week 2 lectures--tentative

10.7 Kinetic-Molecular Theory 420Application to the Gas Laws

10.8 Molecular Effusion and Diffusion 423Graham's Law of EffusionDiffusion and Mean Free Path

10.9 Real Gases: Deviations from Ideal Behavior 427The van der Waals Equation

Chapter 11 Intermolecular Forces, Liquids, and Solids

11.1 A Molecular Comparison of Liquids and Solids 44411.2 Intermolecular Forces 445

Ion-Dipole ForcesDipole-Dipole ForcesLondon Dispersion ForcesHydrogen BondingComparing Intermolecular Forces

11.3 Some Properties of Liquids 453Viscosity Surface Tension

11.4 Phase Changes 455Energy Changes Accompanying Phase ChangesHeating CurvesCritical Temperature and Pressure

11.5 Vapor Pressure 459Explaining Vapor Pressure on the Molecular LevelVolatility, Vapor Pressure, and TemperatureVapor Pressure and Boiling Point

Week 2 lectures--tentative

• Theory developed to explain gas behavior.• Theory based on properties at the molecular level.• Kinetic molecular theory gives us a model for

understanding pressure and temperature at the molecular level.

• Pressure of a gas results from the number of collisions per unit time on the walls of container.

10.7 Kinetic Molecular Theory10.7 Kinetic Molecular Theory

• There is a spread of individual energies of gas molecules in any sample of gas.

• As the temperature increases, the average kinetic energy of the gas molecules increases.

Kinetic Molecular TheoryKinetic Molecular Theory

• Assumptions:– Gases consist of a large number of molecules in constant

random motion.

– Volume of individual molecules negligible compared to volume of container.

– Intermolecular forces (attractive or repulsive forces between gas molecules) are negligible.

– Energy can be transferred between molecules, but total kinetic energy is constant at constant temperature.

– Average kinetic energy of molecules is proportional to temperature.

10.7 Kinetic Molecular Theory10.7 Kinetic Molecular Theory

• Magnitude of pressure given by how often and how hard the molecules strike.

• Gas molecules have an average kinetic energy.

• Each molecule may have a different energy.

• As kinetic energy increases, the velocity of the gas molecules increases.

• Root mean square speed, u, is the speed of a gas molecule having average kinetic energy.

• Average kinetic energy, , is related to root mean square speed:

221 mu

Do you remember how to calculatevxy from vx and vy ?

22yxxy vvv

And how about v from all threecomponents?

222zyx vvvv

Remember these equations!! They’ll popup again in Chap. 11.

Note that the mean value of velocity is

The Maxwell-Boltzmann Distribution of Velocities

The Maxwell Distribution of Speeds

3Speedrms

8SpeedAverage

2SpeedProbaleMost

rmsrms

225.1:128.1:13:8

:2::, 212

rmsmp vvvAnd

ump<u>

The Maxwell-Boltzmann Distribution of Velocities

This is also theform of a Gaussian (normal) distribution,where ump = <u> = urms.

Application to Gas Laws• As volume increases at constant temperature, the average

kinetic of the gas remains constant. Therefore, u is constant. However, as the volume increases the gas molecules have to travel further to hit the walls of the container. Therefore, pressure decreases.

• If temperature increases at constant volume, the average kinetic energy of the gas molecules increases. Therefore, there are more collisions with the container walls and the pressure increases.

Molecular Effusion and Diffusion• As kinetic energy increases, the velocity of the gas

molecules increases.• Average kinetic energy of a gas is related to its mass:

• Consider two gases at the same temperature: the lighter gas has a higher velocity than the heavier gas.

• Mathematically:

221 mu

Molecular Effusion and Diffusion

• The lower the molar mass, M, the higher the urms.

SAMPLE EXERCISE 10.14 Calculating a Root-Mean-Square Speed

Calculate the rms speed, u, of an N2 molecule at 25°C.

Comment: This corresponds to a speed of 1150 mi/hr. Because the average molecular weight of air molecules is slightly greater than that of N2, the rms speed of air molecules is a little slower than that for N2. The speed at which sound propagates through air is about 350 m/s, a value about two-thirds the average rms speed for air molecules.

SolutionAnalyze: We are given the identity of the gas and the temperature, the two quantities we need to calculate the rms speed.Plan: We will calculate the rms speed using Equation 10.22.

Solve: In using Equation 10.22, we should convert each quantity to SI units so that all the units are compatible. We will also use R in units of J/mol-K (Table 10.2) in order to make the units cancel correctly.

(These units follow from the fact that 1 J = 1 kg-m2/s2 )

RTurms

Graham’s Law of Effusion• As kinetic energy increases,

the velocity of the gas molecules increases.

• Effusion is the escape of a gas through a tiny hole.

• The rate of effusion can be quantified.

Graham’s Law of Effusion

• Consider two gases with molar masses M1 and M2, the relative rate of effusion is given by:

• Only those molecules that hit the small hole will escape through it.

• Therefore, the higher the urms the greater the likelihood of a gas molecule hitting the hole.

SAMPLE EXERCISE 10.15 Applying Graham’s Law

An unknown gas composed of homonuclear diatomic molecules effuses at a rate that is only 0.355 times that of O2 at the same temperature. Calculate the molar mass of the unknown, and identify it.

SolutionAnalyze: We are given the rate of effusion of an unknown gas relative to that of O2, and we are asked to find the molar mass and identity of the unknown. Thus, we need to connect relative rates of effusion to relative molar masses.Plan: We can use Graham’s law of effusion, Equation 10.23, to determine the molar mass of the unknown gas. If we let rx and represent the rate of effusion and molar mass of the unknown gas, Equation 10.23 can be written as follows:

Solve: From the information given,

Because we are told that the unknown gas is composed of homonuclear diatomic molecules, it must be an element. The molar mass must represent twice the atomic weight of the atoms in the unknown gas. We conclude that the unknown gas is I2.

SAMPLE EXERCISE 10.15 continued

We now solve for the unknown molar mass,

PRACTICE EXERCISECalculate the ratio of the effusion rates of

Graham’s Law of Effusion

• Consider two gases with molar masses M1 and M2, the relative rate of effusion is given by:

• Only those molecules that hit the small hole will escape through it.

• Therefore, the higher the rms the more likelihood of a gas molecule hitting the hole.

Diffusion and Mean Free Path • Diffusion of a gas is the spread of the gas through space.• Diffusion is faster for light gas molecules.• Diffusion is significantly slower than rms speed (consider

someone opening a perfume bottle: it takes while to detect the odor but rms speed at 25C is about 1150 mi/hr).

• Diffusion is slowed by gas molecules colliding with each other.

Diffusion and Mean Free Path • Average distance of a gas molecule between collisions is

called mean free path.• At sea level, mean free path is about 6 10-6 cm.

• From the ideal gas equation, we have

• For 1 mol of gas, PV/nRT = 1 for all pressures.• In a real gas, PV/nRT varies from 1 significantly and is

called Z.

• The higher the pressure the more the deviation from ideal behavior.

Real Gases: Deviations Real Gases: Deviations from Ideal Behaviorfrom Ideal Behavior

• From the ideal gas equation, we have

• For 1 mol of gas, PV/RT = 1 for all temperatures.• As temperature increases, the gases behave more ideally.• The assumptions in kinetic molecular theory show where

ideal gas behavior breaks down:– the molecules of a gas have finite volume;

– molecules of a gas do attract each other.

• As the pressure on a gas increases, the molecules are forced closer together.

• As the molecules get closer together, the volume of the container gets smaller.

• The smaller the container, the more space the gas molecules begin to occupy.

• Therefore, the higher the pressure, the less the gas resembles an ideal gas.

• As the gas molecules get closer together, the smaller the intermolecular distance.

• The smaller the distance between gas molecules, the more likely attractive forces will develop between the molecules.

• Therefore, the less the gas resembles and ideal gas.• As temperature increases, the gas molecules move faster

and further apart.• Also, higher temperatures mean more energy available to

break intermolecular forces.

• Therefore, the higher the temperature, the more ideal the gas.

The van der Waals Equation• We add two terms to the ideal gas equation one to correct

for volume of molecules and the other to correct for intermolecular attractions

• The correction terms generate the van der Waals equation:

where a and b are empirical constants characteristic of each gas.

The van der Waals Equation

• General form of the van der Waals equation:

nRTnbVV

Corrects for molecular volume

Corrects for molecular attraction

SAMPLE EXERCISE 10.16 Using the van der Waals Equation

If 1.000 mol of an ideal gas were confined to 22.41 L at 0.0°C, it would exert a pressure of 1.000 atm. Use the van der Waals equation and the constants in Table 10.3 to estimate the pressure exerted by 1.000 mol of Cl2(g) in 22.41 L at 0.0°C.

Check: We expect a pressure not far from 1.000 atm, which would be the value for an ideal gas, so our answer seems very reasonable.

SolutionAnalyze: The quantity we need to solve for is pressure. Because we will use the van der Waals equation, we must identify the appropriate values for the constants that appear there.Plan: Using Equation 10.26, we have

Solve: Substituting n = 1.000 mol, R = 0.08206 L-atm/mol-K, T = 273.2 K, V = 22.41 L, a = 6.49 L2-atm/mol2, and b = 0.0562 l/mol:

Comment: Notice that the first term, 1.003 atm, is the pressure corrected for molecular volume. This value is higher than the ideal value, 1.000 atm, because the volume in which the molecules are free to move is smaller than the container volume, 22.41 L. Thus, the molecules must collide more frequently with the container walls. The second factor, 0.013 atm, corrects for intermolecular forces. The intermolecular attractions between molecules reduce the pressure to 0.990 atm. We can conclude, therefore, that the intermolecular attractions are the main cause of the slight deviation of Cl2(g) from ideal behavior under the stated experimental conditions.

SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together

Cyanogen, a highly toxic gas, is composed of 46.2% C and 53.8% N by mass. At 25°C and 751 torr, 1.05 g of cyanogen occupies 0.500 L. (a) What is the molecular formula of cyanogen? (b) Predict its molecular structure. (c) Predict the polarity of the compound.

Because the ratio of the moles of the two elements is essentially 1:1, the empirical formula is CN.To determine the molar mass of the compound, we use Equation 10.11.

SolutionAnalyze: First we need to determine the molecular formula of a compound from elemental analysis data and data on the properties of the gaseous substance. Thus, we have two separate calculations to do.(a) Plan: We can use the percentage composition of the compound to calculate its empirical formula. • (Section 3.5) Then we can determine the molecular formula by comparing the mass of the empirical formula with the molar mass. • (Section 3.5)Solve: To determine the empirical formula, we assume that we have a 100-g sample of the compound and then calculate the number of moles of each element in the sample:

SAMPLE INTEGRATIVE EXERCISE continued

The molar mass associated with the empirical formula, CN, is 12.0 + 14.0 = 26.0 g/mol. Dividing the molar mass of the compound by that of its empirical formula gives (52.0 g/mol)/(26.0 g/mol) = 2.00. Thus, the molecule has twice as many atoms of each element as the empirical formula, giving the molecular formula C 2N2

The Lewis structure shows that each atom has two electron domains. (Each nitrogen has a nonbonding pair of electrons and a triple bond, whereas each carbon has a triple bond and a single bond.) Thus the electron-domain geometry around each atom is linear, causing the overall molecule to be linear.

(c) Plan: To determine the polarity of the molecule, we must examine the polarity of the individual bonds and the overall geometry of the molecule.Solve: Because the molecule is linear, we expect the two dipoles created by the polarity in the carbon–nitrogen bond to cancel each other, leaving the molecule with no dipole moment.

(This structure has zero formal charges on each atom.)

(b) Plan: To determine the molecular structure of the molecule, we must first determine its Lewis structure. • (Section 8.5) We can then use the VSEPR model to predict the structure. • (Section 9.2)Solve: The molecule has 2(4) + 2(5) = 18 valence-shell electrons. By trial and error, we seek a Lewis structure with 18 valence electrons in which each atom has an octet and in which the formal charges are as low as possible. The following structure meets these criteria:

Chapter 11 -- Chapter 11 -- Intermolecular Forces, Intermolecular Forces,

Liquids, and SolidsLiquids, and Solids

In many ways, this chapter is simply acontinuation of our earlier discussion of‘real’ gases.

Remember this nice, regular behavior described by the ideal gas equation.

This plot for SO2 is a morerepresentativeone of real systems!!!

This plot includes a realistic one for Volume as a function of Temperature!

Why do the boiling points vary? Is there anything systematic?

What determines whether a substance existsas a gas, liquid, or solid?

Two primary factors are involved:

Kinetic Energy of the particles.

Strength of attractions betweenthe particles.

What are the important Intermolecular Forces i.e, forces between molecules ?

Note that earlier chapters concentrated on Intramolecular Forces, those within the molecule.

Important ones:

ion-ion similar to atomic systems

ion-dipole (review definition of dipoles)

dipole-dipole

dipole-induced dipole

London Dispersion Forces:(induced dipole-induced dipole) related to polarizability

Hydrogen Bonding

van der Waalsforces

How do you know the relative strengthsof each? Virtually impossible experimentally!!!

Most important though: Establish which are present. London Dispersion Forces: Always All others depend on defining property such as existing dipole for d-d.

It has been possible to calculate therelative strengths in a few cases.

Relative Energies of Various Interactions

d-d d-id disp

Ar 0 0 50

N2 0 0 58

C6H6 0 0 1086

C3H8 0.0008 0.09 528

HCl 22 6 106

CH2Cl2 106 33 570

SO2 114 20 205

H2O 190 11 38

HCN 1277 46 111

Primary factor here is London Dispersion Forces

Ion-Dipole Interactions

• A fourth type of force, ion-dipole interactions are an important force in solutions of ions.

• The strength of these forces are what make it possible for ionic substances to dissolve in polar solvents.

Dipole-Dipole Forces

Ion-dipole interaction

Let’s take a closer look at these interactions:

Dipole-dipole interactions.This is the simple one.

But we also have to consider other shapes.Review hybridization and molecular shapes.

Recall the discussion of sp, sp2, and sp3

hybridization?

Dipole-dipole interactions

London dispersion forcesor induced dipole-induced dipole

A Polarized He atomwith an induced dipole

molecule F2 Cl2 Br2 I2 CH4

polarizability 1.3 4.6 6.7 10.2 2.6

molecular wt. 37 71 160 254 16

Molecular Weight predicts the trends in the boiling points of atoms or molecules without dipole moments because polarizability tends to increase with increasing mass.

But polarizability also depends on shape, as well as MW.

Water provides our best example of Hydrogen Bonding.

These boiling points demonstrate the enormouscontribution of hydrogen bonding.

Water is alsounusual in the relative densities of the liquid and solid phases.

The crystal structure suggests a reason for the unusual

high density of ice.

But hydrogen bonding is not limited to water:

But water isn’t the only substance to show hydrogen bonding!

Viscosity—the resistance to flow of a liquid, such asoil, water, gasoline, molasses, (glass !!!)

Surface Tension – tendency to minimize the surface areacompare water, mercury

Cohesive forces—bind similar molecules together

Adhesive forces – bind a substance to a surface

Capillary action results when these two are not equal

Soap reduces the surface tension, permitting onematerial to ‘wet’ another more easily

11.3 Some Properties of Liquids11.3 Some Properties of Liquids

Viscosity• Resistance of a liquid

to flow is called viscosity.

• It is related to the ease with which molecules can move past each other.

• Viscosity increases with stronger intermolecular forces and decreases with higher temperature.

The SI unit is kg/m-s. Many tables still use the older unit of viscosity, the poise,

which is 1 g/cm-s, with typical values listed as cP = 0.01 P.

Surface Tension

Surface tension results from the net inward force experienced by the molecules on the surface of a liquid.

RationaleforSurfaceTension

Surface Tension• Surface molecules are only attracted inwards towards the

bulk molecules.– Therefore, surface molecules are packed more closely than bulk

molecules.

• Surface tension is the amount of energy required to increase the surface area of a liquid, in J/m2.

• Cohesive forces bind molecules to each other.• Adhesive forces bind molecules to a surface.

Surface Tension• Meniscus is the shape of the liquid surface.

– If adhesive forces are greater than cohesive forces, the liquid surface is attracted to its container more than the bulk molecules. Therefore, the meniscus is U-shaped (e.g. water in glass).

– If cohesive forces are greater than adhesive forces, the meniscus is curved downwards.

• Capillary Action: When a narrow glass tube is placed in water, the meniscus pulls the water up the tube.

• Remember that surface molecules are only attracted inwards towards the bulk molecules.

also called

FUSION

• Sublimation: solid gas.• Vaporization: liquid gas.• Melting or fusion: solid liquid.• Deposition: gas solid.• Condensation: gas liquid.• Freezing: liquid solid.

Phase ChangesPhase Changes

Cp(s):37.62

J/mol-K

ΔHfus:6,010 J/mol

Cp(l):72.24

J/mol-K

ΔHvap:40,670 J/mol

Cp(g):33.12

J/mol-K

SAMPLE EXERCISE 11.4 Calculating H for Temperature and Phase Changes

Calculate the enthalpy change upon converting 1.00 mol of ice at –25°C to water vapor (steam) at 125°C under a constant pressure of 1 atm. The specific heats of ice, water, and steam are 2.09 J/g-K, 4.18 J/g-K and 1.84 J/g-K, respectively. For H2O, Hfus = 6.01 kJ/mol and Hvap = 40.67 kJ/mol.

SolutionAnalyze: Our goal is to calculate the total heat required to convert 1 mol of ice at –25°C to steam at 125°C.Plan: We can calculate the enthalpy change for each segment and then sum them to get the total enthalpy change (Hess’s law, Section 5.6).Solve: For segment AB in Figure 11.19, we are adding enough heat to ice to increase its temperature by 25°C. A temperature change of 25°C is the same as a temperature change of 25 K, so we can use the specific heat of ice to calculate the enthalpy change during this process:

For segment BC in Figure 11.19, in which we convert ice to water at 0°C, we can use the molar enthalpy of fusion directly:

The enthalpy changes for segments CD, DE, and EF can be calculated in similar fashion:

Check: The components of the total energy change are reasonable in comparison with the lengths of the horizontal segments of the lines in Figure 11.19. Notice that the largest component is the heat of vaporization.

The total enthalpy change is the sum of the changes of the individual steps:

11.5 Vapor Pressure• Explaining Vapor Pressure on the Molecular Level, Volatility, • Vapor Pressure, and Temperature; • Vapor Pressure and Boiling Point

11.6 Phase Diagrams• The Phase Diagrams of H2O and CO2

11.7 Structures of Solids• Unit Cells• The Crystal Structure of Sodium Chloride• Close Packing of Spheres

11.8 Bonding in Solids• Molecular Solids• Covalent-Network Solids• Ionic Solids• Metallic Solids

Chapter 13 Properties of Solutions13.1 The Solution Process

• Energy Changes and Solution Formation• Solution Formation, Spontaneity, and Disorder• Solution Formation and Chemical Reactions

The observation of Vapor Pressure:

And, at higher temperatures, the vp is higher.

The samepicture isuseful torationalizeVapor Pressure

With the beaker covered, equilibrium issoon established.

Contrast that with an uncovered beaker!

NormalBP

defined

But recall thedefinition of ANY BoilingPoint.

(Add slide of C-C eq and plot.)

But, recall, we always prefer straight lines!

R TCvap

← Temp. increases

O r w ith P th e va p o r p ressu re a t T a n d P th e va p o r p ressu re a t T

R TC a n d P

su b tra c t to g ive

va p va p

1 1 2 2

2 12 1

Using the Clausius-Clapeyron Equation:

Tabulate P in atm and T in K; Calculate lnP and 1/TPlot lnP vs 1/T

The slope is = - ΔHvap /R

The slope!

week 2 lectures--tentative

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