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Liquids and Solids Polarity of Molecules In the previous chapter, learned how

atoms bond to each other and how these bonds affect the shape of the molecules they form.

We also learned how to determine

the type of bond that exists between two atoms.

The shape of molecules, and the type of bonds they possesses, affect the distribution of electron density within them.

When a molecule has an

unsymmetrical distribution of e- within the molecule, we say that the molecule is POLAR.

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When a molecule has a symmetrical

distribution of e- within the molecule, we say that the molecule is NONPOLAR.

The covalent bonds that make up molecules come in two types:

1) Covalent Bond - a bond resulting in an equal sharing of e-. - the two atoms in the bond have identical electronegativities.

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2) Polar Covalent Bond -a bond resulting in an unequal sharing of e-. - the two atoms in the bond have different electronegativities.

- the more electronegative atom in the bond will have a partial negative charge ( -).

- the less electronegative atom in the bond will have a partial positive charge ( +).

- the existence of +and - in a

bond is called a DIPOLE. When a molecule has an unsymmetrical distribution of e- (it is POLAR), an overall +and - exists

within the molecule. When this occurs, we say that the molecule has a DIPOLE.

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The measure of how much of a dipole a molecule possesses is called the Dipole Moment. The greater the dipole moment, the more polar the molecule.

However, merely having polar bonds does not necessarily mean that the molecule is polar.

The geometry of a molecule must be

considered when determining a molecules’ polar character. Linear Molecules:

H F F Be F

H F

Polar

BeF F

Nonpolar

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Trigonal Planar:

B

Cl

Cl

Cl

B

Cl

F

Cl

PolarNonpolar Tetrahedral Molecules:

CCl ClCl

F

PolarNonpolar

Cl

CCl ClCl

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INTERMOLECULAR FORCES 1) Dipole Forces These interactions occur in polar

molecules. This intermolecular force is only 1%

as strong as a covalent bond.

The +/ - (dipole) of one polar

molecule lines up with +/ - (dipole)

of another polar molecule (opposites attract).

The greater the dipole moments (the

measure of the polarity of a molecule) of the molecules, the stronger the attractive force. This interaction (attraction) really

only works when the molecules are close together.

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When the molecules are in the gas phase, the dipole forces of attraction are negligible.

2) Hydrogen Bonding

This attraction occurs in polar molecules, HOWEVER, only in molecules that have X – H bonds where X = N, O, or F.

This attractive force is an unusually strong dipole force of attraction.

Why is the hydrogen bonding such a powerful attractive force?

2 – Reasons: 1. There is a large difference in the

electronegativity of the X and H H(2.2) F(4.0)

H(2.2) O(3.5) H(2.2) N(3.0)

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The H – atom almost behave as a “naked” proton.

2. The H – atom is very small.

The lone pair of electrons on F, O, and N can get really close to H.

Let’s look at the pattern of boiling points.

bp bp bp

NH3 -33oC H2O 100 oC HF 19oC

PH3 -88oC H2S -60oC HCl -85oC

AsH3 -63oC H2Se -42oC HBr -67oC

SbH3 -18oC H2Te -2oC HI -35oC

*Note the effect of hydrogen bonding in the first row of boiling points.

IMPORTANT Although these intermolecular force are very important, they are very weak compared to a covalent bond.

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3) (London) Dispersion Forces Found in all molecular-nonpolar

substances.

involves a temporary or induced dipole.

Consider the H2 molecule Nonpolar covalent bond

equal sharing of electrons For an instant, the electrons within

the molecule can concentrate closer to one atom in the molecule.

● Produces a +/ - (dipole) within the

molecule. ● This temporary dipole induces a

similar dipole within another molecule.

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● These temporary dipoles result in the two molecules attracting each other. This attraction is the Dispersion Force!

The attraction is dependant on:

1. The # of electrons in the molecules involved.

2. The ease of the electrons in the

molecules to be dispersed within the individual molecules.

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MOLECULAR INTERACTIONS To begin our discussion, let’s talk about the simplest phase of matter – the gas phase.

When these particles pass each other, 3 possible scenarios can occur.

1) They are indifferent to each

other. 2) They are attracted to each

other. 3) They are repelled by each

other. Question: In order for an ice cube to form, which of these options must exist? Answer: ATTRACTION!!!

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3 Types of attractive forces exist. They are (in order of weakest to strongest): 1. Dispersion Forces 2. Dipole/Dipole Forces 3. Hydrogen Bonding Force (a stronger version of Dipole/Dipole Forces) All 3 of these attractive forces can exist in all three phase of matter. Let’s force our discussion on the Liquid and Solid.

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Liquids & Solids Liquid State

Much more ordered than a gas. Greater attractive force than a gas.

Keeps molecules together. Attractive Forces:

- Hydrogen-Bonding - Dipole/Dipole - Dispersion

Fluidity – can flow and assume the shape of the container. Density – most liquids are 1000X more dense than a gas. -They are 10% as dense as a solid.

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Compressibility If 1000 atm of pressure is applied to a gas – the volume can be decreased by 99.9%.

For liquids (1000 atm) will decrease volume by 4%.

Diffusion Liquid can diffuse Much more

slowly than a gas. Affect by Temp. Surface Tension

A force that tends to pull adjacent parts of a liquid’s surface together.

Result: Decrease surface area.

H2O

Air

Surface Why? Intermolecular Forces

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Capillary Action (Related to Surface Tension)

Attraction of a liquid to a surface Water in a tube Upward pull

Evaporation vs. Boiling (Vaporization) Evaporation = Process in which a liquid is converted into a gas by particles leaving the surface of the liquid. (Not boiling) Boiling = A process as above, however, the vapor pressure of the liquid is equal to atmospheric pressure.

*This varies with altitude.

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Solidification (Freezing) Occurs when heat is lost by a liquid. Solids

The particles in a solid are much more ordered than in a liquid

Packed together.

Vibrate in place.

Very dense and is NOT compressible.

Very low rate of diffusion

- If a zinc plate and copper plate are clamped together for a very long time, a few atoms will diffuse into each other.

Diffusion occurs millions of

times slower in solid that liquids.

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Types of Solids 1) Crystalline Solids

Exist as single or group of atoms 3D arrangement of particles in a crystal is called a Crystal Structure

7-types of crystal structures exist:

Triclinic

Hexagonal

Cubic Tetragonal Orthorhombic

Monoclinic Rhombohedral

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4 Types of Crystalline Solids exist: Ionic Molecular Covalent & & Metallic Network Covalent A) Ionic Crystalline Solids - Made of Metal/Nonmetals - Strong bonding forces between + & - ions - Hard & Brittle - High Melting Points B) Metallic Crystalline Solids - Made only of metals - Metals atoms surrounded by a “Sea of Electrons” - Good conductors of electricity - Melting Points vary

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C) Network Covalent Crystalline Solids - Made of single atoms (nonmetals) - Atoms covalently bonded to each other. - Bonding makes up a network -Very hard & brittle - Non-conductor - High melting point D) Molecular Covalent Crystalline Solids - Covalently bonded molecules held to each by intermolecular forces. - Tend to have low M.P. - Vaporize easily - Tend to be soft - Good insulators

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2) Amorphous Solids Not crystalline–no regular shape Very slow flowing Example: foam, glass, plastic

Equilibrium & Phase Changes

Equilibrium A dynamic condition in which 2 opposing changes occur at equal rates in a closed system. Example: A 2-lane highway with vehicles traveling at 55 mph in either direction.

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WaterConsider a dry flask in which room temperature water is added.

One phase exist in the flask.

Phase = any part of a system that has

uniform composition and properties.

When the flask is capped, and the some surface water molecules have enough energy, the liquid water will overcome the intermolecular forces and become a gas.

Liquid + heat energy vapor

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If some of the water vapor falls down to the surface of the water, it will condense.

vapor Liquid + heat energy

After some time, these two process will occur at the same rate – equilibrium.

These two processes can be

summarized with the following equation: Liquid + heat energy vapor

The ideas of equilibrium were first Understood and explained in 1888

by the French chemist, Henri Louis Le Chatelier.

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Le Chatelier develop of the following principle:

Le Chatelier Principle - When a

system at equilibrium is disturbed by a stress, the system will achieve a new equilibrium position in order to counter act the stress.

Consider this principle for this equilibrium process:

Liquid + heat energy vapor

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When all three phases of matter are considered together, the following equilibrium situation must be considered:

Solid Liquid GasVaporization

Freezing

Melting

Sublimation

Deposition

Condensation

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Equilibrium Vapor Pressure The pressure exerted by a vapor in equilibrium with its corresponding liquid at a given temperature. Greater equilibrium vapor pressure when the intermolecular forces decrease. Volatile/Nonvolatile Liquids Volatile Liquid Nonvolatile Liquid

Weak Intermolecular

Forces

StrongIntermolecular

Forces

EvaporateEasily

Do NotEvaporate

Easily

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Heat, Energy & Phase Changes

Energy - the potential or capacity to do work or produce heat.

Kinetic Energy – energy associated

with motion. KE = ½ mv2

Potential Energy – energy of an

object due to its position.

PE = mgh

SI Unit: joule (J)

Non-SI Unit: calorie (cal) 1 cal = 4.184 J

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Heat Flow Heat – energy that is transferred from one object to another due to a difference in their temperatures. Heat flow accompanies all physical

and chemical processes.

All physical and chemical processes occur within the universe.

The universe can be divided into 2 parts: System & Surroundings

System Surroundings

THE UNIVERSE

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Direction & Sign of Heat Flow Heat flow is given the symbol “q”.

If q is “+”, heat flows from the surroundings to the system.

endothermic

Temperature of the system If q is “-”, heat flows to the

surroundings from the system exothermic

Temperature of the system

When a chemical or physical process occurs, we must always be consistent with the Law of Conservation of Mass and Energy.

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Heat flow can be calculated for a chemical or physical process, however, let us first discuss how heat flows to and from a system.

The heat flow, q, can be calculated.

For a pure substance,

q = m x c x T

c = specific heat of the system = amount of heat needed to raise the

temperature of 1 gram of pure substance 1oC Unit = J/g oC

m = mass of the sample

(different for all substances) Unit = g

T = Tfinal - Tinitial

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Unit = oC Specific Heat of Several Pure

Substances at 20oC

Substance Specific Heat (J/g oC)

Air(g) 1.00

Al(s) 0.895

C(graphite) 0.711

C(diamond) 0.502

CO2(g) 0.832

Cu(s) 0.387

Au(s) 0.129

Fe(s) 0.448

Pb(s) 0.128

Ag(s) 0.233

H2O(l) 4.18

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Calorimetry

To measure the heat flow for a chemical or physical process, a device called a calorimeter is used to measure the heat transferred to or from a chemical reaction or physical process by making the following assumption:

qprocess = -qcalorimeter

So,

qprocess = s x m x T

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Example #1: Calculate the quantity of heat (in J) needed to warm water from 25°C to 95°C for making tea. The cup will hold 250 g of water. Answer: Specific Heat (H2O): s = 4.18 J/g°C Mass: m = 250 g ∆T = 95°C - 25°C = 70°C SO, q = (s)(m)(∆T) q = (4.18 J/g°C)(250 g)(70°C) q = 73000 J

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Example #2: 100.00 g of water at 22.4°C is placed in a calorimeter. A 75.25 g sample of aluminum is removed from hot water at a temperature of 99.3°C and quickly placed in the calorimeter. The substances reach a final temperature of 32.9°C. Determine the specific heat of the aluminum. Answer: Step #1: Determine the heat transferred to the water from Al. Step #2: Determine the specific heat of the aluminum.

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Step #1 q = (s)(m)(∆T) s = 4.18 J/g°C m = 100.0 g ∆T = 32.9°C - 22.4°C = 10.5°C q = (4.18 J/g°C)(100.00 g)(10.5°C) q = 4390 J Step #2 q = (s)(m)(∆T) s = ??? m = 75.25 g ∆T = 99.3°C - 32.9°C = 66.4°C q = 4390 J 4390J = (s)(75.25 g)(66.4°C) s = 0.879 J/g°C

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Energy Changes Accompanying a Change in State

During a change in phase, the energy of a substance changes.

Solid Liquid Gas

Freezing Condensation

Melting Evaporation

Heat isabsorbed

Heat isabsorbed

Heat isLost

Heat isLost

What temperature does ice (H2O) melt?

What temperature does H2O freeze?

What temperature does H2O evaporate?

What temperature does H2O condense (liquefy)?

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Regardless of the substance, the melting & the freezing points are the same.

Also, regardless of the substance, the evaporation & the condensing points are the same.

The heat needed to vaporize a liquid into a gas is called the

Heat of Vaporization.

The heat lost when a gas turns into a liquid is called the

Heat of Condensation.

Heatvaporization - Heatcondensation

Similarly, the heat needed to melt a solid into a liquid is called the

Heat of Fusion.

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The heat lost when a liquid turns into a solid is called the

Heat of Solidification.

Heatfusion - Heatsolidification

These relationships can be represented graphically:

Tem

per

atu

re

Boiling

Point

MeltingPoint

Heat Added

Heat of Fusion

Heat of Vaporization

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The heats of fusion & vaporization are expressed in the units of J/g.

Heats of Fusion & Vaporization Substance Heat of

Fusion (J/g) Heat of

Vaporization(J/g)

Cu 205 4726

Ethanol 109 876

Au 64.5 1578

Pb 24.7 858

Ag 88 2300

H2O 334 2260

What is the heat transferred when 20.0 g of ice melts (s = 4.18 J/g oC)?

Let’s use what we have learned:

q = (s)(m)(ΔT), s = 4.18 J/g oC m = 20.0 g ΔT = 0 oC (at melting, no ΔT occurs!) q = (4.18 J/g oC)(20.0 g)(0) = 0 J *** Does this make sense?! ***

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Melting requires no heating??!! *** THAT MAKES NO SENSE! ***

Therefore, calculating the heat transferred during a phase change must be calculated differently.

HeatTransferred

(heat of phase )(mass)

Example #1: Calculate the heat to change a 250.0 g sample of liquid water into steam a 100.0oC. Answer: (Hint: What phase change is happening?) q=(heat of phase change)(m) q=(heat of vaporization)(m) =(2260 J/g)(250.0 g) =565000 J

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Example #2: Calculate the heat needed to melt a 15 g ice cube at 0.0oC to give water, also at 0.0oC. Answer: q=(heat of phase change)(m) q=(heat of fusion)(m) =(334 J/g)(15 g) =5.0 x 103 J = 5.0 kJ