huntington beach high school ap chemistry exam review winter review

Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved

Huntington Beach High School

AP CHEMISTRYAP CHEMISTRYEXAM REVIEWEXAM REVIEW

WINTER REVIEWWINTER REVIEW


Intermolecular ForcesIntermolecular ForcesForces between molecules, Forces between molecules,

between ions, or between between ions, or between molecules and ions.molecules and ions.

Table 13.1:Table 13.1: summary of forces summary of forces and their relative strengths.and their relative strengths.


Intermolecular ForcesIntermolecular ForcesIon-Ion ForcesIon-Ion Forces

Intermolecular ForcesIntermolecular ForcesIon-Ion ForcesIon-Ion Forces

NaNa++ — Cl — Cl-- in salt. in salt.These are the These are the

strongest forces. strongest forces. Solids have high Solids have high

melting melting temperatures.temperatures.

NaCl, mp = 800 NaCl, mp = 800 ooCCMgO, mp = 2800 MgO, mp = 2800 ooCC


Attraction Between Attraction Between Ions and Permanent Ions and Permanent

DipolesDipoles

Attraction Between Attraction Between Ions and Permanent Ions and Permanent

DipolesDipoles

Water is highly polar, Water is highly polar, interacts with positive ions, interacts with positive ions, gives gives hydratedhydrated ions in ions in water.water.

HH

water dipole

••

••

O-

+H

H

water dipole

••

••

O-

+


Attraction Between Ions and

Permanent Dipoles


Permanent Dipoles

Many metal ions are Many metal ions are hydrated.hydrated.

It is the reason It is the reason metal salts metal salts dissolve in water.dissolve in water.

Co(HCo(H22O)O)662+2+



Permanent Dipoles


Permanent Dipoles

Attraction between ions and dipole Attraction between ions and dipole depends on depends on ion chargeion charge and and ion-ion-dipole distancedipole distance..

-1922 kJ/mol-1922 kJ/mol -405 kJ/mol-405 kJ/mol -263 kJ/mol-263 kJ/mol

See Example 13.1, page 588.See Example 13.1, page 588.

OH

H+

-• • • O

H

H+

-• • • O

H

H+

-• • •

Na+Mg2+

Cs+


Dipole-Dipole Dipole-Dipole ForcesForces


Such forces bind molecules Such forces bind molecules having permanent dipoles having permanent dipoles to one another.to one another.

O O O-+

CC+ --+

C




Influence of dipole-dipole forces is seen Influence of dipole-dipole forces is seen in the boiling points of simple in the boiling points of simple molecules.molecules.

CompdCompd Mol. Wt.Mol. Wt. Boil PointBoil Point

NN22 2828 -196 -196 ooCC

COCO 2828 -192 -192 ooCC

BrBr22 160160 59 59 ooCC

IClICl 162162 97 97 ooCC


Hydrogen BondingHydrogen BondingHydrogen BondingHydrogen Bonding

A special form of dipole-dipole A special form of dipole-dipole attraction, which enhances attraction, which enhances dipole-dipole attractions.dipole-dipole attractions.

Hydrogen bonding in HFHydrogen bonding in HF


Hydrogen BondingHydrogen BondingHydrogen BondingHydrogen BondingA special form of dipole-dipole attraction, A special form of dipole-dipole attraction,

which enhances dipole-dipole attractions.which enhances dipole-dipole attractions.

H-bonding is strongest when H-bonding is strongest when H is joined toH is joined to

N, O, or FN, O, or F


H-Bond Between Ethanol and H-Bond Between Ethanol and WaterWater

H-Bond Between Ethanol and H-Bond Between Ethanol and WaterWater

H-bondH-bondH-bondH-bond--

++

--


Hydrogen Bonding in HHydrogen Bonding in H22OOHydrogen Bonding in HHydrogen Bonding in H22OOH-bonding is very strong in water:H-bonding is very strong in water:

• the O—H bond is very polarthe O—H bond is very polar• there are 2 lone there are 2 lone pairs on the O pairs on the O atomatom

Accounts for Accounts for many of water’s many of water’s unique propertiesunique properties


FORCES WITH INDUCED DIPOLESFORCES WITH INDUCED DIPOLESFORCES WITH INDUCED DIPOLESFORCES WITH INDUCED DIPOLES• Non-polar molecules condense to (l) Non-polar molecules condense to (l)

and (s) by “and (s) by “induced dipolesinduced dipoles.”.”

• Example: IExample: I22 dissolving in alcohol. dissolving in alcohol.

• Alcohol INDUCES a Alcohol INDUCES a temporary temporary dipoledipole in I in I22

OH

OH

ROH dipoledistorts orpolarizes theI2 electroncloud

-

+

I-I

R-

++

-

I-I

R

..


FORCES INVOLVING INDUCED FORCES INVOLVING INDUCED DIPOLESDIPOLES

FORCES INVOLVING INDUCED FORCES INVOLVING INDUCED DIPOLESDIPOLES

Dipoles can form in nonpolar Dipoles can form in nonpolar molecules. Electrons in one molecules. Electrons in one molecule REPEL the electrons in molecule REPEL the electrons in another and create TEMPORARY another and create TEMPORARY DIPOLES (van der Waals or London DIPOLES (van der Waals or London Forces)Forces)


FORCES INVOLVING INDUCED DIPOLESFORCES INVOLVING INDUCED DIPOLESFORCES INVOLVING INDUCED DIPOLESFORCES INVOLVING INDUCED DIPOLESThe size of the dipole depends on the The size of the dipole depends on the

tendency to be distorted. tendency to be distorted. Higher molec. weight ---> larger dipoles.Higher molec. weight ---> larger dipoles. MoleculeMolecule Boiling Point Boiling Point ((ooC)C)

CHCH44 (methane) (methane) - 161.5- 161.5

CC22HH66 (ethane) (ethane) - 88.6 - 88.6

CC33HH88 (propane) (propane) - 42.1- 42.1

CC44HH1010 (butane) (butane) - 0.5- 0.5


Boiling Points of HydrocarbonsBoiling Points of HydrocarbonsBoiling Points of HydrocarbonsBoiling Points of Hydrocarbons

Linear relation between bp & molar mass

CHCH44

CC22HH66

CC33HH88

CC44HH1010

LiquidsLiquidsSection 13.3Section 13.3LiquidsLiquids

Section 13.3Section 13.3•• molecules are in molecules are in

constant motionconstant motion

•• almost incompressiblealmost incompressible

•• do not fill the containerdo not fill the container

•• molecules are in molecules are in constant motionconstant motion

•• almost incompressiblealmost incompressible

•• do not fill the containerdo not fill the container•appreciable appreciable intermolecular intermolecular forcesforces

•molecules molecules close togetherclose together



Section 13.3Section 13.3

LIQUID VAPORbreak IM bonds

make IM bonds

Add energy

Remove energy

<---<---condensationcondensation

evaporationevaporation--->--->



Section 13.3Section 13.3

To evaporate, molecules must have To evaporate, molecules must have sufficient energy to break IM forces.sufficient energy to break IM forces.

To evaporate, molecules must have To evaporate, molecules must have sufficient energy to break IM forces.sufficient energy to break IM forces.

Breaking IM forces Breaking IM forces requires energy. requires energy. The process of The process of evaporation is evaporation is endothermicendothermic..



Section 13.3Section 13.3When molecules of liquid When molecules of liquid

evaporate, they exert evaporate, they exert

VAPOR PRESSUREVAPOR PRESSURE

EQUILIBRIUM EQUILIBRIUM VAPOR PRESSUREVAPOR PRESSURE = pressure of the vapor = pressure of the vapor over a liquidover a liquid in a closed in a closed containercontainer when when rate of rate of evaporationevaporation = = rate of rate of condensation.condensation.

See Fig. 13.18See Fig. 13.18

When molecules of liquid When molecules of liquid evaporate, they exert evaporate, they exert

VAPOR PRESSUREVAPOR PRESSURE

EQUILIBRIUM EQUILIBRIUM VAPOR PRESSUREVAPOR PRESSURE = pressure of the vapor = pressure of the vapor over a liquidover a liquid in a closed in a closed containercontainer when when rate of rate of evaporationevaporation = = rate of rate of condensation.condensation.

See Fig. 13.18See Fig. 13.18


Boiling LiquidsBoiling LiquidsBoiling LiquidsBoiling Liquids

Liquid boils when its vapor pressure equals atmospheric pressure.

Liquid boils when its vapor pressure equals atmospheric pressure.


Boiling Point at Lower Boiling Point at Lower PressurePressure

Boiling Point at Lower Boiling Point at Lower PressurePressure

When pressure is lowered, the vapor When pressure is lowered, the vapor pressure can equal the external pressure can equal the external pressure at a lower temperature.pressure at a lower temperature.


T of boiling when P = 1 atm is the T of boiling when P = 1 atm is the NORMAL BOILING POINTNORMAL BOILING POINT

VP of a molecule depends on IM forces. VP of a molecule depends on IM forces. Note IM forces in molecules below.Note IM forces in molecules below.

T of boiling when P = 1 atm is the T of boiling when P = 1 atm is the NORMAL BOILING POINTNORMAL BOILING POINT

VP of a molecule depends on IM forces. VP of a molecule depends on IM forces. Note IM forces in molecules below.Note IM forces in molecules below.

C2H5H5C2 HH5C2 HH

wateralcoholether

increasing strength of IM interactions

extensiveH-bondsH-bonds

dipole-dipole

OOOC2H5H5C2 HH5C2 HH

wateralcoholether

increasing strength of IM interactions

extensiveH-bondsH-bonds

dipole-dipole

OOO

LiquidsLiquidsLiquidsLiquids


LiquidsLiquidsLiquidsLiquids

HEAT OF VAPORIZATIONHEAT OF VAPORIZATION = heat = heat needed (at constant P) to vaporize the needed (at constant P) to vaporize the liquid.liquid.

LIQ + heat ---> VAPLIQ + heat ---> VAP

Compd.Compd. HHvapvap (kJ/mol) (kJ/mol) IM ForceIM Force

HH22OO 40.7 (100 40.7 (100 ooC)C) H-bondsH-bonds

SOSO22 26.8 (-47 26.8 (-47 ooC)C) dipoledipole

XeXe 12.6 (-107 12.6 (-107 ooC)C) induced induced dipoledipole

HEAT OF VAPORIZATIONHEAT OF VAPORIZATION = heat = heat needed (at constant P) to vaporize the needed (at constant P) to vaporize the liquid.liquid.

LIQ + heat ---> VAPLIQ + heat ---> VAP

Compd.Compd. HHvapvap (kJ/mol) (kJ/mol) IM ForceIM Force

HH22OO 40.7 (100 40.7 (100 ooC)C) H-bondsH-bonds

SOSO22 26.8 (-47 26.8 (-47 ooC)C) dipoledipole

XeXe 12.6 (-107 12.6 (-107 ooC)C) induced induced dipoledipole


Types of SolidsTypes of SolidsTable 13.6Table 13.6

Types of SolidsTypes of SolidsTable 13.6Table 13.6

TYPETYPE EXAMPLEEXAMPLE FORCEFORCE

Ionic Ionic NaCl, CaFNaCl, CaF22, ZnS, ZnS Ion-ionIon-ion

MetallicMetallic Na, FeNa, Fe MetallicMetallic

MolecularMolecular Ice, IIce, I22 DipoleDipole

Ind. dipoleInd. dipole

NetworkNetwork DiamondDiamond ExtendedExtendedGraphiteGraphite covalentcovalent

TYPETYPE EXAMPLEEXAMPLE FORCEFORCE

Ionic Ionic NaCl, CaFNaCl, CaF22, ZnS, ZnS Ion-ionIon-ion

MetallicMetallic Na, FeNa, Fe MetallicMetallic

MolecularMolecular Ice, IIce, I22 DipoleDipole

Ind. dipoleInd. dipole

NetworkNetwork DiamondDiamond ExtendedExtendedGraphiteGraphite covalentcovalent


Network SolidsNetwork SolidsNetwork SolidsNetwork Solids

DiamondDiamond

GraphiteGraphite


Properties of SolidsProperties of SolidsProperties of SolidsProperties of Solids1. Molecules, atoms or ions 1. Molecules, atoms or ions

locked in a locked in a CRYSTAL CRYSTAL LATTICELATTICE

2. Particles are CLOSE 2. Particles are CLOSE togethertogether

3. STRONG IM forces3. STRONG IM forces

4. Highly ordered, rigid, 4. Highly ordered, rigid, incompressibleincompressible

1. Molecules, atoms or ions 1. Molecules, atoms or ions locked in a locked in a CRYSTAL CRYSTAL LATTICELATTICE

2. Particles are CLOSE 2. Particles are CLOSE togethertogether

3. STRONG IM forces3. STRONG IM forces

4. Highly ordered, rigid, 4. Highly ordered, rigid, incompressibleincompressible ZnS, zinc sulfideZnS, zinc sulfide


Crystal LatticesCrystal LatticesCrystal LatticesCrystal Lattices

Regular 3-D arrangements of Regular 3-D arrangements of equivalent LATTICE POINTS.equivalent LATTICE POINTS.

The lattice points define The lattice points define UNIT CELLSUNIT CELLS, , the smallest repeating internal unit the smallest repeating internal unit that has the symmetry characteristic that has the symmetry characteristic of the solid. of the solid.

7 basic crystal systems -- we are 7 basic crystal systems -- we are only concerned with only concerned with CUBICCUBIC..

Regular 3-D arrangements of Regular 3-D arrangements of equivalent LATTICE POINTS.equivalent LATTICE POINTS.

The lattice points define The lattice points define UNIT CELLSUNIT CELLS, , the smallest repeating internal unit the smallest repeating internal unit that has the symmetry characteristic that has the symmetry characteristic of the solid. of the solid.

7 basic crystal systems -- we are 7 basic crystal systems -- we are only concerned with only concerned with CUBICCUBIC..


All sidesequal length

All anglesare 90 degrees

All sidesequal length

All anglesare 90 degrees

Cubic Unit CellsCubic Unit CellsCubic Unit CellsCubic Unit Cells


Cubic Unit CellsCubic Unit CellsFigure 13.28Figure 13.28

Cubic Unit CellsCubic Unit CellsFigure 13.28Figure 13.28

•• simple cubic (SC)simple cubic (SC)

•• body centered cubic (BCC)body centered cubic (BCC)

•• face centered cubic (FCC)face centered cubic (FCC)

•• simple cubic (SC)simple cubic (SC)

•• body centered cubic (BCC)body centered cubic (BCC)

•• face centered cubic (FCC)face centered cubic (FCC)


Simple Ionic CompoundsSimple Ionic CompoundsSimple Ionic CompoundsSimple Ionic CompoundsLattices of many simple ionic solids Lattices of many simple ionic solids

are built by taking a SC or FCC are built by taking a SC or FCC lattice of ions of one type and lattice of ions of one type and placing ions of opposite charge in placing ions of opposite charge in the holes in the lattice.the holes in the lattice.

EXAMPLE:EXAMPLE: CsCl has a SC lattice of CsCl has a SC lattice of CsCs++ ions with Cl ions with Cl-- in the center. in the center.


Simple Ionic CompoundsSimple Ionic CompoundsSimple Ionic CompoundsSimple Ionic Compounds1 unit cell has 1 Cl1 unit cell has 1 Cl--

ion plus ion plus

(8 corners)(1/8 Cs(8 corners)(1/8 Cs++ per corner)per corner)

= 1 net Cs= 1 net Cs++ ion. ion.

TRANSITIONS BETWEEN PHASESTRANSITIONS BETWEEN PHASESSection 13.7, Figure 13.37Section 13.7, Figure 13.37

TRANSITIONS BETWEEN PHASESTRANSITIONS BETWEEN PHASESSection 13.7, Figure 13.37Section 13.7, Figure 13.37


TRANSITIONS TRANSITIONS BETWEEN BETWEEN PHASESPHASESSection 13.7Section 13.7

TRANSITIONS TRANSITIONS BETWEEN BETWEEN PHASESPHASESSection 13.7Section 13.7

Lines connect all conditions of T and Lines connect all conditions of T and P where EQUILIBRIUM exists P where EQUILIBRIUM exists between the phases on either side of between the phases on either side of the line. the line. At equilibrium particles move At equilibrium particles move from liquid to gas as fast as they move from liquid to gas as fast as they move from gas to liquid, for example. from gas to liquid, for example. INTERPRET PHASE DIAGRAMS.INTERPRET PHASE DIAGRAMS.

Lines connect all conditions of T and Lines connect all conditions of T and P where EQUILIBRIUM exists P where EQUILIBRIUM exists between the phases on either side of between the phases on either side of the line. the line. At equilibrium particles move At equilibrium particles move from liquid to gas as fast as they move from liquid to gas as fast as they move from gas to liquid, for example. from gas to liquid, for example. INTERPRET PHASE DIAGRAMS.INTERPRET PHASE DIAGRAMS.


TRANSITIONS BETWEEN PHASESTRANSITIONS BETWEEN PHASESTRANSITIONS BETWEEN PHASESTRANSITIONS BETWEEN PHASES

As P and T increase, you reach As P and T increase, you reach CRITICAL TCRITICAL T and and PP

As P and T increase, you reach As P and T increase, you reach CRITICAL TCRITICAL T and and PP

.

LIQUID

GAS

Pcritical

Hig

h P

ress

ure

High Temperature

Tcritical

Note that linegoes straight up

Above critical Above critical T no liquid T no liquid exists no exists no matter how matter how high the high the pressure.pressure.

Above critical Above critical T no liquid T no liquid exists no exists no matter how matter how high the high the pressure.pressure.


Critical T and PCritical T and PCritical T and PCritical T and P

COMPDCOMPD TTcc((ooC)C) PPcc(atm)(atm)

HH22OO 374374 218218

COCO22 3131 7373

CHCH44 -82-82 4646

Freon-12Freon-12 112112 4141(CCl(CCl22FF22))

TTcc and P and Pcc depend on IM forces. depend on IM forces.

COMPDCOMPD TTcc((ooC)C) PPcc(atm)(atm)

HH22OO 374374 218218

COCO22 3131 7373

CHCH44 -82-82 4646

Freon-12Freon-12 112112 4141(CCl(CCl22FF22))

TTcc and P and Pcc depend on IM forces. depend on IM forces.


Solid-Vapor EquilibriumSolid-Vapor EquilibriumSolid-Vapor EquilibriumSolid-Vapor Equilibrium

At P < 4.58 mmHg and T < 0.0098 At P < 4.58 mmHg and T < 0.0098 C C

solid Hsolid H22O can go directly to vapor. O can go directly to vapor.

This process is called This process is called SUBLIMATIONSUBLIMATION

This is how a frost-free refrigerator This is how a frost-free refrigerator works.works.

At P < 4.58 mmHg and T < 0.0098 At P < 4.58 mmHg and T < 0.0098 C C

solid Hsolid H22O can go directly to vapor. O can go directly to vapor.

This process is called This process is called SUBLIMATIONSUBLIMATION

This is how a frost-free refrigerator This is how a frost-free refrigerator works.works.


Some DefinitionsSome DefinitionsSome DefinitionsSome DefinitionsA solution is a A solution is a

HOMOGENEOUSHOMOGENEOUS mixture of 2 or more mixture of 2 or more substances in one substances in one phase. phase.

One substance is the One substance is the SOLVENTSOLVENT and the and the others are the others are the SOLUTESSOLUTES..


Solutions can be Solutions can be unsaturatedunsaturated or or saturatedsaturated..

A A saturatedsaturated solution solution contains the contains the maximum quantity maximum quantity of solute that of solute that dissolves at that dissolves at that temperature.temperature.

DefinitionsDefinitionsDefinitionsDefinitions

Unsaturated Unsaturated solutions have less solutions have less than the maximum than the maximum solute.solute.


SUPERSATURATED SUPERSATURATED SOLUTIONSSOLUTIONS contain more than contain more than is possible (at the is possible (at the ambient temp-ambient temp-erature) and are erature) and are unstable.unstable.

Prepare them at Prepare them at higher temperature higher temperature and allow to cool.and allow to cool.

DefinitionsDefinitionsDefinitionsDefinitions


Colligative PropertiesColligative PropertiesColligative PropertiesColligative PropertiesThe properties of the solvent in a The properties of the solvent in a

solution are changed.solution are changed.

• Vapor pressure Vapor pressure decreasesdecreases

• Melting point Melting point decreasesdecreases

• Boiling point Boiling point increasesincreases

• Osmosis is possible (osmotic Osmosis is possible (osmotic pressure)pressure)

The properties of the solvent in a The properties of the solvent in a solution are changed.solution are changed.

• Vapor pressure Vapor pressure decreasesdecreases

• Melting point Melting point decreasesdecreases

• Boiling point Boiling point increasesincreases

• Osmosis is possible (osmotic Osmosis is possible (osmotic pressure)pressure)


Colligative PropertiesColligative PropertiesColligative PropertiesColligative PropertiesThese changes are called These changes are called

COLLIGATIVE PROPERTIESCOLLIGATIVE PROPERTIES. .

They depend only on the They depend only on the NUMBERNUMBER of of solute particles relative to solvent solute particles relative to solvent particles, not on the particles, not on the KINDKIND of solute of solute particles.particles.

Applies to Applies to IDEAL SOLUTIONS.IDEAL SOLUTIONS. Properties depend only on the Properties depend only on the concentration of solute.concentration of solute.

These changes are called These changes are called COLLIGATIVE PROPERTIESCOLLIGATIVE PROPERTIES. .

They depend only on the They depend only on the NUMBERNUMBER of of solute particles relative to solvent solute particles relative to solvent particles, not on the particles, not on the KINDKIND of solute of solute particles.particles.

Applies to Applies to IDEAL SOLUTIONS.IDEAL SOLUTIONS. Properties depend only on the Properties depend only on the concentration of solute.concentration of solute.


Need concentration Need concentration units to tell us the units to tell us the number of solute number of solute particles per solvent particles per solvent particle.particle.

The unit “molarity” The unit “molarity” does not do this!does not do this!

Need concentration Need concentration units to tell us the units to tell us the number of solute number of solute particles per solvent particles per solvent particle.particle.

The unit “molarity” The unit “molarity” does not do this!does not do this!

Concentration UnitsConcentration UnitsConcentration UnitsConcentration Units


MOLE FRACTION, XMOLE FRACTION, X

For a mixture of A, B, and CFor a mixture of A, B, and C

MOLE FRACTION, XMOLE FRACTION, X

For a mixture of A, B, and CFor a mixture of A, B, and C


A fraction mol XA A fraction mol XA

C mol B mol A mol

Amol



MOLALITY, mMOLALITY, m

MOLALITY, mMOLALITY, m

solvent kg

solute mol solute of m

solvent kg

solute mol solute of m


Calculating ConcentrationsCalculating ConcentrationsCalculating ConcentrationsCalculating Concentrations

250. g H250. g H22O = 13.9 molO = 13.9 mol

X X glycolglycol = 0.0672 = 0.0672

250. g H250. g H22O = 13.9 molO = 13.9 mol

X X glycolglycol = 0.0672 = 0.0672

Dissolve 62.1 g (1.00 mol) of ethylene Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Hglycol in 250. g of H22O. Calculate X, O. Calculate X,

and m of glycol.and m of glycol.

Dissolve 62.1 g (1.00 mol) of ethylene Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Hglycol in 250. g of H22O. Calculate X, O. Calculate X,

and m of glycol.and m of glycol.

Xglycol = 1.00 mol glycol

1.00 mol glycol + 13.9 mol H2OXglycol =

1.00 mol glycol1.00 mol glycol + 13.9 mol H2O


Calculating ConcentrationsCalculating ConcentrationsCalculating ConcentrationsCalculating Concentrations

Calculate molalityCalculate molality

Calculate molalityCalculate molality

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Hof H22O. Calculate X, m, and % of glycol.O. Calculate X, m, and % of glycol.

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Hof H22O. Calculate X, m, and % of glycol.O. Calculate X, m, and % of glycol.

molal 4.00

OH kg 0.250

glycol mol 1.00 (m)conc

2

molal 4.00

OH kg 0.250

glycol mol 1.00 (m)conc

2


Dissolving Gases & Dissolving Gases & Henry’s LawHenry’s Law

Dissolving Gases & Dissolving Gases & Henry’s LawHenry’s Law

Gas solubilityGas solubility

(M) = k(M) = kHH • P • Pgasgas

kkHH for O for O22 = 1.66 x 10 = 1.66 x 10-6-6 M/mmHg M/mmHg

When PWhen Pgasgas drops, solubility drops. drops, solubility drops.

Note M is used for gases since each Note M is used for gases since each gas molecule is independent of gas molecule is independent of others.others.


.

H—O

H

HH—O

H H—O

H—O

H

surface

Understanding Understanding Colligative PropertiesColligative Properties


Colligative properties depend on the Colligative properties depend on the LIQUID-LIQUID-VAPOR EQUILIBRIUMVAPOR EQUILIBRIUM of a solution. of a solution.

Colligative properties depend on the Colligative properties depend on the LIQUID-LIQUID-VAPOR EQUILIBRIUMVAPOR EQUILIBRIUM of a solution. of a solution.




To understand To understand colligative colligative properties, study properties, study the the LIQUID-LIQUID-VAPOR VAPOR EQUILIBRIUMEQUILIBRIUM for a solution.for a solution.

To understand To understand colligative colligative properties, study properties, study the the LIQUID-LIQUID-VAPOR VAPOR EQUILIBRIUMEQUILIBRIUM for a solution.for a solution.




VP of HVP of H22O over a solution depends on O over a solution depends on

the number of Hthe number of H22O molecules per O molecules per

solute molecule.solute molecule.

PPsolventsolvent = X = Xsolventsolvent • P • Poosolventsolvent

RAOULT’S LAWRAOULT’S LAW

The vapor pressure of solvent over a The vapor pressure of solvent over a solution is always solution is always LOWEREDLOWERED!!

VP of HVP of H22O over a solution depends on O over a solution depends on

the number of Hthe number of H22O molecules per O molecules per

solute molecule.solute molecule.

PPsolventsolvent = X = Xsolventsolvent • P • Poosolventsolvent

RAOULT’S LAWRAOULT’S LAW

The vapor pressure of solvent over a The vapor pressure of solvent over a solution is always solution is always LOWEREDLOWERED!!


Changes in Freezing and Boiling Points of SolventChanges in Freezing and Boiling Points of Solvent

See Figure 14.13See Figure 14.13

VP solventafter addingsolute

VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T


VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T


Elevation of Boiling Point Elevation of Boiling Point Elevation of Boiling Point Elevation of Boiling Point Elevation in BP = Elevation in BP = ttBPBP = K = KBP BP • m• m

(K(KBPBP is characteristic of solvent) is characteristic of solvent)


VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T


VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T


Change in Boiling Point Change in Boiling Point Change in Boiling Point Change in Boiling Point

Dissolve 62.1 g of glycol (1.00 mol) Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the BP in 250. g of water. What is the BP of the solution?of the solution?

KKBPBP = +0.512 = +0.512 ooC/molal for water C/molal for water (see (see Table 14.3).Table 14.3).

SolutionSolution

1.1. Calculate solution molality = Calculate solution molality = 4.00 m4.00 m


Change in Boiling Point Change in Boiling Point Change in Boiling Point Change in Boiling Point

2.2. ttBPBP = K = KBPBP • m • m

ttBPBP = =

+0.512 +0.512 ooC/molal (4.00 molal)C/molal (4.00 molal)

ttBPBP = +2.05 = +2.05 ooCC

BP = 102.05 BP = 102.05 ooCC

(100 (100 ooCC + + ttBPBP))


Change in Freezing Point Change in Freezing Point Change in Freezing Point Change in Freezing Point

The freezing point The freezing point of a solution is of a solution is LOWERLOWER than than that of the pure that of the pure solvent.solvent.

FP depression = FP depression = ttFPFP = K = KFPFP•m•m

Pure waterPure waterEthylene glycol + Ethylene glycol + water solutionwater solution


Calculate the FP of a 4.00 molal Calculate the FP of a 4.00 molal glycol/water solution.glycol/water solution.

KKFPFP = -1.86 = -1.86 ooC/molal (Table 14.4)C/molal (Table 14.4)

SolutionSolution

ttFPFP = K = KFPFP • m • m

= (-1.86 = (-1.86 ooC/molal)(4.00 m)C/molal)(4.00 m)

ttFP FP = -7.44 = -7.44 ooCC

Freezing Point DepressionFreezing Point DepressionFreezing Point DepressionFreezing Point Depression


Boiling Point Elevation and Boiling Point Elevation and Freezing Point DepressionFreezing Point Depression

Boiling Point Elevation and Boiling Point Elevation and Freezing Point DepressionFreezing Point Depression

t = K • m • it = K • m • iA generally useful equation A generally useful equation

i = van’t Hoff factor = number of particles i = van’t Hoff factor = number of particles produced per formula unit.produced per formula unit.

CompoundCompound Theoretical Value of iTheoretical Value of i

glycolglycol 11

NaClNaCl 22

CaClCaCl22 33


Osmosis Calculating a Molar MassOsmosis Calculating a Molar MassOsmosis Calculating a Molar MassOsmosis Calculating a Molar Mass

Dissolve 35.0 g of hemoglobin in Dissolve 35.0 g of hemoglobin in water to make 1.00 L of solution. water to make 1.00 L of solution. is is 10.0 torr at 25 10.0 torr at 25 C. Calculate the molar C. Calculate the molar mass of hemoglobin.mass of hemoglobin.

SolutionSolution

(a)(a) Calc. concentration from Calc. concentration from = cRT = cRT

K)(298K)L/moltorr (62.4

torr 10.0 Conc

Conc = 5.39 x 10Conc = 5.39 x 10-4-4 mol/L mol/LConc = 5.39 x 10Conc = 5.39 x 10-4-4 mol/L mol/L


Osmosis Osmosis Calculating a Molar MassCalculating a Molar Mass

Osmosis Osmosis Calculating a Molar MassCalculating a Molar Mass

Conc = 5.39 x 10Conc = 5.39 x 10-4-4 mol/L mol/L

(b)(b)Calc. molar massCalc. molar mass

Molar mass = Molar mass =

35.0 g / 5.39 x 1035.0 g / 5.39 x 10-4-4 mol/L mol/L

Molar mass = 65,100 g/molMolar mass = 65,100 g/mol

Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L

of solution. of solution. measured to be 10.0 mm Hg at 25 measured to be 10.0 mm Hg at 25 C. Calc. C. Calc. molar mass of hemoglobin.molar mass of hemoglobin.

SolutionSolution


• Thermodynamics tells if a reaction is Thermodynamics tells if a reaction is favors products or reactants.favors products or reactants.

• But this gives no info on HOW FAST But this gives no info on HOW FAST reaction goes. reaction goes.

• KINETICSKINETICS — study of — study of REACTION REACTION RATESRATES and their relation to the way and their relation to the way the reaction proceeds, i.e., its the reaction proceeds, i.e., its MECHANISMMECHANISM..

• Thermodynamics tells if a reaction is Thermodynamics tells if a reaction is favors products or reactants.favors products or reactants.

• But this gives no info on HOW FAST But this gives no info on HOW FAST reaction goes. reaction goes.

• KINETICSKINETICS — study of — study of REACTION REACTION RATESRATES and their relation to the way and their relation to the way the reaction proceeds, i.e., its the reaction proceeds, i.e., its MECHANISMMECHANISM..

Chemical KineticsChemical KineticsChemical KineticsChemical Kinetics


Reaction MechanismsReaction MechanismsReaction MechanismsReaction MechanismsThe sequence of events at the molecular level The sequence of events at the molecular level

that control the speed and outcome of a that control the speed and outcome of a reaction.reaction.

Step 1:Step 1: Br + OBr + O33 ---> BrO + O ---> BrO + O22

Step 2:Step 2: Cl + OCl + O33 ---> ClO + O ---> ClO + O22

Step 3:Step 3:

BrO + ClO + light ---> Br + Cl + OBrO + ClO + light ---> Br + Cl + O22

NET: NET: 2 O2 O33 ---> 3 O ---> 3 O22

The sequence of events at the molecular level The sequence of events at the molecular level that control the speed and outcome of a that control the speed and outcome of a reaction.reaction.

Step 1:Step 1: Br + OBr + O33 ---> BrO + O ---> BrO + O22

Step 2:Step 2: Cl + OCl + O33 ---> ClO + O ---> ClO + O22

Step 3:Step 3:

BrO + ClO + light ---> Br + Cl + OBrO + ClO + light ---> Br + Cl + O22

NET: NET: 2 O2 O33 ---> 3 O ---> 3 O22


• Reaction rate = change in Reaction rate = change in concentration of a reactant or concentration of a reactant or product with time.product with time.

• Know about Know about initial rateinitial rate, , average average raterate, and , and instantaneous rateinstantaneous rate. See . See text figure 15.3 and p. 689-690.text figure 15.3 and p. 689-690.

Reaction Rates Reaction Rates Section 15.1Section 15.1

Reaction Rates Reaction Rates Section 15.1Section 15.1


• ConcentrationsConcentrations and and physical statephysical state of reactants and products. Greater of reactants and products. Greater conc. of reactants gives faster rxn.conc. of reactants gives faster rxn.

• TemperatureTemperature Rate incr. w/T Rate incr. w/T

• CatalystsCatalysts Forward and reverse Forward and reverse reactions go faster.reactions go faster.

Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2

Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2


Concentrations and Rates Concentrations and Rates Concentrations and Rates Concentrations and Rates

To postulate a reaction To postulate a reaction mechanism, we studymechanism, we study

•• reaction ratereaction rate and and

•• its its concentration dependenceconcentration dependence

To postulate a reaction To postulate a reaction mechanism, we studymechanism, we study

•• reaction ratereaction rate and and

•• its its concentration dependenceconcentration dependence


Concentrations, Rates, Concentrations, Rates, and Rate Laws and Rate Laws Concentrations, Rates, Concentrations, Rates, and Rate Laws and Rate Laws

In general, for In general, for

a A + b B ---> x Xa A + b B ---> x X with a catalyst Cwith a catalyst C

Rate = k [A]Rate = k [A]mm[B][B]nn[C][C]pp

The exponents m, n, and p The exponents m, n, and p

•• are the reaction orderare the reaction order

•• can be 0, 1, 2 or fractionscan be 0, 1, 2 or fractions

•• must be determined by experiment!must be determined by experiment!

In general, for In general, for

a A + b B ---> x Xa A + b B ---> x X with a catalyst Cwith a catalyst C


The exponents m, n, and p The exponents m, n, and p

•• are the reaction orderare the reaction order

•• can be 0, 1, 2 or fractionscan be 0, 1, 2 or fractions

•• must be determined by experiment!must be determined by experiment!

CisplatinCisplatin


Interpreting Rate Laws Interpreting Rate Laws Interpreting Rate Laws Interpreting Rate Laws


• If m = 1, rxn. is 1st order in AIf m = 1, rxn. is 1st order in A

Rate = k [A]Rate = k [A]11

If [A] doubles, rate goes up by factor of ? If [A] doubles, rate goes up by factor of ? • If m = 2, rxn. is 2nd order in A. If m = 2, rxn. is 2nd order in A. • Rate = k [A]Rate = k [A]22

Doubling [A] increases rate by ?Doubling [A] increases rate by ?


• If m = 1, rxn. is 1st order in AIf m = 1, rxn. is 1st order in A


If [A] doubles, rate goes up by factor of ? If [A] doubles, rate goes up by factor of ? • If m = 2, rxn. is 2nd order in A. If m = 2, rxn. is 2nd order in A. • Rate = k [A]Rate = k [A]22

Doubling [A] increases rate by ?Doubling [A] increases rate by ?

CisplatinCisplatin


Interpreting Rate Laws Interpreting Rate Laws Interpreting Rate Laws Interpreting Rate Laws

If m = 0, rxn. is zero order.If m = 0, rxn. is zero order.


If [A] doubles, rate ?If [A] doubles, rate ?

If m = 0, rxn. is zero order.If m = 0, rxn. is zero order.


If [A] doubles, rate ?If [A] doubles, rate ?

CisplatinCisplatin


Deriving Rate Laws Deriving Rate Laws Deriving Rate Laws Deriving Rate Laws

Derive rate law and k for Derive rate law and k for

CHCH33CHO(g) ---> CHCHO(g) ---> CH44(g) + CO(g)(g) + CO(g)

from experimental data for rate of from experimental data for rate of disappearance of CHdisappearance of CH33CHOCHO

Derive rate law and k for Derive rate law and k for

CHCH33CHO(g) ---> CHCHO(g) ---> CH44(g) + CO(g)(g) + CO(g)

from experimental data for rate of from experimental data for rate of disappearance of CHdisappearance of CH33CHOCHO

CisplatinCisplatin



Expt. Expt. [CH[CH33CHO]CHO] Rate that Rate that

CHCH33CHOCHO

disappearsdisappears(mol/L)(mol/L) (mol/L•sec)(mol/L•sec)

11 0.100.10 0.0200.020

22 0.200.20 0.0810.081

33 0.300.30 0.1820.182

44 0.400.40 0.3180.318

Expt. Expt. [CH[CH33CHO]CHO] Rate that Rate that

CHCH33CHOCHO

disappearsdisappears(mol/L)(mol/L) (mol/L•sec)(mol/L•sec)

11 0.100.10 0.0200.020

22 0.200.20 0.0810.081

33 0.300.30 0.1820.182

44 0.400.40 0.3180.318

CisplatinCisplatin



Rate of rxn = k [CHRate of rxn = k [CH33CHO]CHO]22

Rate goes up by ______ when initial conc. Rate goes up by ______ when initial conc. doubles. This rxn. is ___ order.doubles. This rxn. is ___ order.

Find the value of k (with expt. #3 data)Find the value of k (with expt. #3 data)

0.182 mol/L•s = k (0.30 mol/L)0.182 mol/L•s = k (0.30 mol/L)22

k = 2.0 (L / mol•s)k = 2.0 (L / mol•s)

Using k you can calc. rate at other values of Using k you can calc. rate at other values of [CH[CH33CHO] at same T.CHO] at same T.

Rate of rxn = k [CHRate of rxn = k [CH33CHO]CHO]22

Rate goes up by ______ when initial conc. Rate goes up by ______ when initial conc. doubles. This rxn. is ___ order.doubles. This rxn. is ___ order.

Find the value of k (with expt. #3 data)Find the value of k (with expt. #3 data)

0.182 mol/L•s = k (0.30 mol/L)0.182 mol/L•s = k (0.30 mol/L)22

k = 2.0 (L / mol•s)k = 2.0 (L / mol•s)

Using k you can calc. rate at other values of Using k you can calc. rate at other values of [CH[CH33CHO] at same T.CHO] at same T.

CisplatinCisplatin


Concentration/Time Concentration/Time Relations Relations

Concentration/Time Concentration/Time Relations Relations

Conc. of reactant as function of time. Conc. of reactant as function of time.

FIRST ORDER REACTIONSFIRST ORDER REACTIONSFor 1st order reactions, the rate law isFor 1st order reactions, the rate law is

Conc. of reactant as function of time. Conc. of reactant as function of time.

FIRST ORDER REACTIONSFIRST ORDER REACTIONSFor 1st order reactions, the rate law isFor 1st order reactions, the rate law is

CisplatinCisplatin

[A] = - k tln[A]o

naturallogarithm [A] at time = 0

[A] = - k tln[A]o

naturallogarithm [A] at time = 0


Concentration/Time Relations Concentration/Time Relations Concentration/Time Relations Concentration/Time Relations

Sucrose decomposes to Sucrose decomposes to simpler sugarssimpler sugars

Rate of disappearance Rate of disappearance of sucrose of sucrose

= k [sucrose]= k [sucrose]

k = 0.21 hrk = 0.21 hr-1-1

Sucrose decomposes to Sucrose decomposes to simpler sugarssimpler sugars

Rate of disappearance Rate of disappearance of sucrose of sucrose

= k [sucrose]= k [sucrose]

k = 0.21 hrk = 0.21 hr-1-1

SucroseSucrose

Initial [sucrose] = 0.010 MInitial [sucrose] = 0.010 M

How long to drop 90% (to 0.0010 M)?How long to drop 90% (to 0.0010 M)?

Initial [sucrose] = 0.010 MInitial [sucrose] = 0.010 M

How long to drop 90% (to 0.0010 M)?How long to drop 90% (to 0.0010 M)?


Concentration/Time Concentration/Time RelationshipsRelationshipsRate of disappear of sucrose = k Rate of disappear of sucrose = k [sucrose], k = 0.21 hr[sucrose], k = 0.21 hr-1-1.. If initial [sucrose] If initial [sucrose] = 0.010 M, how long to drop 90% or to = 0.010 M, how long to drop 90% or to 0.0010 M?0.0010 M?

Use the first order integrated rate lawUse the first order integrated rate law

ln (0.100) = - 2.3 = - (0.21 hrln (0.100) = - 2.3 = - (0.21 hr-1-1) • time) • time

time = 11 hourstime = 11 hours


[N2O5] vs. time

time

1

0

0 5

[N2O5] vs. time

time

1

0

0 5

l n [N2O5] vs. time

time

0

-2

0 5

l n [N2O5] vs. time

time

0

-2

0 5

Using the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate Law

2 N2 N22OO55(g) ---> 4 NO(g) ---> 4 NO22(g) + O(g) + O22(g) Rate = k [N(g) Rate = k [N22OO55]]

Conc. vs. time Conc. vs. time plot do not fit plot do not fit straight line.straight line.

Plot of ln [NPlot of ln [N22OO55] ]

vs. time is a vs. time is a straight line!straight line!


Using the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate LawUsing the Integrated Rate Law

All 1st order reactions All 1st order reactions have straight line have straight line plot for ln [A] vs. plot for ln [A] vs. time. time.

2nd order gives 2nd order gives straight line for plot straight line for plot of 1/[A] vs. time.of 1/[A] vs. time.

0 order gives straight 0 order gives straight line for plot of [A] line for plot of [A] vs. time.vs. time.

ln [N2O5] vs. time

time

0

-2

0 5

ln [N2O5] vs. time

time

0

-2

0 5


MECHANISMSMECHANISMSA Microscopic View of ReactionsA Microscopic View of Reactions

Sections 15.5 and 15.6Sections 15.5 and 15.6

MECHANISMSMECHANISMSA Microscopic View of ReactionsA Microscopic View of Reactions

Sections 15.5 and 15.6Sections 15.5 and 15.6

How are reactants converted to How are reactants converted to products at the molecular level?products at the molecular level?

Want to connect the Want to connect the

RATE LAW ----> RATE LAW ----> MECHANISMMECHANISM

experiment ---->experiment ----> theorytheory

How are reactants converted to How are reactants converted to products at the molecular level?products at the molecular level?

Want to connect the Want to connect the

RATE LAW ----> RATE LAW ----> MECHANISMMECHANISM

experiment ---->experiment ----> theorytheory


Energy involved in conversion of Energy involved in conversion of trans to cis butenetrans to cis butene

Energy involved in conversion of Energy involved in conversion of trans to cis butenetrans to cis butene

trans

cis

energy ActivatedComplex

26.9 kJ/mol

30.2 kJ/mol3.9 kJ/mol

233 kJ 229 kJ

MECHANISMSMECHANISMSMECHANISMSMECHANISMS

See Figure 15.15See Figure 15.15



Reaction Reaction

passes thru apasses thru a

TRANSITIONTRANSITION

STATE STATE where where

there is an there is an

activated complexactivated complex that has sufficient that has sufficient

energy to become a product. energy to become a product.

Reaction Reaction

passes thru apasses thru a

TRANSITIONTRANSITION

STATE STATE where where

there is an there is an

activated complexactivated complex that has sufficient that has sufficient

energy to become a product. energy to become a product.

trans

cis


26.9 kJ/mol


233 kJ 229 kJ

trans

cis


26.9 kJ/mol


233 kJ 229 kJ



trans

cis


26.9 kJ/mol


233 kJ 229 kJ

ACTIVATION ENERGY, EACTIVATION ENERGY, Eaa = energy = energy req’d to form activated complex. form activated complex.

Here EHere Eaa = 233 kJ/mol = 233 kJ/mol

ACTIVATION ENERGY, EACTIVATION ENERGY, Eaa = energy = energy req’d to form activated complex. form activated complex.

Here EHere Eaa = 233 kJ/mol = 233 kJ/mol


Activation EnergyActivation EnergyActivation EnergyActivation Energy

In general, In general, differences in differences in activation energyactivation energy are the reason are the reason reactions vary from fast to slow.reactions vary from fast to slow.

In general, In general, differences in differences in activation energyactivation energy are the reason are the reason reactions vary from fast to slow.reactions vary from fast to slow.



1. Why is reaction of 1. Why is reaction of cis ---> trans cis ---> trans observed to observed to be 1st order? be 1st order?

As [trans] doubles, number of molecules with As [trans] doubles, number of molecules with enough E also doubles.enough E also doubles.

2. Why is the reaction faster at higher 2. Why is the reaction faster at higher temperature?temperature?

Fraction of molecules with sufficient Fraction of molecules with sufficient activation energy increases with T.activation energy increases with T.

1. Why is reaction of 1. Why is reaction of cis ---> trans cis ---> trans observed to observed to be 1st order? be 1st order?

As [trans] doubles, number of molecules with As [trans] doubles, number of molecules with enough E also doubles.enough E also doubles.

2. Why is the reaction faster at higher 2. Why is the reaction faster at higher temperature?temperature?

Fraction of molecules with sufficient Fraction of molecules with sufficient activation energy increases with T.activation energy increases with T.


Collision TheoryCollision TheoryCollision TheoryCollision Theory

Reactions require Reactions require

(a) activation energy and (a) activation energy and

(b) correct geometry. (b) correct geometry.

OO33(g) + NO(g) ---> O(g) + NO(g) ---> O22(g) + NO(g) + NO22(g)(g)

Reactions require Reactions require

(a) activation energy and (a) activation energy and

(b) correct geometry. (b) correct geometry.

OO33(g) + NO(g) ---> O(g) + NO(g) ---> O22(g) + NO(g) + NO22(g)(g)


Collision TheoryCollision TheoryCollision TheoryCollision Theory

2. Activation 2. Activation energy and energy and geometrygeometry

1. Activation1. Activation energy energy



OO33 + NO reaction occurs in a single + NO reaction occurs in a single

ELEMENTARYELEMENTARY step. Most others involve a step. Most others involve a sequence of elementary steps.sequence of elementary steps.

Adding elementary steps gives NET reaction.Adding elementary steps gives NET reaction.

OO33 + NO reaction occurs in a single + NO reaction occurs in a single

ELEMENTARYELEMENTARY step. Most others involve a step. Most others involve a sequence of elementary steps.sequence of elementary steps.

Adding elementary steps gives NET reaction.Adding elementary steps gives NET reaction.



2 I2 I-- + H + H22OO22 + 2 H + 2 H++ ---> I ---> I22 + 2 H + 2 H22OO

Rate = k [IRate = k [I--] [H] [H22OO22]]

1 — slow HOOH + I1 — slow HOOH + I-- --> HOI + OH --> HOI + OH--

2 — fast2 — fast HOI + I HOI + I-- --> I --> I22 + OH + OH--

3 — fast3 — fast 2 OH 2 OH- - + 2 H + 2 H++ --> 2 H --> 2 H22OO

Rate of rxn. controlled by slow step Rate of rxn. controlled by slow step

RATE DETERMINING STEPRATE DETERMINING STEP (rds) (rds)

Rate can be no faster than rds!Rate can be no faster than rds!

2 I2 I-- + H + H22OO22 + 2 H + 2 H++ ---> I ---> I22 + 2 H + 2 H22OO

Rate = k [IRate = k [I--] [H] [H22OO22]]

1 — slow HOOH + I1 — slow HOOH + I-- --> HOI + OH --> HOI + OH--

2 — fast2 — fast HOI + I HOI + I-- --> I --> I22 + OH + OH--

3 — fast3 — fast 2 OH 2 OH- - + 2 H + 2 H++ --> 2 H --> 2 H22OO

Rate of rxn. controlled by slow step Rate of rxn. controlled by slow step

RATE DETERMINING STEPRATE DETERMINING STEP (rds) (rds)

Rate can be no faster than rds!Rate can be no faster than rds!



Step 1Step 1 involves I involves I-- and HOOH. This and HOOH. This predicts the rate law should bepredicts the rate law should be

Rate = k [IRate = k [I--] [H] [H22OO22] — as observed!!] — as observed!!

The species HOI and OHThe species HOI and OH-- are are reaction reaction intermediates.intermediates.

Step 1Step 1 involves I involves I-- and HOOH. This and HOOH. This predicts the rate law should bepredicts the rate law should be

Rate = k [IRate = k [I--] [H] [H22OO22] — as observed!!] — as observed!!

The species HOI and OHThe species HOI and OH-- are are reaction reaction intermediates.intermediates.

2 I2 I-- + H + H22OO22 + 2H + 2H++ ---> I ---> I22 + 2H + 2H22O Rate = k[IO Rate = k[I--][H][H22OO22]]

Step 1 — slowStep 1 — slow HOOH + IHOOH + I-- --> HOI + OH --> HOI + OH--

Step 2 — fastStep 2 — fast HOI + IHOI + I-- --> I --> I22 + OH + OH--

Step 3 — fastStep 3 — fast 2 OH2 OH- - + 2 H + 2 H++ --> 2 H --> 2 H22OO


CATALYSISCATALYSISCATALYSISCATALYSIS

Catalysts speed up reactions by Catalysts speed up reactions by altering the mechanism to lower altering the mechanism to lower the activation energy barrier.the activation energy barrier.

Catalysts speed up reactions by Catalysts speed up reactions by altering the mechanism to lower altering the mechanism to lower the activation energy barrier.the activation energy barrier.


Iodine-Catalyzed Reaction of Iodine-Catalyzed Reaction of Hydrogen PeroxideHydrogen Peroxide

Figure 15.18


Properties of an Equilibrium Properties of an Equilibrium Properties of an Equilibrium Properties of an Equilibrium

Equilibrium systems areEquilibrium systems are

• DYNAMIC (in constant DYNAMIC (in constant motion)motion)

• REVERSIBLE REVERSIBLE

• can be approached from can be approached from either directioneither direction

Equilibrium systems areEquilibrium systems are

• DYNAMIC (in constant DYNAMIC (in constant motion)motion)

• REVERSIBLE REVERSIBLE

• can be approached from can be approached from either directioneither direction

Pink to blue Co(HPink to blue Co(H22O)O)66ClCl22 ---> Co(H ---> Co(H22O)O)44ClCl22 + 2H + 2H22OO

Blue to pink Co(HBlue to pink Co(H22O)O)44ClCl22 + 2H + 2H22O ---> Co(HO ---> Co(H22O)O)66ClCl22


Chemical EquilibriumChemical EquilibriumFeFe3+3+ + SCN + SCN-- FeSCN FeSCN2+2+

• After some time, the After some time, the concentrations of concentrations of reactants and reactants and products are constant. products are constant.

• The forward and The forward and reverse reactions reverse reactions continue after continue after equilibrium is equilibrium is reached.reached.


THE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANT

a A + b B a A + b B c C + d Dc C + d D

K is a CONSTANT (at a given T)K is a CONSTANT (at a given T)

K =[C]c [D]d

[A]a [B]b

conc. of products

conc. of reactantsequilibrium constantWith K -- predict concs. of products or reactantsWith K -- predict concs. of products or reactants


Determining KDetermining KDetermining KDetermining K2 NOCl(g) 2 NOCl(g) 2 NO(g) + 2 NO(g) +

ClCl22(g)(g)

Place 2.00 mol of NOCl in a 1.00 L Place 2.00 mol of NOCl in a 1.00 L flask. At equilibrium you find 0.66 flask. At equilibrium you find 0.66 mol/L of NO. Calculate K.mol/L of NO. Calculate K.

We need a table showingWe need a table showing

The The reactionreaction and concentrations and concentrations at 3 conditions:at 3 conditions:

InitialInitialChangeChange EquilibriumEquilibrium


Determining KDetermining KDetermining KDetermining K2 NOCl(g) 2 NOCl(g) 2 NO(g) + Cl 2 NO(g) + Cl22(g)(g)

Place 2.00 mol of NOCl is a 1.00 L flask. At Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. equilibrium you find 0.66 mol/L of NO. Calculate K.Calculate K.

SolutionSolution

[NOCl][NOCl] [NO][NO] [Cl[Cl22]]

InitialInitial 2.002.00 00 00

ChangeChange -0.66-0.66 +0.66+0.66 +0.33+0.33

EquilibriumEquilibrium 1.341.34 0.660.66 0.330.33


Determining KDetermining KDetermining KDetermining K2 NOCl(g) 2 NOCl(g) 2 NO(g) + Cl 2 NO(g) + Cl22(g)(g)

Write the K expression and substitute Write the K expression and substitute the the EQUILIBRIUMEQUILIBRIUM values into it. values into it.



K [NO]2[Cl2 ]

[NOCl]2 K

[NO]2[Cl2 ]

[NOCl]2


Determining KDetermining KDetermining KDetermining K2 NOCl(g) 2 NOCl(g) 2 NO(g) + Cl2 NO(g) + Cl22(g)(g)

Write the K expression and substitute the Write the K expression and substitute the EQUILIBRIUMEQUILIBRIUM values into it. values into it.



K [NO]2[Cl2 ]

[NOCl]2 K

[NO]2[Cl2 ]

[NOCl]2

K [NO]2[Cl2 ]

[NOCl]2 =

(0.66)2(0.33)

(1.34)2 = 0.080K

[NO]2[Cl2 ]

[NOCl]2 =

(0.66)2(0.33)

(1.34)2 = 0.080


K [NH4

+][OH- ][NH3 ]

K [NH4

+][OH- ][NH3 ]

Writing and Manipulating K Writing and Manipulating K ExpressionsExpressions


Solids and liquids NEVER Solids and liquids NEVER appear in equilibrium appear in equilibrium expressions.expressions.

NHNH33(aq) + H(aq) + H22O(liq) NHO(liq) NH44++

(aq) + OH(aq) + OH--(aq)(aq)




Adding equations for reactionsAdding equations for reactions

S(s) + OS(s) + O22(g) (g) SOSO22(g) K(g) K11 = [SO = [SO22]/[O]/[O22]]

SOSO22(g) + 1/2 O(g) + 1/2 O22(g) (g) SO SO33(g)(g)

K K22 = [SO = [SO33] / [SO] / [SO22][O][O22]]1/21/2

Net eqn:Net eqn: S(s) + 3/2 O S(s) + 3/2 O22(g) (g) SO SO33(g)(g)

Knet [SO3]

[O2 ]3/2 = K1 • K2 Knet [SO3]

[O2 ]3/2 = K1 • K2


The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of K

1.1. Can tell if a reaction favors Can tell if a reaction favors products or reactants.products or reactants.

For NFor N22(g) + 3 H(g) + 3 H22(g) (g) 2 NH 2 NH33(g)(g)

Conc. of products is Conc. of products is much greatermuch greater than than that of reactants at equilibrium. that of reactants at equilibrium. Products are Products are favoredfavored..

Kc = [NH3 ]2

[N2 ][H2]3 = 3.5 x 108Kc =

[NH3 ]2

[N2 ][H2]3 = 3.5 x 108


The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of K2. Can tell if rxn. is at equilibrium, and 2. Can tell if rxn. is at equilibrium, and

how it moves to reach equilibrium.how it moves to reach equilibrium.

H

H

H

H

H

H

H

H

H H H

CH

HHH H

H—C—C—C—C—H H—C—C—C—H

K =

n-butane iso-butane

[iso]

[n] = 2.5

H

H

H

H

H

H

H

H

H H H

CH

HHH H


K =

n-butane iso-butane

[iso]

[n] = 2.5


The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of K

If [iso] = 0.35 M and [n] = 0.15 M, are If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? you at equilibrium?

Which way does the reaction “shift” Which way does the reaction “shift” to approach equilibrium?to approach equilibrium?

H

H

H

H

H

H

H

H

H H H

CH

HHH H


K =

n-butane iso-butane

[iso]

[n] = 2.5


The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of KIn general, all reacting chemical systems In general, all reacting chemical systems

are characterized by their are characterized by their REACTION REACTION QUOTIENT, QQUOTIENT, Q..

If Q = K, then system is at equilibrium.If Q = K, then system is at equilibrium.

Q = product concentrationsreactant concentrations



The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of KREACTION QUOTIENT, QREACTION QUOTIENT, Q..

If Q = K, the system is at equilibrium.If Q = K, the system is at equilibrium.[iso] = 0.35 M and [n] = 0.15 M, Is this [iso] = 0.35 M and [n] = 0.15 M, Is this

equilibrium?equilibrium?

Q (2.3) < K (2.5). Q (2.3) < K (2.5).



Q = conc. of isoconc. of n

= 0.350.15

= 2.3Q = conc. of isoconc. of n

= 0.350.15

= 2.3


The Meaning of KThe Meaning of KThe Meaning of KThe Meaning of KREACTION QUOTIENT, QREACTION QUOTIENT, Q..

If Q = K, then system is at equilibrium.If Q = K, then system is at equilibrium.

Q (2.33) < K (2.5). Q (2.33) < K (2.5).

Reaction is NOT at equilibrium, so [Iso] Reaction is NOT at equilibrium, so [Iso] must become larger and [n] must must become larger and [n] must become smaller. Reaction runs become smaller. Reaction runs FORWARD. FORWARD.



Q = conc. of isoconc. of n

= 0.350.15

= 2.3Q = conc. of isoconc. of n

= 0.350.15

= 2.3


Typical CalculationsTypical CalculationsTypical CalculationsTypical Calculations

PROBLEM: 1.00 mol each of HPROBLEM: 1.00 mol each of H22 and I and I22 in a 1.00 L flask. Calc. equilibrium in a 1.00 L flask. Calc. equilibrium concentrations.concentrations.

HH22(g) + I(g) + I22(g) (g) ¸̧ 2 HI(g) 2 HI(g)

Kc = [HI]2

[H2 ][I2 ] = 55.3Kc =

[HI]2

[H2 ][I2 ] = 55.3


H2(g) + I2(g) 2 HI(g), Kc = 55.3

Step 1. Set up table to define Step 1. Set up table to define EQUILIBRIUM concentrations.EQUILIBRIUM concentrations.

[H[H22]] [I[I22]] [HI][HI]

Initial Initial 1.001.00 1.001.00 00

ChangeChange -x-x -x-x +2x+2x

EquilibEquilib 1.00-x1.00-x 1.00-x1.00-x 2x2x


H2(g) + I2(g) 2 HI(g), Kc = 55.3

Step 2. Put equilibrium Step 2. Put equilibrium concentrations into Kconcentrations into Kcc

expression.expression.

Kc = [2x]2

[1.00 - x][1.00 - x] = 55.3Kc =

[2x]2

[1.00 - x][1.00 - x] = 55.3


H2(g) + I2(g) 2 HI(g), Kc = 55.3

7.44 = 2x

1.00 - x7.44 =

2x1.00 - x

Step 3. Solve KStep 3. Solve Kcc expression - take square expression - take square

root of both sides.root of both sides.

x = 0.79x = 0.79Therefore, at equilibriumTherefore, at equilibrium

[H[H22] = [I] = [I22] = 1.00 - x = 0.21 M] = 1.00 - x = 0.21 M

[HI] = 2x = 1.58 M[HI] = 2x = 1.58 M

huntington beach high school ap chemistry exam review winter review

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