rms

.

Kinetic Molecular Theory

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

.

Kinetic Molecular Theory

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:   1

mu

2 2 • Consider two gases at the same temperature: the lighter gas has a higher velocity than the heavier gas.

• Mathematically:

u

 3

RT

M

Kinetic Molecular Theory

Molecular Effusion and Diffusion

• The lower the molar mass,

M

, the higher the u rms .

3

RT M

1 2 

u rms

SAMPLE EXERCISE 10.14

Calculating a Root-Mean-Square Speed

Calculate the rms speed,

u

, of an N 2 molecule at 25°C.

Solution Analyze:

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.

u rms

 3

RT M

1 2

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-m 2 /s 2 )

Comment:

This corresponds to a speed of 1150 mi/hr. Because the average molecular weight of air molecules is slightly greater than that of N 2 , the rms speed of air molecules is a little slower than that for N 2 . 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.

Kinetic Molecular Theory

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.

Kinetic Molecular Theory

Graham’s Law of Effusion

• Consider two gases with molar masses

M 1

and

M 2

, the relative rate of effusion is given by:

r

1

r

2 

M M

2 1 • Only those molecules that hit the small hole will escape through it.

• Therefore, the higher the u rms 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 O 2 at the same temperature. Calculate the molar mass of the unknown, and identify it.

Solution Analyze:

We are given the rate of effusion of an unknown gas relative to that of O 2 , 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

r x

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, Thus,

SAMPLE EXERCISE 10.15

continued We now solve for the unknown molar mass, 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 I 2 .

PRACTICE EXERCISE

Calculate the ratio of the effusion rates of

Kinetic Molecular Theory

Graham’s Law of Effusion

• Consider two gases with molar masses

M 1

and

M 2

, the relative rate of effusion is given by:

r r

1 2 

u

1

u

2  3

RT M

1 3

RT M

2 

M M

2 1 • 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.

Kinetic Molecular Theory

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 25  C is about 1150 mi/hr).

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

Kinetic Molecular Theory

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.

Real Gases: Deviations from Ideal Behavior

• From the ideal gas equation, we have

PV



n RT or PV nRT

 1 • For 1 mol of gas,

PV

/n

RT

= 1 for all pressures.

• In a real gas,

PV

/n

RT

varies from 1 significantly and is called Z.

Z



PV nRT

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

Real Gases: Deviations from Ideal Behavior

• From the ideal gas equation, we have

PV RT



n

• 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.

Real Gases: Deviations from Ideal Behavior

• 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.

Real Gases: Deviations from Ideal Behavior

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

Real Gases: Deviations from Ideal Behavior

• 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.

Real Gases: Deviations from Ideal Behavior

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

Real Gases: Deviations from Ideal Behavior

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:

P



V nRT



nb



n

2

a V

2 where

a

and each gas.

b

are empirical constants characteristic of

Real Gases: Deviations from Ideal Behavior

The van der Waals Equation

P



V nRT



nb



n

2

a V

2 Corrects for molecular volume Corrects for molecular attraction • General form of the van der Waals equation:    

P



n

2

a V

2     

V



nb

 

nRT

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 Cl 2 (

g

) in 22.41 L at 0.0°C.

Solution Analyze:

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 L 2 atm/mol 2 , and

b

= 0.0562 l/mol:

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.

SAMPLE EXERCISE 10.16

continued

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 Cl 2 (

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.

Solution Analyze:

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: 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.

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 2 N 2

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

Chapter 11 - Intermolecular Forces, Liquids, and Solids

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

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

This plot for SO 2 is a more representative one 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 exists as a gas, liquid, or solid?

Two primary factors are involved:

Kinetic Energy of the particles

.

Strength of attractions between the 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 ion-dipole similar to atomic systems (review definition of

dipoles

) dipole-dipole dipole-induced dipole London Dispersion Forces: (induced dipole-induced dipole) related to polarizability van der Waals forces Hydrogen Bonding

How do you know the relative strengths of 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 the relative strengths in a few cases.

Relative

Energies of Various Interactions Ar N 2 C 6 H 6 C 3 H 8 HCl CH 2 Cl 2 SO 2 H 2 O HCN 0 0

d-d

0 0.0008

22 106 114 190 1277 0 0

d-id

0 0.09

6 33 20 11 46

disp

50 58 1086 528 106 570 205 38 111

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

Let’s take a closer look at these interactions:

Ion-dipole interaction

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, sp 2 , and sp 3 hybridization?

Dipole-dipole interactions

A Polarized He atom with an induced dipole

London dispersion forces or induced dipole-induced dipole

molecule polarizability molecular wt.

F 2 1.3

37 Cl 2 4.6

71 Br 2 6.7

160 I 2 10.2

254 CH 4 2.6

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 enormous contribution of hydrogen bonding.

Water is also unusual 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!

11.3 Some Properties of Liquids

Viscosity

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

Surface Tension

– tendency to minimize the surface area compare 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 one material to ‘wet’ another more easily

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.

Rationale for

Surface Tension

• •

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/m 2 .

Cohesive forces Adhesive forces

bind molecules to each other.

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

Phase Changes

• • • • • •

Sublimation

: solid  gas.

Vaporization

: liquid  gas.

Melting

or

fusion

: solid 

Deposition

: gas  solid.

liquid.

Condensation

: gas 

Freezing

: liquid  liquid.

solid.

ΔH fus : 6,010 J/mol C p (s): 37.62 J/mol-K ΔH vap : 40,670 J/mol C p (l): 72.24 J/mol-K C p (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 H 2 O, 

H

fus = 6.01 kJ/mol and 

H

vap = 40.67 kJ/mol.

Solution Analyze:

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:

SAMPLE EXERCISE 11.4

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

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.

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 H 2 O and CO 2 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 Solutions

13.1 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 same picture is useful to rationalize Vapor Pressure

With the beaker covered, equilibrium is soon established.

Contrast that with an uncovered beaker!

Normal BP defined But recall the definition of

ANY

Boiling Point.

But, recall, we always prefer straight lines!

ln

P vap

  

H vap R

  1

T

  

C

(Add slide of C-C eq and plot.) ← Temp. increases

Using the Clausius-Clapeyron Equation: Tabulate P in atm and T in K; Calculate lnP and 1/T Plot lnP vs 1/T The slope is = - ΔH vap /R

Or with P the vapor pressure at T

1 1

and P the vapor pressure at T

2 2 ln

P

2   

H vap R

1 

T

2 

subtract to give C and

ln

P

1   

H vap R

1 

T

1 

C

ln

P

2  ln

P

1   

H vap R

1 

T

2  1

T

1 

or

ln 

P P

1 2  

H vap R

1 

T

2  1

T

1 

The slope!