1 Chapter 10 Liquids and Solids. 2 3 Van Der Waals Forces These are intermolecular forces of...

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Chapter 10

Liquids and Solids

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Van Der Waals Forces

• These are intermolecular forces of attraction between neutral molecules.

• The Nobel Prize in Physics 1910 (Johannes van der Waals)

• "for his work on the equation of state for gases and liquids"

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intER vs. intRA molecular forces

• Intramolecular forces are the forces within a molecule or ionic compound

NaCl Ionic bond between atom of Na and atom of Cl

• Intermolecular forces are the forces between molecules or ions and molecules

Example: Solid liquid gas

.

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Intermolecular Force

Model Basis of Attraction

Energy (kJ/mol)

Example

Ion-dipole                     Ion and polar

molecule600 – 40 Na+ &

H2O

Dipole-dipole                     Partial charges of polar molecules

25 – 5 HCl & HCl

Hydrogen bond

                   H bonded to N, O, or F, and another N, O, or F

40 – 10 H2O &

NH3

London dispersion

                      Induced dipoles of polarizable molecules

40 – 0.05 Xe & Xe

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3 Types of van der Waals Forces

• Dipole-Dipole forces

• London Dispersion forces

• Hydrogen bonding

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DIPOLE-DIPOLE FORCES

• These are forces of attraction that occur between polar molecules. (big difference in electron negativity)

• These forces are effective only when polar molecules are very close. As distance increase strength of bond decreases.

• For molecules of approximately equal mass and size, the strength of force of attraction increases as the polarity increases.

• Radius have an effect on strength of dipole.

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DIPOLE-DIPOLE FORCES

+

_

_

_ +

+

+_

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DIPOLE-DIPOLE FORCES

• Molecules with larger dipole moments have higher melting and boiling points (hard to break) than those with small dipole moments.

• Dipole attractions are relatively weak and tend to be liquids or gas at room temperature.

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HYDROGEN BONDING

• A special type of dipole-dipole interaction between the hydrogen atom in a polar bond and an unshared electron pair of an element that is very electronegative usually a F, O, or N atom on another molecule

• (note that all of these have very high EN’s and small atomic radii).

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14

FYI

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HYDROGEN BONDING

• These types of bonds are super-humanly strong.

• (4X stronger that diopole dipole)

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HYDROGEN BONDING IN WATER

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HYDROGEN BONDING

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WHY HYDROGEN BONDING IS EFFECTIVE

• F, O, & N are extremely small and very electronegative atoms.

• Hydrogen atoms have no inner core of electrons, therefore, the positive side of the bond dipole has the concentrated charge of the partially exposed, nearly bare proton of the nucleus

• …in other words, the atoms have a large difference in electronegativity and their nuclei can get really close.

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IMPORTANCE OF HYDROGEN BONDING

• Are important biologically, in stabilizing proteins and keeping DNA together.

• Also explains why ice is less dense than water (see text).

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LONDON DISPERSION FORCES

• Fritz London

• These are forces that arise as a result of temporary dipoles induced in the atoms or molecules.( it’s a temporary accident!)

• All molecules have some degree of LD forces

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LONDON DISPERSION FORCES

• LD forces occur between neutral non-polar molecules. (nobles gases and nonpolar compounds)

• LD forces are weak

• The greater the number of electrons the greater the LD force. (ie the greater the melting and boiling pt.)

• LD force molecules have Low melting and boiling pts

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See Graphic on next slide

• The motion of electrons in an atom or molecule can create an instantaneous dipole moment.

• EX: in a collection of He (g) the average distribution of electrons about a nucleus is spherical, the molecules are non-polar and there is no attraction.

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INSTANTANEOUS AND INDUCED DIPOLES

Pg 454- 455 in text

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LONDON DISPERSION FORCES (CONT)

• These forces tend to increase in strength with an increase in molecular weight (The size of the molecule generally increases with mass and the electrons are less tightly held…allows the electron cloud to be more easily distorted.

• These forces are stronger in linear molecules than comparable “bunched up” molecules.

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LONDON DISPERSION FORCES

LD forces are generally the WEAKEST intermolecular forces.

Molecules with more electrons will experience more LD forces

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n-pentane vs neopentane

• BP = 309.4 K BP = 282.7 K

• Same atomic masses different structure

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Generalizations Regarding Relative Strengths of IM Forces

• If molecules have comparable molecular weights and shapes, dispersion forces are approximately equal. Any difference in attractive forces is due to dipole-dipole attractions.

• If molecules differ widely in molecular weight, dispersion forces are the decisive factor. The most massive molecule has the strongest attractions.

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Because melting points (MPs) and boiling points (BPs) of covalent molecules increase with the strengths of the forces holding them together, it is common to use MPs and BPs as a way to compare the strengths of intermolecular forces.

This is shown below, with the molecular formulas, molar masses and normal BPs of the first five straight-chain hydrocarbons.

Molecular Formula Molar Mass Normal BP (C)CH4 16 - 161.5C2H6 30 - 88.5C3H8 44 - 42.1C4H10 58 - 0.5C5H12 72 36.1

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Which noble gas element has the lowest boiling point?

He

Ne

Ar

Kr

Xe

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The chemical forces between HCl is/are

• Dispersion

• Covalent bond

• Hydrogen bond

• Dipol-dipole

• Two of the above

All Molecules Have

Not symmetrical Polar

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Consider the following list of compounds. How many of these have hydrogen bonding as their

principle IMF

HCl

NH3

CH3OH

H2S

CH4

PH3

Hydrogen Bonding is between H and highley EN

atoms such as N, O, F, and H

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Which of the following statements are false or correct and why?

O2 is dipole dipole

HCl is hydrogen bonding

CO2 is dipole dipole

NH3 is hydrogen

FALSE Dipole dipole not symmetric/polar

FALSE London Dispersion

symmetric/nonpolar

TRUE H + N,O, or F

FALSE London Dispersion

symmetric/nonpolar

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ION-DIPOLE FORCES

• Attraction between an ion and the partial charge on the end of a polar molecule.

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ANSWER

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A.Identify the types of bonds in

1. Glucose

2. Cyclohexane

B.Glucose is soluble in water but cyclohexane is not. Why?

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A.Identify the types of bonds in

1. Glucose

H, LD, VanderWal, Dip-dip

2. Cyclohexane

LD only

B.Glucose is soluble in water but cyclohexane is not. Why?

Glucose is polar and cyclohexane is nonpolar. Polar compounds are soluble in polar solvent and visversa.

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ION-DIPOLE FORCES

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ION-DIPOLE FORCES (CONT)

• The magnitude of attraction increases as either the charge of the ion increases or magnitude of the dipole moment increases.

• Ion-dipole forces are important in solutions of ionic substances in polar liquids (e.g. water)

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ION-DIPOLE FORCES AND THE SOLUTION PROCESS

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Homework

• Pg 504-505

#’s : 35, 36, 37, 39 (you may need to read 10.1 for this part esp. LD portion)

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10.2 Properties of liquids

• Liquids are vital to our lives.

• Water is a means of food preparation

• Cooling machines n industrial processes

• Recreation

• Cleaning

• Transportation

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CHARACTERISTICS OF LIQUIDS

• Surface tension

• Capillary action

• Viscosity

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COHESIVE FORCES

• Intermolecular forces that bind like molecules to one another (e.g. hydrogen bonding).

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ADHESIVE FORCES

• Intermolecular forces that bind a substance to a surface.

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SURFACE TENSION

• A measure of the inward forces that must be overcome in order to expand the surface area of a liquid.

• The greater the forces of attraction between molecules of the liquid, the greater the surface tension.

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Surface Tension Cont.

• Surface tension of a liquid decreases with increasing temperature.

• The stronger the intermolecular forces the stronger the surface tension.

Water has a high surface tension do to hydrogen

bonding.

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CAPILLARY ACTION

• Another way surface tension manifests.

• The rise of liquids up very narrow tubes. This is limited by adhesive and cohesive forces.

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Formation of meniscus

• Water : adhesive forces are greater than cohesive forces

• Mercury: Cohesive are greater than adhesive forces.

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VISCOSITY

• The resistance of a liquid to flow.

• The less “tangled” a molecule is expected to be, the less viscous it is.

Water = less Viscosity

syrup = high Viscosity

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Viscosity Cont.

• Viscosity decreases with increasing temperature (molecules gain kinetic energy and can more easily overcome forces of attraction).

• Viscosity Increases as pressure increases.

• Liquids with strong IMF have a higher viscosity.

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Homework

• Pg 505

#’s 43-45 all

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10.3 Structure of Solids

• Two ways to categorize solids– Crystalline – Amorphous

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Crystalline Solid

• Ridged and long range order of its atoms.

• Solids have flat surfaces

• Sharp melting points

• EX: Quartz, diamond, sodium Chloride.

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Amorphous Solid

• Lack a well defined arrangement

• No long range order

• IMF vary in strength

• DO NOT have sharp melting points.

EX: rubber, glass

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Unit Cell

• The smallest part of a crystal that will reproduce the crystal when repeated in a three dimensions.

• Three types– Simple /primitive cubic cell– Face centered cubic cell– Body Centered

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X-Ray Diffraction

• X-Ray crystallography lead to the discovery of DNA

Beams of light shot at DNA and scattered to reveal the double

helix structure.Watson, Franklin, Crick

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X-ray Diffraction

• Derived by the English physicists Sir W.H. Bragg and his son Sir W.L. Bragg in 1913 to explain why the faces of crystals appear to reflect X-ray beams at certain angles of incidence (theta, θ).

• d is the distance (Ǻ) between atomic layers in a crystal

• lambda λ is the wavelength (Ǻ) of the incident X-ray beam

• n is the order of the diffraction• Bragg’s Equation: nλ = 2d sinθ

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Bragg’s Equation Demo

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Example

• The first order diffraction of x-rays from crystal planes separated by 2.81 Ǻ occurs at 11.8°.

a. what is the wavelength of the x-ray

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The work

• n = 1 (first order)• d = 2.81 Ǻ • Θ = 11.8°• λ = ?nλ = 2d sinθ

1(λ) = 2(2.81 Ǻ) sin 11.8°Sin-1 = 24.14 °λ = 24.14 °

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Homework

• Pg 505

• #’s :47, 63, 64,

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Changes of state

• Transformation from one state to another

Vaporization

AKA: steam

Condensation

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Changes in state

• Liquid Gas Vaporization Endothermic

• Gas Liquid Condensation Exothermic

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• Solid Gas Sublimation Endothermic

• Gas Solid Deposition Exothermic

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• Solid Liquid Melting Endothermic

• Liquid Solid Freezing Exothermic

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Changes of state

• The energy involved it phase changes is calculated using

– Heat of fusion (solid liquid or liquid solid)

– Heat of vaporization (liquid gas or gas liquid)

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Energy Changes and Phase Changes

Heat of Vaporization: Vaporization is an endothermic process ( it requires heat). Energy is required to overcome intermolecular forces to turn liq to gas.

Hvap is an Indicator of strength of IMF

CH4 = 9.2 kJ/mol C3H8 = 18.1 kJ/mol

Larger molecule…greater IMF…greater Hvap

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Question

How much energy does it take to vaporizer 111 g of water?

Given: Hvap water= 40.67 kJ/mol

111 g H2O 1 mol x 40.6kJ = 250kJ 18g 1mol

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• Heat of Fusion: the enthalpy change associated with melting. (Solid to liquid.)

• Hfusion water= 6.01 kJ/mol

• NOTE: heat of fusion is always smaller than heat of vaporization. This makes sense think about the level of “order” in the molecules in these phases.

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Heating Curve• A plot of the temperature versus time

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Heat of Vaporization

Heat of Fusion

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Example

Calculate the enthalpy change associated with converting 1.00 mole of ice -25ºC to water 150ºC at 1 atm. Specific heat of ice, water, and steam are 2.09 J/g ºC and 4.184 J/g ºC, 1.84 J/g ºC . The heat of fusion of ice is 6.01 kJ/mol and heat of vaporization of water is 40.67 kJ/mol

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ICE -25 °C

q1

Ice water 0°C

q2

Water 0°C

q3

Water vapor 100°C

q4

Vapor

100°C

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1 mol ice 1 mol ice 1 mol water 1 mol water 1 mol steam

T= -25ºC 0ºC 0ºC 100ºC 100ºC

qtotal = q1 + q2 + q3 + q4

1.)q = 2.09(18g)(-25-0)2.) q = 6.02 KJ/mol (convert heat of fusion) 3.) q = 4.184(18g) (100-0)4.) q = 40.7 KJ/mol (convert heat of

vaporization)5. ) q = 1.84(18 g) (100-0)

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Critical Stuff

• Critical Temperature: The temperature above which it is impossible to liquefy the gas under study no matter how high the applied pressure.

• Critical Pressure: The pressure required to liquefy a gas as at its critical temperature

NOTE: the critical temp of a gas gives an indication of the strength of the IMF of that gas. A substance with weak attractive forces would have a low critical temp.

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Which gas can be liquefied at 25ºC

Gas Critical Temp

ºC

Critical Pressure

atm

Ammonia 132 112

Ethanol 158 78

Argon -186 6

Critical Temp above 25ºC

Critical Temp under 25ºC

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Vapor Pressure (vp)Vapor Pressure: Pressure

exerted by molecules that have enough energy to escape the surface.

As T ↑ VP ↑evaporation ↑

Liquids with high VP are volatile (alcohol evaporates easily)

Liquids that have strong IMF have

low vapor pressures. (take a lot of energy to overcome

IMF so it can evaporate)

Kinetic energy

%

of

Molecules

T2

• At higher temperature more molecules have enough energy

• Higher vapor pressure.

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• Liquids with high VP are volatile (alcohol evaporates easily)

• Liquids that have

strong IMF have low vapor pressures.

• (take a lot of energy to overcome IMF so it can evaporate)

substance vapor pressure at

25oC

diethyl etherC4H10O

0.7 atm

BromineBr2

0.3 atm

ethyl alcoholC2H5OH

0.08 atm

WaterH2O

0.03 atm

Evaporation

• Molecules at the surface break away and become gas.

• Only those with enough KE

escape

• Evaporation is a cooling process.

• It requires heat.

• Endothermic.

Condensation

Change from gas to liquid

Achieves a dynamic equilibrium with vaporization in a closed system.

What the heck is a “dynamic equilibrium?”

When first sealed the molecules gradually escape the surface of the liquid

As the molecules build up above the liquid some condense back to a liquid.

Dynamic equilibrium

As time goes by the rate of vaporization remains constant

but the rate of condensation increases because there are more molecules to condense.

Equilibrium is reached when Rate of Vaporization = Rate of Condensation

Dynamic equilibrium

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VP example

In a closed container the number of partials changing from liquid vapor will eventually equal the number changing form vapor liquid.

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Boiling Point

Note: The normal boiling point of water is 100oC. The term normal refers to standard pressure or 1 atm, or also 101.3

kPa.

The vapor pressure of the liquid = air pressure above the liquid

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Boiling Pts. of H2O at Various Elevations

Altitude compared to Sea Level

(m)

Boiling Point

(°C)

1609 98.3

177 100.3

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How to make something boil

1. Increase the VP of the liquid (heat it) so that the VP of the liquid is > that of the atmosphere.

2. Lower the atmospheric pressure (pressure above the liquid)

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Boiling Point

↑ boiling pt by

↑ in IMF

Or

↓ VP

At high altitudes (low air pressure) water boils at a lower temperature

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Normal Boiling Point

• Temperature at which something boils when the vp =1 atm

• Note the lower the external pressure the lower the boiling point.

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Freezing point/melting point

• They are the same but in opposite directions.

• When heated the particles vibrate more rapidly until they shake themselves free of each other.

• Ionic solids have strong intermolecular forces so a high mp.

• Covalent/molecular solids have weak intermolecular forces so a low mp.

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Phase Diagram

• A graphical way to summarize the conditions under which equilibrium exists between different states of matter.

• Allows you to predict the phase of a substance that is stable at a given temperature and pressure

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Triple point = three phase are in equilibrium with each other at the same time

Boiling Point Melting Point

1 atm

Critical point

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Critical Point: The temp beyond which the ,molecules of a substance have to much kinetic

energy to stick together to form a liquid.

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Phase diagrams of substances other than water the slope of the solid liquid line slopes forward. (positive)

In water the slope of the solid-liquid lines slopes downward. (negative)

Water Not Water

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Homework

• Pg 508

• #’s : 85, 87,89, 91,