sat ii chem prep ppt mrs. gupta modified from mark rosengartens powerpoint

SAT II CHEM PREP PPT

Mrs. Gupta

Modified from Mark Rosengarten’s Powerpoint

Setup of the SAT II Chem Exam

• 85 total questions, 1 hour (about 42 s/question)• - All multiple choice, 1/4th point taken off for every incorrect

answer• - if you can narrow down to two choices, then guess

otherwise leave blank• - scoring scale from 200-800

What to Bring to the Exam

• #2 pencil, eraser• No calculators allowed (brush up on your basic math skills)• Your brain. Please don’t leave it at home.:)

How To Prepare

• DO NOT CRAM. Get your studying done with by the night before. Get a good night’s sleep and have breakfast the morning of the exam.

• Actively participate in any and all review classes and activities offered by your teacher.

Matter

1) Properties of Phases2) Types of Matter3) Phase Changes

Properties of Phases

• Solids: Crystal lattice (regular geometric pattern), vibration motion only

• Liquids: particles flow past each other but are still attracted to each other.

• Gases: particles are small and far apart, they travel in a

straight line until they hit something, they bounce off without losing any energy, they are so far apart from each other that they have effectively no attractive forces and their speed is directly proportional to the Kelvin temperature (Kinetic-Molecular Theory, Ideal Gas Theory)

Solids

The positive and negative ions alternate in the ionic crystal latticeof NaCl.

Liquids

When heated, the ions movefaster and eventuallyseparate from each other to form a liquid. The ions areloosely held together by theoppositely charged ions, butthe ions are moving too fastfor the crystal lattice to staytogether.

GasesSince all gas molecules spread outthe same way, equal volumes of gas under equal conditions of temperature and pressure will contain equal numbers of molecules of gas. 22.4 L of any gas at STP (1.00 atm and 273K)will contain one mole (6.02 X 1023) gas molecules.

Since there is space between gasmolecules, gases are affected bychanges in pressure.

Types of Matter

• Substances (Homogeneous)– Elements (cannot be decomposed by chemical

change): Al, Ne, O, Br, H– Compounds (can be decomposed by chemical

change): NaCl, Cu(ClO3)2, KBr, H2O, C2H6

• Mixtures– Homogeneous: Solutions (solvent + solute)– Heterogeneous: soil, Italian dressing, etc.

Elements

• A sample of lead atoms (Pb). All atoms in the sample consist of lead, so the substance is homogeneous.

• A sample of chlorine atoms (Cl). All atoms in the sample consist of chlorine, so the substance is homogeneous.

Compounds

• Lead has two charges listed, +2 and +4. This is a sample of lead (II) chloride (PbCl2). Two or more elements bonded in a whole-number ratio is a COMPOUND.

• This compound is formed from the +4 version of lead. This is lead (IV) chloride (PbCl4). Notice how both samples of lead compounds have consistent composition throughout? Compounds are homogeneous!

Mixtures

• A mixture of lead atoms and chlorine atoms. They exist in no particular ratio and are not chemically combined with each other. They can be separated by physical means.

• A mixture of PbCl2 and PbCl4 formula units. Again, they are in no particular ratio to each other and can be separated without chemical change.

The Atom

1) Nucleons2) Isotopes3) Natural Radioactivity4) Half-Life5) Nuclear Power6) Electron Configuation7) Development of the Atomic Model

Nucleons

• Protons: +1 each, determines identity of element, mass of 1 amu, determined using atomic number, nuclear charge

• Neutrons: no charge, determines identity of isotope of an element, 1 amu, determined using mass number - atomic number (amu = atomic mass unit)

• 3216S and 33

16S are both isotopes of S

• S-32 has 16 protons and 16 neutrons• S-33 has 16 protons and 17 neutrons• All atoms of S have a nuclear charge of +16 due to the 16

protons.

Isotopes

• Atoms of the same element MUST contain the same number of protons.

• Atoms of the same element can vary in their numbers of neutrons, therefore many different atomic masses can exist for any one element. These are called isotopes.

• The atomic mass on the Periodic Table is the weight-average atomic mass, taking into account the different isotope masses and their relative abundance.

• Rounding off the atomic mass on the Periodic Table will tell you what the most common isotope of that element is.

Weight-Average Atomic Mass• WAM = ((% A of A/100) X Mass of A) + ((% A of B/100) X Mass of B) + …

• What is the WAM of an element if its isotope masses and abundances are:– X-200: Mass = 200.0 amu, % abundance = 20.0 %– X-204: Mass = 204.0 amu, % abundance = 80.0%

– amu = atomic mass unit (1.66 × 10-27 kilograms/amu)

Most Common Isotope

• The weight-average atomic mass of Zinc is 65.39 amu. What is the most common isotope of Zinc? Zn-65!

• What are the most common isotopes of: C, H and O?

• FACT: one atomic mass unit (1.66 × 10-27 kilograms) is defined as 1/12 of the mass of an atom of C-12.

Natural Radioactivity

• Alpha Decay • Beta Decay• Positron Decay• Gamma Decay• Charges of Decay Particles

• Natural decay starts with a parent nuclide that ejects a decay particle to form a daughter nuclide which is more stable than the parent nuclide was.

Alpha Decay

• The nucleus ejects two protons and two neutrons. The atomic mass decreases by 4, the atomic number decreases by 2.

• 23892U

Beta Decay

• A neutron decays into a proton and an electron. The electron is ejected from the nucleus as a beta particle. The atomic mass remains the same, but the atomic number increases by 1.

• 146C

Positron Decay

• A proton is converted into a neutron and a positron. The positron is ejected by the nucleus. The mass remains the same, but the atomic number decreases by 1.

• 5326Fe

Gamma Decay

• The nucleus has energy levels just like electrons, but the involve a lot more energy. When the nucleus becomes more stable, a gamma ray may be released. This is a photon of high-energy light, and has no mass or charge. The atomic mass and number do not change with gamma. Gamma may occur by itself, or in conjunction with any other decay type.

Charges of Decay Particles

Half-Life

• Half life is the time it takes for half of the nuclei in a radioactive sample to undergo decay.

• Problem Types:– Going forwards in time– Going backwards in time– Radioactive Dating

Going Forwards in Time

• How many grams of a 10.0 gram sample of I-131 (half-life of 8 days) will remain in 24 days?

• #HL = t/T = 24/8 = 3• Cut 10.0g in half 3 times: 5.00, 2.50, 1.25g

Going Backwards in Time

• How many grams of a 10.0 gram sample of I-131 (half-life of 8 days) would there have been 24 days ago?

• #HL = t/T = 24/8 = 3• Double 10.0g 3 times: 20.0, 40.0, 80.0 g

Radioactive Dating

• A sample of an ancient scroll contains 50% of the original steady-state concentration of C-14. How old is the scroll?

• 50% = 1 HL• 1 HL X 5730 y/HL = 5730y

Nuclear Power

• Artificial Transmutation• Particle Accelerators• Nuclear Fission• Nuclear Fusion

Artificial Transmutation

• 4020Ca + _____ -----> 40

19K + 11H

• 9642Mo + 2

1H -----> 10n + _____

• Nuclide + Bullet --> New Element + Fragment(s)

• The masses and atomic numbers must add up to be the same on both sides of the arrow.

Nuclear Fission

• 23592U + 1

0n 9236Kr + 141

56Ba + 3 10n + energy

• The three neutrons given off can be reabsorbed by other U-235 nuclei to continue fission as a chain reaction

• A tiny bit of mass is lost (mass defect) and converted into a huge amount of energy.

Chain Reaction

Nuclear Fusion

• 21H + 2

1H 42He + energy

• Two small, positively-charged nuclei smash together at high temperatures and pressures to form one larger nucleus.

• A small bit of mass is destroyed and converted into a huge amount of energy, more than even fission.

Electron Configuration

• Basic Configuration• Valence Electrons• Electron-Dot (Lewis Dot) Diagrams• Excited vs. Ground State• Rules for Electron Filling• Para and Diamagnetic• Lewis Structures and Hybridization

Basic Configuration

• The number of electrons is determined from the atomic number.

• Look up the basic configuration below the atomic number on the periodic table. (PEL: principal energy level = shell)

• He: 2 (2 e- in the 1st PEL)• Na: 2-8-1 (2 e- in the 1st PEL, 8 in the 2nd and 1 in the 3rd)• Br: 2-8-18-7 (2 e- in the 1st PEL, 8 in the 2nd, 18 in the 3rd and

7 in the 4th)

Valence Electrons

• The valence electrons are responsible for all chemical bonding.

• The valence electrons are the electrons in the outermost PEL (shell).

• He: 2 (2 valence electrons)• Na: 2-8-1 (1 valence electron)• Br: 2-8-18-7 (7 valence electrons)

• The maximum number of valence electrons an atom can have is EIGHT, called a STABLE OCTET.

Electron-Dot Diagrams

• The number of dots equals the number of valence electrons.• The number of unpaired valence electrons in a nonmetal tells

you how many covalent bonds that atom can form with other nonmetals or how many electrons it wants to gain from metals to form an ion.

• The number of valence electrons in a metal tells you how many electrons the metal will lose to nonmetals to form an ion. Caution: May not work with transition metals.

• EXAMPLE DOT DIAGRAMS

(c) 2006, Mark Rosengarten

Example Dot Diagrams

Carbon can also have this dot diagram, which ithas when it forms organic compounds.

Excited vs. Ground State

• Configurations on the Periodic Table are ground state configurations.

• If electrons are given energy, they rise to higher energy levels (excited state).

• If the total number of electrons matches in the configuration, but the configuration doesn’t match, the atom is in the excited state.

• Na (ground, on table): 2-8-1• Example of excited states: 2-7-2, 2-8-0-1, 2-6-3

Ways to Represent Electron Configuration1.Expanded Electron Configuration

2.Condensed Electron Configurations

3.Orbital Notation

4.Electron Dot Structure

Write the above four electron configurations for Zinc, Zinc ion and Cu ion.

Electron Configuration of Ions

• -Group configurations: s block ns1-2, p block ns2 np 1-6, d block ns0-2 (n-1) d 1-10, f block ns 0-

2 (n-1) d 1 (n-2) f 1-14

• - Remember that outmost electrons are lost first (which means that it will always be s or p electrons lost, never d or f). Ex. Sc+ or Sc3+ electron configuration would be:

Rules for Electron Filling

• - Afbau’s Principle: Electrons tend to occupy the lowest energy orbitals first.

• - Hund’s Rule: Pairing of e in the degenerate orbitals does not take place till every orbital has one e.

• - Pauli’s Exclusion Principle: No two electrons can have all four same quantum numbers.

Diamagnetism, Paramagnetism

• Diamagnetism: does not show magnetic properties in external magnetic field. No unpaired electrons.

• Paramagnetism: shows magnetic properties in external magnetic field. Has unpaired electrons.

• Best way to predict dia or paramagnetism is by drawing orbital diagrams.

Writing Lewis Structures• Lewis structures are used to

depict bonding pairs and lone pairs of electron in the molecule.

Step 1• Total number of valence

electrons in the system: Sum the number of valence electrons on all the atoms . Add the total negative charge if you have an anion. Subtract the charge if you have a cation.Example: CO3

2- Step 2• Number of electrons if each

atom is to be happy: Atoms in our example will need 8 e (octet rule) or 2 e ( hydrogen). So, for the ex.

•

• Step 3

Step 3• Calculate number of bonds in the system:

Covalent bonds are made by sharing of e. You need 32 and you have 24. You are 8 e deficient. If you make 4 bonds ( with 2 e per bond) , you will make up the deficiency. Therefore,

• # of bonds= ( e in step 2- e in step 1)/2 =(32-

24)/2= 4 bondsStep 4• Draw the structure: The central atom is C

( usually the atom with least electro negativity will be in the center). The oxygens surround it . Because there are four bonds and only three atoms, there will be one double bond.

Step 5 • Double check your answer by counting total

number of electrons.

© 2009, Prentice-Hall, Inc.

Formal Charges: Writing Lewis Structures

• Then assign formal charges.– For each atom, count the electrons in lone pairs and half

the electrons it shares with other atoms.– Subtract that from the number of valence electrons for that

atom: the difference is its formal charge.


Writing Lewis Structures

• The best Lewis structure…– …is the one with the fewest charges.– …puts a negative charge on the most

electronegative atom.


Resonance

• One Lewis structure cannot accurately depict a molecule like ozone.

• We use multiple structures, resonance structures, to describe the molecule.


Resonance

Just as green is a synthesis of blue and yellow…

…ozone is a synthesis of these two resonance structures.


Molecular Shapes

• The shape of a molecule plays an important role in its reactivity.

• By noting the number of bonding and nonbonding electron pairs we can easily predict the shape of the molecule.


What Determines the Shape of a Molecule?

• Simply put, electron pairs, whether they be bonding or nonbonding, repel each other.

• By assuming the electron pairs are placed as far as possible from each other, we can predict the shape of the molecule.


Electron Domains

• We can refer to the electron pairs as electron domains.

• In a double or triple bond, all electrons shared between those two atoms are on the same side of the central atom; therefore, they count as one electron domain.


• The central atom in this molecule, A, has four electron domains.

Valence Shell Electron Pair Repulsion Theory (VSEPR)

“The best arrangement of a given number of electron domains is the one that minimizes the repulsions among them.”


Electron-Domain Geometries

These are the electron-domain geometries for two through six electron domains around a central atom.


Hybridization

• Refers to mixing of orbitals.

• Atomic orbitals of central atom undergo change to accommodate incoming atoms.

• Hybridization could be sp, sp2, sp3, sp3d and sp3d2.

• How do you tell the hybridization on the central atom?

9.1 – 9.2: V.S.E.P.R.Valence-shell electron-pair repulsion theory– Because e- pairs repel, molecular shape adjusts so the valence e-

pairs are as far apart as possible around the central atom.– Electron domains: areas of valence e- density around the central

atom; result in different molecular shapes• Includes bonding e- pairs and nonbonding e- pairs• A single, double, or triple bond counts as one domain

Summary of LmABn (Tables 9.1 - 9.3):

L = lone or non-bonding pairs

A = central atom

B = bonded atomsBond angles notation used here:

< xº means ~2-3º less than predicted

<< xº means ~4-6º less than predicted

Tables 9.1 - 9.3# of e-

domains&

# and type of hybrid orbitals

e- domain geometry

Formula &

Molecular geometry

Predicted bond

angle(s)

Example(Lewis

structure with

molecular shape)

2

Two sp hybrid orbitals

Linear

AB2

Linear

180ºBeF2

CO2

A|X

X

A|B

B

3

Three sp2 hybrid orbitals

Trigonal planar

AB3

Trigonal planar

120º

BF3

Cl-C-Cl<< 120º

Cl2CO

LAB2

Bent

< 120º

NO21-

A|X

XX

A|B

BB

A|B

:B

Example: CH4

Molecular shape = tetrahedral

Bond angle = 109.5º

H|

H—C—H|H

109.5º

109.5º

109.5º

109.5º

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4

Four sp3

hybrid orbitals

or

Tetrahedral

AB4

Tetrahedral

109.5º

CH4

LAB3

Trigonal pyramidal

< 109.5º

Ex: NH3 = 107º NH3

L2AB2 Bent

<<109.5º

Ex: H2O = 104.5º

H2O

X

A

X

XX

X

A

X

XX

B

A

B

BB

:

A

B

BB

:

A

B

B:

PCl5

Molecular shape = trigonal bipyramidal

Bond angles

equatorial = 120º

axial = 90º

:Cl: :Cl:\ /

:Cl—P—Cl:|

:Cl::

:::

:: :

120º

120º

90º90º

90º

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5

Five sp3d

hybrid orbitals

Trigonal bipyramida

l

AB5

Trigonal bipyramida

l

Equatorial = 120º

Axial = 90º

PCl5

LAB4

Seesaw

Equatorial < 120º

Axial< 90º

SF4

X

X

X

A

X

|

X

B

B

B

A

B

|

B

:

B - A - B

B B

5

Five sp3d

hybrid orbitals

Trigonal bipyramidal

L2AB3

T-shaped

Axial<< 90º

ClF3

L3AB2

Linear

Axial = 180º

XeF2

X

X

X

A

X

|

X

B

:

:

A

B

|

B

:

:

:

A

B

|

B

6

Six sp3d2

hybrid orbitals

or

Octahedral

AB6

Octahedral

90º

SF6

LAB5

Square pyramidal

< 90º

BrF5

X

X

A

X

|

X

X

X

B

B

A

B

|

B

B

B

B

B

A

B

|..

B

BA

X

|

X

XX

X

X

6

Six sp3d2 hybrid orbitals

or

L2AB4

Square planar

90º

XeF4

or

L3AB3

T-shaped

<90º

KrCl31-

A

B

|

B

BB

..

..B

B

A

..

|..

B

B

B

B

A

..

|..

..

B

A

..

|

B

BB

..

..

What Is Light?

• Light is formed when electrons drop from the excited state to the ground state.

• The lines on a bright-line spectrum come from specific energy level drops and are unique to each element.

• Ex. Emission and Absorption Spectra ( line spectra)

EXAMPLE SPECTRUM

This is the bright-line spectrum of hydrogen. The topnumbers represent the energy level transition change that produces the light with that color and the bottom number is thewavelength of the light (in nanometers, or 10-9 m).

No other element has the same bright-line spectrum ashydrogen, so these spectra can be used to identifyelements or mixtures of elements.

Development of the Atomic Model

• Thompson Model• Rutherford Gold Foil Experiment and Model• Bohr Model• Quantum-Mechanical Model

Thompson Model

• The atom is a positively charged diffuse mass with negatively charged electrons stuck in it.

Rutherford Model

• The atom is made of a small, dense, positively charged nucleus with electrons at a distance, the vast majority of the volume of the atom is empty space.

Alpha particles shotat a thin sheet of goldfoil: most go through(empty space). Somedeflect or bounce off(small + chargednucleus).

Bohr Model

• Electrons orbit around the nucleus in energy levels (shells). Atomic bright-line spectra was the clue.

Quantum-Mechanical Model

• Electron energy levels are wave functions.• Electrons are found in orbitals, regions of space where an

electron is most likely to be found.• You can’t know both where the electron is and where it is

going at the same time.• Electrons buzz around the nucleus like gnats buzzing around

your head.

Orbital Quantum NumbersSymbol Name Description Meaning Equations

nPrinciple

Q.N.

Energy level

(i.e. Bohr’s theory)

Shell number

n = 1, 2, 3, 4, 5, 6, 7

n = 1, 2, 3, …

lAngular

Momentum Q.N.

General probability

plot (“shape” of the

orbitals)

Subshell number

l = 0, 1, 2, 3

l = 0 means “s”

l = 1 means “p”

l = 2 means “d”

l = 3 means “f”

l = 0, 1, 2, …, n – 1

Ex: If n = 1, l can only be 0; if n = 2, l can be 0 or 1.

Symbol Name Description Meaning Equations

mlMagnetic

Q.N.3-D orientation of the orbital

s has 1

p has 3

d has 5

f has 7

ml = -l, -l +1, …,

0, l, …, +l

There are

(2l + 1) values.

ms Spin Q.N.Spin of the electron

Parallel or antiparallel

to field

ms = +½ or

-½

* s, p, d, and f come from the words sharp, principal, diffuse, and fundamental.

Permissible Quantum Numbers

(4, 1, 2, +½)

(5, 2, 0, 0)

(2, 2, 1, +½)

76

Not permissible; if l = 1, ml = 1, 0, or –1 (p orbitals only have 3 subshells)

Not permissible; ms = +½ or –½

Not permissible; if n = 2, l = 0 or 1 (there is no 2d orbital)

Phase Changes

• Phase Change Types• Phase Change Diagrams• Heat of Phase Change• Evaporation

Phase Change Types

Phase Change Diagrams

AB: Solid PhaseBC: Melting (S + L)CD: Liquid PhaseDE: Boiling (L + G)EF: Gas Phase

Notice how temperature remains constant during a phase change? That’s because the PE is changing, not the KE.

Heat of Phase Change

• How many joules would it take to melt 100. g of H2O (s) at 0oC?

• q=mHf = (100. g)(334 J/g) = 33400 J• How many joules would it take to boil 100. g of H2O (l) at

100oC?• q=mHv = (100.g)(2260 J/g) = 226000 J

Evaporation

• When the surface molecules of a gas travel upwards at a great enough speed to escape.

• The pressure a vapor exerts when sealed in a container at equilibrium is called vapor pressure, and can be found on Table H.

• When the liquid is heated, its vapor pressure increases.• When the liquid’s vapor pressure equals the pressure

exerted on it by the outside atmosphere, the liquid can boil.

• If the pressure exerted on a liquid increases, the boiling point of the liquid increases (pressure cooker). If the pressure decreases, the boiling point of the liquid decreases (special cooking directions for high elevations).

Reference Table H: Vapor Pressure of Four Liquids


83

Phase diagrams: CO2 Lines: 2 phases exist in

equilibrium

Triple point: all 3 phases exist together in equilibrium (X on graph)

Critical point, or critical temperature & pressure: highest T and P at which a liquid can exist (Z on graph)

For most substances, inc P will cause a gas to condense (or deposit), a liquid to freeze, and a solid to become more dense (to a limit.)

Temp (ºC)

84

Phase diagrams: H2O• For H2O, inc P

will cause ice to melt.

The Periodic Table

• Metals• Nonmetals• Metalloids• Chemistry of Groups• Electronegativity• Ionization Energy

Metals

• Have luster, are malleable and ductile, good conductors of heat and electricity

• Lose electrons to nonmetal atoms to form positively charged ions in ionic bonds

• Large atomic radii compared to nonmetal atoms• Low electronegativity and ionization energy• Left side of the periodic table (except H)

Nonmetals

• Are dull and brittle, poor conductors• Gain electrons from metal atoms to form negatively

charged ions in ionic bonds

• Share unpaired valence electrons with other nonmetal atoms to form covalent bonds and molecules

• Small atomic radii compared to metal atoms• High electronegativity and ionization energy• Right side of the periodic table (except Group 18)

Metalloids

• Found lying on the jagged line between metals and nonmetals flatly touching the line (except Al and Po).

• Share properties of metals and nonmetals (Si is shiny like a metal, brittle like a nonmetal and is a semiconductor).

Chemistry of Groups

• Group 1: Alkali Metals• Group 2: Alkaline Earth Metals• Groups 3-11: Transition Elements• Group 17: Halogens• Group 18: Noble Gases

• Diatomic Molecules

Group 1: Alkali Metals

• Most active metals, only found in compounds in nature

• React violently with water to form hydrogen gas and a strong base: 2 Na (s) + H2O (l) 2 NaOH (aq) + H2 (g)

• 1 valence electron• Form +1 ion by losing that valence electron• Form oxides like Na2O, Li2O, K2O

Group 2: Alkaline Earth Metals

• Very active metals, only found in compounds in nature

• React strongly with water to form hydrogen gas and a base: – Ca (s) + 2 H2O (l) Ca(OH)2 (aq) + H2 (g)

• 2 valence electrons• Form +2 ion by losing those valence electrons• Form oxides like CaO, MgO, BaO

Groups 3-11: Transition Metals

• Many can form different possible charges of ions• If there is more than one ion listed, give the charge as a

Roman numeral after the name• Cu+1 = copper (I) Cu+2 = copper (II)• Compounds containing these metals can be colored.

Group 17: Halogens

• Most reactive nonmetals• React violently with metal atoms to form halide

compounds: 2 Na + Cl2 2 NaCl

• Only found in compounds in nature• Have 7 valence electrons• Gain 1 valence electron from a metal to form -1

ions• Share 1 valence electron with another nonmetal

atom to form one covalent bond.

Group 18: Noble Gases

• Are completely nonreactive since they have eight valence electrons, making a stable octet.

• Kr and Xe can be forced, in the laboratory, to give up some valence electrons to react with fluorine.

• Since noble gases do not naturally bond to any other elements, one atom of noble gas is considered to be a molecule of noble gas. This is called a monatomic molecule. Ne represents an atom of Ne and a molecule of Ne.

Diatomic Molecules

• Br, I, N, Cl, H, O and F are so reactive that they exist in a more chemically stable state when they covalently bond with another atom of their own element to make two-atom, or diatomic molecules.

• Br2, I2, N2, Cl2, H2, O2 and F2

• The decomposition of water: 2 H2O 2 H2 + O2

Electronegativity

• An atom’s attraction to electrons in a chemical bond.• F has the highest, at 4.0• Fr has the lowest, at 0.7• If two atoms that are different in EN (END) from each other

by 1.7 or more collide and bond (like a metal atom and a nonmetal atom), the one with the higher electronegativity will pull the valence electrons away from the atom with the lower electronegativity to form a (-) ion. The atom that was stripped of its valence electrons forms a (+) ion.

• If the two atoms have an END of less than 1.7, they will share their unpaired valence electrons…covalent bond!

Ionization Energy

• The energy required to remove the most loosely held valence electron from an atom in the gas phase.

• High electronegativity means high ionization energy because if an atom is more attracted to electrons, it will take more energy to remove those electrons.

• Metals have low ionization energy. They lose electrons easily to form (+) charged ions.

• Nonmetals have high ionization energy but high electronegativity. They gain electrons easily to form (-) charged ions when reacted with metals, or share unpaired valence electrons with other nonmetal atoms.

Ions

• Ions are charged particles formed by the gain or loss of electrons.– Metals lose electrons (oxidation) to form (+) charged

cations.– Nonmetals gain electrons (reduction) to form (-) charged

anions.• Atoms will gain or lose electrons in such a way that they end

up with 8 valence electrons (stable octet).– The exceptions to this are H, Li, Be and B, which are not

large enough to support 8 valence electrons. They must be satisfied with 2 (Li, Be, B) or 0 (H).

Metal Ions (Cations)

• Na: 2-8-1

• Na+1: 2-8

• Ca: 2-8-8-2

• Ca+2: 2-8-8

• Al: 2-8-3

• Al+3: 2-8

Note that when the atom loses its valence electron, the next lower PEL becomes the valence PEL.

Notice how the dot diagrams for metal ions lack dots! Place brackets around the element symbol and put the charge on the upper right outside!

Nonmetal Ions (Anions)

• F: 2-7• F-1: 2-8

• O: 2-6• O-2: 2-8

• N: 2-5• N-3: 2-8

Note how the ions all have 8 valence electrons. Also note the gained electrons as red dots. Nonmetal ion dot diagrams show 8 dots, with brackets around the dot diagram and the charge of the ion written to the upper right side outside the brackets.

Chemical Bonding

•Intermolecular Bonding: Ionic, Covalent, Metallic and Covalent Network Bonds•Intermolecular Bonding: H bond, dipole-dipole interaction, LDFs

Ionic Bonding

• If two atoms that are different in EN (END) from each other by 1.7 or more collide and bond (like a metal atom and a nonmetal atom), the one with the higher electronegativity will pull the valence electrons away from the atom with the lower electronegativity to form a (-) ion. The atom that was stripped of its valence electrons forms a (+) ion.

• The oppositely charged ions attract to form the bond. It is a surface bond that can be broken by melting or dissolving in water.

• Ionic bonding forms ionic crystal lattices, not molecules.

Example of Ionic Bonding

Covalent Bonding

• If two nonmetal atoms have an END of 1.7 or less, they will share their unpaired valence electrons to form a covalent bond.

• A particle made of covalently bonded nonmetal atoms is called a molecule.

• If the END is between 0 and 0.4, the sharing of electrons is equal, so there are no charged ends. This is NONPOLAR covalent bonding.

• If the END is between 0.5 and 1.7, the sharing of electrons is unequal. The atom with the higher EN will be - and the one with the lower EN will be + charged. This is a POLAR covalent bonding. (means “partial”)

Examples of Covalent Bonding

Sigma and Pi bondsSigma () bond:• Covalent bond that results from axial overlap of orbitals between atoms in

a molecule• Lie directly on internuclear axis• “Single” bonds, could form between s-s orbital or s-p orbital or p-p orbital

by axial overlappingEx: F2

Pi () bond:• Covalent bond that results from side-by-side overlap of orbitals between

atoms in a molecule.• Are “above & below” and “left & right” of the internuclear axis and

therefore have less total orbital overlap, so they are weaker than bonds. Forms between two p orbitals (py or pz)

• Make up the 2nd and 3rd bonds in double & triple bonds.Ex: O2 N2

Metallic Bonding

• Metal atoms of the same element bond with each other by sharing valence electrons that they lose to each other.

• This is a lot like an atomic game of “hot potato”, where metal kernals (the atom inside the valence electrons) sit in a crystal lattice, passing valence electrons back and forth between each other).

• Since electrons can be forced to travel in a certain direction within the metal, metals are very good at conducting electricity in all phases.

Types of Compounds

• Ionic: made of metal and nonmetal ions. Form an ionic crystal lattice when in the solid phase. Ions separate when melted or dissolved in water, allowing electrical conduction. Examples: NaCl, K2O, CaBr2

• Molecular: made of nonmetal atoms bonded to form a distinct particle called a molecule. Bonds do not break upon melting or dissolving, so molecular substances do not conduct electricity. EXCEPTION: Acids [H+A- (aq)] ionize in water to form H3O+ and A-, so they do conduct.

• Network: made up of nonmetal atoms bonded in a seemingly endless matrix of covalent bonds with no distinguishable molecules. Very high m.p., don’t conduct.

Ionic Compounds


Ionic Crystal Structure, then adding heat (or dissolving in water) to break up the crystal into a liquid composed of free-moving ions.

Molecular Compounds


Network Solids

Network solids are made of nonmetal atoms covalently bonded together to form large crystal lattices. No individual molecules can be distinguished. Examples include C (diamond) and SiO2 (quartz). Corundum (Al2O3) also forms these, even though Al is considered a metal. Network solids are among the hardest materials known. They have extremely high melting points and do not conduct electricity.

Intermolecular Forces

The attractions between molecules are not nearly as strong as the intramolecular attractions that hold compounds together.



They are, however, strong enough to control physical properties such as boiling and melting points, vapor pressures, and viscosities.



These intermolecular forces as a group are referred to as van der Waals forces.


van der Waals Forces

• Dipole-dipole interactions• Hydrogen bonding• London dispersion forces


Ion-Dipole Interactions

• Ion-dipole interactions (a fourth type of force), are important in solutions of ions.

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


Dipole-Dipole Interactions

• Molecules that have permanent dipoles are attracted to each other.– The positive end of one is

attracted to the negative end of the other and vice-versa.

– These forces are only important when the molecules are close to each other.


Dipole-Dipole Interactions

The more polar the molecule, the higher is its boiling point.


London Dispersion Forces

While the electrons in the 1s orbital of helium would repel each other (and, therefore, tend to stay far away from each other), it does happen that they occasionally wind up on the same side of the atom.



At that instant, then, the helium atom is polar, with an excess of electrons on the left side and a shortage on the right side.



Another helium nearby, then, would have a dipole induced in it, as the electrons on the left side of helium atom 2 repel the electrons in the cloud on helium atom 1.



London dispersion forces, or dispersion forces, are attractions between an instantaneous dipole and an induced dipole.



• These forces are present in all molecules, whether they are polar or nonpolar.

• The tendency of an electron cloud to distort in this way is called polarizability.


Factors Affecting London Forces

• The shape of the molecule affects the strength of dispersion forces: long, skinny molecules (like n-pentane tend to have stronger dispersion forces than short, fat ones (like neopentane).

• This is due to the increased surface area in n-pentane.


Factors Affecting London Forces

• The strength of dispersion forces tends to increase with increased molecular weight.

• Larger atoms have larger electron clouds which are easier to polarize.


Which Have a Greater Effect?Dipole-Dipole Interactions or Dispersion Forces

• If two molecules are of comparable size and shape, dipole-dipole interactions will likely the dominating force.

• If one molecule is much larger than another, dispersion forces will likely determine its physical properties.


How Do We Explain This?

• The nonpolar series (SnH4 to CH4) follow the expected trend.

• The polar series follows the trend from H2Te through H2S, but water is quite an anomaly.


Hydrogen Bonding

• The dipole-dipole interactions experienced when H is bonded to N, O, or F are unusually strong.

• We call these interactions hydrogen bonds.


Hydrogen Bonding

• Hydrogen bonding arises in part from the high electronegativity of nitrogen, oxygen, and fluorine.


Also, when hydrogen is bonded to one of those very electronegative elements, the hydrogen nucleus is exposed.

Attractive Forces

• Molecules have partially charged ends. The + end of one molecule attracts to the - end of another molecule.

• Ions are charged (+) or (-). Positively charged ions attract other to form ionic bonds, a type of attractive force.

• Since partially charged ends result in weaker attractions than fully charged ends, ionic compounds generally have much higher melting points than molecular compounds.

• Determining Polarity of Molecules • Hydrogen Bond Attractions

Determining Polarity ofMolecules


-----------------------------------------------------------------------------

Hydrogen BondAttractions

A hydrogen bond attraction is a very strong attractive force between the H end of one polar molecule and the N, O or F end of another polar molecule. This attraction is so strong that water is a liquid at a temperature where most compounds that are much heavier than water (like propane, C3H8) are gases. This also gives water its surface tension and its ability to form a meniscus in a narrow glass tube.

Summarizing Intermolecular Forces


Compounds

1) Types of Compounds2) Formula Writing3) Formula Naming4) Empirical Formulas5) Molecular Formulas6) Types of Chemical Reactions7) Balancing Chemical Reactions8) Attractive Forces

Formula Writing

• The charge of the (+) ion and the charge of the (-) ion must cancel out to make the formula. Use subscripts to indicate how many atoms of each element there are in the compound, no subscript if there is only one atom of that element.

• Na+1 and Cl-1 = NaCl• Ca+2 and Br-1 = CaBr2

• Al+3 and O-2 = Al2O3

• Zn+2 and PO4-3 = Zn3(PO4)2

• Try these problems!

Formulas to Write

• Ba+2 and N-3

• NH4+1 and SO4

-2

• Li+1 and S-2

• Cu+2 and NO3-1

• Al+3 and CO3-2

• Fe+3 and Cl-1

• Pb+4 and O-2

• Pb+2 and O-2

Formula Naming

• Compounds are named from the elements or polyatomic ions that form them.

• KCl = potassium chloride• Na2SO4 = sodium sulfate

• (NH4)2S = ammonium sulfide

• AgNO3 = silver nitrate

• Notice all the metals listed here only have one charge listed? So what do you do if a metal has more than one charge listed? Take a peek!

The Stock System

• CrCl2 = chromium (II) chloride Try

• CrCl3 = chromium (III) chloride Co(NO3)2 and

• CrCl6 = chromium (VI) chloride Co(NO3)3

• FeO = iron (II) oxide MnS = manganese (II) sulfide• Fe2O3 = iron (III) oxide MnS2 = manganese (IV) sulfide

• The Roman numeral is the charge of the metal ion!

Math of Chemistry

1) Formula Mass2) Percent Composition3) Mole Problems4) Gas Laws5) Neutralization6) Concentration7) Significant Figures and Rounding8) Metric Conversions9) Calorimetry

Formula Mass

• Gram Formula Mass = sum of atomic masses of all elements in the compound

• Round given atomic masses to the nearest tenth• H2O: (2 X 1.0) + (1 X 16.0) = 18.0 grams/mole

• Na2SO4: (2 X 23.0)+(1 X 32.1)+(4 X 16.0) = 142.1 g/mole

• Now you try:– BaBr2

– CaSO4

– Al2(CO3)3

Percent Composition

What is the % composition, by mass,of each element in SiO2?

%Si = (28.1/60.1) X 100 = 46.8%%O = (2 X 16.0 = 32.0), (32.0/60.1) X 100 = 53.2%

The mass of part is the number of atoms of that element in the compound. The mass of whole is the formula mass of the compound. Don’t forget to take atomic mass to the nearest tenth! This is a problem for you to try.

Practice PercentComposition Problem

• What is the percent by mass of each element in Li2SO4?

Mole Problems

• Grams <=> Moles• Molecular Formula• Stoichiometry

Grams <=> Moles

• How many grams will 3.00 moles of NaOH (40.0 g/mol) weigh?

• 3.00 moles X 40.0 g/mol = 120. g

• How many moles of NaOH (40.0 g/mol) are represented by 10.0 grams?

• (10.0 g) / (40.0 g/mol) = 0.250 mol

Molecular Formula• Molecular Formula = (Molecular Mass/Empirical Mass) X Empirical Formula

• What is the molecular formula of a compound with an empirical formula of CH2 and a molecular mass of 70.0 grams/mole?

• 1) Find the Empirical Formula Mass: CH2 = 14.0

• 2) Divide the MM/EM: 70.0/14.0 = 5• 3) Multiply the molecular formula by the result:

5 (CH2) = C5H10

Stoichiometry

• Moles of Target = Moles of Given X (Coefficent of Target/Coefficient of given)

• Given the balanced equation N2 + 3 H2 2 NH3, How many moles of H2 need to be completely reacted with N2 to yield 20.0 moles of NH3?

• 20.0 moles NH3 X (3 H2 / 2 NH3) = 30.0 moles H2

Limiting Reactant

• controls the amount of product formed.CO(g) + 2H2 (g) Ch3OH

a. If 500 mol of CO react with 750 mol of H2, which is the limiting reactant?

1.Use either given amount to calculate required amount of other.

2.Compare calculated amount to amount given

b. How many moles of excess reactant remain unchanged?

147

H2

125 mol CO

Percent yield= (actual yield/ theoretical yield)*100

• Theoretical yield is the maximum amount of product that can be produced from a given amount of reactant

• Actual yield is the measured amount of a product obtained from a reaction

148

Theoretical yield= 117.5 g SnF2

Actual yield = 113. 4g SnF2

Percent yield = 113.4 g SnF2

117.5 g SnF2

*100

Determining empirical formula from combustion data

When a compound containing C,H and O undergoes combustion, it forms CO2 and H2O. Then from the mass of CO2 and H2O, we can calculate the mass of C and Hand then find the mass of O by subtracting the sum of masses of C and H from total g present of that substance. From the mass of C,H and O, we can calculate the moles of C,H and O.Then the smallest whole number ratios of these moles will give the empirical formula.

Ex. A 0.6349 g sample of the unknown produced 1.603 g of CO2 and 0.2810 g of H2O. Determine the empirical formula of the compound. Ans. C7H6O2

Empirical Formulas

• Ionic formulas: represent the simplest whole number mole ratio of elements in a compound.

• Ca3N2 means a 3:2 ratio of Ca ions to N ions in the compound.

• Many molecular formulas can be simplified to empirical formulas– Ethane (C2H6) can be simplified to CH3. This is the

empirical formula…the ratio of C to H in the molecule.• All ionic compounds have empirical formulas.

Molecular Formulas

• The count of the actual number of atoms of each element in a molecule.

• H2O: a molecule made of two H atoms and one O atom covalently bonded together.

• C2H6O: A molecule made of two C atoms, six H atoms and one O atom covalently bonded together.

• Molecular formulas are whole-number multiples of empirical formulas:– H2O = 1 X (H2O)

– C8H16 = 8 X (CH2)

• Calculating Molecular Formulas


Calculating Empirical Formulas

One can calculate the empirical formula from the percent composition.



The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.



Assuming 100.00 g of para-aminobenzoic acid,

C: 61.31 g x = 5.105 mol C

H: 5.14 g x = 5.09 mol H

N: 10.21 g x = 0.7288 mol N

O: 23.33 g x = 1.456 mol O

1 mol12.01 g

1 mol14.01 g

1 mol1.01 g

1 mol16.00 g


Calculating Empirical FormulasCalculate the mole ratio by dividing by the smallest number of moles:

C: = 7.005 7

H: = 6.984 7

N: = 1.000

O: = 2.001 2

5.105 mol0.7288 mol

5.09 mol0.7288 mol

0.7288 mol0.7288 mol

1.458 mol0.7288 mol



These are the subscripts for the empirical formula:

C7H7NO2


Combustion Analysis

• Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this.– C is determined from the mass of CO2 produced.

– H is determined from the mass of H2O produced.

– O is determined by difference after the C and H have been determined.


Stoichiometric Calculations

The coefficients in the balanced equation give the ratio of moles of reactants and products.



Starting with the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant).



Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6…use the coefficients to find the moles of H2O…and then turn the moles of water to grams.

C6H12O6 + 6 O2 6 CO2 + 6 H2O

Molecular FormulaActual ratio of atoms in a compound. Ex. H2O, C6H12O6

To determine the molecular formula, divide the molar mass by empirical formula mass. This will give the number of empirical formula units (n) in actual molecule.

n= Molar Mass/ Empirical Formula Mass

Ex. Determine the empirical and molecular formula of each of the following: 1.Ethylene glycol, the substance used as antifreeze has 38.70 % C, 9.70 % H and 51.60 % O , mm= 62.10 g2.Caffeine, a stimulant in coffee has the following percent composition:49.50 % C, 5.15% H, 28.90 % N and 16.50 % O , molar mass= 195.00g

Types of Chemical Reactions

• Redox Reactions: driven by the loss (oxidation) and gain (reduction) of electrons. Any species that does not change charge is called the spectator ion.– Synthesis– Decomposition– Single Replacement

• Ion Exchange Reaction: driven by the formation of an insoluble precipitate. The ions that remain dissolved throughout are the spectator ions.– Double Replacement

Synthesis

• Two elements combine to form a compound• 2 Na + O2 Na2O

• Same reaction, with charges added in:– 2 Na0 + O2

0 Na2+1O-2

• Na0 is oxidized (loses electrons), is the reducing agent• O2

0 is reduced (gains electrons), is the oxidizing agent

• Electrons are transferred from the Na0 to the O20.

• No spectator ions, there are only two elements here.

Decomposition

• A compound breaks down into its original elements.• Na2O 2 Na + O2

• Same reaction, with charges added in:– Na2

+1O-2 2 Na0 + O20

• O-2 is oxidized (loses electrons), is the reducing agent• Na+1 is reduced (gains electrons), is the oxidizing agent

• Electrons are transferred from the O-2 to the Na+1.

• No spectator ions, there are only two elements here.

Single Replacement• An element replaces the same type of element in a

compound.• Ca + 2 KCl CaCl2 + 2 K• Same reaction, with charges added in:

– Ca0 + 2 K+1Cl-1 Ca+2Cl2-1 + 2 K0

• Ca0 is oxidized (loses electrons), is the reducing agent• K+1 is reduced (gains electrons), is the oxidizing agent

• Electrons are transferred from the Ca0 to the K+1.

• Cl-1 is the spectator ion, since it’s charge doesn’t change.

Double Replacement

• The (+) ion of one compound bonds to the (-) ion of another compound to make an insoluble precipitate. The compounds must both be dissolved in water to break the ionic bonds first.

• NaCl (aq) + AgNO3 (aq) NaNO3 (aq) + AgCl (s)

• The Cl-1 and Ag+1 come together to make the insoluble precipitate, which looks like snow in the test tube.

• No species change charge, so this is not a redox reaction.• Since the Na+1 and NO3

-1 ions remain dissolved throughout the reaction, they are the spectator ions.

• How do identify the precipitate?

Identifying the Precipitate

• The precipitate is the compound that is insoluble. AgCl is a precipitate because Cl- is a halide. Halides are soluble, except when combined with Ag+ and others.

Balancing Chemical Reactions

• Balance one element or ion at a time• Use a pencil• Use coefficients only, never change formulas• Revise if necessary

• The coefficient multiplies everything in the formula by that amount– 2 Ca(NO3)2 means that you have 2 Ca, 4 N and 12 O.

• Examples for you to try!

Reactions to Balance

• ___NaCl ___Na + ___Cl2

• ___Al + ___O2 ___Al2O3

• ___SO3 ___SO2 + ___O2

• ___Ca + ___HNO3 ___Ca(NO3)2 + ___H2

• __FeCl3 + __Pb(NO3)2 __Fe(NO3)3 + __PbCl2

Writing Net Ionic Equations

• Cancel all the spectator ions.• Dissociate all dissociable ionic compounds

(refer to solubility rules)• All gases and liquids NEVER dissociate. • Write the net ionic equation.

Manometers: measure P of a gas1. Closed-end: difference in Hg levels (h) shows P of

gas in container compared to a vacuum2. http://www.chm.davidson.edu/ChemistryApplets/GasLaws/Pressure.html

172

closed

2. Open-end:• Difference in Hg levels (h) shows P of gas in container compared to Patm

173

Gas Laws

• Make a data table to put the numbers so you can eliminate the words.

• Make sure that any Celsius temperatures are converted to Kelvin (add 273).

• Rearrange the equation before substituting in numbers. If you are trying to solve for T2, get it out of the denominator first by cross-multiplying.

• If one of the variables is constant, then eliminate it.• Try these problems!

Gas Law Problem 1

• A 2.00 L sample of N2 gas at STP is compressed to 4.00 atm at constant temp-erature. What is the new volume of the gas?

• V2 = P1V1 / P2

• = (1.00 atm)(2.00 L) / (4.00 atm)

• = 0.500 L

Gas Law Problem 2

• To what temperature must a 3.000 L sample of O2 gas at 300.0 K be heated to raise the volume to 10.00 L?

• T2 = V2T1/V1

• = (10.00 L)(300.0 K) / (3.000 L) = 1000. K

Gas Law Problem 3

• A 3.00 L sample of NH3 gas at 100.0 kPa is cooled from 500.0 K to 300.0 K and its pressure is reduced to 80.0 kPa. What is

the new volume of the gas? • V2 = P1V1T2 / P2T1

• = (100.0 kPa)(3.00 L)(300. K) / (80.0 kPa)(500. K)• = 2.25 L

Gay Lussac’s Law of Combining Volumes

N2

1 volume+

+

3 H2

3 volumes

→→

2 NH3

2 volumes

2

2

H

N

V 3=

V 13

2

NH

H

V 2=

V 33

2

NH

N

V 2=

V 1

When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.

Gay Lussac’s Law of Combining Volumes When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.

Avogadro’s Law

Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.

• Volume of one mole of any gas at STP = 22.4 L.

• 22.4 L at STP is known as the molar volume of any gas.

Mole-Mass-Volume Relationships

V PnTnRT

V = P

PV = nRT

atmospheres

Determination of Density Using the Ideal Gas Equation

• Density = mass/volume

gRTM =

PV

D = MP/ RT

Mole fraction (X):• Ratio of moles of one component to the total moles in

the mixture (dimensionless, similar to a %)

184

Ex: What are the mole fractions of H2 and He in the previous example?

29.02.10

0.60X

2H 714.02.10

1.50XHe

Collecting Gases “over Water”• When a gas is bubbled through water, the vapor

pressure of the water (partial pressure of the water) must be subtracted from the pressure of the collected gas:

PT = Pgas + PH2O

∴ Pgas = PT – PH2O

185

See Appendix B for vapor pressures of water at different temperatures.

Graham’s Law of Effusion

• The rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses.

• Rate of effusion of A = MB

Rate of effusion of B MA

Neutralization

• 10.0 mL of 0.20 M HCl is neutralized by 40.0 mL of NaOH. What is the concentration of the NaOH?

• #H MaVa = #OH MbVb, so Mb = #H MaVa / #OH Vb

• = (1)(0.20 M)(10.0 mL) / (1) (40.0 mL) = 0.050 M

• How many mL of 2.00 M H2SO4 are needed to completely neutralize 30.0 mL of 0.500 M KOH?

Concentration

• Molarity• Parts per Million• Percent by Mass• Percent by Volume

Molarity

• What is the molarity of a 500.0 mL solution of NaOH (FM = 40.0) with 60.0 g of NaOH (aq)?– Convert g to moles and mL to L first!– M = moles / L = 1.50 moles / 0.5000 L = 3.00 M

• How many grams of NaOH does it take to make 2.0 L of a 0.100 M solution of NaOH (aq)?– Moles = M X L = 0.100 M X 2.0 L = 0.200 moles– Convert moles to grams: 0.200 moles X 40.0 g/mol = 8.00 g

Parts Per Million

• 100.0 grams of water is evaporated and analyzed for lead. 0.00010 grams of lead ions are found. What is the concentration of the lead, in parts per million?

• ppm = (0.00010 g) / (100.0 g) X 1 000 000 = 1.0 ppm• If the legal limit for lead in the water is 3.0 ppm, then the water

sample is within the legal limits (it’s OK!)

Percent by Mass

• A 50.0 gram sample of a solution is evaporated and found to contain 0.100 grams of sodium chloride. What is the percent by mass of sodium chloride in the solution?

• % Comp = (0.100 g) / (50.0 g) X 100 = 0.200%

Percent By Volume

• Substitute “volume” for “mass” in the above equation.

• What is the percent by volume of hexane if 20.0 mL of hexane are dissolved in benzene to a total volume of 80.0 mL?

• % Comp = (20.0 mL) / (80.0 mL) X100 = 25.0%

Colligative Properties• Vapor Pressure Lowering• B.P. Elevation DTf= m. kf (or D Tb= m. kb)• F.P. Depression• Osmotic Pressure• Colligative properties depend upon # of

particles (ions, atoms, molecule= particle)• Which will have lowest B.P. 1M NaCl, 1 M

C6H12O6 or 1M Na3PO4?

How many Sig Figs?

• Start counting sig figs at the first non-zero.• All digits except place-holding zeroes are sig figs.

Measurement # of Sig Figs

234 cm 3

67000 cm 2

_ 45000 cm

4

560. cm 3

560.00 cm 5

Measurement # of Sig Figs

0.115 cm 3

0.00034 cm 2

0.00304 cm 3

0.0560 cm 3

0.00070700 cm 5

What Precision?

• A number’s precision is determined by the furthest (smallest) place the number is recorded to.

• 6000 mL : thousands place• 6000. mL : ones place• 6000.0 mL : tenths place• 5.30 mL : hundredths place• 8.7 mL : tenths place• 23.740 mL : thousandths place

Rounding with addition and subtraction

• Answers are rounded to the least precise place.

1) 4.732 cm 2) 17.440 mL 3) 32.0 MW 16.8 cm 3.895 mL + 0.0059 MW + 0.781 cm + 16.77 mL --------------- ---------- -------------- 22.313 cm 38.105 mL 32.0059 MW 22.3 cm 38.11 mL 32.0 MW

Rounding with multiplicationand division

• Answers are rounded to the fewest number of significant figures.1) 37.66 KW 2) 14.922 cm 3) 98.11 kg x 2.2 h x 2.0 cm x 200 m ---------- ----------- ---------- 82.852 KWh 29.844 cm2 19 622 kgm 83 KWh 30. cm2 20 000 kgm

Metric Conversions

• Determine how many powers of ten difference there are between the two units (no prefix = 100) and create a conversion factor. Multiply or divide the given by the conversion factor.

How many kg are in 38.2 cg?

(38.2 cg) /(100000 cg/kg) = 0.000382 km

How many mL in 0.988 dL?

(0.988 dg) X (100 mL/dL) = 98.8 mL

Calorimetry

• This equation can be used to determine any of the variables here. You will not have to solve for C, since we will always assume that the energy transfer is being absorbed by or released by a measured quantity of water, whose specific heat is given above.

• Solving for q• Solving for m• Solving for T

Solving for q

• How many joules are absorbed by 100.0 grams of water in a calorimeter if the temperature of the water increases from 20.0oC to 50.0oC?

• q = mCT = (100.0 g)(4.18 J/goC)(30.0oC) = 12500 J

Solving for m

• A sample of water in a calorimeter cup increases from 25oC to 50.oC by the addition of 500.0 joules of energy. What is the mass of water in the calorimeter cup?

• q = mCT, so m = q / CT = (500.0 J) / (4.18 J/goC)(25oC) = 4.8 g

Solving for T

• If a 50.0 gram sample of water in a calorimeter cup absorbs 1000.0 joules of energy, how much will the temperature rise by?

• q = mCT, so T = q / mC = (1000.0 J)/(50.0 g)(4.18 J/goC) = 4.8oC

• If the water started at 20.0oC, what will the final temperature be?– Since the water ABSORBS the energy, its temperature will INCREASE

by the T: 20.0oC + 4.8oC = 24.8oC

Reaction Rate

• Reactions happen when reacting particles collide with sufficient energy (activation energy) and at the proper angle.

• Anything that makes more collisions in a given time will make the reaction rate increase.– Increasing temperature– Increasing concentration (pressure for gases)– Increasing surface area (solids)

• Adding a catalyst makes a reaction go faster by removing steps from the mechanism and lowering the activation energy without getting used up in the process.

Heat of Reaction

• Reactions either absorb PE (endothermic, +H) or release PE (exothermic, -H)

Exothermic, PEKE, Temp

Endothermic, KEPE, Temp

Rewriting the equation with heat included:

4 Al(s) + 3 O2(g) 2 Al2O3(s) + 3351 kJ

N2(g) + O2(g) +182.6 kJ 2 NO(g)

5.3: Enthalpy, H• Since most reactions occur in containers open to the air,

w is often negligible. If a reaction produces a gas, the gas must do work to expand against the atmosphere. This mechanical work of expansion is called PV (pressure-volume) work.

• Enthalpy (H): change in the heat content (qp) of a reaction at constant pressure

H = E + PVH = E + PV (at constant P)H = (qp + w) + (-w)H = qp

205

• Sign conventionsH > 0 Heat is gained from surroundings

+ H in endothermic reaction

H < 0 Heat is released to surroundings- H in exothermic reaction

206

5.4: Enthalpy of Reaction (Hrxn)• Also called heat of reaction:

1. Enthalpy is an extensive property (depends on amounts of reactants involved).

Ex: CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

Hrxn = - 890. kJ

– Combustion of 1 mol CH4 produces 890. kJ

… of 2 mol CH4 → (2)(-890. kJ) = -1780 kJ

What is the H of the combustion of 100. g CH4?

207

2. Hreaction = - Hreverse reaction

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

H = - 890. kJ

CO2 (g) + 2 H2O (l) CH4 (g) + 2 O2 (g)

H = +890. kJ

208

5.6: Hess’ Law• If a rxn is carried out in a series of steps,

Hrxn = (Hsteps) = H1 + H2 + H3 + …

209

Ex. What is Hrxn of the combustion of propane?C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

3 C (s) + 4 H2 (g) C3 H8 (g) H1 = -103.85 kJ

C (s) + O2 (g) CO2 (g) H2 = -393.5 kJ

H2 (g) + ½ O2 (g) H2O (l) H3 = -285.8 kJHrxn = 103.85 + 3(- 393.5) + 4(- 285.8) = - 2219.8 kJ

3[ ] 3( )

4[ ] 4( )

Germain Hess(1802-1850)

C3H8 (g) 3 C (s) + 4 H2 (g) H1 = +103.85 kJ

5.7: Enthalpy of Formation (Hf)• Formation: a reaction that describes a substance formed

from its elementsNH4NO3 (s)

• Standard enthalpy of formation (Hf): forms 1 mole of compound from its elements in their standard state (at 298 K)

C2H5OH (l)Hf = - 277.7 kJ

– Hf of the most stable form of any element equals zero.H2, N2 , O2 , F2 , Cl2 (g)Br2 (l), Hg (l)C (graphite), P4 (s, white), S8 (s), I2 (s) 210

Ex: 2 N2 (g) + 4 H2 (g) + 3 O2 (g) 2

2 C (graphite) + 3 H2 (g) + ½ O2 (g)

Hess’ Law (again)

211

Ex. Combustion of propane:C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

Given: Compound Hrxn (kJ/mol)C3H8 (g) -103.85CO2 (g) -393.5H2O (l) -285.8H2O (g) -241.82

Hrxn = [3(- 393.5) + 4(- 285.8)] – [1(-103.85) + 5(0)]

= - 2219.8 kJ

5.5: Calorimetry• Measurement of heat flow• Heat capacity, C: amount of heat required to raise T of an

object by 1 K

q = C T• Specific heat (or specific heat capacity, c): heat capacity

of 1 g of a substance

q = m c T

Ex: How much energy is required to heat 40.0 g of iron (c = 0.45 J/(g K) from 0.0ºC to 100.0ºC?

q = m c T = (40.0 g)(0.45 J/(g K))(100.0 – 0.0 ºC)= 1800 J 212

Potential Energy Diagrams

• Steps of a reactions:– Reactants have a certain amount of PE stored in their

bonds (Heat of Reactants)– The reactants are given enough energy to collide and

react (Activation Energy)– The resulting intermediate has the highest energy that

the reaction can make (Heat of Activated Complex)– The activated complex breaks down and forms the

products, which have a certain amount of PE stored in their bonds (Heat of Products)

– Hproducts - Hreactants = H EXAMPLES

Making a PE Diagram

• X axis: Reaction Coordinate (time, no units)• Y axis: PE (kJ)• Three lines representing energy (Hreactants, Hactivated complex, Hproducts)

• Two arrows representing energy changes:

– From Hreactants to Hactivated complex: Activation Energy

– From Hreactants to Hproducts : H

• ENDOTHERMIC PE DIAGRAM• EXOTHERMIC PE DIAGRAM

Endothermic PE Diagram

If a catalyst is added?

Endothermic with Catalyst

The red line represents the catalyzed reaction.

Exothermic PE Diagram

What does it look like with a catalyst?

Exothermic with a Catalyst

The red line represents the catalyzed reaction. Lower A.E. and faster reaction time!

19.1: Spontaneous Processes• Reversible reaction: can proceed forward and backward

along same path (equilibrium is possible)Ex: H2O freezing & melting at 0ºC

• Irreversible reaction: cannot proceed forward and backward along same pathEx: ice melting at room temperature

Spontaneous reaction: an irreversible reaction that occurs without outside interventionEx: Gases expand to fill a container, ice melts at room temperature (even though endothermic), salts dissolve in water

219

EntropyEntropy (S): a measure of molecular randomness or

disorder– S is a state function: S = Sfinal - Sinitial

+ S = more randomness- S = less randomness

– For a reversible process that occurs at constant T:

– Units: J/K220

T

q S rev

system

Examples of spontaneous reactions:

• Gases expand to fill a container:

• Ice melts at room temperature:

• Salts dissolve in water:

221

Particles are more evenly distributed

Particles are no longer in an ordered crystal latticeIons are not locked in crystal lattice

19.3: 3rd Law of Thermodynamics• The entropy of a crystalline solid at 0 K is 0.

How to predict S:• Sgas > Sliquid > Ssolid

• Smore gas molecules > Sfewer gas molecules

• Shigh T > Slow T

Ex: Predict the sign of S for the following:1. CaCO3 (s) → CaO (s) + CO2 (g)2. N2 (g) + 3 H2 (g) → 2 NH3 (g)

3. N2 (g) + O2 (g) → 2 NO (g) 222

+, solid to gas

-, fewer moles produced?

19.5: Gibbs free energy, G• Represents combination of two

forces that drive a reaction:H (enthalpy) and S (disorder)

• Units: kJ/mol G = H - TS G° = H° - TS°

(absolute T)

223

reactantsfproductsf mG - Gn G Called “free energy” because G represents maximum useful work that can be done by the system on its surroundings in a spontaneous reaction. (See p. 708 for more details.)

Josiah Willard Gibbs(1839-1903)

Determining Spontaneity of a Reaction

If G is: reaction is spontaneous (proceeds in the forward direction

• PositiveForward reaction is non-spontaneous; the reverse reaction is spontaneous

• Zero The system is at equilibrium

224

19.6: Free Energy & Temperature• G depends on enthalpy, entropy, and temperature:

G = H - TSH S G and reaction outcome- + Always (-); spontaneous at all T

2 O3 (g) → 3 O2 (g)+ - Always +; non-spontaneous at all T

3 O2 (g) → 2 O3 (g)- - Spontaneous at low T; non-spontaneous at high T

H2O (l) → H2O (s)+ + Spontaneous only at high T ; non-spontaneous at

low TH2O (s) → H2O (l)

225

Solubility Curves• Solubility: the maximum quantity of solute that can be

dissolved in a given quantity of solvent at a given temperature to make a saturated solution.

• Saturated: a solution containing the maximum quantity of solute that the solvent can hold. The limit of solubility.

• Supersaturated: the solution is holding more than it can theoretically hold OR there is excess solute which precipitates out. True supersaturation is rare.

• Unsaturated: There are still solvent molecules available to dissolve more solute, so more can dissolve.

• How ionic solutes dissolve in water: polar water molecules attach to the ions and tear them off the crystal.

Solubility

Solubility: go to the temperature and up to the desired line, then across to the Y-axis. This is how many g of solute are needed to make a saturated solution of that solute in 100g of H2O at that particular temperature.

At 40oC, the solubility of KNO3 in 100g of water is 64 g. In 200g of water, double that amount. In 50g of water, cut it in half.

Supersaturated

If 120 g of NaNO3 are added to 100g of water at 30oC:

1) The solution would be SUPERSATURATED, because there is more solute dissolved than the solubility allows

2) The extra 25g would precipitate out

3) If you heated the solution up by 24oC (to 54oC), the excess solute would dissolve.

UnsaturatedIf 80 g of KNO3 are added to 100g of water at 60oC:

1) The solution would be UNSATURATED, because there is less solute dissolved than the solubility allows

2) 26g more can be added to make a saturated solution

3) If you cooled the solution down by 12oC (to 48oC), the solution would become saturated

How Ionic Solutes Dissolve in Water

Water solvent molecules attach to the ions (H end to the Cl-, O end to the Na+)

Water solvent holds the ions apart and keeps the ions from coming back together

Formulas, Naming and Properties of Acids

• Arrhenius Definition of Acids: molecules that dissolve in water to produce H3O+ (hydronium) as the only positively charged ion in solution.

• HCl (g) + H2O (l) H3O+ (aq) + Cl-

• Properties of Acids• Naming of Acids• Formula Writing of Acids

Properties of Acids

• Acids react with metals above H2 on Table J to form H2(g) and a salt.

• Acids have a pH of less than 7.• Dilute solutions of acids taste sour.• Acids turn phenolphthalein CLEAR, litmus RED

and bromthymol blue YELLOW.• Acids neutralize bases.• Acids are formed when acid anhydrides (NO2,

SO2, CO2) react with water for form acids. This is how acid rain forms from auto and industrial emissions.

Naming of Acids

• Binary Acids (H+ and a nonmetal)– hydro (nonmetal) -ide + ic acid

• HCl (aq) = hydrochloric acid• Ternary Acids (H+ and a polyatomic ion)

– (polyatomic ion) -ate +ic acid• HNO3 (aq) = nitric acid

– (polyatomic ion) -ide +ic acid• HCN (aq) = cyanic acid

– (polyatomic ion) -ite +ous acid• HNO2 (aq) = nitrous acid

Formula Writing of Acids

• Acids formulas get written like any other. Write the H+1 first, then figure out what the negative ion is based on the name. Cancel out the charges to write the formula. Don’t forget the (aq) after it…it’s only an acid if it’s in water!

• Hydrosulfuric acid: H+1 and S-2 = H2S (aq)

• Carbonic acid: H+1 and CO3-2 = H2CO3 (aq)

• Chlorous acid: H+1 and ClO2-1 = HClO2 (aq)

• Hydrobromic acid: H+1 and Br-1 = HBr (aq)• Hydronitric acid:• Hypochlorous acid:• Perchloric acid:

Formulas, Naming and Properties of Bases

• Arrhenius Definition of Bases: ionic compounds that dissolve in water to produce OH- (hydroxide) as the only negatively charged ion in solution.

• NaOH (s) Na+1 (aq) + OH-1 (aq)

• Properties of Bases• Naming of Bases• Formula Writing of Bases

Properties of Bases

• Bases react with fats to form soap and glycerol. This process is called saponification.

• Bases have a pH of more than 7.• Dilute solutions of bases taste bitter.• Bases turn phenolphthalein PINK, litmus BLUE and

bromthymol blue BLUE.• Bases neutralize acids.• Bases are formed when alkali metals or alkaline earth metals

react with water. The words “alkali” and “alkaline” mean “basic”, as opposed to “acidic”.

Naming of Bases

• Bases are named like any ionic compound, the name of the metal ion first (with a Roman numeral if necessary) followed by “hydroxide”.

Fe(OH)2 (aq) = iron (II) hydroxide

Fe(OH)3 (aq) = iron (III) hydroxide

Al(OH)3 (aq) = aluminum hydroxide

NH3 (aq) is the same thing as NH4OH:

NH3 + H2O NH4OH

Also called ammonium hydroxide.

Formula Writing of Bases

• Formula writing of bases is the same as for any ionic formula writing. The charges of the ions have to cancel out.

• Calcium hydroxide = Ca+2 and OH-1 = Ca(OH)2 (aq)

• Potassium hydroxide = K+1 and OH-1 = KOH (aq)• Lead (II) hydroxide = Pb+2 and OH-1 = Pb(OH)2 (aq)

• Lead (IV) hydroxide = Pb+4 and OH-1 = Pb(OH)4 (aq)

• Lithium hydroxide =• Copper (II) hydroxide =• Magnesium hydroxide =

Neutralization

• H+1 + OH-1 HOH• Acid + Base Water + Salt (double replacement)• HCl (aq) + NaOH (aq) HOH (l) + NaCl (aq)

• H2SO4 (aq) + KOH (aq) 2 HOH (l) + K2SO4 (aq)

• HBr (aq) + LiOH (aq) • H2CrO4 (aq) + NaOH (aq) • HNO3 (aq) + Ca(OH)2 (aq) • H3PO4 (aq) + Mg(OH)2 (aq)

pH

• A change of 1 in pH is a tenfold increase in acid or base strength.

• A pH of 4 is 10 times more acidic than a pH of 5.• A pH of 12 is 100 times more basic than a pH of 10.

16.2: Dissociation of Water• Autoionization of water:

H2O (l) ↔ H+ (aq) + OH- (aq)

241

][

]][[

2OH

OHHKc

M 55.6 g 18.0

mol 1

wKOHH -1410 x 0.1]][[

KW = ion-product constant for water

H3O+ (aq) or H+ (aq) = hydronium

L 1

g 1000 (l)] O[H2

Indicators

At a pH of 2:

Methyl Orange = red

Bromthymol Blue = yellow

Phenolphthalein = colorless

Litmus = red

Bromcresol Green = yellow

Thymol Blue = yellow

Methyl orange is red at a pH of 3.2 and below and yellow at a pH of 4.4 and higher. In between the two numbers, it is an intermediate color that is not listed on this table.

Alternate Theories

• Arrhenius Theory: acids and bases must be in aqueous solution.

• Alternate Theory: Not necessarily so!– Acid: proton (H+1) donor…gives up H+1 in a reaction.– Base: proton (H+1) acceptor…gains H+1 in a reaction.

• HNO3 + H2O H3O+1 + NO3-1

– Since HNO3 lost an H+1 during the reaction, it is an acid.

– Since H2O gained the H+1 that HNO3 lost, it is a base.

16.11: Lewis Acids & Bases• Lewis acid: “e- pair acceptor”– Brønsted-Lowry acid = H+ donor– Arrhenius acid = produces H+

• Lewis base: “e- pair donor”– B-L base = H+ acceptor– Arrhenius base = produces OH-

Ex:NH3 + BF3 → NH3BF3

Lewis base Lewis acid Lewis salt

6 CN- + Fe3+ → Fe(CN)63-

Lewis base Lewis acid Coordination compound244

Gilbert N. Lewis(1875 – 1946)

15.1: Chemical Equilibrium

• Occurs when opposing reactions are proceeding at the same rate– Forward rate = reverse rate of reaction

Ex:• Vapor pressure: rate of vaporization = rate of condensation• Saturated solution: rate of dissociation = rate of

crystallization

• Expressing concentrations:– Gases: partial pressures, PX

– Solutes in liquids: molarity, [X]

245

Reversible Reactions and Rate

246

Reac

tion

Rate

TimeBackward rate

Forward rate

Equilibrium is established:

Forward rate = Backward rate

When equilibrium is achieved:[A] ≠ [B] and kf/kr = Keq

15.2: Law of Mass Action• Derived from rate laws by Guldberg andWaage (1864)– For a balanced chemical reaction

in equilibrium:

a A + b B ↔ c C + d D– Equilibrium constant expression (Keq):

247

ba

dc

c [B] [A]

[D] [C] K b

Ba

A

dD

cC

p )(P)(P

)(P)(PK

Keq is strictly based on stoichiometry of the reaction (is independent of the mechanism).

Units: Keq is considered dimensionless (no units)

Cato Guldberg Peter Waage (1836-1902) (1833-1900)

or

Relating Kc and Kp• Convert [A] into PA:

248

ba

dc

([B]RT)([A]RT)

([D]RT)([C]RT)b

Ba

A

dD

cC

p )(P)(P

)(P )(PK

baba

dcdc

(RT)[B][A]

(RT)[D][C]

(RT) K K b)(a - d)(ccp

where n == change in coefficents of products – reactants (gases only!)= (c+d) - (a+b)

(RT) K nc

RT

P

V

nM RTAPA ][

Magnitude of Keq

• Since Keq [products]/[reactants], the magnitude of Keq

predicts which reaction direction is favored:

– If Keq > 1 then [products] > [reactants]and equilibrium “lies to the right”

– If Keq < 1 then [products] < [reactants]and equilibrium “lies to the left”

249

Relationship Between Q and K

• Reaction Quotient (Q): The particular ratio of concentration terms that we write for a particular reaction is called reaction quotient.

• For a reaction, A B, Q= [B]/[A]• At equilibrium, Q= K Reaction Direction: Comparing Q and K• Q<K, reaction proceeds to right, until equilibrium

is achieved (or Q=K)• Q>K, reaction proceeds to left, until Q=K

250

Value of KFor the referencerxn, A>B,

For the reverse rxn, B >A,

For the reaction,2A > 2B

For the rxn,A > CC > B

K(ref)= [B]/[A]

K= 1/K(ref) K= K(ref)2 K (overall)= K1 X K2

251

15.3: Types of Equilibria

• Homogeneous: all components in same phase (usually g or aq)

N2 (g) + H2 (g) ↔ NH3 (g)

252

3H

1N

2NH

P )(P)(P

)(PK

22

3

bB

aA

dD

cC

P )(P)(P

)(P)(PK

3 21

Fritz Haber(1868 – 1934)

• Heterogeneous: different phasesCaCO3 (s) ↔ CaO (s) + CO2 (g)

Definition: What we use:

253

][CaCO

)(P [CaO] K

3

COeq

22COp P K

Even though the concentrations of the solids or liquids do not appear in the equilibrium expression, the substances must be present to achieve equilibrium.

Concentrations of pure solids and pure liquids are not included in Keq expression because their concentrations do not vary, and are “already included” in Keq (see p. 548).

15.4: Calculating Equilibrium Constants

Steps to use “ICE” table:1. “I” = Tabulate known initial and equilibrium

concentrations of all species in equilibrium expression

2. “C” = Determine the concentration change for the species where initial and equilibrium are known

• Use stoichiometry to calculate concentration changes for all other species involved in equilibrium

3. “E” = Calculate the equilibrium concentrations

254

• Ex: Enough ammonia is dissolved in 5.00 L of water at 25ºC to produce a solution that is 0.0124 M ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that [OH1-] is 4.64 x 10-4 M. Calculate Keq at 25ºC for the reaction:

NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)

255

NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)

Initial

Change

Equilibrium

256

][NH

]][OH[NH K

3

-114

c

0.0124 M

- x

0.0119 M

0 M 0 M

+ x + x

4.64 x 10-4 M 4.64 x 10-4 M

5-2-4

101.81x 0.0119

)10(4.64

NH3 (aq)H2O (l)

NH41+ (aq) OH1- (aq)

XXX

x = 4.64 x 10-4 M

Equilibrium


When the rate of the forward reaction equals the rate of the reverse reaction.

Examples of Equilibrium

• Solution Equilibrium: when a solution is saturated, the rate of dissolving equals the rate of precipitating.– NaCl (s) Na+1 (aq) + Cl-1 (aq)

• Vapor-Liquid Equilibrium: when a liquid is trapped with air in a container, the liquid evaporates until the rate of evaporation equals the rate of condensation.– H2O (l) H2O (g)

• Phase equilibrium: At the melting point, the rate of solid turning to liquid equals the rate of liquid turning back to solid.– H2O (s) H2O (l)

Le Châtelier’s Principle

• If a system at equilibrium is stressed, the equilibrium will shift in a direction that relieves that stress.

• A stress is a factor that affects reaction rate. Since catalysts affect both reaction rates equally, catalysts have no effect on a system already at equilibrium.

• Equilibrium will shift AWAY from what is added• Equilibrium will shift TOWARDS what is removed.• This is because the shift will even out the change in reaction

rate and bring the system back to equilibrium

»NEXT

Steps to Relieving Stress• 1) Equilibrium is subjected to a STRESS.• 2) System SHIFTS towards what is removed from the system

or away from what is added.• The shift results in a CHANGE OF CONCENTRATION for both

the products and the reactants.– If the shift is towards the products, the concentration of

the products will increase and the concentration of the reactants will decrease.

– If the shift is towards the reactants, the concentration of the reactants will increase and the concentration of the products will decrease.

» NEXT

Examples

• For the reaction N2(g) + 3H2(g) 2 NH3(g) + heat – Adding N2 will cause the equilibrium to shift RIGHT, resulting in an

increase in the concentration of NH3 and a decrease in the concentration of N2 and H2.

– Removing H2 will cause a shift to the LEFT, resulting in a decrease in the concentration of NH3 and an increase in the concentration of N2 and H2.

– Increasing the temperature will cause a shift to the LEFT, same results as the one above.

– Decreasing the pressure will cause a shift to the LEFT, because there is more gas on the left side, and making more gas will bring the pressure back up to its equilibrium amount.

– Adding a catalyst will have no effect, so no shift will happen.

Oxidation Numbers• Rules for Assigning Oxidation States • The oxidation state of an atom in an uncombined element is 0.• The oxidation state of a monatomic ion is the same as its charge.• Oxygen is assigned an oxidation state of –2 in most of its

covalent compounds. Important exception: peroxides (compounds containing the O2 2- group), in which each oxygen is assigned an oxidation state of –1)

• In its covalent compounds with nonmetals, hydrogen is assigned an oxidation state of +1

• For a compound, sum total of ON s is zero.

• For an ionic species (like a polyatomic ion), the sum of the oxidation states must equal the overall charge on that ion.

16.6: Weak Acids

• Weak acids partially ionize in water (equilibrium is somewhere between ions and molecules). HA (aq) ↔ A- (aq) + H+ (aq)

263

eqa HA

AHK

][

]][[

Ka = acid-dissociation constant in water

Weak acids generally have Ka < 10-3

See Appendix D for full listing of Ka values


Dissociation Constants

• For a generalized acid dissociation,

the equilibrium expression would be

• This equilibrium constant is called the acid-dissociation constant, Ka.

[H3O+] [A-][HA]

Kc =

HA (aq) + H2O (l) A- (aq) + H3O+ (aq)


Dissociation ConstantsThe greater the value of Ka, the stronger is the acid.


Calculating Ka from the pH

The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is 2.38. Calculate Ka for formic acid at this temperature.

We know that

[H3O+] [COO-][HCOOH]

Ka =



The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is 2.38. Calculate Ka for formic acid at this temperature.

To calculate Ka, we need the equilibrium concentrations of all three things.

We can find [H3O+], which is the same as [HCOO-], from the pH.



pH = -log [H3O+]

2.38 = -log [H3O+]

-2.38 = log [H3O+]

10-2.38 = 10log [H3O+] = [H3O+]

4.2 10-3 = [H3O+] = [HCOO-]


Calculating Ka from pH

Now we can set up a table…

[HCOOH], M [H3O+], M [HCOO-], M

Initially 0.10 0 0

Change - 4.2 10-3 + 4.2 10-3 + 4.2 10-3

At Equilibrium 0.10 - 4.2 10-3

= 0.0958 = 0.10

4.2 10-3 4.2 10-3


Calculating Ka from pH

[4.2 10-3] [4.2 10-3][0.10]

Ka =

= 1.8 10-4


Calculating Percent Ionization

• Percent Ionization = 100• In this example

[H3O+]eq = 4.2 10-3 M

[HCOOH]initial = 0.10 M

[H3O+]eq

[HA]initial

Percent Ionization = 1004.2 10-3

0.10= 4.2%


Calculating pH from Ka

Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25C.

HC2H3O2 (aq) + H2O (l) H3O+ (aq) + C2H3O2- (aq)

Ka for acetic acid at 25C is 1.8 10-5.



The equilibrium constant expression is

[H3O+] [C2H3O2-]

[HC2H3O2]Ka =



We next set up a table…

[C2H3O2], M [H3O+], M [C2H3O2-], M

Initially 0.30 0 0

Change -x +x +x

At Equilibrium 0.30 - x 0.30 x x

We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.



Now,

(x)2

(0.30)1.8 10-5 =

(1.8 10-5) (0.30) = x2

5.4 10-6 = x2

2.3 10-3 = x



pH = -log [H3O+]pH = -log (2.3 10-3)pH = 2.64


Polyprotic Acids……have more than one acidic proton

If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.


Weak Bases

Bases react with water to produce hydroxide ion.


Weak Bases

The equilibrium constant expression for this reaction is

[HB] [OH-][B-]Kb =

where Kb is the base-dissociation constant.


Weak Bases

Kb can be used to find [OH-] and, through it, pH.


pH of Basic Solutions

What is the pH of a 0.15 M solution of NH3?

[NH4+] [OH-]

[NH3]Kb = = 1.8 10-5

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)



Tabulate the data.

[NH3], M [NH4+], M [OH-], M

Initially 0.15 0 0

At Equilibrium 0.15 - x 0.15 x x



(1.8 10-5) (0.15) = x2

2.7 10-6 = x2

1.6 10-3 = x2

(x)2

(0.15)1.8 10-5 =



Therefore,[OH-] = 1.6 10-3 MpOH = -log (1.6 10-3)pOH = 2.80pH = 14.00 - 2.80pH = 11.20


Ka and Kb

Ka and Kb are related in this way:

Ka Kb = Kw

Therefore, if you know one of them, you can calculate the other.


Factors Affecting Acid Strength

• The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound.

• So acidity increases from left to right across a row and from top to bottom down a group.



In oxyacids, in which an -OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid.



For a series of oxyacids, acidity increases with the number of oxygens.



Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic.

16.9: Salt Solutions as Acids & Bases

• Hydrolysis: acid/base reaction of ion with water to produce H+ or OH-

– Anion (A-) = a conjugate baseA- (aq) + H2O (l) ↔ HA (aq) + OH- (aq)

– Cation (B+) = a conjugate acidB+ (aq) + H2O (l) ↔ BOH (aq) + H+ (aq)

290

17.1: Common Ion Effect

• Addition of a “common ion”: solubility of solids decrease because of Le Châtelier’s principle.Ex: AgCl (s) ↔ Ag+ (aq) + Cl- (aq)

– Addition of Cl- shifts equilibrium toward solid

291

17.4: Solubility Equilibria• Dissolving & precipitating of salts– Solubility rules discussed earlier are generalized

qualitative observations of quantitative experiments.Ex: PbCl2 (s) ↔ Pb2+ (aq) + 2 Cl- (aq)

Ksp = [Pb2+][Cl-]2 = 1.6 x 10-5

Ksp = solubility-product constant (found in App. D)

– Recall that both aqueous ions and solid must be present in solution to achieve equilibrium

– Changes in pH will affect the solubility of salts composed of a weak acid or weak base ion.

292


Solubility Products

Consider the equilibrium that exists in a saturated solution of BaSO4 in water:

BaSO4(s) Ba2+(aq) + SO42−(aq)


Solubility Products

The equilibrium constant expression for this equilibrium is

Ksp = [Ba2+] [SO42−]

where the equilibrium constant, Ksp, is called the solubility product.


Solubility Products

• Ksp is not the same as solubility.

• Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).

Calculating Ksp from solubility

1. Calculate Ksp for Ag2CrO4, if its solubility is 0.022 g/L.

(Ans: 6.6 X 10^-5)

Calculating Solubility given Ksp2. Ksp for MgF2 is 6.4 X 10^-9 at 250C. Calculate its solubility in mol/L and g/L. (Ans: 1.2 X 10^-3 M, 7.3 X 10^-2 g/L)


Factors Affecting Solubility

• The Common-Ion Effect– If one of the ions in a solution equilibrium is

already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease.

BaSO4(s) Ba2+(aq) + SO42−(aq)



• pH– If a substance has a

basic anion, it will be more soluble in an acidic solution.

– Substances with acidic cations are more soluble in basic solutions.



• Amphoterism– Amphoteric metal oxides

and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.

– Examples of such cations are Al3+, Zn2+, and Sn2+.

17.2: Buffers:• Solutions that resist drastic changes in pH upon

additions of small amounts of acid or base.– Consist of a weak acid and its conjugate base

(usually in salt form)Ex: acetic acid and sodium acetate:

HC2H3O2 + NaC2H3O2

– Or consist of a weak base and its conjugate acid (usually in salt form)Ex: ammonia and ammonium chloride:

NH3 + NH4Cl 301


Buffers

• Buffers are solutions of a weak conjugate acid-base pair.

• They are particularly resistant to pH changes, even when strong acid or base is added.


Buffers

If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.


Buffers

Similarly, if acid is added, the F− reacts with it to form HF and water.


Titration

In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base).


Titration

A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.


Titration of a Strong Acid with a Strong Base

From the start of the titration to near the equivalence point, the pH goes up slowly.



Just before (and after) the equivalence point, the pH increases rapidly.



At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.



As more base is added, the increase in pH again levels off.


Titration of a Weak Acid with a Strong Base

• Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.

• At the equivalence point the pH is >7.

• Phenolphthalein is commonly used as an indicator in these titrations.

Practice Problem on Titration:

If 7.3 mL of 1.25 M HNO3 is required to neutralize 25.00 mL of a potassiumhydroxide solution, what is the molarity of the potassium hydroxide? 0.044 M KOH

Titration of a Weak Base and Strong Acid

- Half Equivalence Point , pH= pKa

- pka or pkb of weak acid or base in a buffer should be clsoe to the desired pH of the buffer solution.

313

14

pH 7

0

0 30Volume of HCl added (mL)

Redox: Reduction occurs when an atom gains one or more electrons. Ex: Oxidation occurs when an atom or ion loses one or more electrons. Ex: LEO goes GERCopper metal reacts with silver nitrate to form silver metal and copper nitrate: Cu + 2 Ag(NO3) 2 Ag + Cu(NO3)2.

Identifying OX, RD, SI Species

• Ca0 + 2 H+1Cl-1 Ca+2Cl-12 + H2

0

• Oxidation = loss of electrons. The species becomes more

positive in charge. For example, Ca0 Ca+2, so Ca0 is the species that is oxidized.

• Reduction = gain of electrons. The species becomes more

negative in charge. For example, H+1 H0, so the H+1 is the species that is reduced.

• Spectator Ion = no change in charge. The species does not gain or lose any electrons. For example, Cl-1 Cl-1, so the Cl-1 is the spectator ion.

Oxidizing Agent and Reducing Agent:

Oxidizing agent gets reduced itself and reducing agent gets oxidized itself, so a strong oxidizing agent should have a great tendency to accept e and a strong reducing agent should be willing to lose e easily. What are strong oxidizing agents- metals or non metals? Why?

Which is the strongest oxidizing agent and which is the strongest reducing agent?

Agents

• Ca0 + 2 H+1Cl-1 Ca+2Cl-12 + H2

0

• Since Ca0 is being oxidized and H+1 is being reduced, the electrons must be going from the Ca0 to the H+1.

• Since Ca0 would not lose electrons (be oxidized) if H+1 weren’t there to gain them, H+1 is the cause, or agent, of Ca0’s oxidation. H+1 is the oxidizing agent.

• Since H+1 would not gain electrons (be reduced) if Ca0 weren’t there to lose them, Ca0 is the cause, or agent, of H+1’s reduction. Ca0 is the reducing agent.

Steps for Balancing a Redox Reaction: Half Reaction MethodIn half reaction method, oxidation and reduction half- reactions are written and balanced separately before combining them into a balanced redox reaction. It is a good method for balancing redox reactions because this method can be used both for reactions carried out in acidic and basic medium .

Steps for Balancing Redox Reaction Using Half Reaction Method IN ACIDIC MEDIUM:Step 1: Write unbalanced equation in ionic form.Step 2: Write separate half reactions for the oxidation and reduction processes. (Use Oxidation Numbers for identifying oxidation and reduction reactions)Step 3: Balance atoms in the half reactions•First, balance all atoms except H and O•Balance O by adding H2O•Balance H by adding H+

Step 4: Balance Charges on each half reaction, by adding electrons.Step 5: Multiply each half reaction by an appropriate number to make the number of electrons equal in both half reactions.Step 6: Add two half reactions and simplify where possible by canceling species appearing in both sides.Step 7: Check equation for same number of atoms and charges on both sides.

Writing Half-Reactions

• Ca0 + 2 H+1Cl-1 Ca+2Cl-12 + H2

0

• Oxidation: Ca0 Ca+2 + 2e- • Reduction: 2H+1 + 2e- H2

0

The two electrons lost by Ca0 are gained by the two H+1 (each H+1 picks up an electron).

PRACTICE SOME!

Practice Half-Reactions

• Don’t forget to determine the charge of each species first!

• 4 Li + O2 2 Li2O

• Oxidation Half-Reaction:• Reduction Half-Reaction:

• Zn + Na2SO4 ZnSO4 + 2 Na

• Oxidation Half-Reaction:• Reduction Half-Reaction:

Steps for Balancing Redox Reaction Using Half Reaction Method IN BASIC MEDIUM:For balancing redox reactions in basic solutions, all the steps are the same as acidic medium balancing, except you add one more step to it. The H+ ions can then be “neutralized” by adding an equal number of OH- ions to both sides of the equation. Ex.

Standard Cell Potential

Just as the water tends to flow from a higher level to a lower level, electrons also move from a higher “potential” to a lower potential. This potential difference is called the electromotive force (EMF) of cell and is written as Ecell.

The standard for measuring the cell potentials is called a SHE (Standard Hydrogen Electrode).

Description of SHE (Standard Hydrogen Electrode)Reaction 2H+

(aq, 1M)+ 2e - H2(g, 101kPa) E0= 0.00 V

Standard Reduction PotentialsMany different half cells can be paired with the SHE and the standard reduction potentials for each half cell is obtained. Check the table for values of reduction potential for various substances:Would substances with high reduction potential be strong oxidizing agents or strong reducing agents? Why?

Activity Series

• For metals, the higher up the chart the element is, the more likely it is to be oxidized. This is because metals like to lose electrons, and the more active a metallic element is, the more easily it can lose them.

• For nonmetals, the higher up the chart the element is, the more likely it is to be reduced. This is because nonmetals like to gain electrons, and the more active a nonmetallic element is, the more easily it can gain them.

Metal Activity

• Metallic elements start out with a charge of ZERO, so they can only be oxidized to form (+) ions.

• The higher of two metals MUST undergo oxidation in the reaction, or no reaction will happen.

• The reaction 3 K + FeCl3 3 KCl + Fe WILL happen, because K is being oxidized, and that is what Table J says should happen.

• The reaction Fe + 3 KCl FeCl3 + 3 K will NOT happen.

3 K0 + Fe+3Cl-13

REACTION

Fe0 + 3 K+1Cl-1

NO REACTION

Voltaic Cells (Galvanic Cells)

A voltaic cell converts chemical energy from a spontaneous redox reaction into electrical energy. Ex: Cu and Zn voltaic cell (More positive reduction potential is the cathode)Key Words:•Cathode•Anode•Salt Bridge

How a Voltaic Cell Works: An Ox, Red Cat

Representing Electrochemical Cells

Voltaic Cells

• Produce electrical current using a spontaneous redox reaction• Used to make batteries!• Materials needed: two beakers, piece of the metals (anode, -

electrode and cathode + electrode), solution of each metal, porous material (salt bridge), solution of a salt that does not contain either metal in the reaction, wire and a load to make use of the generated current!

• Use Reference Table J to determine the metals to use– Higher = (-) anode (lower reduction potential)– Lower = (+) cathode (higher reduction potential)

Making Voltaic Cells

Electrolytic Cells

• Use electricity to force a nonspontaneous redox reaction to take place.

• Uses for Electrolytic Cells:– Decomposition of Alkali Metal Compounds– Decomposition of Water into Hydrogen and Oxygen– Electroplating

• Differences between Voltaic and Electrolytic Cells:– ANODE: Voltaic (-) Electrolytic (+)– CATHODE: Voltaic (+) Electrolytic (-)– Voltaic: 2 half-cells, a salt bridge and a load– Electrolytic: 1 cell, no salt bridge, IS the load

Decomposing AlkaliMetal Compounds

2 NaCl 2 Na + Cl2

The Na+1 is reduced at the (-) cathode, picking up an e- from the battery

The Cl-1 is oxidized at the (+) anode, the e- being pulled off by the battery (DC)

Decomposing Water

2 H2O 2 H2 + O2

The H+ is reduced at the (-) cathode, yielding H2 (g), which is trapped in the tube.

The O-2 is oxidized at the (+) anode, yielding O2 (g), which is trapped in the tube.

Electroplating

The Ag0 is oxidized to Ag+1 when the (+) end of the battery strips its electrons off.

The Ag+1 migrates through the solution towards the (-) charged cathode (ring), where it picks up an electron from the battery and forms Ag0, which coats on to the ring.

Spontaneity of Redox Reactions:

E0 = E0red( reduction process-cathode) – E0

red (oxidation process-anode)

A positive value of E0 indicates a spontaneous process and a negative value of E0 indicates a nonspontaneous value.

Steps for Predicting Spontaneity of Redox Reactions•First write the reaction as oxidation and reduction half reactions.•Then plug standard reduction potential values in the equation given above.•Check for the spontaneity by a positive or a negative value of E0

Ex:

Hydrocarbons

• Molecules made of Hydrogen and Carbon• Carbon forms four bonds, hydrogen forms one bond• Hydrocarbons come in three different homologous series:

– Alkanes (single bond between C’s, saturated)– Alkenes (1 double bond between 2 C’s, unsaturated)– Alkynes (1 triple bond between 2 C’s, unsaturated)

• These are called aliphatic, or open-chain, hydrocarbons.• Count the number of carbons and add the appropriate suffix!

Alkanes

• CH4 = methane

• C2H6 = ethane

• C3H8 = propane

• C4H10 = butane

• C5H12 = pentane

• To find the number of hydrogens, double the number of carbons and add 2.

Methane

Meth-: one carbon

-ane: alkane

The simplest organic molecule, also known as natural gas!

Ethane

Eth-: two carbons

-ane: alkane

Propane

Prop-: three carbons

-ane: alkane

Also known as “cylinder gas”, usually stored under pressure and used for gas grills and stoves. It’s also very handy as a fuel for Bunsen burners!

Butane

But-: four carbons

-ane: alkane

Liquefies with moderate pressure, useful for gas lighters. You have probably lit your gas grill with a grill lighter fueled with butane!

Pentane

Pent-: five carbons

-ane: alkane

Your Turn!!!

Draw Hexane:

Draw Heptane:

Alkenes

• C2H4 = Ethene

• C3H6 = Propene

• C4H8 = Butene

• C5H10 = Pentene

• To find the number of hydrogens, double the number of carbons.

EtheneTwo carbons, double bonded. Notice how each carbon has four bonds? Two to the other carbon and two to hydrogen atoms.

Also called “ethylene”, is used for the production of polyethylene, which is an extensively used plastic. Look for the “PE”, “HDPE” (#2 recycling) or “LDPE” (#4 recycling) on your plastic bags and containers!

Propene

Three carbons, two of them double bonded. Notice how each carbon has four bonds?

If you flipped this molecule so that the double bond was on the right side of the molecule instead of the left, it would still be the same molecule. This is true of all alkenes.

Used to make polypropylene (PP, recycling #5), used for dishwasher safe containers and indoor/outdoor carpeting!

ButeneThis is 1-butene, because the double bond is between the 1st and 2nd carbon from the end. The number 1 represents the lowest numbered carbon the double bond is touching.

This is 2-butene. The double bond is between the 2nd and 3rd carbon from the end. Always count from the end the double bond is closest to.

ISOMERS: Molecules that share the same molecular formula, but have different structural formulas.

Pentene

This is 1-pentene. The double bond is on the first carbon from the end.

This is 2-pentene. The double bond is on the second carbon from the end.

This is not another isomer of pentene. This is also 2-pentene, just that the double bond is closer to the right end.

Alkynes

C2H2 = Ethyne C3H4 = Propyne C4H6 = Butyne C5H8 = Pentyne

To find the number of hydrogens, double the number of carbons and subtract 2.

Ethyne

Now, try to draw propyne! Any isomers? Let’s see!

Also known as “acetylene”, used by miners by dripping water on CaC2 to light up mining helmets. The “carbide lamps” were attached to miner’s helmets by a clip and had a large reflective mirror that magnified the acetylene flame.

Used for welding and cutting applications, as ethyne burns at temperatures over 3000oC!

Propyne

This is propyne! Nope! No isomers.

OK, now draw butyne. If there are any isomers, draw them too.

Butyne

Well, here’s 1-butyne!

And here’s 2-butyne!

Is there a 3-butyne? Nope! That would be 1-butyne. With four carbons, the double bond can only be between the 1st and 2nd carbon, or between the 2nd and 3rd carbons.

Now, try pentyne!

Pentyne

1-pentyne

2-pentyne

Now, draw all of the possible isomers for hexyne!

Naming: Check this link outhttp://www.chem.ucalgary.ca/courses/351/orgnom/javanom/nomenclature2StartPage.htm

IsomersIsomers are compounds that have same molecular formula (same number of atoms) but a different structure.

There are three types of isomers:1.Structural Isomers: Same number of atoms, arranged differently.2.Geometric Isomers (Cis- trans-): Happens in = or triple bonded compounds since these are inflexible bonds. Ex.3.Optical Isomers ( D- and L-): Need a central atom that is “Chiral” (all four groups attached to it are different). These are non super imposable mirror images. Usually this central atom is C.

Substituted Hydrocarbons

• Hydrocarbon chains can have three kinds of “dingly-danglies” attached to the chain. If the dingly-dangly is made of anything other than hydrogen and carbon, the molecule ceases to be a hydrocarbon and becomes another type of organic molecule.– Alkyl groups– Halide groups– Other functional groups

• To name a hydrocarbon with an attached group, determine which carbon (use lowest possible number value) the group is attached to. Use di- for 2 groups, tri- for three.

Alkyl Groups

Halide Groups

Organic Families• Each family has a functional group to identify it.

– Alcohol (R-OH, hydroxyl group)– Organic Acid (R-COOH, primary carboxyl group)– Aldehyde (R-CHO, primary carbonyl group)– Ketone (R1-CO-R2, secondary carbonyl group)– Ether (R1-O-R2)– Ester (R1-COO-R2, carboxyl group in the middle)– Amine (R-NH2, amine group)– Amide (R-CONH2, amide group)

• These molecules are alkanes with functional groups attached. The name is based on the alkane name.

Alcohol

On to DI and TRIHYDROXY ALCOHOLS

Di and Tri-hydroxy Alcohols

Positioning of Functional Group

PRIMARY (1o): the functional group is bonded to a carbon that is on the end of the chain.

SECONDARY (2o): The functional group is bonded to a carbon in the middle of the chain.

TERTIARY (3o): The functional group is bonded to a carbon that is itself directly bonded to three other carbons.

Organic Acid

These are weak acids. The H on the right side is the one that ionized in water to form H3O+. The -COOH (carboxyl) functional group is always on a PRIMARY carbon.

Can be formed from the oxidation of primary alcohols using a KMnO4 catalyst.

Aldehyde

Aldehydes have the CO (carbonyl) groups ALWAYS on a PRIMARY carbon. This is the only structural difference between aldehydes and ketones.

Formed by the oxidation of primary alcohols with a catalyst. Propanal is formed from the oxidation of 1-propanol using pyridinium chlorochromate (PCC) catalyst.*

Ketone

Ketones have the CO (carbonyl) groups ALWAYS on a SECONDARY carbon. This is the only structural difference between ketones and aldehydes.

Can be formed from the dehydration of secondary alcohols with a catalyst. Propanone is formed from the oxidation of 2-propanol using KMnO4 or PCC catalyst.*

Ether

Ethers are made of two alkyl groups surrounding one oxygen atom. The ether is named for the alkyl groups on “ether” side of the oxygen. If a three-carbon alkyl group and a four-carbon alkyl group are on either side, the name would be propyl butyl ether. Made with an etherfication reaction.

Ester

Esters are named for the alcohol and organic acid that reacted by esterification to form the ester. If the alcohol was 1-propanol and the acid was hexanoic acid, the name of the ester would be propyl hexanoate. Esters contain a COO (carboxyl) group in the middle of the molecule, which differentiates them from organic acids.

Amine

- Component of amino acids, and therefore proteins, RNA and DNA…life itself!

- Essentially ammonia (NH3) with the hydrogens replaced by one or more hydrocarbon chains, hence the name “amine”!

Amide

Synthetic Polyamides: nylon, kevlar

Natural Polyamide: silk!

For more information on polymers, go here.

Organic Reactions

• Combustion• Fermentation• Substitution• Addition• Dehydration Synthesis

– Etherification– Esterification

• Saponification• Polymerization

Combustion

• Happens when an organic molecule reacts with oxygen gas to form carbon dioxide and water vapor. Also known as “burning”.

Substitution

• Alkane + Halogen Alkyl Halide + Hydrogen Halide• The halogen atoms substitute for any of the hydrogen atoms in the alkane. This happens one atom at a time. The halide generally replaces an H on the end of

the molecule.

C2H6 + Cl2 C2H5Cl + HCl

The second Cl can then substitute for another H:

C2H5Cl + HCl C2H4Cl2 + H2

Addition

• Alkene + Halogen Alkyl Halide• The double bond is broken, and the halogen adds at either

side of where the double bond was. One isomer possible.


Etherification*

• Alcohol + Alcohol Ether + Water

• A dehydrating agent (H2SO4) removes H from one alcohol’s OH and removes the OH from the other. The two molecules join where there H and OH were removed.

Note: dimethyl ether and diethyl ether are also produced from this reaction, but can be separated out.

Esterification

• Organic Acid + Alcohol Ester + Water• A dehydrating agent (H2SO4) removes H from the organic acid and

removes the OH from the alcohol. The two molecules join where there H and OH were removed.

Saponification

The process of making soap from glycerol esters (fats).

Glycerol ester + 3 NaOH soap + glycerol

Glyceryl stearate + 3 NaOH sodium stearate + glycerol

The sodium stearate is the soap! It emulsifies grease…surrounds globules with its nonpolar ends, creating micelles with - charge that water can then wash away. Hard water replaces Na+ with Ca+2 and/or other low solubility ions, which forms a precipitate called “soap scum”.

Water softeners remove these hardening ions from your tap water, allowing the soap to dissolve normally.

Polymerization

• A polymer is a very long-chain molecule made up of many monomers (unit molecules) joined together.

• The polymer is named for the monomer that made it.– Polystyrene is made of styrene monomer– Polybutadiene is made of butadiene monomer

• Addition Polymers• Condensation Polymers• Rubber

Addition PolymersJoining monomers together by breaking double bonds

Polyvinyl chloride (PVC): vinyl siding, PVC pipes, etc.

Vinyl chloride polyvinyl chloride

n C2H3Cl -(-C2H3Cl-)-n

Polytetrafluoroethene (PTFE, teflon):

TFE PTFE

n C2F4 -(-C2F4-)-n

Condensation Polymers

Condensation polymerization is just dehydration synthesis, except instead of making one molecule of ether or ester, you make a monster molecule of polyether or polyester.

Rubber

The process of toughing rubber by cross-linking the polymer strands with sulfur is called...

THE END


sat ii chem prep ppt mrs. gupta modified from mark rosengartens powerpoint

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