Chapter 12 Liquids, Solids, and Intermolecular Forces

Chapter 12

Liquids, Solids, and Intermolecular Forces

Homework

• Assigned Problems (odd numbers only)• Sections 12.4 to 12.6 • “Problems” (41-49), (51-61), (63-69)• “Cumulative Problems” (87 to 93), (95-97)• “Challenge Questions” 105 only (page 443)

Interactions between Molecules

• Kinetic molecular theory of matter states that the particles present in any phase of matter are in constant, random motion: kinetic energy

• Kinetic (thermal) energy is temperature dependent and increases with increasing temperature

• The particles interact together through attractions and repulsions that creates potential energy within the particles


• Kinetic energy gives the particles their motion and tends to move the particles away from each other (disruptive force)

• Potential energy is stored energy that matter possesses mainly due to electrostatic interactions from positive and negatively charged particles (cohesive force)These are attractions and repulsions in which

positively and negatively charged particles attract and repel each other


• Temperature influences the kinetic energy of the particles and this plays an important role in determining the physical state of a system (i.e. solid, liquid, or gas)

• The charged particles interact through attractions and repulsions and therefore contain potential energy– Particles of opposite electrical charge will attract– Particles of identical charge will repel


• The relative influence of kinetic and potential energy is the main consideration when KM theory is used to explain the general properties of the gas, liquid, or solid states of matter

• The type of energy that dominates will influence a substance’s physical state


• Intermolecular Forces are forces that act between a molecule and another molecule

• They are electrostatic forces that are similar to the intramolecular forces involved in ionic and covalent bonding

• It is the strength of the intermolecular forces that determine the physical state of molecular substances at room temperature

Properties of Liquids and Solids

• The solid state is predominated by potential energy rather than by (thermal) kinetic energy

• Particles are in a fixed position by strong electrostatic attractions but vibrate due to kinetic energy

• Definite volume and shape and do not assume the shape of their container

• High density: The particles are located as close as possible

Properties of Liquids and Solids• The liquid state is not dominated by potential energy

or by kinetic energy• Particles are not in a fixed position due to kinetic

energy which provides just enough motion energy for the particles to slide over each other.

• The potential energy (cohesive force) is strong enough prevent total separation and retain a fixed volume– Assumes the shape of the container it occupies– Definite volume and indefinite shape– High density: The particles are not widely

separated but located relatively close together

Properties of Liquids and Solids• The gaseous state is dominated completely by kinetic

energy• Particles of a gas are independent of each other and

move in a totally random manner due to their kinetic energy

• The attractive forces between particles have been overcome by kinetic energy which allows particles to travel in all directions

• Assumes both the volume and shape of the container it occupies

• Indefinite volume and indefinite shape• Low density: The particles are widely separated and

relatively few particles per unit volume

Properties of Liquids and Solids• Liquid and solid phases have many similar

characteristics but gases are very different• The average distance between the particles is

only slightly different in the solid and liquid but vastly different in the gaseous state

Evaporation and Condensation• Evaporation is the process by which molecules

escape from the liquid phase to the gas phase• According to KM theory, at any given instant, not all

molecules will have the same kinetic energy• The molecules with above average KE can overcome

the attractive forces that are holding them in the liquid’s surface and escape into the gas phase

Evaporation and Condensation• Liquids are constantly evaporating and molecules at the

surface break away from the liquid and enter into the gas phase

• An increase in temperature will give molecules the minimum needed kinetic energy to escape from the intermolecular attractions of other molecules

• So, the rate of evaporation always increases as the temperature increases.

Vapor Pressure • Liquids are constantly evaporating and

molecules at the surface break away from the liquid and enter the gas phase

• A substance that readily evaporates (at room temp.) is described as a volatile substance

• The molecules that escape from an evaporating liquid are referred to as a vapor

• Vapor: The gaseous state of a substance at a temperature and pressure at which the substance is normally a liquid

Vapor Pressure

• In a closed container, the molecules continue to escape the liquid phase increasing their concentration in the gas phase

• As the concentration increases in the gas phase (vaporization) , some of the particles return back to the liquid phase (condensation)

• Since vaporization and condensation are occurring simultaneously, eventually a condition of equilibrium is established

Vapor Pressure• At equilibrium, the rates of evaporation and

condensation are the same • The concentration of molecules in the liquid

phase and gas phases are no longer changing• The vapor pressure is the pressure exerted by a

vapor above a liquid when the liquid and vapor are in equilibrium

Boiling• Boiling is a special form of

evaporation in which conversion from the liquid to the vapor state occurs within the body of a liquid through bubble formation

• It occurs when a liquid is heated and the vapor pressure of the liquid reaches a value of the outside pressure

• The normal boiling point of a liquid is the temp. at which its vapor pressure equals atmospheric pressure of 1 atm

Energetics of Evaporation and Condensation

• Evaporation is the process by which molecules escape from the liquid phase to the gas phase

• It is an endothermic process and requires the absorption of heat energy

• The rate of evaporation always increases as a liquid’s temperature increases

• This increases the number of molecules that possess the minimum KE needed to overcome the attractive forces (escape to vapor phase)

Energetics of Evaporation and Condensation

• Condensation is the process by which a gas is changed to a liquid

• This change of state is the reverse process of evaporation

• It is an exothermic process and requires the release of heat energy

• Energy is released as the intermolecular forces increase in number

Heating CurveHeating CurveA heating curve is a plot of temperature versus time A heating curve is a plot of temperature versus time

with a constant amount of heat addedwith a constant amount of heat addedIllustrates the steps involved in changing a solid to a Illustrates the steps involved in changing a solid to a

gasgasHeat added is shown on the x-axisHeat added is shown on the x-axisTemperature is shown on the y-axisTemperature is shown on the y-axis

Energy required to undergo a series of phase Energy required to undergo a series of phase changes depends on the (three) property values of changes depends on the (three) property values of the substance:the substance:Specific HeatSpecific HeatHeat of FusionHeat of FusionHeat of VaporizationHeat of Vaporization

Heating CurveHeating Curve

No temp change

No temp change

Heating Curve during Boiling• Boiling is a phase

change and the process occurs at a constant temperature

• The energy absorbed is only used to overcome the intermolecular forces

• The energy is also released as intermolecular forces are increased

Heating Curve for Water

VaporizationVaporization/Condensation/CondensationHeat of Vaporization

Heat energy required to vaporize one mole of a substance (e.g. water)

Heat energy that must be removed to condense one mole of a substance (e.g. water)

molkJ40.7onvaporizatiofheat water

)()( 22 gOHOH

)()( 22 OHgOH

gJ2260 onvaporizatiofheat water

Change of State ProblemChange of State ProblemCalculate the heat neededCalculate the heat needed(in Joules) to heat 15 g of(in Joules) to heat 15 g ofwater from 75 °C to 100 °C,water from 75 °C to 100 °C,and to convert it to steamand to convert it to steamat 100 °Cat 100 °C

Two parts to the problemTwo parts to the problemHeat the waterHeat the water (use (use

specific heat for water)specific heat for water)Convert water to steamConvert water to steam

(use(use heat of vaporization heat of vaporization for water)for water)

Change of State ProblemChange of State Problem• Heat the waterHeat the waterCalculate the heat energy needed to warm the water from 75 °C Calculate the heat energy needed to warm the water from 75 °C

to 100 °Cto 100 °C

J 1570 g15.0

Cg

J4.184SH OH2

Cg

J 4.184

)( CC 75100

Δt(g)massSH q OH21

×25 °C

Given: 15.0 g H2O Find: kJ

)()( 22 OHOH Ti = 75 °C Tf = 100 °C

1q

Change of State ProblemChange of State ProblemCalculate the heat required to vaporize 15.0 g of liquid Calculate the heat required to vaporize 15.0 g of liquid

water to steam at 100 °Cwater to steam at 100 °C

No change in temperature for a phase changeNo change in temperature for a phase change

kJ33.9 J10

kJ3

OH g 15.0 2

g

J2260 ΔHvap


OH g

J 2260

2

)()( 22 gOHOH Ti = 100 °C Tf = 100 °C

2q

Change of State ProblemChange of State ProblemCalculate the heat required to vaporize 15.0 g of liquid Calculate the heat required to vaporize 15.0 g of liquid

water to steam at 100 °Cwater to steam at 100 °C


kJ33.9 OH mol

kJ40.7

2

OH g 15.0 2

mol

kJ40.7ΔHvap


OH g18.02

OH mol 1

2

2

)()( 22 gOHOH Ti = 100 °C Tf = 100 °C

1 mol H2O = 18.02 g H2O

2q

Change of State ProblemChange of State ProblemCombining Energy CalculationsCombining Energy Calculations

Calculate the total heatCalculate the total heatHeat the water (qHeat the water (q11))

Convert liquid water to steam (qConvert liquid water to steam (q22)

J35,500qtotal

total21 qqq

J 1,570 q1

J33,900q2

Energetics of Melting and Freezing• Melting is the conversion of a solid to a liquid• It is an endothermic process that requires the

absorption of heat• The heat energy absorbed is used to partially

disrupt the intermolecular attractions, allowing the solid to melt

• Freezing is the reverse process of melting and converts a liquid to a solid

• It is an exothermic process that requires the release of heat

MeltingMelting/Freezing/FreezingReversible, change of state processesHeat of Fusion

Heat energy required to melt 1 mole of a substance (e.g. water)

Heat energy that must be removed to freeze 1 mole of a substance (e.g. water)

Heat energy (to melt) 1 g of ice (water at 0 °C):

molkJ6.02 fusionofheat water

)()( 22 OHsOH

)()( 22 sOHOH

gJ334 fusionofheat water

Change of State ProblemChange of State Problem

• Two parts to the problemMelt ice (use heat

of fusion for water)Heat water

(use specific heat of water)

Calculate the heat required (in Joules) to melt 15 Calculate the heat required (in Joules) to melt 15 g of ice at 0°C, and to heat the water to 75 °Cg of ice at 0°C, and to heat the water to 75 °C

Change of State ProblemChange of State ProblemCalculate the heat required to melt 15.0 g of Calculate the heat required to melt 15.0 g of

ice to liquid water at 0 °C ice to liquid water at 0 °C


kJ5.01 3 J10kJOHg 15.0

2

g

J334ΔHfus


OH g

J 334

2

)()( 22 OHsOH Ti = 0 °C Tf = 0 °C

Change of State ProblemChange of State ProblemCalculate the heat required to melt 15.0 g of Calculate the heat required to melt 15.0 g of

ice to liquid water at 0 °C ice to liquid water at 0 °C


kJ5.01 mol

kJ6.02OHg 15.0

2

mol

kJ6.02ΔH fus


OH g18.02

OH mol 1

2

2

)()( 22 OHsOH Ti = 0 °C Tf = 0 °C

1 mol H2O = 18.02 g H2O

Change of State ProblemChange of State Problem• Heat the waterHeat the waterCalculate the heat energy needed to warm the water from 0°C to 75 °CCalculate the heat energy needed to warm the water from 0°C to 75 °C

J 4707 g15.0

Cg

J4.184SH OH2

Cg

J 4.184

)5( CC 07

Δt(g)massSH OH2

×75 °C


)()( 22 OHOH Ti = 0 °C Tf = 75 °C

Change of State ProblemChange of State ProblemCombining Energy CalculationsCombining Energy Calculations

Calculate the total heatCalculate the total heatMelt the ice (qMelt the ice (q11))

Heat the liquid water (qHeat the liquid water (q22)

J9,717qtotal

total21 qqq

J 5,010 q1

J4,707q2

SublimationSublimation• Sublimation: A phase change from solid to gas without Sublimation: A phase change from solid to gas without

going through the liquid stategoing through the liquid state• For example, the sublimation of water and carbon dioxideFor example, the sublimation of water and carbon dioxide

Requires the absorption of heatRequires the absorption of heatNo temperature change occurs during this processNo temperature change occurs during this processDeposition is the reverse process (heat is released)Deposition is the reverse process (heat is released)

)()( 22 gOHsOHHeat

)()( 22 gCOsCOHeat

Intermolecular Forces• The strength of the attractive forces that exist

between the molecules when they are close together will determine whether a substance is a solid, liquid, or a gas (e.g. water is a liquid, methane is gas, and glucose is a solid)

• Intermolecular forces are the attractive forces that act between a molecule and another molecule

• The type and magnitude of these forces will determine the characteristics of a compound:

Relative Strengths of the Intermolecular Forces:covalent bonds >>> hydrogen bonds > dipole-dipole forces > London Forces

intramolecular intermolecular

Intermolecular ForcesIntermolecular forces influence:

The physical state of a substance at room temperatureThe boiling and freezing points of a substance

• There are two factors that cause intermolecular forces:

• The polarity of the bonds within molecules • The unsymmetrical movement of the electrons about the

nuclei

Dispersion ForceDispersion ForceDispersion (London) forces: Short-lived dipoles caused by

uneven shifts in electron densityThe uneven shift causes one end of the molecule to be

slightly positive and one end slightly negativeThis induces the same electron shift in adjacent molecules

which creates a network of attractive forcesAll molecules can form these instantaneous dipoles but it

is the only type of intermolecular force possible in non-polar substances

The magnitude of this force increases with increasing molar mass

Dipole-dipole ForceDipole-dipole ForceDipole-dipole: Attractive forces between molecules that have

permanent dipoles (polar molecules)The nonsymmetrical distribution of the charge causes the molecules

to line up Positive end of one directed (attracted) toward negative end of

otherOnly molecules that possess permanent dipoles can engage in this

type of interactionThe magnitude of this interaction increases as the polarity of the

molecule increases

Hydrogen BondingHydrogen BondingThe strongest of the three attractive forces is hydrogen bonding: A

special type of dipole-dipole interactionOccurs between molecules that have a H atom bonded to F, O, or NHighly polar, covalent bonds with a H atom bonded with a small,

highly electronegative atom leaving a significant partial positive charge on hydrogen

The small size of the hydrogen atom allows close proximity to lone pair electrons on an F, O, or N of another molecule

O H

H

O

H

H

Heats of Fusion and Vaporization vs. Intermolecular ForcesHeats of Fusion and Vaporization vs. Intermolecular Forces

Dispersionforces

Dipole-dipole

H-bonding Ionic forces

• End

Documents

Chapter 12 Liquids, Solids, and Intermolecular Forces