Chem 106, Prof. T. L. Heise 1 CHE 106: General Chemistry CHAPTER ELEVEN Copyright © Tyna L. Heise...

Chem 106, Prof. T. L. Heise 11

CHE 106: General Chemistry

CHAPTER ELEVEN

Copyright © Tyna L. Heise 2001

All Rights Reserved


All elements can exist in three different phases of matter

– Chemical properties remain the same

– Physical properties are different

–Many of the substances to be considered in solid and liquid phase are molecular

» Virtually all liquids are molecular

» Physical properties examined are due to the forces of attractions NOT within a molecule BUT between molecules

Molecular ComparisonMolecular Comparison of Solids and Liquids of Solids and Liquids

Chapt. 11.1


Solids, Liquids, and Gases

Chapt. 11.1


Solids, Liquids, and Gases

Chapt. 11.1

Substances can be changed from one phase to another by -heating and cooling to change the average kinetic energy

-adding and decreasing pressure changes the distance between molecules which leads to an alteration in intermolecular forces

- increasing pressure pushes molecules closer together, increasing molecular attraction

- decreasing pressure allows molecules to spread farther apart, decreasing molecular attraction


Intermolecular ForcesIntermolecular Forces

The strengths of intermolecular forces of different substances vary greatly, however they are much weaker than ionic or covalent bonds:

You are not actually breaking apart a molecule, just separating molecules from each other as you move from solid to liquid and the to a gas phase.

Physical Properties reflect the IM forces

- boiling points

- melting points

Chapt. 11.2



Three types of IM forces:

dipole - dipole forces

ion - dipole forces

London dispersion forces

hydrogen bonding

Chapt. 11.2


Ion - Dipole forcesIon - Dipole forces

Important in aqueous solutions

- important between an ion and the partial charge on the end of a polar compound

negative ion attracts to +

positive ion attracts to -

Chapt. 11.2


Dipole - Dipole forcesDipole - Dipole forces

Important in aqueous solutions

- important between two dipoles

negative end attracts to +

positive end attracts to -

**for molecules of equal size, the strengths of dipole forces increases with electronegativity

Chapt. 11.2


London Dispersion forcesLondon Dispersion forces

Important between two nonpolar molecules

- important to look at instantaneous electron arrangements

- at an instant, the position of an electron can force a temporary dipole moment.

- the temporary dipole moment can induce its’ neighbor to experience a temporary dipole moment, causing an attraction

- this force is only significant when the molecules are very close to each other

Chapt. 11.2



Chapt. 11.2



The amount of polarizability increases with size of molecule.

- boiling point increases with molecular weight

The more contact a molecule has with another also increases the polarizability

- n-pentane has higher boiling point than neopentane due to larger available surface area

Chapt. 11.2


Hydrogen BondingHydrogen Bonding

A special intermolecular force which exists between the H of one molecule and the O, F, or N of a neighboring polar molecule

Chapt. 11.2


Hydrogen BondingHydrogen Bonding

Chapt. 11.2



Chapt. 11.2

Summary: Intermolecular Forces can be determined using 1. Composition

2. Structure

Dispersion forces are in ALL molecules

•Strengths increase with increased weight, and depend on shape

Dipole-dipole forces add to dispersion forces and are found ONLY in polar molecules

Hydrogen bonds are found between the H of one molecule and the F, O, or N of another

None of these are as strong as ionic or covalent bonds, but of these, hydrogen bonding is the strongest


Properties of LiquidsProperties of Liquids

Chapt. 11.3

Two important properties of Liquids:

Viscosity - the resistance of a liquid to flowlow viscosity flows easilyhigh viscosity flows slowly

Surface Tension - the energy required to increase the surface area of a liquid by a unit amount.

Molecules at surface all are attracted inwards instead of in all directions, thispacks surface molecules closer together

The measurement of the inward forces that must be overcome is the surface tension



Chapt. 11.3

Viscosity



Chapt. 11.3

Surface Tension


Phase ChangesPhase Changes

Chapt. 11.4

Many important properties of liquids and solids relate to the ease in which they change from one state to another:



Chapt. 11.4

Each phase change is accompanied by a change in energy of the system.

Energy must be added to overcome intermolecular forces and achieve a less ordered state.

Energy must be released as intermolecular forces begin to form and a molecule achieves a more ordered state



Chapt. 11.4

Heating Curves: plot of the process of changing phase of substance

Heat of Fusion: the energy needed to change a solid to a liquid

Heat of Vaporization: the amount of energy needed to change a liquid to a gas



Chapt. 11.4

Special Notes:

Supercooling: reducing the heat of the liquid so fast it’s temperature falls below its

freezing point , but it remains a liquid. This occurs because the liquid never has a chance to form an ordered solid.

Critical Temperature: Highest temperature at which a substance can exist as a liquid

Critical Pressure: The pressure needed to bring about liquefaction at critical temp.


Vapor PressureVapor Pressure

Chapt. 11.5

Molecules can escape from the surface of liquid into the gas phase by vaporization or evaporation.



Chapt. 11.5

Why?

•Molecules of liquids move at various speeds

•At any instant a molecule on the surface may possess enough energy to overcome its intermolecular forces and escape into the gas phase

•The movement of molecules between liquid and gas phases goes on continuously, eventually only so many molecules can escape in a closed container and a dynamic equilibrium will be attained



Chapt. 11.5

Open Containers

•As liquid evaporates, the gas molecules move away from surface of liquid, causing the possibility of recapture by the liquid to be very small

•Equilibrium never occurs, and the vapor continues to form until the liquid is gone

•Liquids that easily evaporate due to low intermolecular forces, creating a high vapor pressure in a closed container, are considered volatile



Chapt. 11.5

Boiling Point: the temperature at which the vapor pressure equals the external pressure acting on the surface of the liquid.

Normal Boiling Point: the boiling point of a liquid at 1 atm


Phase DiagramsPhase Diagrams

Chapt. 11.6


Structures of SolidsStructures of Solids

Chapt. 11.7

Crystalline Solid: particles are in a well ordered arrangement

- flat surfaces or surfaces at definite angles to each other- orderly stacking of particles causes them to have orderly shapes- examples - quartz and diamonds- melts at specific temperature



Chapt. 11.7

Amorphous Solid: particles have no orderly structure

- mixtures of molecules that do not stack well together.- composed of large complicated molecules- examples: rubber and glass- intermolecular forces vary in strength from one part of solid to another due to irregularities in solid, melting point varies



Chapt. 11.7

Unit Cells: the repeating unit of a crystalline solid that builds the definite patterns- three dimensional array called a lattice- generally parallelpipeds, can be described using two terms, length and angle between edges



Chapt. 11.7

Unit Cells



Chapt. 11.7

Sample exercise: The body-centered cubic cell off a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron.



Chapt. 11.7


1) D = m V



Chapt. 11.7


1) D = m V

Mass = 55.845 amu



Chapt. 11.7


1) D = m V

Mass = 55.845 amu 1 g 6.02 x 1023 amu



Chapt. 11.7

Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 angstroms on each side. Calculate the density of this form of iron.

1) D = m V

Mass = 2(55.8) amu 1 g = 1.85 x 10-22 g 6.02 x 1023 amu



Chapt. 11.7


1) D = m V 3

Volume = (2.8664 angstroms) (10-10 m) 1 angstroms



Chapt. 11.7


1) D = m V 3

Volume = (2.8664 angstroms) (10-10 m) = 2.36 x 10-29

1 angstroms



Chapt. 11.7


1) D = m V

Volume = 2.36 x 10-29 m3 1 L

10-3 m3



Chapt. 11.7


1) D = m V

Volume = 2.36 x 10-29 m3 1 L = 2.36 x 10-26 L

10-3 m3



Chapt. 11.7

Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 ≈ on each side. Calculate the density of this form of iron.

1) D = m = 1.85 x 10-22 g V 2.36 x 10-26 L



Chapt. 11.7

Sample exercise: The body-centered cubic cell of a particular crystalline form of iron is 2.8664 ≈ on each side. Calculate the density of this form of iron.

1) D = m = 1.85 x 10-22 g = 7,838 g V 2.36 x 10-26 L L



Chapt. 11.7

Close packing of spheres: structures adopted by crystalline solids are those that bring particles in closest contact to maximize intermolecular attractions

* Each sphere has 12 neighbors, known as the coordination number



Chapt. 11.7

Sample exercise: Aluminum metal crystallizes in a cubic close-packed structure (face-centered cell)

(a) How many aluminum atoms are in the unit cell



Chapt. 11.7


(a) How many aluminum atoms are in the unit cell

1/2 atom at each of 6 faces = 3 atoms

1/8 atom at each of 8 corners = 1 atom

4 atoms total



Chapt. 11.7


(b) What is the coordination number of each aluminum atom?



Chapt. 11.7


(b) What is the coordination number of each aluminum atom?

12



Chapt. 11.7


(c) If each aluminum atom has a radius of 1.43 angstroms, what is the length of each side?



Chapt. 11.7


(c) If each aluminum atom has a radius of 1.43 angstroms, what is the length of each side?

4 radius on diagonal=4(1.43 angstroms)= 5.72 angstroms



Chapt. 11.7


(c) If each aluminum atom has a radius of 1.43 A, what is the length of each side?

4 radius on diagonal = 4(1.43 A) = 5.72 A

a2 + b2 = c2

2a2 = c22a2 = (5.72 A)2

2a2 = 32.72 A a2 = 16.36 A

a = 4.04 A



Chapt. 11.7


(d) Calculate the density of the of aluminum metal



Chapt. 11.7



1) D = m V



Chapt. 11.7



1) D = m V

Mass = 26.98 amu



Chapt. 11.7



1) D = m V

Mass = 26.98 amu 1 g 6.02 x 1023 amu



Chapt. 11.7



1) D = m V

Mass = 4(26.98) amu 1 g = 1.79 x 10-22 g 6.02 x 1023 amu



Chapt. 11.7



1) D = m V 3

Volume = (4.04 A) (10-10 m) 1 L 1 A 10-3 m3



Chapt. 11.7



1) D = m V

Volume = 6.59 x 10-26 L



Chapt. 11.7



1) D = m = 1.79 x 10-22 g V 6.59 x 10-26 L



Chapt. 11.7



1) D = m = 1.79 x 10-22 g = 2,720 g/L

V 6.59 x 10-26 L


Bonding in SolidsBonding in Solids

Chapt. 11.7

The physical properties of crystalline solids, such as melting point and hardness, depend both on the arrangements off particles and on the attractive forces between particles in solids.

4 Types : Molecular Covalent & Network Ionic Metallic



Chapt. 11.7

Molecular - made up of atoms or molecules - London dispersion forces, dipole-dipole

forces, hydrogen bonding - soft, low to moderately high melting point,

poor thermal and electrical conductivity

Network Covalent - atoms connected in networks - covalent bonding - very hard, very high melting point, often poor thermal and electrical

conductivity



Chapt. 11.7

Ionic - made up of positively and negatively charged ions- electrostatic attractions- hard, high melting point, poor thermal and electrical conductivity

Metallic - atoms - metallic bonding - soft to very hard, low to very high melting point, excellent thermal and electrical conductivity, malleable and ductile


Three types of Intermolecular ForcesThree types of Intermolecular Forces ViscosityViscosity Surface TensionSurface Tension Phase changesPhase changes Heats of Fusion and VaporizationHeats of Fusion and Vaporization Vapor PressureVapor Pressure Phase Diagrams, triple pointPhase Diagrams, triple point Crystalline solidsCrystalline solids Amorphous solidsAmorphous solids Unit CellsUnit Cells Bonding in SolidsBonding in Solids

Chapter Eleven; ReviewChapter Eleven; Review

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Chem 106, Prof. T. L. Heise 1 CHE 106: General Chemistry CHAPTER ELEVEN Copyright © Tyna L. Heise...

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