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Vanessa N. Prasad-PermaulValencia Community College

CHM 1046

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In a gas, the particles of matter are far apart and are uniformly distributed throughout the container.

Gases have an indefinite shape and assume the shape of their container.

Gases can be compressed and have an indefinite volume.

Gases have the most energy of the three states of matter.

Gases are in constant random motion (Kinetic-Molecular Theory).

STATES OF MATTER

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In a liquid, the particles of matter are loosely packed and are free to move past one another.

Liquids have an indefinite shape and assume the shape of their container.

Liquids cannot be compressed and have a definite volume.

Liquids have less energy than gases but more energy than solids.

Liquids are also in constant random motion, but are more tightly packed (less free space).

STATES OF MATTER

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In a solid, the particles of matter are tightly packed together.

Solids have a definite, fixed shape.

Solids cannot be compressed and have a definite volume.

Solids have the least energy of the three states of matter.

Solids do NOT move about freely, but oscillate or vibrate about fixed sites, explaining the rigidity.

STATES OF MATTER

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STATES OF MATTER

RECALL that gases normally follow theIdeal Gas Law: PV = nRT

Van der Waals Equation: (P + n2a)*(V-nb) = nRT

V2

Where: a is intermolecular forces b is molecular size

Liquids and solids are different from gases in that they have strong attractive forces

between particles6

STATES OF MATTER

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PHASE TRANSITIONS

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When a solid changes to a liquid, the phase change is called melting.

A substance melts as the temperature increases.

H2O(s) H2O(l) (melting,

fusion)

When a liquid changes to a solid, the phase change is called freezing.

A substance freezes as the temperature decreases.

H2O(l) H2O(s) (freezing)

PHASE TRANSITIONS

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When a liquid changes to a gas, the phase change is called vaporization.

H2O(l) H2O(g) (vaporization)

PHASE TRANSITIONS

A substance vaporizes as the temperature increases.

When a gas changes to a liquid, the phase change is called condensation.

H2O(l) H2O(g) (condensation)

A substance condenses as the temperature decreases.

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When a solid changes directly to a gas, the phase

change is called sublimation.H2O(s) H2O(g) (sublimation)

A substance sublimes as the temperature increases.

When a gas changes directly to a solid, the phasechange is called deposition.H2O(g) H2O(s) (deposition)

A substance undergoes deposition as the temperature decreases.

PHASE TRANSITIONS

Physical form changes but chemical identity does not change :

Fusion (melting) solid liquidFreezing liquid solidVaporization liquid gasCondensation gas liquidSublimation solid gas

Deposition gas solid

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PHASE TRANSITIONS

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PHASE TRANSITIONS

Vapor pressure of a liquid is the partial pressure of the vapor over the liquid, measured at equilibrium at a given

temperature.

Numerical value of Vapor Pressure depends on:

a)      Magnitude of intermolecular forcesThe smaller the forces the higher

the vapor pressure, loosely held molecules escape easily

 b)      Temperature

The higher the temperature, the higher the vapor pressure, larger fraction of molecules have sufficient kinetic energy to escape

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PHASE TRANSITIONS

The higher the temperature and the lower the boiling point of the substance the greater the fraction of molecules in the sample that have sufficient kinetic energy to break free from the surface of the liquid and escape into the vapor.

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PHASE TRANSITIONS

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PHASE TRANSITIONS

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PHASE TRANSITIONS

Boiling Point: The temperature at which the vaporpressure of liquid equals the pressure exerted on theliquid .

H2O(l) H2O(g)

Freezing Point: The temperature at which a pure liquid changes to a crystalline solid, or freezes.

Solid Liquid

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PHASE TRANSITIONS

Heat of Fusion: the heat needed for melting of a solid, (Hfus).

H2O(s) H2O(l) ; Hfus = 6.01kj/mol

Heat of Vaporization: the heat needed for the evaporation of liquid, (Hvap).

H2O(s) H2O(l) ; Hvap = 40.7kj/mol

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PHASE TRANSITIONS

HEATING CURVE FOR WATER

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EXAMPLE 11.1: A particular refrigerator cools by evaporating liquefied dichlorodifluoromethane. How many kg of this liquid must be evaporated to freeze a tray of water @ 0oC. The mass of the water is 525g, the heat of fusion of ice is 6.01kj/mol, and the heat of vaporization of CCl2F2 is 17.4kj/mol.

525g H2O x 1mol H2O x -6.01kj = -175kj 18.0g H2O 1 mol H2O

175kj x 1 mol CCl2F2 x 121g CCl2F2 = 1.22 x 103g CCl2F2

1.22kg of CCl2F2 must be evaporated in order to freeze 525g of water

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EXERCISE 11.1: The heat of vaporization of ammonia is 23.4kj/mol. How much heat is required to vaporize 1.00kg of ammonia? How many grams of water @ 0oC could be frozen to ice at 0oC by the evaporation of this amount of ammonia?

The Clausius-Clapeyron Equation: Relating vapor pressure and liquid temperature

ln Pvap = - Hvap + C RT

Y = m x + bWhere:m is the slopeb is the y-interceptR is the gas constant 8.314 J/K molC is a constant characteristic of each specific

substanceT temperature in Kelvin

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PHASE TRANSITIONS

The Clausius-Clapeyron Equation

ln P2 = Hvap 1 - 1

P1 R T1 T2

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PHASE TRANSITIONS

This equation makes it possible to calculate the heat of vaporization of a liquid by measuring its vapor pressure at several temperatures and then plotting the results to obtain the slope

Once the heat of vaporization and the vapor pressure at one temperature are known, the vapor pressure of the liquid at any other temperature can be calculated.

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PHASE TRANSITIONS

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EXAMPLE 11.2: Estimate the vapor pressure of water @ 85oC. Note that the normal boiling point of water is 100oC and it’s heat of vaporization is 40.7kj/mol.

Using the Clausius-Clapeyron Equation: ln P2 = Hvap 1 - 1 P1 R T1 T2

ln P2 = 40.7kj/mol 1 - 1 760mmHg 8.314J/(K.mol) 373K 358K

= (4898K)(-1.123x10-41/K)

= -0.550

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EXAMPLE 11.2 cont…

P2 = antiln (-0.550) 760mmHg

REMEMBER e-0.550 = 0.577

P2 = 0.577 760mmHg

P2 = 0.577 x 760mmHg

= 439mmHg

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EXERCISE 11.2: Carbon disulfide has a normal boiling point of 46oC and a heat of vaporization of 28.6kJ/mol. What is the vapor pressure of CS2 at 35oC?

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EXAMPLE 11.3: Calculate the heat of vaporization of diethyl ether C4H10O, from the following vapor pressures: 400mmHg @ 18oC and 760mmHg @ 35oC.

Using the Clausius-Clapeyron Equation: ln P2 = Hvap 1 - 1 P1 R T1 T2

ln 760mmHg = Hvap 1 - 1 400mmHg 8.314J/(K.mol) 291K 308K

0.642 = (-2.28 x 10-5) x Hvap/(J/mol) Hvap = 2.82 x 104 J/mol

= 28.2kJ/mol

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EXERCISE 11.3: Selenium tetrafluoride is a colorless liquid. It has a vapor pressure of 757mmHg @ 105oC and 522mmHg @ 95oC. What is the heat of vaporization of SeF4?

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Phase Diagrams

PHASE DIAGRAM: a graphical way to summarize the conditions under which the different states of a substance are stable.

Phase Diagrams

Shows which phase is stable at different combinations of pressure and temperature.

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Phase Diagrams

Triple Point: The only condition under which all three phases can be in equilibrium with one another.

Critical Temperature (Tc): The temperature at and above which vapor of the substance cannot be liquefied, no matter how much pressure is applied.

Critical Pressure (Pc) : The minimum pressure required to liquefy a gas at its critical temp.

Supercritical Fluid: Neither true liquid nor true gas

Normal boiling and melting point always at 1 atm 31

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EXAMPLE 11.4: The critical temperatures of ammonia and nitrogen are 132oC and -147oC, respectively. Explain why ammonia can be liquefied at room temperature by merely compressing the gas to a high enough pressure, whereas the compression of nitrogen requires a low temperature as well.

Above the critical temperature, a gas cannot be liquefied, so the critical temperature of NH3 is way above room temperature (~25oC) if NH3 is compressed @ room temperature, it will liquefy. Nitrogen, however, has a critical temperature well below room temperature, so it cannot liquefy at room temperature simply by applying pressure.

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EXERCISE 11.4: Describe how you could liquefy the following gases:

A.Methyl chloride (CH3Cl - critical point 144oC, 66atm)

B. Oxygen (O2 - critical point -119oC, 50atm)

Properties of Liquids

Surface Tension

1. The resistance of a liquid to spread out and increase its surface area

2. Caused by differences in intermolecular forces experienced by molecules at the surface and the interior

3. Surface molecules feel attractive forces on only one side and are drawn in toward the liquid

4. Interior molecules are drawn equally in all directions

5. Higher in liquids that have stronger intermolecular forces

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Viscosity

1. The measure of a liquids resistance to flow

2. Related to the ease with which individual molecules move around in the liquid and thus to the intermolecular forces present

3. Substances with small non-polar molecules have weak intermolecular forces and low viscosities (free flowing)

4. More polar substances have stronger intermolecular forces and have higher viscosities

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Properties of Liquids

Van der Waals forces – intermolecular forces as a whole, all are electrical in origin and result from the mutual attraction of unlike charge or mutual repulsion of like charges.

4 main types Dipole-dipole Ion-dipole London Dispersion forces Hydrogen bonding

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Intermolecular Forces

Intermolecular Forces

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Polar Molecules

Polar Molecules

Just as bonds can be polar, molecules as a whole can be polar

Net sum of individual bond polarities and lone-pair contributions

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Polar Covalent Bonds

Polar Covalent Bonds Form between a non-metal/non-metal of

different electronegativities

EN = ENA – ENB

When EN 1.7 ionic bondWhen EN < 1.7 polar covalent bondWhen EN < .5 non-polar covalent bond

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Dipole-dipole

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a) Neutral but polar molecules experience dipole-dipole forces as a result of electrical interactions among dipoles on neighboring molecules.

b) Forces can be attractive or repulsive, depending on the orientation of the molecules.

c) These forces are weak 3-4 kJ/mol and only significant if molecules are close

Ion-dipole

a) Result of electrical interactions between an ion and the partial charges on a polar molecule

b) Particularly important in aqueous solutions of ionic substances such as NaCl, in which polar water molecules surround the ions

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Na+

OHH

OH H

OH

H

O

H

HCl

OHH

OH H

London Dispersion Forces

a) Result from the motion of electrons

b) At any given time more electrons may be in a particular area of the molecule

c)  This gives the molecule an instantaneous dipole

d)  This short lived dipole can affect the electron distribution in neighboring molecules and induce temporary dipoles in them

e) More electrons a molecule has the stronger the dispersion forces

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Hydrogen Bonding

a) Attractive interaction between a hydrogen atom bonded to a very electronegative atom (O, N, F) and an unshared electron pair on another electronegative atom

b) Hydrogen bonds arise because O-H, N-H, and F-H bonds are highly polar with partial positive charge on the hydrogen and partial negative on the electronegative atom.

c) Hydrogen has no core electrons to shield its nucleus and it is small so it can be approached closely by other molecules

d) The dipole-dipole attraction between the hydrogen and an unshared electron pair on a nearby atom is usually strong 43

Hydrogen Bonding

e) Water is able to form a vast 3D network of hydrogen bonds because each H2O molecule has two hydrogens and two electron pairs

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INTERMOLECULAR FORCES

• Vapor pressure depends on intermolecular

forces

• l g depends on the attraction to

other molecules

• When intermolecular forces are strong in a

liquid, vapor pressure will be low

• Van der Waals forces• London forces exist in every molecule• Dipole dipole forces are usually

appreciable only in small polar molecules or large molecules having very large dipole moments

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INTERMOLECULAR FORCES

• London Forces increase with increasing molecular mass• More electrons, more polarizable

• With molecules of the same molecular mass, but

different arrangements (isomers)•More compact, less polarizable•London forces are smaller•Lower vapor pressure•Lower Hvap

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EXAMPLE 11.5: What kind of intermolecular forces (London, dipole-dipole, hydrogen bonding) are expected in the following substances?

A.CH4, methaneB. CHCl3, tricholoromethaneC.CH3CH2CH2CH2OH, butanol

A.Methane is a nonpolar molecule so the only intermolecular forces are the London Dispersion Forces.B.Tricholormethane is an unsymmetrical molecule with polar bonds, so there will be dipole-dipole and London forces.C.Butanol has a hydrogen atom attached to an oxygen atom, there will be hydrogen bonding. There will also be dipole-dipole and London forces.

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EXERCISE 11.5: List the different intermolecular forces you would expect for each of the following compounds:

A.Propanol, CH3CH2CH2OHB.Carbon Dioxide, CO2C.Sulfur Dioxide, SO2

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EXAMPLE 11.6: For each of the following pairs, choose the substance you expect to have a lower vapor pressure at a given temperature:

A.CO2 or SO2B.Dimethyl ether CH3OCH3 or ethanol CH3CH2OH

The molecular mass of CO2 is 44amu and the molecule is non-polar, so there are no dipole-dipole forces, onlyLondon forces exist.

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The molecular mass of SO2 is 64amu and the molecule

is polar, so dipole-dipole forces as well as London

forces exist.

Sulfur dioxide has a lower vapor pressure Experimental values confirm that the vapor

pressure of CO2 @ 20oC is 56.3atm and SO2 is 3.3atm.

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Dimethyl Ether

Ethanol

The molecular masses are equal (46amu) so the London

forces are approximately the same. There is a strong

hydrogen bonding in ethanol ethanol should have a

lower vapor pressure.

Experimental values confirm that the vapor pressure @

20oC is 4.88atm for dimethyl ether and 0.056atm for

Ethanol.

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EXERCISE 11.6: Arrange the following hydrocarbons in order of increasing vapor pressure:

Ethane, C2H6

Propane, C3H8 Butane, C4H10

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EXERCISE 11.7: At the same temperature, methyl chloride CH3Cl, has a vapor pressure of 1490mmHg. Ethanol has a vapor pressure of 42mmHg. Explain why this is to be expected.

Types of solids

1. Molecular solid -held together by intermolecular forces -H2O(s), CO2(s)

2. Metallic solid -positively charged atomic cores surrounded by delocalized electrons -Fe, Cu, Ag

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SOLID STATE

Types of solids3. Ionic solid

-cations and anions held together by electrical attraction of opposite

charges -NaCl

4. Covalent network solid -atoms held together in large networks by covalent bonds -diamonds

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SOLID STATE

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EXAMPLE 11.7: Which of the four basic types of solids would you expect the following substances to be?

A.Solid ammoniaB.CesiumC.Cesium iodideD.Silicon

A. Ammonia NH3 has a molecular structure, it freezes as a molecular solid.B.Cesium (Cs) is a metal, it is a metallic solid.C.Cesium Iodide, CsI, is n ionic substance, so it exists as an ionic solid.D.Silicon (Si) atoms might be expected to form covalent bonds to other silicon atoms, as carbon does in diamond. A covalent network solid would result.

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EXERCISE 11.8: Classify each of the following solids according to the forces of attraction that exist between the structural units:

A.ZincB.Sodium iodideC.Silicon carbideD.Methane

PHYSICAL PROPERTIES:

MELTING POINT AND STRUCTURE The stronger the intermolecular forces, the

higher the melting points

HARDNESS AND STRUCTURE Molecular crystals with weak intermolecular

forces are soft compared to ionic crystals in which the attractive forces are stronger

Generally brittle whereas metallic crystals are malleable

ELECTRICAL CONDUCTIVITY AND STRUCTURE Metals are good electrical conductors Covalent and ionic solids are non-conductors

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SOLID STATE

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EXAMPLE 11.9: Arrange the following elements in order of increasing melting point: silicon, hydrogen and lithium. Explain your reasoning.

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EXERCISE 11.9: Decide what type of solid is formed for each of the following substances:C2H5OH, CH4, CH3Cl, MgSO4. On the basis of the type of solid and the expected magnitude of intermolecular forces (for molecular crystals), arrange these substances in order of increasing melting point. Explain your reasoning.

Example 1

Chloromethane

a) Calculate the dipole moment

b) Calculate the % ionic character of the bond

Experimentally measured dipole moment = 1.87 D

C-Cl bond distance = 178 pm = 178 x 10-12 m

If we assume that the contributions of the nonpolar C-H bonds are small, then most of the chloromethane dipole moment is due to the C-Cl bond 61

H

Cl

HH

Example 2

Hydrochloric acid

Calculate the % ionic charcater1.      Distance between atoms is 127 pm2.      Experimentally measured dipole moment = 1.03 D

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Cl H

Example 3

Tell which of the following compounds are likely to have a dipole moment and show the direction of each. a) SF6 b) CHCl3

c) CH2Cl2 d) CH2CH2

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Example 4

Identify the likely kinds of intermolecular forces in the following

A)   HCl

B) CH3CH3

C) CH3NH2

D) Kr

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ClHH

H

H

H

H

H

N

H

H

H

H

H

Example 5

Of the substances Ar, Cl2, CCl4 and HNO3 which has:

a)  The largest dipole-dipole forces?

b)   The largest hydrogen-bond forces?

c) The smallest dispersion forces?

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Example 6

Chloroform has Hvap = 29.2 kJ/mol and Svap = 87.5 J/K mol. What is the boiling point of chloroform?

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Example 7

The vapor pressure of ethanol at 34.7C is 100.0 mm Hg, and the heat of vaporization of ethanol is 38.6 kJ/mol. What is the vapor pressure of ethanol in mm Hg at 65.0C?

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Example 8

The normal boiling point of benzene is 80.1 C and the heat of vaporization is 30.8 kJ/mol. What is the boiling point of benzene (in C) on top of Mt. Everest where P = 260 mm Hg?

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Example 9

Can you label the following?

a) solid regionb) Liquid regionc)  Gas regiond)  Normal boiling pointe)  Normal melting pointf)   Triple pointg)  Supercritical fluid

regionh) Critical point, what is

the critical pressure and temperature

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