Segment 1 PowerPoint Lessons

Segment 1: Lesson 1—Slide 1 Heating (Review Ch. 9)

• The heat required to heat an object is proportional to:Mass of the sample (m)

Heat for one gram for 1 degree (Specific Heat Capacity)Temperature change required. (T)

q = (m)(S.H.)(T) • The product, mass times specific heat is the heat capacity. cp

q = cp T (calorimetry)• When a hot sample is in contact with a colder one, the heat gained by the colder sample must equal the heat lost by the hot sample.• The heat required to vaporize one mole of a liquid is its molar heat of vaporization. Hvap• The heat removed when one mole of a liquid is frozen is the molar heat of fusion. Hfusion• The heat to absorbed when a mole of solid passes directly from the

solid into to the gas phase is called the molar heat of sublimation.Hsub

Segment 1: Lesson 1—Slide 2 Heating Problems

1. A system (Cp = 20.8 J/ oC) was heated at a constant 1 atm pressure from 25 oC to 50 oC. If the volume increased from 24.4 L to 26.5 L, calculate: q, w, E, and H

Solution: q = Cp T = (20.8 J/ oC) (50-25) oC = + 520 J (Sign?)

Since heating was at constant pressure, q = H H = +520 Jw = -P V = - (1 atm)(26.5 - 24.4) L = -2.1 L atm= - 213 J(sign?)

H = E + (PV)

+520 J = E + (1atm)(26.5-24.4)L

+520 J = E + 213 J

E = +307 J (sign?)

Is the comparison between H and E correct?

(Should more heat be required at constant pressure?)

Segment 1: Lesson 1—Slide 3 Heating Problems

2. How much heat is needed to heat 150 g of Fe (SH = 0.45 J/g oC) from 0.00 oC to 300 K?

Solution:q = m x SH x T = (150 g)(0.45 J/ g oC) (26.85 - 0.00) oC

q = 1810 J---SIGN?

3. What will be the final temperature if 50.0 g of Fe at 100 oC is dropped into 150 g of water (SH = 4.18 J / g oC) at 0.00 OC?Solution:

Heat lost = Heat gained

(50.0 g)(0.45 J / g oC)(100 - T) = (150 g)(4.18 J/ g oC)(T - 0.00)

2250 - 22.5 T = 627 T - 0.00 649.5 T = 2250T = 3.46 oC (LOGICAL?) checkpoint

Segment 1: Lesson 2—Slide 4 INTERMOLECULAR FORCES

(Chapter 6)• ALL particles: atoms, ions, and molecules attract each other due to LONDON DISPERSION FORCE.

• LONDON DISPERSION FORCE INCREASES WITH:

1. Molecular weight of the molecule

2. Volume of the molecule

3. Number of electrons in the molecule

4. Distance of electrons from nuclei.


• LONDON DISPERSION FORCE IS CAUSED BY:

1. Electrons move much faster than nuclei and try to avoid other electrons on nearby molecules.2. Electrons on the side of a molecule that is adjacent to another molecule will cause the electrons of the adjacent molecule to move away.3. As electrons of the second molecule move away, positive charge is exposed causing an attraction for the electrons of the first molecule.

4. Electrons of both molecules continue to oscillate in phase resulting in a permanent attraction between the two molecules.

Segment 1: Lesson 3—Slide 6 (non covalent interactions)

Molecular Polarity• Polar molecules attract ions and each other.

1. Polar molecules must contain bonds between atoms that have different electronegativities.2. Polar molecules must have polar bonds that are arranged unsymmetrically.

• Polar molecules will have a dipole moment.1. If a molecule is polar, the center of the positive charge (from nuclei) and the center of the negative charge (from electrons) will be separated by a distance “r”.2. The dipole moment, µ, is the product of the charge separated, Q, and the distance of separation, r. (µ=Qr)

Segment 1: Lesson 4—Slide 7INTERMOLECULAR FORCES

• HYDROGEN BONDING IS THE ATTRACTION OF A HYDROGEN ATOM ATTACHED TO A VERY ELECTRONEGATIVE ATOM ON ONE MOLECULE FOR A PAIR OF ELECTRONS ON AN ATOM OF AN ADJACENT MOLECULE.

1. The hydrogen atom must be attached to F, O, or N (the three most electronegative atoms.)

2. Any atom of an adjacent molecule that has a lone pair of electrons can complete the hydrogen bond.3. The hydrogen atom “bridges” the two molecules.

4. The hydrogen bridge is linear (180o)

Segment 1: Lesson 4—Slide 8 INTERMOLECULAR FORCES• HYDROGEN BONDING (BRIDGING) IS THE STRONGEST INTERMOLECULAR FORCE.

IF PRESENT AT ALL!!

1. Each hydrogen bridge is much weaker than a single covalent bond.2. However, a mole of hydrogen bonds can amount to TENS of kJ.

3. The hydrogen bonding in water is extremely strong. The energy of atomic bombs is required to vaporize 1 inch of water over a 1 square mile area!!!

Paper

2.5


PROBLEMS

1. List all the types of INTERMOLECULAR FORCES that exist between molecules of:

Cl2 CHCl3 PH3 AsH3 SbH3 HCF3 CH3OCH3 CH3CN

CH3NH2 ONF CH3CH2CF3 CH3NHCH3CH3COCH3

2. Arrange the 2 hydrogen bonded molecules order of LDF.

3. Arrange the 9 non hydrogen bonded molecules in order of LDF.

4. Arrange the 2 non-polar molecules in order of LDF.

checkpoint

Segment 1: Lesson 5—Slide 10SPONTANEITY OF SOLUTIONS

(Chatper 12 pgs 506-516)• Natural processes are favored by: (DRIVING FORCES FOR SPONTANEOUS CHANGE)

1. An INCREASE in randomness

2. An INCREASE in the amount of matter on which ENERGY is distributed.

(DISORDER)

Segment 1: Lesson 5—Slide 11 SPONTANEITY OF SOLUTIONS

(Chatper 12 pgs 506-516)• SPONTANEOUS MIXING (MISCIBILITY)

1. The environment of a molecule in the solution should be similar to the environment it was in before mixing.

2. The intermolecular forces between solute and solvent molecules should be similar to those between molecules in the pure solvent and those between pure solute molecules.3. The act of mixing generates disorder and favors forming a solution.

4. Strong attraction between solute and solvent molecules liberates energy and favors forming a solution.

SPONTANEOUS MIXINGMiscibility

IMF

IMF

If IMF = IMF

Remove partition




Random location for all particlesMiscible

Each particle experiences same force regardless of type of particle near it.


• PROBLEMS:

1. Place the following under H2O or CCl4 indicating which of solvent you think the following is soluble.

H2O CCl4

NaCl CH3OH CH3OCH3 SO2 Br2 HCl

NaCl

CH3OH

CH3OCH3

SO2

Br2

HCl

Segment 1: Lesson 6—Slide 17BULK PROPERTIES AND IMF

(pgs 470-478)• The equilibrium VAPOR PRESSURE of a liquid is the pressure in a sealed container when the rate at which molecules escape the liquid surface is the same as the rate at which they return.

1. RECALL: The average kinetic energy of a collection of molecules is fixed by the temperature. (KE = 3/2 RT)

2. At any temperature, there is a distribution of molecular energies. (BOLTZMAN DISTRIBUTION)3. At any temperature, there is a fraction of molecules with a kinetic energy greater than that needed to escape the surface.

4. These molecules escape into the gas phase and create the vapor pressure.


• The vapor pressure of a liquid increases with temperature. the mathematical relationship is:

lnP = a ∆Hvap

R1T( )

• A plot of ln P vs 1/T should be linear with a slope of -∆Hvap/R

lnP

1/T

run

rise

Slope = riserun

Or ln(P2/P1) = - ∆Hvap /R (1/T2 - 1/T1)


Problem: The heat of vaporization of water is 40.65 kJ/mol and the vapor pressure of water is 92.6 Torr at 50 oC. Calculate the expected boiling point of water.

SOLUTION:The vapor pressure is 760 torr at the boiling point.And 50 oC = 323.15 K.

ln(760/92.6) = - 40.65 x 103/8.314 (1/T2 - 1/323.15)

2.105 = - 4889.34(1/T2 - 0.0030945)-4.3053 x 10-4 + 0.0030945 = 1/T2

0.002664 = 1/T2

T2 = 375.37 K = 102.2 oC checkpoint

Segment 1: Lesson 6—Slide 20 BULK PROPERTIES AND IMF• VAPOR PRESSURE

1. If the intermolecular force is HIGH, few molecules can escape the liquid surface and the vapor pressure is LOW.

2. INCREASING the temperature increases the fraction of the molecules that can escape and INCREASES vapor pressure.

3. The BOILING POINT is the temperature at which the vapor pressure is equal to the applied pressure. (at the NORMAL boiling point, the vapor pressure is 1.00 atm.)

4. At the melting point, the solid and the liquid have the same vapor pressure. checkpoint

Segment 1: Lesson 6—Slide 21 BULK PROPERTIES AND IMF

(pgs 254-256)• A PHASE DIAGRAM is a plot of pressure (Y-axis) vs. Temperature (X-axis) showing the pressure at which two phases coexist at any given temperature.

1. The solid, liquid and gas phases are open areas on the diagram.

2. Lines represent the temperature and pressure where two phases exist in equilibrium.

3. The Critical point is the end point of the liquid-vapor line.

4. At the triple point all three phases are in equilibrium.

Segment 1: Lesson 6—Slide 22 BULK PROPERTIES AND IMF

PROBLEMS:

1. What is the composition at each of the points numbered:

1, 2, 3, 4, 5, 6, 7, 8, 9.

2. Describe what occurs in moving from point 1 to point 5 and from point 6 to point 9.3. Identify the: Critical point, Triple point,

Supercritical fluid region.

• Refer this phase diagram:

checkpoint

pressure

Temperature

Solid

Liquid

Gas

1

2

3

4

5

6 7 8 91 atm

Segment 1: Lesson 7—Slide 23 CRYSTALLINE SOLIDS

(Chapter 18)• CRYSTALLINE SOLIDS have an ordered, 3-dimensional arrangement of particles. Amorphous solids have a random arrangement of particles.

1. The array of points representing the location of particles in crystalline solids is called the CRYSTAL LATTICE.

2. The smallest repeating unit of a crystal lattice is called the UNIT CELL.

A. The unit cell has the same density as the bulk sample.

B. The unit cell has the same ratio of atoms (formula) as the bulk sample.C. Only 14 crystal lattices can exist and only 3 of them are cubic unit cells (same length, width, and height.)


• SIMPLE CUBIC UNIT CELL

1. Particles are located at the 8 corners of a cube.

2. Each of the 8 corners is shared by 8 unit cells.

3. Each unit cell has only one particle in it.

4. Since a particle is located at each corner of the unit cell, the edge length of the unit cell is twice the radius of the particles. l = 2r

5. The density of the unit cell (mass / volume) is:

Density = (1 x atomic wt./ 6.02 x 1023) l 3


• BODY-CENTERED CUBIC UNIT CELL

1. Particles are located at the 8 corners and at the center of the unit cell.

2. The corners again contribute 1 particle to the unit cell and the one in the center contributes another one.3. Particles touch across the diagonal from the front-top-left to the back-bottom-right. There are four radii across this length. 3 l = 4r4. The density (mass / volume) is:


Segment 1: Lesson 7—Slide 26 CRYSTALLINE SOLIDS F

• FACE-CENTERED UNIT CELL

1. The 8 corners again contribute one particle to the unit cell and each of the 6 face contribute ½ a particle to each of two unit cells. The total particles per unit cell is 4.

2. Particles touch each other across the diagonal of each face of the unit cell. This distance spans 4 radii.

2 l = 4r

3. The density (mass / volume) is:


4. Copper forms a face-centered cubic unit cell with an edge length of 361 pm. What is the radius of the copper atom? What is the density of copper?

PROBLEMS:

2 L = 4R

Body Centered Cubic Lattice

R = (1.414)(361 pm)

4 R = 128 pmRadius:

Density =

4 (At.Wt.)6.02 x 1023

L3

L = 361 x 10-12 m 10 2 cm1 m

L = 3.61 x 10-8 cm

Density = (4)(63.54)

(6.02 x 1023)(3.61 x 10-8)3

= 8.97 g/cm3

checkpoint

Equilibrium Vapor PressureEnergy Input

Gas molecules Rate of escape

Rate ofreturn

EQUALS

Boltzman Distribution of Energy

N(E)

Energy(E)

Most Probable Energy

Average Energy

Vaproization Energy

Molecules with enoughenergy to escape andcause vapor pressure

Simple Cubic Crystal Lattice

Unit Cell

Length of unit cell(L) = 2R

1 234

Each particle includedin 8 unit cells (1/8)

8 corners(patticles)1/8 x 8 = 1 particle per unit cell

Density = m/V=

(1) At.Wt.

6.02 x 1023

L3 (2R)3

R R

L

1 R

Face Centered Cubic Lattice

L

L

L

L2 L = 234

All 6 faces are identicalonly front face is shown

Particles touch eachother across diagonal

Each face particleshared in twounit cells. Six facescontribute 3particles

Each cornerparticle in 8unit cells

Eight cornerscontribute1 particle

Total particles per unit cell = 4

Density =

4 (At.Wt.)6.02 x 1023

L3


Number of particles per unit cell = (8 x 1/8 + 1)

2 Particles per unit cell




L

2 L

3 L

3 L = 4 R

Density =

2 x Atomic Mass6.02 x 1023

L 3

Phase Diagrams

pressure

Temperature

Solid

Liquid

Gas

1

2

3

4

5

6 7 8 91 atm

HYDROGEN BONDING(BRIDGING)

X H

Hydrogen bonded to an atom with same electronegaivity:

Hydrogen bonded to O, N or F:Adjacent atom with lone pair:

X H

YX HX + Y

180o

Attraction Pair of electrons

LONDON DISPERSION FORCE

+++

SINGLE MOLECULE

++ ++

ADJACENT MOLECULE

Attr.+++ ++ ++ ++ +

ELECTRONEGATIVITY

Li Be B C N O F

Cl

Br

I

1.0 1.5 2.0 2.5 3.0 3.5 4.0

P

H2.1INCREASE:

INCREASE:

OBSERVATION:When comparing atoms on different rowson the periodic table, the element closestto the noble gas on its row will have the higher electronegativity.

DIPOLE MOMENTS

+

Spherical electron distribution

Center of negative charge Center of positive charge

No charge separation

r = 0 and = 0

+ Charges separated

r

= q x r

Polarity and shape of molecules

Carbonate, CO32-, is NOT polar but contains polar bonds,

The bond dipole moments are in symmetrically arranged becauseall angles are 120o. The bond dipolemoments all cancel, resulting in no dipole moment for the ion .

Nitrite, NO21-,IS polar because polar bond dipole moments don’t cancel.

Bond dipole moments are not symmetrical and add to produce a net dipole moment through thenitrogen atom bisecting the ONO angle.

O

OOC

2-

O

ON

1-

UNSYMMETRICAL MOLECULESVSEPR:

AXE Nonsymmetrical(N)AX2 Symmetrical(S)2.

3. AX3 S AX2E (N) AXE2(N)

4. AX4 S AX3E (N) AX2E2 (N) AXE3 (N)

5. AX5 S AX4E (N) AX3E2 (N) AX2E3 (S) AXE4 (N)

6. AX6 S AX5E (N) AX4E2 (S) AX3E3 (N) AX2E4 (S) AXE5 (N)

EXAMPLE: SF3- anion Valence electrons = 6 + 3 x7 +1 = 28

2883

244

42

2

VSEPR = A X E 3 2

(m+n)

SF3- anion is Non symmetricaland POLAR.

Hydrogen Bonding (Bridging)

H

H

OH

H

O -----

This is a H-Bond

This is a H-O covalent bond

N

H H

H

N

HH

H

These are H-Bonds

O---H O angle is 180o

Hydrogen Bonding (Bridging)

Segment 1—Check point #1

Your two problems for 2 points.

1. Given that 23.6 kJ of hear are required to completely vaporize 60.0 g of benzene, (C6H6(l)) at 80.1 oC, calculate the molar Enthalpy of vaporization of benzene.

2. A 50.0 g sample of Cu (Cp = 24.4 J/mol/oC) at 98.6 oC is dropped into 125 g of water (Cp = 75.3 J/mol/oC) at 25.3 oC. What is the final temperature of the Cu and water?

return

NCl3 PCl3 PCl5 HCF3 HNCl2

Consider the molecules below:

Your 4 Problems:

1. Which two molecule(s) are not polar?

2. Which molecule(s) have hydrogen bonding?

3. Which molecule has the most polar BOND?

4. Which molecule has the highest London Force?

5 parts

Segment 1—Check point #2

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Segment 1—Check Point #3

Your Problem for 1 point

Problem: The heat of vaporization of water is 40.65 kJ/mol and the vapor pressure of water is 92.6 Torr at 50 oC. Calculate the vapor pressure of water at 25.0 oC.

NOTE: The measured vapor pressure of water at 25oC is 23.8 Torr

return


Your problem:

Draw the Lewis bonding structure for CH4O, CH3CN, and CH3F

Rank each molecule according to the criteria on each columnIn the table below: Use ranking 1,2,3 for largest to smallest. (0 if the molecule does not have the property to be ranked.)

H-bonding London Dispersion Most polar bond boiling point

CH4O

CH3CN

CH3F

15 parts return


1. What do points a, b, c, d represent?

2. What phase changes occur in going along the arrow from point 1 to point 2?

Your 2 problems:

Consider this phase diagram:

7 pointsreturn

P(atm)

1.0

a b c d1

2

T(oC)

Segment 1—Point #6

Your problem:

Platinum (at. mass = 195.08) has a density of 21.45 g/cm3

and forms a face centered cubic crystal. What is the diameter of the platinum atom?

1 part

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Segment 1 PowerPoint Lessons