusion

- 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.

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