chemistry 14 outline guide

Table of ContentsAtomic Structure

The Development of the Different Atomic Models 1The Subatomic Particles 2Quantum Mechanical Model and Quantum Numbers 3The Periodic Table 6

Periodicity of Properties 7The Chemical Bond

Ionic or Electrovalent Bond 8The Covalent Bond 9

Properties of Covalent Bonds 14Theories on Covalent Bonds 14

Changes in MatterChemical Changes 16

The Mole Concept 16Empirical and Molecular Formula 18Stoichiometry 19

Nuclear Changes 22Nuclear vs. Chemical Change 22Types of Radioactive Decay 23Types of Nuclear Changes 24

Phases of MatterIntermolecular Forces of Attraction 25Comparison of the Phases of Matter 26Phase Changes and Phase Diagrams 26The Gaseous State 27

Kinetic Molecular Theory 27The Gas Laws 28Dalton's Law of Partial Pressures 30Deviations from Ideal Behavior 30

The Liquid State 30The Solid State 31

SolutionsThe Dissolution Process 31Factors Affecting Solubility 32Types of Solutions 33Concentration 33

Concentration Units 33Dilution 34

Colligative Properties 34Acids and Bases

Definitions� 35Types of Acids 36Titration and Neutralization 36

Chemical ThermodynamicsBasic Concepts 37The First Law of Thermodynamics 38

Heats of Reaction 39The Second Law of Thermodynamics 41

CHEMISTRY 14The Fundamentals of General Chemistry IProf. Noel S. QuimingDepartment of Physical Sciences and Mathematics

Section TFC || Tuesdays and Fridays: 10:00-11:30AMGAB 105

ii | Chemistry 14: Fundamentals of General Chemistry I

Entropy 41Gibb’s�Free�Energy�(ΔG)� 41

Third Law of Thermodynamics 43Chemical Kinetics

Theories on Reaction Rates 44Factors Affecting Rates of Reactions 44

Chemical EquilibriumMolecular Equilibrium 45

Definitions� 45Factors Affecting Equilibrium 46

Ionic Equilibrium 48

Atomic Structure

The Development of the Different Atomic Models

Democritus- proposed that matter is made up of small indivisible “atoms”� -�small,�hard,�infinite�and�has�different�shapes- found when something is divided and divided: Ends up with an atom of gold- said that they are always moving and capable of joining- nobody believed him - Aristotle was more believed, where he said that the world is made up of 4 elements

John Dalton - Billiard Ball ModelPotulates: - matter is made up of tiny indivisible atoms - all atoms are identical - different elements have different atoms - compounds are made by joining atoms- was believed for 100 years until the discovery of subatomic particles

J.J. Thompson - Plum Pudding/Raisin Bread Model- he discovered the electron using a cathode ray tube - seen through a neutral gas acting like its negatively charged- this proved that the atom was divisible� -�he�showed�that�there�were�negatively�charged�“corpuscles”�(electrons)�in�matter- he proposed that there must be another positively charged particle to neutralize the charge- was believed for only 5 years

Dalton’s model was similar to Thompson’s model by both believing that the atom is solid.

Ernest Rutherford - Nuclear Model-�first�to�propose�a�nucleus�in�the�atom- performed the gold foil experiment - a stream of alpha rays were aimed perpendicularly at a gold foil� � -�most�light�passed�through�the�film�and�some�were�reflected� -�proved�that�the�atom�is�not�solid�(mostly�empty�space) - it must have a very small dense area of positive charge - the positive particle proton negated the electron and is 2000 times more massive- no results would give a complete description of the electron - he just assumed they were scattered around the nucleus

Limitations of Rutherford- What holds the nucleus?- Why don’t the electrons collapse into the nucleus?- Wrong proportion of masses with number of particles. - suggested that there was another particle

James Chadwick- he discovered the neutron- showed that the nucleus contained protons and neutrons- nucleons: protons and neutrons

Modern Model (Rutherford) vs. Dalton- atoms are divisible- atoms of the same element may not be identical- atoms of an element can be changed into another element by nuclear reactions

| 1Prof. Noel S. Quiming

The Subatomic Particles

Atomic Mass Unit (amu)- mass of 1/12 of a 12C atom

Isotopic Notation

or

AXZ

where: A is the mass no. and Z is the atomic no.

Atomic�No.�(Z)�� =�No.�of�protons� �� =�No.�of�electrons�(if�neutral)

Mass�No.�(A)� =�No.�of�protons�+�No.�of�neutrons� � =�No.�of�electons�+�No.�of�neutrons�(if�neutral)� � =�Z�+�No.�of�Neutrons

For�Cations�(+):��No.�of�protons�>�No.�of�electronsFor�Anions�(-):�� No.�of�Protons�<�No.�of�electrons�Charge�=�algebraic�sum�of�No.�of�protons�and�No.�of�electrons

No. of protons No. of electrons No. of neutrons

12Mg2+ 12 10 12

16S2- 16 18 16

30Zn2+30 28 33

Isotopes- atoms of the same element that have the same No. of protons but different No. of neutrons- same Z but different A

Examples: 1H 2H 3H; 12C 13C 14C; 235U 238U

No. of protons No. of neutrons Mass Number

1H�(protium) 1 0 1

2H�(deuterium) 1 1 2

3H�(tritium) 1 2 3

H+�=�proton

13C�-�used�in�NMR�(Nuclear�Magnetic�Resonance)14C�-�used�in�carbon�dating�t½�=�5700�years

Atomic Mass- averaged from the masses of isotopes- uses the relative abundance of isotopes

Example:

Given:35Cl : relative abundance = 75.77% Atomic Mass = 34.968 amu37Cl : relative abundance = 24.23% Atomic Mass = 36.965 amu

2 | Chemistry 14: Fundamentals of General Chemistry I

Z XA

24

32

63

Average Atomic Mass = (relative abundance 1)(atomic mass 1) +

(relative abundance 2)(atomic mass 2)

Average Atomic Mass = (0.7577)(34.968) + (0.2423)(36.965)

= 35.45 amu

Isobars�-�atoms�of�different�elements�but�with�the�same�Mass�No.�(same�A)Isotones - atoms with the same No. of neutrons

Niels Bohr - Planetary Model-�electrons�move�in�ciruclar�orbits�(orbitals)-�energies�are�quantized�meaning�they�have�definite�values- electrons need to absorb energy to jump across orbitals� -�minimum�energy�needed�=�energy�difference�(E�=ΔE)

E = −Rh

(1

n2

)

Rh = Rydberg constant

= 2.1× 10−18J

∆E = E2 − E1

= −Rh

(1

n22

)−−Rh

(1

n21

)

= −Rh

(1

n22

− 1

n21

)

- after jumping into the excited state, electrons need to go back to the original orbital releas-ing light energy.

E =hc

λ

h = Planck′s constant

= 6.63× 10−34 Js

c = Speed of Light

= 3.0× 108 m/s

Example:n = 5 → n = 1

∆E = −Rh

(1

n2f

− 1

n2i

)

= −2.1× 10−18

(1

12− 1

52

)

= −2.016× 10−18 J

λ =hc

E=

((6.63× 10−34 Js)(3.0× 108 m/s)

2.016× 10−18 J

)(1× 104 nm

1 m

)

= 98.66 nm

Application: Flame Tests - wavelengths resulting from energy release result in different colors of light

Limitations of Bohr- cannot describe a multi-electron model

Quantum Mechanical Model and Quantum NumbersErwin Schrödinger - Quantum Mechanical Model-�electrons�move�in�a�3D�space�(electron�cloud)

OrbitalsNode�-�a�region�in�space�where�there�is�zero�probability�of�finding�an�electron

s orbitals: sp orbitals: px, py, pz

d orbitals: dxy, dxz, dyz, dz^2, dx^2-y^2

Quantum Numbers- sets of numbers that are used to describe electrons



Symbol Possible Values Comments

Principal

Higher n: - higher energy - bigger atom/ electron cloud

n

n�=�1�(K�shell)��=�2�(L�shell)��=�3�(M�shell) and so on...

positive integers(1,�2,�3,�...)

- describes the energy level of an electron- describes the size of the electron cloud- main energy level/shell

Angular Momentumor Azimuthal

l

l�=�1�(s�shell)��=�2�(p�shell)��=�3�(d�shell)��=�4�(f�sublevel)

n�=�1n�=�2n�=�3n�=�4

l�=�0l�=�0,�1l�=�0,�1,�2l�=�0,�1,�2,�3

- describes the shape of the electron cloud- describes the sub-shell/sublevel

Magnetic m or ml l�=�1l�=�2

l�=�3

m�=�-1,�0,�+1m�=�-2,�-1,�0,��+1,�+2m�=�-3,�-2,�-1,��0,�+1,�+2,�+3

- determines the ori-entation of space of the electron cloud- describes the orbital

Spin s or ms +½,�-½ - describes the direc-tion of the rotation of the orbital

Pauli’s Exclusion Principle- no two electrons can have the same set of quantum numbers - same n: same shell - same l: same subshell - same m: same orbital - electrons in the same orbital cannot have the same s- orbitals can only accomodate two electrons

No. of orbitals No. of electrons

n�=�1 1 2

n�=�2 4 8

n�=�3 9 32

Electron ConfigurationAUFBAU Building Principle-�lower�energy�levels�are�filled�up�first

n+l rule-�the�higher�the�n+l��value,�the�higher�the�energy�of�the�subshell- determine the energy level of subshells

Example:� 4s:�n+l�=�4+0�=�4� 3d:�n+l�=�3+2�=�5� 4p:�n+l�=�4+1�=�5�

-�If�both�n+l�values�are�equal,�you�describe�the�one�with�the�higher�shell�first�

Examples:

1H: 1s1

2He: 1s2


6C: 1s22s22p2

2p: ↓_ ↓_ __ 2s:↓↑Ground state: 1s:↓↑

-�orbitals�in�the�same�subshell�are�degenerate�with�one�another�(have�the�same�energy) - p orbitals: ↓_ ↓_ __�=�↓_ __ ↓_�=�__�↓_ ↓_

Hund's Rule of maximum Multiplicity- one electron is filled in each orbital first before pairing with an unpaired one

Atoms at ground state can undergo excitation into an excited state:

6C: 1s22s22p2

2p: ↓_ ↓_ __ 2s:↓↑Ground state: 1s:↓↑

2p: ↓_ ↓_ ↓_

2s:↓_Excited state: 1s:↓↑

Quantum Numbers for electrons in an orbital:

1s: ↓↑�� (1,�0,�0,�-½),�� (1,�0,�0,�+½)2s: ↓↑�� (2,�0,�0,�-½),�� (2,�0,�0,�+½)2p: ↓↑↓↑↓↑ (2,�1,�-1,�-½),�� (2,�1,�-1,�+½)� � (2,�1,�0,�-½),�� (2,�1,�0,�+½)� � (2,�1,�+1,�-½),� (2,�1,�+1,�+½)

Differentiating Electron: last entering electron-�for�C:�(2,�1,�0,�-½)

paramagnetic - if an atom has an unpaired electrondiamagnetic - if there are no unpaired electrons

Shortcut: Using noble gases Mn: [Ar]4s23d5

Valence electrons - electrons occupying the outermost energy level C: 5 e-

Examples:

7N: 1s22s22p3

2p: ↓_ ↓_ ↓_ 2s:↓↑Ground state: 1s:↓↑

Differentiating�Electron:�(2,�1,�+1,�-½)� � Valence�Electrons:�5

8O: 1s22s22p4

2p: ↓↑ ↓_ ↓_ 2s:↓↑Ground state: 1s:↓↑


25Mn: 1s22s22p63s23p64s23d5

Ground state: 1s:↓↑ 2s:↓↑ 2p: ↓↑ ↓↑ ↓↑ 3s: ↓↑ 3p: ↓↑↓↑↓↑4s: ↓↑3d: ↓_ ↓_ ↓_ ↓_ ↓_�Differentiating�Electron:�(3,�2,�+2,�-½)� � Valence�Electrons:�2

51Sb: 1s22s22p63s23p64s23d104p65s24d105p3�=�[Kr]5s24d105p3



73Ta: 1s22s22p63s23p64s23d104p65s24d105p66s24p145d3�=�[Xe]6s24p145d3

Differentiating�Electron:�(5,�2,�0,�-½)� � Valence�Electrons:�2

38Sr: 1s22s22p63s23p64s23d104p65s2=�[Kr]5s2

Differentiating�Electron:�(5,�0,�0,�+½)� � Valence�Electrons:�2

Given:�Differentiating�Elctron:�(4,�1,�0,�+½)Answer: 1s22s22p63s23p64s23d104p5�=�[Ar]4s23d104p5 =�Bromine

Stability - if an atom has no incompletely filled orbitals, then it is stable-�completely�filled�>�half�filled�>�random�incompletely�filled- exceptions in AUFBAU principle because of this

18Ar: [He]2p63s23p6 3p: ↓↑ ↓↑ ↓↑�� (Stable)

15P: [He]2p63s23p3 3p: ↓_ ↓_ ↓_

16S: [He]2p63s23p4 3p: ↓↑ ↓_ ↓_

Stability:�Ar�>�P�>�S

29Cu: [Ar]4s23d9 4s: ↓↑ 3d: ↓↑↓↑↓↑↓↑↓_� (LESS�STABLE) [Ar]4s13d10 4s: ↓_ 3d: ↓↑ ↓↑ ↓↑ ↓↑ ↓↑� (MORE�STABLE)

Pseudo-Noble Gases

30Zn: [Ar]4s23d10 - completely filled, a relatively stable element - still less stable than a noble gas

30Zn2+: [Ar]4s03d10 - psuedo noble gas ion - also stable since all orbitals are filled

26Fe: [Ar]4s23d6 26Fe2+: [Ar]4s03d6

26Fe3+: [Ar]4s03d5� (most�stable�ion)

29Cu: [Ar]4s13d10 29Cu+: [Ar]4s03d10 (most�stable�ion)

29Cu2+: [Ar]4s03d9 (from 4s23d9)

The Periodic TableAnton Laurent Lavosier- tabulated 33 elements in his time- only a simple table with no arrangement

Dalton's Chemical Symbols- made symbols for 36 different elements

Dobereiner's Law of Triads- groupings with three elements each- based on similar chemical and physical properties

Newland's Law of Octaves- arranged the elements into increasing atomic mass horizontally- every 8th element would have same properties as the previous 8th- new rows would start at the 8th element

Lothar Meyer- observed similarities in properties in relation with atomic masses

Mendeleev and Meyer's Periodic Law- the physical and chemical properties of elements are periodic functions of their atomic weights


- discrepancies with elements such as K and Ar - atomic weight is not the bast basis for arranging the elements

Moseley's Periodic Law- the physical and chemical properties of elements are periodic functions of their atomic numbers- the modern periodic table is based from this

Location of Elements- group no: column- period no: row-�electron�configuration�can�be�used�to�tell�the�locations�of�elements

Old System- Family A: Representative Elements - if the differentiang electron occupies an s or p orbital - group no: # of valence electrons - period no: highest n value- Family B: Transition Elements - if the differentiating electron occupies a d or f orbital� -�group�no:�#�of�electrons�in�ns�+�(n-1)d�orbitals�(roman�numerals) - period no: highest n value

Examples:

25Mn: [Ar]4s23d5� Family�B� Group�No�VIIB� Period�No�4

17Cl: [Ne]3s23p5� Family�A� Group�No�VIIA� Period�No�3

IUPAC System- groups numbered 1-18-�group�No�� =�no.�of�electrons�in�ns�(s�block)�� =�10�+�no�of�valence�electrons�(p�block)� � =�no�of�electrons�in�ns�+�(n-1)d�(d�block)- period no: highest n value

Example:

15P: [He]2p63s23p3 Group 15 Period 3

Periodicity of PropertiesPeriodic Properties- mostly obsered at family A, the d and f blocks only differ in the inner electrons

Atomic Size/Radius- one-half of the nuclear distance ⇓ : increasing size� � -�increase�in�n�=�higher�number�of�energy�levels ⇒:decreasing size - same energy level but increase in p+and e-

- higher electrostatic attraction

Ionization Potential/Energy- amount of energy required to remove an electron from a neutral atom to convert it to a positively charged ion- 1st electron: 1st ionization potential, 2nd electron: 2nd ionization potential ⇓ : decreasing potential� � -�bigger�atom�=�less�attraction�from�nucleus,�easier�to�remove ⇒:increasing potential - higher electrostatic atractionExceptions:-�Group�3�<�2

4Be: 1s22s2 and 5B: 1s22s22p1


- since the electron at 2p is farther from the nucleus, it is easier to remove-�Group�6�<�5

7Be: 1s22s22p3 and 8B: 1s22s22p4

- N is already stable and there is an e- - e- repulsion that decreases IP

- noble gases have high IP because of stability

Electron Affinity- amount of enegry released when a neutral atom accepts an electron in its outermost shell to becme a negatively charged ion- it is expected to have a negative value ⇓ : decreasing affinity ⇒:increasing affinity - metals have less negative electron affinity while nonmetals have more negative ones

Electronegativity- the tendency of an atom to attract electons to itself when it is chemically combined to another atom ⇓ : decreasing electronegativity ⇒:increasing electronegativity� � -�group�VIIA/17�(halogens)�have�high�electronagavity�values�(F�=�highest) - noble gases do not need to attract electrons because they are stable

Effective Nuclear Charge- the positive charge that electrons actually experience- 4Be (1s22s2)�has�a�+4�charge� -�the�electrons�in�1s�experience�a�+4�charge� -�the�electrons�in�2s�do�not�actually�experience�a�+4�charge�because�of�shielding� -�the�1s�electrons�experience�a�+4�charge�while�the�2s�electrons�experience�a�� +2�charge- 5B: 1s22s22p1

� -�1s�=�+5,�2s�=�+3,�2p�=�+1- this is the reason why the electron in the outermost level is the easiest to remove- electrons in the same orbital cannot shield each other

The Chemical Bond- result of the atomic quest for stability� -�products�have�the�same�electron�configuration�as�noble�gases - He: 1s2, others: ns2 np6�(stable�octet)

Ionic or Electrovalent Bond-�electron�transfer:�metal/metalloid�+�nonmetal-�metal(cation):�low�ionization�potential;�nonmetal(anion):�high�electron�affinity- due to electrostatic attraction- forms an ionic bond-�electronegativity�difference�>�1.7

ex.

11Na: 1s22s22p63s1�(attains�stability�by�losing�one�electron) ⇒ Na+: 1s22s22p6�(isoelectronic�with�Ne)

17Cl: [Ne]3s23p5 ⇒ Cl-: [Ne]3s23p6

48Cd: [Kr]5s23d10 ⇒ Cd2+:[Kr]5s03d10�(pseudo-noble�gas�ion)

p/e ratio - determines the trend of size between elements and their ions-�higher�p/e�=�smaller�size-�lower�p/e�=�bigger�size- the atom shrinks because of increased attraction of electrons to protons


#p #e p/e

Fe 26 26 1

Fe2+ 26 24 1.08

Fe3+ 26 23 1.13

Fe�>�Fe2+�>�Fe3+

#p #e p/e

Na+ 11 10 1.1

Mg2+ 12 10 1.2

Al3+ 13 10 1.3

Na+�>�Mg2+�>�Al3+

The Covalent Bond- between nonmetals, neutral particles/molecules- overlapping of orbitals must occur before electron sharing - forming molecular orbitals- head-on overlap: forms a sigma molecular orbital/bond� -�occurs�on�s+s,�px+px, hybrid orbitals- lateral overlap: sideways, forms a pi molecular orbital/bond - py+py, pz+pz

- creates a node at the center of the bond

Polarity- nonpolar covalent bond (ex. H2) - forms between similar atoms - equal electron sharing� -�EN�difference�≤�0.4-�polar�covalent�bond�(ex.�C-H) - forms between dissimilar atoms - unequal sharing of electrons - EN difference from 0.4 to 1.7 - creates a dipole moment - pointed to the more electronegative atom

Covalency No. - number of bonds that the atom has form with other atoms - may be regardless of normal covalency number

Normal Covalency No. - maximum number of covalent bonds that an atom can form� -�CN�=�8�-�No.�of�valence�electrons Examples:� � � C:�8-4=4� � � O:�8-6=2� � � N:�8-5=3

Coordinate Covalent Bond - a bond between two atoms in which the shared electrons come from one atom

Formal Charge�=�Group�No.�-�(No.�of�unshared�electrons�+�No.�of�bonds)

Lewis Dot StructureSteps: (ex: NCl3)� 1. No. of valence electrons needed� � N:�� 1x8�=�8� � Cl:�� 3x8�=�24

#p #e p/e

N3- 7 10 0.7

O2- 8 10 0.8

F- 9 10 0.9

N3->O2->F-


TOTAL: 32 valence electrons 2. No. of valence electrons available� � N:� 1x5�=�5� � Cl:� 3x7�=�21 TOTAL: 26 valence electrons� 3.�No.�of�shared�electrons:�� 32-26�=�6�electrons� 4.�No.�of�bonds:�� 6/2�=�3�bonds

Formal Charges: FCN:�5-3-2�=�0� � � FCCl:�7-1-6�=�0

Central atom: highest covalency number

NH4+

1. No. of valence electrons needed� N:�� 1x8�=�8� H:�� 4x2�=�8 TOTAL: 16 valence electrons2. No. of valence electrons available� N:� 1x5�=�5� H:� 4x1�=�4� TOTAL:�9�valence�electrons�-�1�electron(from�charge)�=�8�electrons3.�No.�of�shared�electrons:�� 16-8�=�8�electrons4.�No.�of�bonds:�� 8/2�=�4�bonds

Formal Charges:FCN:�5-4-0�=�+1�� FCH:�1-1-0�=�0

CO2

1. No. of valence electrons needed� C:�� 1x8�=�8� O:�� 2x8�=�16 TOTAL: 24 valence electrons2. No. of valence electrons available� C:� 1x4�=�4� O:� 2x6�=�12 TOTAL: 16 valence electrons3.�No.�of�shared�electrons:�� 24-16�=�8�electrons4.�No.�of�bonds:�� 8/2�=�4�bonds

Formal Charges:FCC:�4-4-0�=�0� � � FCO:�6-4-2�=�0

CO32-

1. No. of valence electrons needed� C:�� 1x8�=�8� O:�� 3x8�=�24 TOTAL: 32 valence electrons2. No. of valence electrons available� C:� 1x4�=�4� O:� 3x6�=�18� TOTAL:�22�valence�electrons�+�2�=�243.�No.�of�shared�electrons:�� 32-24�=�8�electrons4.�No.�of�bonds:�� 8/2�=�4�bonds

Formal Charges:FCC:�4-4-0�=�0� � � FCO1:�6-4-2�=�0� � � FCO2:�6-6-1�=�-1

Contributing Structures based on Formal Charges-�structures�with�small�formal�charges�(+2,�-2,�or�less)�are�more�likely


- formal charges of opposite signs are usually adjacent to each other- highly electronegative atoms should have negative rather than positive formal charges- most stable structures have the largest sum of the EN differences for adjacent atoms- ones with fewer formal charges are more stable

Resonance Structures - different lewis structures for the same compound- connected by the same framework of sigma bonds

FNO2

1. No. of valence electrons needed� F:�� 1x8�=�8� N:�� 1x8�=�8� O:�� 1x8�=�8 TOTAL: 24 valence electrons2. No. of valence electrons available� F:�� 1x7�=�7� N:�� 1x5�=�5� O:�� 1x6�=�6 TOTAL: 16 valence electrons3.�No.�of�shared�electrons:�� 24-16�=�8�electrons4.�No.�of�bonds:�� 8/2�=�4�bonds

Structure 1: Double bond to O1

FCF: 0 FCN:�+1�� FCO1: 0 FCO2: -1Structure 2: Double bond to O2

FCF: 0 FCN:�+1�� FCO1: -1 FCO2: 0Structure 3: Double bond to FFCF:�+1�� FCN:�+1�� FCO1: -1 FCO2: -1

- the third structure is not contributing because - F, which is the most electronegative has a positive formal charge - adjacent atoms have charges of the same sign

SCN-

1. No. of valence electrons needed� S:�� 1x8�=�8� C:�� 1x8�=�8� N:�� 1x8�=�8 TOTAL: 24 valence electrons2. No. of valence electrons available� S:�� 1x6�=�6� C:�� 1x4�=�4� N:�� 1x5�=�5� TOTAL:�15�valence�electrons�+�1�=�16�electrons3.�No.�of�shared�electrons:�� 24-16�=�8�electrons4.�No.�of�bonds:�� 8/2�=�4�bonds

Structure 1: Double bond to both S and NFCs:�6-4-2�=�0� � FCc:�4-4-0�=�0� � FCN:�5-4-2�=�-1Structure 2: Triple bond to SFCs:�6-2-3�=�+1�� FCc:�4-4-0�=�0� � FCN:�5-6-1�=�-2�Structure 3: Triple bond to NFCs:�6-6-1�=�-1� � FCc:�4-4-0�=�0� � FCN:�5-2-3�=�0

- structure 2 is the least contributing among the structures

Exceptions to the Octet Rule1. Molecules with an odd number of electronsex: NO2

1. No. of valence electrons needed: 24 valence electrons


2. No. of valence electrons available: 17 valence electrons� 3.�No.�of�shared�electrons:�� 24-17�=�7�electrons� 4.�No.�of�bonds:�� 7/2�=�3.5�bonds

- the resulting structure will have a lone electron- compounds with odd numbers of electrons are called free radicals

2. Needs less than an octet to be stableex: BF3

- following to octet rule there will be the following formal charges FCB:�3-4-0�=�-1�FCF1:�7-1-6�=�0� FCF2:�7-1-6�=�0� �FCF3:�7-2-4�=�+1 - to remove the formal charges, B should only have 3 bonds to it - BF3 then is very reactive and a Lewis acid

3. Expanded octet - only for elements at period 3 and belowex: PCl5 - 5 bonds attached to P ICl4

- - 3 bonds and 2 lone pairs attached to I

- to expand an octet, remove the formal charge by adding bonds to the central atom SNF3

1. No. of valence electrons needed: 40 valence electrons 2. No. of valence electrons available: 32 valence electrons� � 3.�No.�of�shared�electrons:�� 40-32�=�8�electrons� � 4.�No.�of�bonds:�� 8/2�=�4�bonds FCS:�6-4-0�=�+2�� FCN:�5-1-6�=�-2 - by adding two more S-N bonds, you can remove the formal charges HClO4

1. No. of valence electrons needed: 42 valence electrons 2. No. of valence electrons available: 32 valence electrons� � 3.�No.�of�shared�electrons:�� 42-32�=�10�electrons� � 4.�No.�of�bonds:�� 5/2�=�5�bonds FCCl:�7-4-0�=�+3�� FCO:�6-1-6�=�-1 - by adding a Cl-O bond to each oxygen except the one bonded to H, you remove the formal charges

Molecular Geometry (VSEPR Theory)- based on bonding and nonbonding electrons (valence e-)- search for the geometry with the most minimized electron repulsion between atoms- repulsion: lone e- pair-lone e-�pair�>�lone�e-�pair-bonding�>�bonding-bonding

hybridization - mixing of two ro more nonequivalnt atomic orbitals to for a new set of equivalent degenerate orbitals

1. HgCl2

80Hg: [Xe]6s24f145d10

GS: 6s: ↓↑ ES: 6s: ↓_ 6p: ↓_ __ __ HS: sp: ↓_ ↓_�� (forms�2�covalent�bonds)

sp: ↓↑Cl ↓↑Cl LINEAR molecular geometry bond angle: 180º2. BF3

5B: 1s22s22p1

GS: 2s: ↓↑ 2p: ↓_ __ __ ES: 2s: ↓_ 2p: ↓_ ↓_ __ HS: sp2: ↓_ ↓_ ↓_�� (forms�3�covalent�bonds)

sp2: ↓↑F ↓↑F ↓↑F TRIGONAL PLANAR molecular geometry bond angle: 120º3. CH4

6C: 1s22s22p2

GS: 2s: ↓↑ 2p: ↓_ ↓_ __ ES: 2s: ↓_ 2p: ↓_ ↓_ ↓_


HS: sp3: ↓_ ↓_ ↓_ ↓_� (forms�4�covalent�bonds)

sp3: ↓↑H ↓↑H ↓↑H ↓↑H TETRAHEDRAL molecular geometry bond angle: 109.5º4. PCl5

15P: [Ne]3s23p3

GS: 3s: ↓↑ 3p: ↓_ ↓_ ↓_ ES: 3s: ↓_ 3p: ↓_ ↓_ ↓_ 3d: ↓_ __ __ __ __ HS: sp3d: ↓_ ↓_ ↓_ ↓_ ↓_� (forms�5�covalent�bonds)

sp3d: ↓↑Cl ↓↑Cl ↓↑Cl ↓↑Cl ↓↑Cl TRIGONAL BIPYRAMIDAL molecular geometry bond�angle:�120º�(equatorial)�90º�(axial)5. SF6

16S: [Ne]3s23p4

GS: 3s: ↓↑ 3p: ↓↑ ↓_ ↓_ ES: 3s: ↓_ 3p: ↓_ ↓_ ↓_ 3d: ↓_ ↓_ __ __ __ HS: sp3d2:↓_ ↓_ ↓_ ↓_ ↓_ ↓_�(forms�6�covalent�bonds)

sp3d2: ↓↑F ↓↑F ↓↑F↓↑F ↓↑F ↓↑F OCTAHEDRAL molecular geometry bond angle: 90º

Geometries with nonbonding electrons6. NH3

7N: 1s22s22p3

GS: 2s: ↓↑ 2p: ↓_ ↓_ ↓_ HS: sp3: ↓↑ ↓_ ↓_ ↓_� (forms�3�covalent�bonds�and�1�lone�pair)

sp3: ↓↑ ↓↑H ↓↑H ↓↑H TRIGONAL PYRAMIDAL molecular geometry bond�angle:�<109.5º�(≈�107)� or�TRIPOD7. H2O

8O: 1s22s22p4

GS: 2s: ↓↑ 2p: ↓↑ ↓_ ↓_ HS: sp3: ↓↑ ↓↑ ↓_ ↓_� (forms�2�covalent�bonds�and�2�lone�pairs)

sp3: ↓↑ ↓↑ ↓↑H ↓↑H BENT�or�V-SHAPED�molecular�geometry bond�angle:�<104.5º8. SF4

16S: [Ne]3s23p4

GS: 3s: ↓↑ 3p: ↓↑ ↓_ ↓_ ES: 3s: ↓↑ 3p: ↓_ ↓_ ↓_ 3d: ↓_ __ __ __ __ HS: sp3d: ↓↑ ↓_ ↓_ ↓_ ↓_� (forms�4�covalent�bonds�and�1�lone�pair)

sp3d: ↓↑ ↓↑F ↓↑F ↓↑F ↓↑F DISTORTED TETRAHEDRON molecular geometry or SEESAW- lone pair will be found on the equatorial plane to minimize electron repulsion9. ClF3

17Cl: [Ne]3s23p5

GS: 3s: ↓↑ 3p: ↓↑ ↓↑ ↓_ ES: 3s: ↓↑ 3p: ↓↑ ↓_ ↓_ 3d: ↓_ __ __ __ __ HS: sp3d: ↓↑ ↓↑ ↓_ ↓_ ↓_� (forms�3�covalent�bonds�and�2�lone�pairs)

sp3d: ↓↑ ↓↑ ↓↑F ↓↑F ↓↑F T-SHAPED molecular geometry- lone pairs will be found on the equatorial plane to minimize electron repulsion10. XeF2

54Xe: [Kr]5s24d105p6

GS: 5s: ↓↑ 5p: ↓↑ ↓↑ ↓↑ ES: 5s: ↓↑ 5p: ↓↑ ↓↑ ↓_ 5d: ↓_ __ __ __ __ HS: sp3d: ↓↑ ↓↑ ↓↑ ↓_ ↓_� (forms�2�covalent�bonds�and�3�lone�pairs)

sp3d: ↓↑ ↓↑ ↓↑ ↓↑F ↓↑F LINEAR molecular geometry- lone pairs will be found on the equatorial plane to minimize electron repulsion

11. BrF5

35Br: [Ar]4s23d104p5

GS: 4s: ↓↑ 4p: ↓↑ ↓↑ ↓_ ES: 4s: ↓↑ 4p: ↓_ ↓_ ↓_ 4d: ↓_ ↓_ __ __ __


HS: sp3d2:↓↑ ↓_ ↓_ ↓_ ↓_ ↓_�(forms�5�covalent�bonds�and�1�lone�pair)

sp3d2: ↓↑ ↓↑F ↓↑F↓↑F ↓↑F ↓↑F SQUARE PYRAMIDAL molecular geometry12. XeF4

54Xe: [Kr]5s24d105p6

GS: 5s: ↓↑ 5p: ↓↑ ↓↑ ↓↑ ES: 5s: ↓↑ 4p: ↓↑ ↓_ ↓_ 5d: ↓_ ↓_ __ __ __ HS: sp3d2:↓↑ ↓↑ ↓_ ↓_ ↓_ ↓_(forms�4�covalent�bonds�and�2�lone�pairs)

sp3d2: ↓↑ ↓↑ ↓↑F↓↑F ↓↑F ↓↑F SQUARE PLANAR molecular geometry

# e- pairs hybrid. # bonding # nonbonding geometry example

2 sp 2 0 linear HgO2, CO2

3 sp23 0 trigonal planar BF3

2 1 bent�(v-shaped) NO2-

4 sp3

4 0 tetrahedral ICH4

3 1 trigonal pyramid NH2

2 2 bent�(v-shaped) H2O

5 sp3d

5 0 trigonal bipyr. PCl54 1 seesaw SF4

3 2 t-shaped ClF3

2 3 linear XeF2

6 sp3d2

6 0 octahedral SF6

5 1 square pyramid BrCl54 2 square planar XeF4

Properties of Covalent Bonds1. Bond polarity� -�nonpolar�bond:�EN�difference�≤�0.4� -�polar�bond:�0.4�≤�EN�difference�≤�1.7� -�ionic�bond:�1.7�≤�EN�difference� � ex:�H-F�>�H-Cl�>�H-Br�>�H-I�(decreasing�pattern)

2. Bond dissociation energy - energy meeded to destroy a covalent bond ↑ bond polarity ↑ bond dissociation energy

3. Bond order - number of bonds present between two atoms ↑ bond order ↑ bond dissociation energy

4. Bond length - internuclear distance ↑ size of atom ↑ bond length ↑ bond polarity ↓ bond length

5. Bond angle - becomes smaller due to the presence of lone pairs - deviations from the normal value - comparing different bond angles - difference in central atom ↑ size of central ↓ bond angle formed - difference in ligand atoms ↑ size of ligand ↑ bond angle formed

Theories on Covalent Bondsin C2H4: All H in s orbitals and all C are sp2 hybridized C: GS: 2s: ↓↑ 2p: ↓_ ↓_ __ ES: 2s: ↓_ 2p: ↓_ ↓_ ↓_


HS: sp2: ↓_ ↓_ ↓_ 2p: ↓_

C-C bond: 4 C-H bonds: 1 sigma bond: sp2 - sp2 1 sigma bond: sp2 - s 1 pi bond: p - p

single bond - sigma bonds 1 sigma bond - sp2 - sp2

double bond - pi bonds 1 pi bond - p - p

- only unhybridized p orbitals can form pi bonds

in C2H2: All H in s orbitals and all C are sp hybridized

C: HS: sp: ↓_ ↓_ 2p: ↓_ ↓_

C-C bond: 2 C-H bonds: 1 sigma bond: sp - sp 1 sigma bond: sp - s 1 pi bond: py - py

1 pi bond: pz - pz

For N and O: the atom follows the hybridization of an adjacent carbon atom if one is present

- the unhybridized p orbital is almost always used as an orbital for pi bonds - the exceptions are with compounds such as aniline, where there are adjacent C and N molecules in which due to N following the hybridization of C, forms � an�unhybridized�p�orbital�not�used�for�bonding�(used�for�lone�pairs)�this�N�atom�� has no pi bonds around it and the electrons in its lone pair is able to delocalize

delocalization - movement of pi electrons/lone pairs

polycentric molecular orbital - is obtained by having pi electrons delocalize around the atom continuously in loops � (continuous�delocalization)

Molecular Orbital Theory- scope only on homonuclear diatomic atoms (H2, O2,�etc)- bonding molecular orbital - result of constructive overlap of molecular orbitals- antibonding molecular orbital - result of destructive overlap of molecular orbitals

2 possible results of orbital overlapping - s orbitals: σ bonding and σ antibonding (σ*)�MO - p orbitals: σ bonding, σ antibonding (σ*),�p bonding, p antibonding (p*)�MO

Molecular Orbital Energy Diagram1.�No.�of�MO�formed�=�No.�of�atomic�orbitals�combined2. Bonding MO will have a lower energy than the original atomic orbital while the antibonding MO will have a higher energy than the original atomic orbital3. The magnitude of decrease of energy for the bonding MO is equal to the magnitude of increase of the antibonding MO

Bond�Order�=�½(#�e- in BMO - # e-�in�ABMO)

- you can also determine if the molecule is paramagnetic or diamagnetic


O2, F2, Ne2 B2, C2, N2

Changes in MatterChemical Changes The Mole ConceptFormula Mass/Weight - refers to the mas of ionic compoundsMolecular mass/Weight - refers to covalent compounds- both are computed by the summation of the atomic weights of the component elements

FW of KCl = AW K + AW Cl

= 39.098 + 25.45

= 74.55 amu

MW of C6H12O6 : C : 6× 12.01 = 72.06

H : 12× 1.01 = 12.12

O : 6× 16.00 = 96.00

= 180.18 amu

FW of Ca3(PO4)2 : Ca : 3× 40.01 = 120.03

P : 2× 30.97 = 61.94

O : 8× 16.00 = 128.00

= 309.97 amu

Mole - amount of substance that contains as many elementary particles as there are atoms in 12 g of 12C-�1�mole�=�6.022145x1023 particles = NA (Avogadro's Number)-�symbolizes�a�collection�of�particles�(like�dozen,�etc.)- of you collect one mole of atoms, its weight is equal to its atomic mass- molar mass: weight of 1 mol of a substance

#�of�particles� ��->� ÷� ->� Moles� ��->� � x� ->� Mass�in�g Avogadro's Number Molar Mass#�of�particles� ��<-� x� <-� Moles� ��<-� � ÷� <-� Mass�in�g

p: __ __

1s 1s

σ: __ σ: __

σ: __ σ: __

σ: __

σ: __

σ*: __ σ*: __

σ*: __ σ*: __

σ*: __ σ*: __

p*: __ __ p*: __ __

p: __ __

2s 2s

2p 2p2p 2p

1s 1s

2s 2s

INC

RE

AS

ING

EN

ER

GY


Examples:Needed: atoms in 4.21 mol of Pt

(4.21 mol Pt)

(6.02× 1023 atoms Pt

1 mol Pt

)= 2.53× 1024 atoms

Needed: molecules in 0.286 mol CH4

(0.286 mol CH4)

(6.02× 1023 molecules CH4

1 mol CH4

)= 1.72× 1023 molecules CH4

same number of C atoms, multiply by 4 for number of H atoms

Needed: formula units in 0.50 mol KBr

(0.50 mol KBr)

(6.02× 1023 formula units KBr

1 mol KBr

)= 3.01× 1023 formula units KBr

ionic compounds exist in a crystal lattice while covalent compounds exist as molecules

Needed: moles in 6.07g of CH4

MW : C : 12.01 +H : 4.04 = 16.05 g/mol

(6.07g CH4)

(1 mol CH4

16.05g CH4

)= 0.38 mol CH4

Needed: moles in 198g of CHCl3

MW : C : 12.01 +H : 1.01 + Cl : 35.45× 3 = 119.37 g/mol

(198g CHCl3)

(1 mol CHCl3119.37g CHCl3

)= 1.66 mol CHCl3

Needed: atoms of H in 25.6g of (NH2)2CO

(25.6g (NH2)2CO)

(1 mol (NH2)2CO

60.06g (NH2)2CO

)(6.02× 1023 molecules (NH2)2CO

1 mol (NH2)2CO

)(4 atoms H

1 molecule (NH2)2CO

)

= 1.02× 1024 atoms H

Needed: grams of 3.14x1019molecules of HCN

(3.14x1019 molecules HCN)

(1 mol HCN

6.02× 1023 molecules HCN

)(27.03g HCN

1 mol HCN

)= 1.41× 10−3 g HCN

Percent Composition- percent by mass of each element in a compound

mass% =(molar mass)(no. of moles)

(mass of compound)

NaNO3 : 22.99 + 14.01 + 3× 16.00 = 85.00 g/mol

%Na :22.99

85.00× 100 = 27.05% Na

%N :14.01

85.00× 100 = 16.48% N

%O :3× 16.00

85.00× 100 = 56.47% O

H2SO4 : 2× 1.01 + 32.07 + 4× 16.00 = 98.09 g/mol

%H :2× 1.01

98.09× 100 = 2.06% H

%S :32.07

98.09× 100 = 32.69% S

%O :4× 16.00

98.09× 100 = 65.25% O


C18H36O2 : 12× 12.01 + 36× 1.01 + 2× 16.00 = 284.54 g/mol

%C :12× 12.01

284.54× 100 = 75.98% C

%H :36× 1.01

284.54× 100 = 12.78% H

%O :2× 16.00

284.54× 100 = 11.25% O

Needed: g Al in 371g Al2O3

Al2O3 : 2× 26.98 + 3× 16.00 = 101.96 g/mol

%Al :2× 26.98

101.96× 100 = 52.92% Al

(371g Al2O3)(0.5292) = 196.33g Al

Empirical and Molecular FormulaEmpirical Formula- formula that is determined experimentally-�lowest�integer�form�(simplest)

Given: 0.540g S and 0.538g O

mol S : (0.540g S)

(1 mol S

32.07g S

)= 0.017 mol S ÷ 0.017 = 1

mol O : (0.538g O)

(1 mol O

16.00g O

)= 0.034 mol O ÷ 0.017 = 2

Empirical Formula : SO2

Empirical Formula based on % Composition

Given: 40.96%C 4.58%H 54.50%O

Basis: 100g

mol C : (40.92g C)

(1 mol C

12.01g C

)= 3.41 mol C ÷ 3.41 = 1× 3 = 3

mol H : (4.58g H)

(1 mol H

1.01 H

)= 4.53 mol H ÷ 3.41 = 1.33× 3 = 4

mol O : (54.50g O)

(1 mol O

16.00g O

)= 3.41 mol O ÷ 3.41 = 1× 3 = 3

Empirical Formula : C3H4O3

Given: 24.75%K 34.77%Mn 40.51%O

Basis: 100g

mol K : (24.75g K)

(1 mol K

39.10g K

)= 0.633 mol K ÷ 0.633 = 1

mol Mn : (34.77g Mn)

(1 mol Mn

59.94 Mn

)= 0.633 mol Mn÷ 0.633 = 1

mol O : (40.51g O)

(1 mol O

16.00g O

)= 2.53 mol O ÷ 0.633 = 4

Empirical Formula : KMnO4

Experimental Determination (Combustion Analysis)- CO2 produced: C content - H2O produced: H content


Given: from 11.5g ethanol, 22.0g CO2 and 13.5g H2O produced

mol C : (22.0g CO2)

(1 mol CO2

44.02g CO2

)(1 mol C

1 mol CO2

)= 0.500 mol C ÷ 0.25 = 2

mol H : (13.5g H2O)

(1 mol H2O

18.02g H2O

)(2 mol H

1 mol H2O

)= 1.498 mol H ÷ 0.25 = 6

g C : (0.500 mol C)

(12.01 g C

1 mol C

)= 6.005g C

g H : (1.498 mol H)

(1.01 g H

1 mol H

)= 1.513g H

g O : 11.5− 6.005− 1.513 = 3.998g O

mol C : (3.998g O)

(1 mol O

16.00 O

)= 0.25 mol C ÷ 0.25 = 1

Empirical Formula : C2H6O

Molecular Formula1.�Determine�the�molar�mass�of�the�empirical�formula�(EW)2. Find the ratio of the molar mass of the compound to the molar mass of the empirical ��formula�(factor�=�MW/EW)3. Determine the number of units in the empirical formula of the compound

Examples1. Given: EF : CH2O MM : 60g/mol

CH2O : 12.01 + 2.02 + 16.00 = 30.03g/mol

f =20

30.03≈ 2

MF : (CH2O)2 or C2H4O2

2. Given: 14.6%C 39.0%O 46.3%F MM : 82g/mol

basis : 100g

146gC

(1mol

12.01gC

)= 1.217molC ÷ 1.217 = 1

390gO

(1mol

16.00gO

)= 2.438molO ÷ 1.217 = 2

463gF

(1mol

19.00gF

)= 2.437molF ÷ 1.217 = 2

EF : CO2F2

CO2F2 : 12.01 + (2× 16.00) + (2× 19.00) = 82.01g/mol

f =82

82.01≈ 1

MF : CO2F2

Stoichiometry- calculations involving balanced chemical reactions

Recipe: balanced chemical reactionExample:�2Na�+�Cl2 ⇒�2NaCl�(2mol�Na�+�1mol�Cl2 ⇒�2mol�NaCl)


1. Given: 2H2 +O2 → 2H2O

Moles of reactants needed? 2 molH2and 1 molO2

Moles of reactants needed to produce 4 moles of H2O? 4 molH2and 2 molO2

Moles of hydrogen needed and water produced from 3 moles of oxygen?

3 molO2

(2 molH2

1 molO2

)= 6 molH2

3 molO2

(2 molH2O

1 molO2

)= 6 molH2O

Mole-Mole Conversions1. Given: 2Na+ Cl2 → 2NaCl 2.6 molCl2

2.6 molCl2

(2 molNaCl

1 molCl2

)= 5.2 molNaCl

2. Given: 4NH3 + 5O2 → 4NO + 6H2O || 25.00 molNH3

molO2= 25.00 molNH3

(5 molO2

4 molNH3

)= 31.25 molO2

molNO = 25.00 molNH3

(4 molNO

4 molNH3

)= 25.00 molNO

molH2O = 25.00 molNH3

(6 molH2O

4 molNH3

)= 37.5 molH2O

Mole-Mass Conversions1. Given: 2Na+ Cl2 → 2NaCl || 5 molNa

gCl2 = 5.00 molNa

(1 molCl2

5 molNa

)(70.90 gCl2

1 molCl2

)= 177.25 gCl2

2. Given: CoCl2 +HF → CoF2 + 2HCl || 15 molCoF2

gCoCl2 = 15.00 molCoF2

(1 molCoCl2

1 molCoF2

)(129.84 gCoCl2

1 molCoCl2

)= 1947.6 gCoCl2

gHF = 15.00 molCoF2

(2 molHF

1 molCoF2

)(20.01 gHF

1 molHF

)= 600.3 gHF

Mass-Mole Conversions1. Given: 2C2H6 + 7O2 → 4CO2 + 6H20 || 10.0 gH2O

molC2H6= 10.0 gH2O

(1 molH2O

18.02 gH2O

)(2 molC2H6

6 molH2O

)= 0.185 molC2H6

2. Given: CH4 + 2O2 → CO2 + 2H2O || 24.0 gCH4

molO2= 24.0 gCH4

(1 molCH4

16.05 gCH4

)(2 molO2

1 molCH4

)= 2.99 molO2

Mass-Mass Conversions1. Given: N2 + 3H2 → 2NH3 || 2.00 gN2

gNH3= 2.00 gN2

(1 molN2

28.02 gN2

)(2 molNH3

1 molN2

)(17.04 gNH3

1 molNH3

)= 2.43 gNH3

2. Given: 2Na+ I2 → 2NaI || 93.25 gI2

gNaI = 93.25 gI2

(1 molI2253.8 gI2

)(2 molNaI

1 molI2

)(149.89 gNaI

1 molNaI

)= 110.14 gNaI


3. Given: 2KClO3 → 2KCl + 3O2 || 46.0 gKClO3

gO2= 46.0 gKClO3

(1 molKClO3

122.55 gKClO3

)(3 molO2

2 molKClO3

)(32.00 gO2

1 molO2

)= 18.02 gO2

Excess Reactant-�said�to�be�in�excess�(there�is�too�much)

Limiting Reactant- limits how much product we get. Once it runs out, the reaction stops

1. Given: 2Al + 3Cl2 → 2AlCl3 || 10.0 gAl || 35.00 gCl2

Limiting Reactant:

If Al: gAlCl3 = 10.0 gAl

(1 molAl

26.98 gAl

)(2 molAlCl3

2 molAl

)(133.33 gAlCl3

1 molAlCl3

)= 49.42 gAlCl3

If Cl: gAlCl3 = 35.00 gCl2

(1 molCl2

70.9 gCl2

)(2 molAlCl3

3 molCl2

)(133.33 gAlCl3

1 molAlCl3

)= 43.38 gAlCl3

The LR is the reactant that produces less product, so in this reaction, Cl2 is the LR

and Al is in excess. The reaction produced 43.38 gAlCl3

Excess: (Law of Conservation of Mass)

35.0gAl + 10.0gCl2 = 45.0greactant − 43.48gproduct = 1.12gAl

2. Given: 2Al + Fe2O3 → Al2O3 + 2Fe|| 124 gAl || 601 gFe2O3

a. Limiting Reactant:

If Al : gAl2O3= 124 gAl

(1 molAl

26.98 gAl

)(1 molAl2O3

2 molAl

)(101.96 gAl2O3

1 molAl2O3

)= 234.30 gAl2O3

If Fe2O3 : gAl2O3 = 601 gFe2O3

(1 molFe2O3

159.7 gFe2O3

)(1 molAl2O3

1 molFe2O3

)(101.96 gAl2O3

1 molAl2O3

)= 383.71g

Al is the LR

The reaction produced 234.30 gAl2O3

b. Excess:

gFe2O3= 124 gAl

(1 molAl

26.98 gAl

)(1 molFe2O3

2 molAl

)(159.7 gFe2O3

1 molFe2O3

)= 366.99 gFe2O3

601gFe2O3 − 366.99gFe2O3 = 234.01gFe2O3

Percent Yield- the amount of product made in a chemical reaction

1. actual yield - experimental yield, what you got in the lab2. theoretical yield - what the balanced equation tells what the yield should be

% yield =

(actual yield

theoretical yield

)× 100

1. Given: 2Al + 3CuSO4 → Al2(SO4)3 + 3Cu || 3.92gAl produced 6.78gCu

a. actual yield: 6.78gCu

b. theoretical yield:

gCu = 3.92gAl

(1 molAl

26.98 gAl

)(3 molCu

2 molAl

)(62.55 gCu

1 molCu

)= 13.85 gCu

c. % yield =

(6.78 gCu

13.85 gCu

)× 100 = 48.95%


2. Given: TiCl4 + 2Mg → Ti+ 2MgCl2 || actual yield: 7.91× 106 gTi

3.54× 107 gTiCl4 || 1.13× 107 gMg

a. theoretical yield:

gTi = 3.54× 107 gTiCl4

(1 molTiCl4

189.67 gTiCl4

)(1 molTi

1 molTiCl4

)(47.87 gTi

1 molTi

)= 8.93× 106 gTi = LR

gTi = 1.13× 107 gMg

(1 molMg

24.31 gMg

)(1 molTi

2 molMg

)(47.87 gTi

1 molTi

)= 1.113× 107 gTi

b. % yield =

(7.91× 106 gTi

8.93× 106 gTi

)× 100 = 88.49%

3. Given: CaSO4 → CaO + SO3 || at 85% yield: 33.0 gCaSO4

136 gCaSO4produces 80.07 gSO3

If 100 % complete: 33.0 gCaSO4÷ 0.85 = 38.82 gCaSO4

a. actual yield:

gCaO = 38.82 gCaSO4

(1 molCaSO4

196.15 gCaSO4

)(1 molCaO

1 molCaSO4

)(56.01 gCaO

1 molCaO

)= 15.98 gCaO

b. gSO3 = 38.82 gCaSO4

(80.07 gSO3

136 gCaSO4

)= 22.83gSO3

% purity =wt. pure

wt. unpure× 100

4. Given: 2KNO3 → 2KNO2 +O2 || gO = 1.42gO2|| gKNO3

= 10gKNO3

gKNO3= 1.42 gO2

(1 molO2

32.00 gO2

)(2 molKNO3

1 molO2

)(101.11 gKNO3

1 molKNO3

)= 8.97 gKNO3

% purity =8.97 g

10.0 g× 100 = 89.7%

5. Given: 2C4H10 + 13O2 → 8CO2 + 10H2O

a. 1.0 kgC4H10

kgO2= 1000 gC4H10

(1 molC4H10

58.14 gC4H10

)(13 molO2

2 molC4H10

)(32.00 gO2

1 molO2

)(1 kgO2

1000 gO2

)= 3.58 kgO2

b. 89% yield

Unused O2 : 3.58 kgO2− 89%

(3.58 kgO2)(0.11) = 0.39 kgO2

c. 85% pureO2|| 1 kgO2

Unburned C4H10 :

100% pureO2 :

kgC4H10 = 1000 gO2

(1 molO2

32.00 gO2

)(2 molC4H10

13 molO2

)(58.14 gC4H10

1 molC4H10

)= 279.52 gC4H10

85% pureO2 :

kgC4H10= 850 gO2

(1 molO2

32.00 gO2

)(2 molC4H10

13 molO2

)(58.14 gC4H10

1 molC4H10

)= 237.59 gC4H10

Unburned C4H10 = 279.52− 237.59 = 41.93 gC4H10

Nuclear ChangesNuclear ChemistryNuclear vs. Chemical Change


Nuclear Change Chemical Change

involve the nucleus and inner electrons involve only valence electrons

nucleus opens and p+ and n0 rearrange valence electron transfer/sharing

elements transform into other elements rearrangement of atoms and bonds

needs/releases a high amount of energy needs/releases a small amount of energy

rates usually not affected by P, T, catalyst rates usually affected by P, T, catalyst

Radioactivity- discovered by Marie Curie-�the�spontaneous�disintegration�(ionizing�radiation)�of�unstable�elements�into�smaller�pieces- these elements that can do this are called radionuclides

Types of Radioactive Decay

1. Alpha Decay - Loss of an a-particle�(a�helium�nucleus):�42He or 4

2α

- ↓ 2 n0 and ↓ 2 p+

23892 U →234

90 Th+42 He

2. Beta Decay - Loss of a b-particle�(a�high�energy�electron):�0−1e or 0

−1β

- ↓ 1 n0 and ↑ 1 p+ ( 10n →1

1 p+0−1 e�)

13153 I →131

54 Xe+0−1 e

3. e+ emision - Loss of a positron (a particle that has the same mass as but opposite

� � charge�than�an�electron):�01e or 0

1β

- ↑ 1 n0 and ↓ 1 p+�(11p →0

1 β +10 n)

116 C →11

5 B +01 e

4. g emission - Loss of a g-ray (high-energy radiation that almost always accompanies the

� � loss�of�a�nuclear�particle):�00γ

5. e- capture - Addition of an electron to a proton in the nucleus. As a result, a proton is

transformed into a neutron. 11p+

0−1 e →1

0 n

- ↑ 1 n0 and ↓ 1 p+

3718Ar +

0−1 e →37

17 Cl

Balancing Nuclear Reactions1. Conserve Mass No.2. Conserve atomic no./nuclear charge

Examples:

1. 212Po : α decay21284 Po →4

2 He+20882 Pb

2. 60Co : β decay6027Co →0

−1 e+6028 Ni

Neutron-proton ratios- elements with more than one proton have repulsions between the protons in the nucleus- a strong nuclear force helps keep the nucleus from flying apart- Neutrons play a key role stabilizing the nucleus.


-�For�(Z�≤�20)�stable�neutron-to-proton�ratio�is�≈�1:1.- larger nuclei takes a greater number of neutrons to stabilize the nucleus.

Stable Nuclei- belt of stability shows what nuclides would be stable in certain ratios-�nuclei�above�this�belt�have�too�many�neutrons�->�tend�to�beta�decay-�Z�>�83�are�unstable�->�tend�to�alpha�decay

TrendsMagic NumbersNuclei with 2, 8, 20, 28, 50, or 82 protons or tend to be more stable 2, 8, 20, 28, 50, 82, or 126 neutrons

Stability (protons-neutrons): even-even > even-odd ≈ odd-even > odd-odd

Radioactive Series- Radioactive decay usually is not a one-step nuclear transformation.-�series�of�decays�until�they�form�a�stable�nuclide�(often�a�nuclide�of�lead).

Half-Life- time it takes for a substance to become 50% of its original mass- constant and independent of initial composition

Mass DefectMD�=�theoretical�mass�-�actual�mass-�some�of�the�mass�can�be�converted�into�energy�(E=mc2)�called� nuclear binding energy (BE) - energy required to break up a nucleus into its component protons and neutrons.

Given: Mn0 = 1.00867 amu || Mp+ = 1.00728 amu || M6027Co = 59.933819 amu

theoretical : (27p+ × 1.00728 amup+) + (33n0 × 1.00867 amun0) = 60.48267 amu

∆M = 60.48267 amu− 59.933819 amu = 0.548851 amu

Types of Nuclear Changes- nucleus undergoes change as a result of bombardment by elementary particles/nuclei

Nuclear Fission- a heavy nucleus dicides and form smaller nuclei- followed by high amount of energy

Nuclear Chain Reaction- starts with bombardment and splitting of a radioactive parent- the neutron products bombard other possible nuclei-�continues�until�not�enough�radioactive�nuclides�are�available�(the�chain�reaction�will�die�out)- Critical Mass - minimum amount of fissionable material present for the chain reaction to be sustained- Supercritical Mass - too much mass that creates an uncontrolled accelerated process that may explode

Nuclear Fusion- combination of nuclei-�can�occur�only�in�very�high�temperatures�(like�at�stars)- not radioactive- possible to perform in a Tokamak apparatus that uses magnetic fields to generate heat

Radiation on Matter


Does it pass through? Paper 0.5 cm lead 10 cm lead

alpha particle X X X

beta particle O X X

gamma ray O O X

Phases of MatterIntermolecular Forces of Attraction- forces of attractions that exist between molecules- determine: - state - melting and boiling points - solubilities

van der Waals or London dispersion forces - weakest of the intermolecular forces - due to uneven electron distributions in neighboring molecules- weak attraction of the nuclei in a molecule for electrons in another-�attraction�of�the�+�side�of�a�nonpolar�molecule�to�the�-�side�of�another- ↑ size (and ↑�electrons),�↑ LDF due to higher polarizability- only type of IFA between nonpolar molecules, present in ALL molecules- only temporary, due to temporary dipole moments- induced dipole-induced dipole attraction

nonpolar�molecules:�µ�total�=�0polar�molecules:�µ�total�=�>0

CCl4 - nonpolar because there is a cancellation of dipole moments - electrically neutral

Examples: H2, Cl2, CO2, CH4�(only�held�by�LDF)

Dipole-Dipole Attractions-�molecules�with�permanent�net�dipoles�(polar�molecules)- partial positive charge on one molecule is electrostatically attracted to the partial negative charge of another- stronger than LDFExamples: SCl2 PCl3, CH3Cl

Hydrogen-Bond- special dipole-dipole- H atoms are attracted to F, O and N atoms- stronger than dipole-dipole and LDF

volume contraction - occurs when you mix 50mL water and 50mL ethanol

Ion-Dipole Attraction- attraction between an ion and a polar molecule- accounts for the solubility of most ionic compounds in polar solvents- the larger the charge the stronger the force

What�type(s)�of�intermolecular�forces�exist�between�each�of�the�following�molecules?HBr�(polar):�� dipole-dipole,�LDFCH4�(nonpolar):��LDFSO2�(polar):� dipole-dipole,�LDF

Summary of IFAs: (strongest to weakest)-�ion-dipole�� (ion�&�polar�molecule)-�H-bonding�� (N,�O,�F�attached�to�H)


-�dipole-dipole�� (polar�molecules)-�ion-induced�dipole�� (ion�&�nonpolar�molecule)-�dipole-induced�dipole�� (polar�&�nonpolar�molecule)-�london�dispersion�forces/Van�der�Waals�� (nonpolar�molecules)

Indicators of Strength of IFA1. Molar Heat of Vaporization (Liquid-Gas)- Heat energy required to evaporate one mole of a given substance at a given temperature- ↑ IFA, ↑ΔHvapExamples: CH4�<�SO2�<�H2O

2. Boiling Point-�vapor�pressure�=�atmospheric�pressure- ↑ IFA, ↑ BPt

3. Vapor Pressure- pressure exerted by a gas on a liquid in which it is in equilibrium- ↑ volatility, ↑�VP,�↓ BPt- ↑ IFA, ↓�VP

4. Molar Heat of Fusion (Solid-Liquid)- heat energy required to melt one mole of a solid- ↑ IFA, ↑ΔHfus

5. Melting Point- temperature at which the liquid and solid phases coexist in equilibrium- ↑ IFA, ↑ MPt

6. Viscosity-�measure�of�a�fluid’s�resistance�to�flow.- measured in Nsm-2�(SI�Units)�or�poise�(P)�or�centipoise�(cP)- ↑ IFA, ↑�Viscosity

7. Surface Tension- amount of energy required to stretch or increase the surface of a liquid by a unit area- due to the unbalanced force experience by molecules at the surface of a liquid-�drops�are�spherical�because�the�sphere�offers�the�smallest�area�for�a�definite�volume.

8. Solubility- max amount of solute that will dissolve in a solvent at a given temperature- ↑ interaction, ↑ solubility

Comparison of the Phases of Matter- solid, liquid and gas-�interaction:�Solid>Liquid>Gas - gases - total disorder and freedom� -�liquids�� -�"gliding"�movement�on�top�of�each�other�(flow) - solids - vibrations- shape and volume:� -�solids�� -�definite�shape�and�volume� -�liquids�� -�indefinite�shape,�definite�volume� -�gases�� -�indefinite�shape�and�volume- only gases are compressible

Phase Changes and Phase DiagramsMolar Heat of Vaporization- Heat energy required to evaporate one mole of a given substance at a given temperature- Energy related to evaporation and condensation� +ΔHvap:�evaporation�� (endothermic)� –ΔHvap:�condensation� (exothermic)


boiling point� � -�vapor�pressure�=�atmospheric�pressurenormal boiling point - boiling point at 1 atm

Molar Heat of Fusion- is the energy that must be absorbed to melt one mole of a substance

melting point/freezing point - the temperature at which the solid and liquid phases coexist in equilibrium

Molar heat of sublimation - the energy required to sublime 1 mole of a solid� � � � -�ΔHsub�=�ΔHfus�+�ΔHvap

Phase Diagram - a graphical way to summarize the conditions under which equilibrium exist between the different states of matter. It allows us to predict the phase of a substance that is stable at any given temperature and pressure

Triple Point: where all three phases are in equilibriumCritical Point: point where critical Temp. and Pressure is achievedCritical Temperature: highest temperature at which a distinct liquid phase can formCritical Pressure: pressure required to bring about liquefaction at critical temperatureSupercritical fluid:� fluid�existing�in�critical�temperatures

negative fusion slope of H2O: causes the lower density of ice than water

The Gaseous State-�gases�are�either�monoatomic�(Ne),�diatomic�(N2)�or�polyatomic�(CH4) 1. have mass 2. compressible� 3.�fill�containers 4. diffusible 5. exert pressure - force on a gas on container walls

Kinetic Molecular Theory-�attraction�(IFA)�vs.�dispersion�(KE)- basis for Gas Laws- predicts IDEAL GAS behavior

Postulate 1: Tiny particles in all matter are in constant motion - constant random straight motion; collide against other molecules or container walls - movement is changed by collidingPostulate 2: Colliding is perfectly elastic - kinetic energy of the system is preservedPostulate 3: Molecular motion is greater at higher temperature.Postulate 4: Volume of gas molecules is negligible compared to the total volume of the container (Vgas<<Vcontainer) - empty spaces are in between particlesPostulate 5: No attractive or repulsive force occur. - negligible IFA

Gas Variables1. n: amount of moles - amount of gas - 6.02x1023 particles/mol2. V: volume � -�assume�Vgas�≈�Vcontainer� -�SI�unit:�L�(1L�=�1dm3)3. T: temperature - average kinetic energy


� -�in�Kelvin:�K�=�Cº�+�2734. P: pressure - exerted on the wall of the container - atm, mmHg and torr� -�1�atm�=�760�torr�=�760�mmHg

Standard Temperature and PressureST:�273K�(0ºC)� � � � SP:�1�atm- because the gas behavior varies in different T and P ranges

The Gas Laws

Boyle's Law P ∝ 1

V k = T, n PV = k P1V1 = P2V2

-�the�P�vs�V�graph�is�logarithmic

1. Given: V1 = 0.55L || P1 = 1atm || P2 = 0.40atm || V2 =?

P1V1 = P2V2

V2 =P1V1

P2=

(1 atm)(0.55 L)

0.40 atm= 1.375 L

2. Given: V1 = 946mL || P1 = 726mmHg || V2 = 154mL || P2 =?

P1V1 = P2V2

P2 =P1V1

V2=

(726 mmHg)(946 mL)

154 mL= 4459.71 mmHg

Charles' Law V ∝ T k = P, n V

T= k

V1

T1=

V2

T2

-�the�V�vs�T�graph�is�linear

1. Given: V1 = 452mL || T1 = 295K || T2 = 460K || V2 =?

V1

T1=

V2

T2

V2 =V1T2

T1=

(452 mL)(460 K)

295 K= 704.81 mL

2. Given: V1 = 3.20L || T1 = 398K || V2 = 1.54L || T2 =?

V1

T1=

V2

T2

T2 =T1V2

V1=

(398 K)(1.54 L)

3.20 L= 191.54 K

Gay-Lussac's Law P ∝ T k = V, n P

T= k

P1

T1=

P2

T2

- the P vs T graph is linear

1. Given: P1 = 3.0atm || T1 = 400K || T2 = 500K || P2 =?

P1

T1=

P2

T2

P2 =P1T2

T1=

(3.0 atm)(500 K)

400 K= 3.75 atm

Combined Gas Law P1V1

T1=

P2V2

T2


1. Given: T1 = 281K || P1 = 6.4atm || V1 = 2.1mL || T2 = 298K || P2 = 1atm || V2 =?

V2 =P1V1T2

T1P2=

(6.4 atm)(2.1 mL)(298 K)

(281 K)(1 atm)= 14.25 mL

2. Given: V1 = 4.0L || P1 = 1.2atm || T1 = 339K || V2 = 1.7L || T2 = 315K || P2 =?

P2 =P1V1T2

T1V2=

(1.2 atm)(4.0 L)(315 K)

(339 K)(1.7 L)= 2.62 atm

Ideal Gas LawPV = nRT

R = 0.08206Latm

molK

1. Given: n = 1.82mol || V = 5.43L || T = 342.5K || P =?

P =nRT

V=

(1.82 mol)(0.08206LatmmolK )(342.5 K)

(5.43 L)= 9.42 atm

2. Given: n = 2.12mol || P = 6.54atm || T = 349K || P =?

V =nRT

P=

(2.12 mol)(0.08206LatmmolK )(349 K)

(6.54 atm)= 9.28 L

Density Formula Molar Mass Formula

ρ =PM

RT M =

ρRT

P

1. Given: P = 752mmHg || T = 328K || MMNH3= 17.04g/mol || ρ =?

ρ =PM

RT=

(752 mmHg)(

1 atm760 mmHg

)(17.04 g

mol )

(0.08206LatmmolK )(328 K)

= 0.63 g/L

2. Given: P = 779mmHg || T = 335K || MMNH3= 352.03g/mol || ρ =?

ρ =PM

RT=

(779 mmHg)(

1 atm760 mmHg

)(352.03 g

mol )

(0.08206LatmmolK )(335 K)

= 13.13 g/L

3. Given: P = 288atm || T = 309K || ρ = 7.71g/L || MM =?

MM =ρRT

P=

(7.71 gL )(0.08206

LatmmolK )(309 K)

(288 atm)= 67.88 g/mol

Through trial and error (MM Cl: 35.45g/mol O: 16.00g/mol) the molecular formula is ClO2

4. Given: 33%Si 67%F || P = 1.20atm || T = 308K || V = 0.210L || mass = 2.38g

basis: 100g

molSi = 33gSi

(1 molSi

28.09 gSi

)= 1.17 molSi ÷ 1.17 = 1

molF = 67gF

(1 molF19.00 gF

)= 3.53 molF ÷ 1.17 = 3

EF : SiF3

EW : 28.09 + (3× 19) = 85.09 g/mol

MM =ρRT

P=

gRT

V P=

(2.38 g)(0.08206LatmmolK )(308 K)

(0.210 L)(1.20 atm)= 168.5 g/mol

factor :168.5

85.04≈ 2

MF : Si2F6


Avogadro's Law V ∝ n k = P, T 1mol = 22.4L at STP� � � � � � � (molar�vol.�of�ideal�gas�at�STP)

1. Given: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l) || VC2H2= 2.64LC2H2

VO2= 2.64 LC2H2

(1 molC2H2

22.4 LC2H2

)(5 molO2

2 molC2H2

)(22.4 LO2

1 molO2

)= 6.6 LO2

2. Given: 2C4H10(g) + 13O2(g)8CO2(g) → +10H2O(l) || VC4H10= 14.9LC4H10

VO2= 14.9 LC4H10

(1 molC4H10

22.4 LC4H10

)(13 molO2

2 molC4H10

)(22.4 LO2

1 molO2

)= 96.85 LO2

3. Given: 2NaN3(s) → 2Na(s) + 3N2(g) || gNaN3= 60gNaN3

|| T = 294K || P = 823mmHg

VN2 = 160 gNaN3

(1 molNaN3

65.02 gNaN3

)(3 molN2

2 molNaN3

)

(0.08206LatmmolK )(294K)

(823mmHg)(

1 atm760 mmHg

)

= 30.74 LN2

4. Given: C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) || gC6H12O6= 5.60gC6H12O6

|| T = 310K || P = 1atm

VCO2 = 5.60 gC6H12O6

(1 molC6H12O6

180.18 gC6H12O6

)(6 molCO2

1 molC6H12O6

)((0.08206Latm

molK )(310K)

(1atm)

)

= 4.74 LCO2

Dalton's Law of Partial Pressures

PT = PA + PB + PC + ...+ Pn Pn = XnPT Xn =

nn

nT

Ideal Behavior- possible in ↑T and ↓P-�if�n=1,�then�PV/RT�=�1

Deviations from Ideal Behavior-�real�gases�have�volume�(ideal�gases�assume�no�volume)- real gases have IFA - will cause the pressure on the container wall to decrease

Van der Waals Equation for Real Gases

(P +

an2

V 2

)(V − nb) = nRT

(P +

an2

V 2

): correct P

(V − nb) : correct V

a: correction due to IFA b: correction due to container walls

The Liquid StateProperties of a Liquid- Liquids are almost incompressible, assume the shape but not the volume of container:- Liquids molecules are held closer together than gas molecules, but not so rigidly that the molecules cannot slide past each other.

1. Viscosity -�liquid’s�resistance�to�flow�due�to�IFA�which�impedes�movement- ↑ SA ↓ viscosity - ↑IFA ↑ viscosity - ↑ T ↓ viscosity

2. Surface Tension- amount of energy required to increase the surface area of a liquid


- measure of the “cohesive force that must be overcome”-�liquid�surfaces�tend�to�have�the�smallest�area�(spherical)- ↑ IFA ↑ Surface TensionExample:�Hg�>�water�>�ethanol

3. Capillarity/ Capillary Action- rising of a liquid through a narrow space against the pull of gravity Cohesive Force�–�binds�molecules�to�each�other�(IFA) Adhesive Force – binds molecules to a surface meniscus - the shape of the liquid surface. �� -�adhesive�forces�>�cohesive�forces - the liquid surface is attracted to its container more than the bulk molecules. Therefore, the meniscus is U-shaped Example: water� � ��-�cohesive�forces��>�adhesive�forces - the meniscus is curved downwards Example: mercury

The Solid StateTypes of Solidcrystalline - solids with an ordered atomic arrangementpolycrystalline - solids with an many ordered atomic arrangementsamorphous - solids without an ordered atomic arrangement

Solutions- homogenous mixtures-�solvent:�the�component�with�the�higher�amount�(dissolver)-�solute:�the�component�with�the�lower�amount�(dissolved)- aqueous solution: water is the solvent

Examples:�air,�antifreeze�in�water,�brass�(and�other�alloys),�soda,�seawater

solid-solid are only called homogenous mixtures

The Dissolution ProcessSolvation/Dissolution: the surrounding of solvent molecules to solute molecules

In ionic compounds: - involves attraction of ions to positive and negative parts of the solvent.- the solvent surrounds the ions -�ΔH�changes-�IFA�(ionic):�ion-dipole


Polar MoleculesExample: - CH3CH2OH is soluble in water due to formation of H-bonds- sucrose is also soluble in water due to formation of H-bonds

Substances insoluble in water- oil: oil is immiscible in water because they cannot form H-bonds and oil is nonpolar - the water-water H-bonds are more stronger than the attraction of oil and water

Energy Changes in Solvation1.�Separation�of�Solute�Particles�(endothermic,�+ΔH)2.�Separation�of�Solvent�Particles�(endothermic,�+ΔH)3.�Formation�of�interaction�between�solute�and�solvent�(exothermic�-ΔH)

In�a�Solution:�� sol-sol�interaction�>�sol-sol�and�solv-solv�interactions

Heat�of�Solution:�� ΔHsoln�=�ΔHsol-sol�+�ΔHsolv-solv�+�ΔH�sol-solv

Physical manifestation: exothermic: increase in temperature felt endothermic: decrease in temperature felt

Entropy Changes in Solvation1.�Separation�of�Solute�Particles�(+ΔS)2.�Separation�of�Solvent�Particles�(+ΔS)3.�Formation�of�interaction�between�solute�and�solvent�(-ΔS)In Total: +∆S in solvation

Factors Affecting Solubility1. Nature of Solute/Solvent - dependent on the polarity of solute and solvent - "like dissolves like" (↑IFA, ↑�solubility) - polar sol, polar solv - nonpolar sol, nonpolar solv - ↑size, ↑ LDF, ↑IFA, ↑ solubility Example: C6H12 and glucose - cyclohexane is nonpolar, it doesn't dissolve in water - glucose is polar, it dissolves in water - gases are slightly soluble in water ↑ solubility, ↑ mass

2. Pressure - only affects gases ↑P, ↑ solubility Henry's Law: SG�=�kPG

3. Temperature solid-liquid: ↑T, ↑solubility�(endothermic)�for�salts Cs2(SO4)3 : exothermic ↑T, ↓ solubility

gas-liquid: ↑T, ↓ solubility

4. Surface Area ↑ SA, ↑ Solubility

Dissolution vs. Chemical Reaction- just because a solid disappears doesn't mean it was dissolved- a chemical reaction may have occurred- dissolution is a physical change while chemical reactions are chemical changes- to check: remove the solvent - if there was remaining solid, it was dissolution - if there was nothing remaining, it was a reaction

Solubility: the maximum amount of solute that can be dissolved in a solvent at a certain temperature


Types of SolutionsDegree of Saturation- qualitative description of concentration1. saturated - holds as much solute as possible at a certain temperature- the dissolved solute is in equilibrium with the undissolved solute

2. unsaturated - less than max amount of solute is held at a given temperature3. supersaturated - solvent holds more than the normal possible amount

ConcentrationConcentration Units

Molarity (M) M =molsolLsoln

M =gsol

(MMsol)(Lsoln)- temperature dependent

Molality (m) m =

molsolkgsolv

Normality (N) N =# equivalent weights

Lsoln

N = M × f

# equivalent weights = molsol × f

Factor1. Acids f = # of replaceable hydronium ion (H+)Examples:�HCl�(f�=�1)� � H2SO4�(f�=�1,�2;�diprotic�acid)� � CH3COOH�(f�=�1)

Polyproptic acid -�stepwise�(one�at�a�time)�donation�of�H+

- If ionization equation is not given: assume max factor

H3PO4 (triprotic)

H3PO4 → H+ +H2PO4− : f = 1

H2PO4− → H+ +HPO4

2− : f = 2

HPO42− → H+ + PO4

3− : f = 3

2. Bases f = # of replaceable OH-

Examples:�� NaOH�(f�=�1)� � � � Mg(OH)2�(f�=�2,�1) Al2(OH)3�(f�=�1,�2,�3)� � � NH3�(f�=�1;�NH4OH)

3. Salts f = total (+) charges or total (-) chargesExamples:� NaCl�(f�=�1)� � � � MgO�(f�=�2) Ca3(PO4)2�(f�=�2)

4. Oxidizing Agents f = # of e- gainedExamples: MnO4

− acidic−−−→ Mn2+ : f = 5

MnO4− basic/neutral−−−−−−−−→MnO2 : f = 3

5. Reducing Agents f = # of e- lostExamples: Fe2+ → Fe3+ : f = 1

Mass % %(w/w) =gsolgsoln

× 100

%(w/v) =gsol

mLsoln× 100

%(v/v) =mLsol

mLsoln× 100


Proof Strength proof = 2×% alcohol

Parts per Million (ppm) ppm =gsol

gsoln or mLsoln× 106

=mgsol

kgsoln or Lsoln

Parts per Billion (ppb) ppb =gsol

gsoln or mLsoln× 109

=µgsol

kgsoln or Lsoln

p-Scale p = −log[solute]� (used�if�concentration�is�very�low)

Mole Fraction χsol =nsol

nsoln

χsolv =nsolv

nsoln

nsoln = nsol + nsolv Xsol +Xsolv = 1

Mole % χ× 100

Dilution C1V1 = C2V2

Example:

Colligative Properties- dependent only on the amount of particles present

non-electrolyte - does not dissociate in solution ex. 1M C6H12O6

electrolyte - dissociates in solution ex. 1M NaCl ⇒ Na+�+�Cl-

For nonvolatile, non-electrolyte solutes:

1. Vapor Pressure Lowering - due to solute particles on the surface - also due to IFA that prevents vaporization of solvent

Pf = χsolvP◦

∆P = χsolP◦

2. Boiling Point Elevation 3. Freezing Point Depression

∆tb = mKb

Kb(water) = 0.52◦C/m

m = molality

∆tf = mKf

Kf (water) = 1.86◦C/m

m = molality

4. Osmotic PressureOsmosis - movement of solvent to a high concentration of solvent to a low concentration of solvent- movement of solvent to a low concentration of solute to a high concentration of solute- passes through a semi-permeable layer that only allows solvent to pass through

Π = CRT C = molar concentration

R = 0.08206Latm

molKT = absolute temperature(K)


van't Hoff factor (i)- considers amount of ionization of electrolyte solutes in solution

i =

∆tb observed

∆tb theoretical (nonelectrolyte)

=∆tb observed

mkb

� ex.�� NaCl��(i�=�2) CaCl2�(i�=�3)

For Electrolyte Solutes:

∆P = i(χsolP◦) ∆tf = i(mKf )

∆tb = i(mKb) Π = i(CRT )

Unless�specified,�we�assume�100%�ionization�of�electrolyte�solute�(i�=�#�of�ions)

Debye–Hückel theory- Due to the ionic atmosphere surrounding an ion, the expected i value becomes non-ideal

Degree of Dissociation (a) α =ireal − 1

iideal − 1

Acids and BasesDefinitions1. Arrhenius Theory - acid: yields H3O

+ / H+�in�solution�(hydronium�ion) - base: yields OH- in solution Example: Acid Base

HCl → H+ + Cl− NaOH → Na+ +OH−

HCl +H2O → H3O+ + Cl− NH3 +H2O → NH+

4 +OH−

2. Brønsted-Lowry Theory - acid: proton donor (proton: H+) - base: proton acceptor - includes conjugate acid-base pairs - strong acids/bases produces a weak conjugate and vice versa Example: Pairs: HCl︸︷︷︸

acid

+H2O︸︷︷︸base

→ H3O+

︸︷︷︸c.base

+ Cl−︸︷︷︸c.acid

HCl/Cl-

H2O/H3O+

CH3COOH︸︷︷︸acid

+H2O︸︷︷︸base

→ CH3COO−︸︷︷︸

c.base

+H3O+

︸︷︷︸c.acid

H2O︸︷︷︸acid

+NH3︸︷︷︸base

→ OH−︸︷︷︸c.base

+NH+4︸︷︷︸

c.acid

Arrhenius�bases�≠�Brønsted�Lowry�Bases - metal hydroxides are only Arrhenius bases

3. Lewis Theory - acid: electron pair acceptor - base: electron pair donor - forms a coordinate covalent bond Example: H+

︸︷︷︸acid

+NH3︸︷︷︸base

→ NH4

Acidity: The ease of losing the H+ ion


Types of AcidsBinary Acids- consists only of H and another atom- acidity is dependent on electronegativity, bond strength and bond polarityExamples: HF, HCl, HBr, HI, H2S

↑ bond polarity, ↑ bond strength, ↓ acidity

Example:�� Acidity:�HF�<�HCl�<�HBr�<�HI�� (HF�is�already�considered�a�weak�acid)

Oxyacids- consists of an ionizable H attached to an O- oxygen-containing acids

Acidity in Difference of central atoms↑ EN of central atom, ↑ Acidity↓ size of central atom, ↑ Acidity

Example: HClO4 is more acidic than HBrO4

Acidity in Difference of the number of Oxygen Atoms↑ O atoms, ↑ Acidity

� Example:� Acidity:�HClO�<�HClO2�<�HClO3�<�HClO4

Examples of Strong Acids:

Polyproptic Acids - ionizes through a series of stepsExample:

H2SO4�(diprotic):��1st ionization : H2SO4 +H2O → H3O

+ +HSO−4 (complete)

2nd ionization : HSO−4 +H2O � H3O

+ + SO2−4 (incomplete)

H3PO4�(triprotic):��

1st ionization : H3PO4 +H2O � H3O+ +H2PO−

4 (incomplete)

2nd ionization : H2PO−4 +H2O � H3O

+ +HPO2−4 (incomplete)

3rd ionization : HPO2−4 +H2O � H3O

+ + PO3−4 (incomplete)

- degree of ionization decreases after every ionization step

Titration and Neutralization- volumetric process to determine the concentration of a sample

Acid-Base Titration:� analyte:�acid� � � titrant:�base�(solution�of�known�concentration) indicator: changes color at the endpoint

- for 1:1 mole ratio, use M1V1�=�M2V2 ex. HCl +NaOH → NaCl +H2O

- for other ratios, you can use normality or stoichiometryExample: H2SO4 + 2NaOH → Na2SO4 + 2H2O analyte: 10.0 mL, H2SO4 titrant: 0.1M, 25.0 mL, NaOH

Ma =

(0.1 molb1 Lb

)(0.025 Lb

1

)(1 mola2 molb

)(1

0.01 La

)= 0.125Macid

Chemical Thermodynamics- measure of energy changes, spontaneity, etc.


Basic ConceptsSystem- The part of the universe on which we wish to focus attention, or any part we want to study.

Surroundings- Everything outside the system.

Types of systems1. Open Systems - Has a boundary or wall that allows exchange of heat and matter with its surroundings. - Also known as NONCONSERVATIVE system.

2. Closed Systems - Has a boundary or wall that allows the exchange of heat but NOT matter with its surroundings. - The wall or boundary is known as the diathermal wall.

3. Isolated Systems - Has a boundary or wall that does NOT allow the exchange of heat and matter with its surroundings. - The wall or boundary is known as the adiabatic wall.

Properties of a SystemExtensive Properties- They are dependent on the amount of matter, that is, the properties changes as the quantity changes. Examples: mass, moles, volume, etc.

Intensive Properties- They are independent on the amount of matter. Examples: density, molecular mass, boiling point, etc.

Intensive Property =Extensive Property 1

Extensive Property 2 Examples:�MM�(g/mol),�density�(mass/vol)

State Functions- The property of a system that depends only on its present state. ��(dependent�on�initial/final�states)- They do not depend on the path taken or how the system arrived to its present state. ��(path-independent) Examples:�ΔU,�ΔH,�ΔS,�ΔG,�ΔA,�V,�P,�T

Non-state functions - Path-dependent: The path should be taken into account when evaluating them.-�One�needs�to�determine�the�change�involved�and�not�on�the�initial�and�final�states. Examples:�Work�and�heat�(w�and�q)

State functions can be related to one another using an equation of state. � Example:�ideal�gas�equation��PV�=�nRT

Non-state functions, can be only evaluated if we know the path or type of change.

Variables and Sign Conventions in Thermodynamics � ΔU�� =�Change�in�Internal�Energy� � q�� =�Heat� ΔH�� =�Change�in�Enthalpy� � � w�� =�Work� ΔS�� =�Change�in�Entropy


� ΔG�� =�Change�in�Gibbs�free�energy

Unit�of�energy:�Joule�(J)Other�conversions:�1�calorie�=�4.184�J

By convention, using the BANK system [(+) - deposit and (-) - withdrawal]

(+)�q�� =�heat�is�absorbed�by�the�system� (-)�q�� =�heat�is�released�by�the�system

(+)�w�� =�work�is�done�on�the�system� � (-)�w�� =�work�is�done�by�the�system

(+)�ΔH�� =�the�process�is�endothermic� � (-)�ΔH�� =�the�process�is�exothermic

The First Law of Thermodynamics- Law of Conservation of Energy-�The�internal�energy�(U):��the�sum�of�the�kinetic�and�potential�energies�of�the�system.-�The�internal�energy�of�the�system�can�be�changed�by�flow�of�work,�heat�or�both

∆U = q + w

A common type of work associated with chemical processes is work done by a gas (through expansion)�or�done�to�a�gas�(compression).�For�instance,�in�automobile�engines,�the�heat�from combustion of fuels expands the gases in the cylinder to push back the piston. This work translates into the motion of the car.

For gases, w = −P∆V � ΔV�=�change�in�volume�upon�expansion�or�compression.

Expansion: V2 > V1

w = −P∆V

= −P (V2 − V1)

= (−)(+)

w = (−)

Compression: V2 < V1

w = −P∆V

= −P (V2 − V1)

= (−)(−)

w = (+)

Example: Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant pressure of 15 atm.

1. Given: V1 = 46L ; V2 = 64L ; P = 15atm ; w =?

w = −P∆V = −P (V2 − V1)

= −(15)(64− 46)

= −270Latm

Converting to Joules:

= −270Latm

(molK

0.08206Latm

)(0.314J

molK

)= −27357.75J = −27.36kJ

Note: R = 0.08206

Latm

molK= 0.314

J

molk 1Latm = 101.325J

Example:�A�balloon�is�being�inflated�to�its�full�extent�by�heating�the�air�inside�it.�In�the�final�stages of this process, the volume of the balloon changes from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J as heat. Assuming that the gas expands against a constant pressure�of�1.0�atm.�Calculate�the�ΔU�for�the�process.


2. Given: V1 = 4.00× 106L ; V2 = 4.50× 106L ; P = 1.0atm ; q = 1.3× 108J

∆U = q + w = q − P∆V = q − P (V2 − V1)

= 1.3× 108J − (1.0atm)(4.50× 106L− 4.00× 106L)

= 1.3× 108J − (5.0× 105Latm)

(J

101.325 Latm

)

= 1.79995× 108J

Applications of the First Law of Thermodynamics

at constant pressure: ∆H = q

The�ΔH�is�commonly�evaluated�instead�of�the�internal�energy�(ΔU).�

1. Obtaining ∆H from ∆U ∆H = ∆U +∆(PV )

Considering a chemical reaction, ∆Hrxn = ∆Urxn +∆(PV )

Assuming that all gaseous species are ideal gases:

PV = nRT

∆(PV ) = ∆(nRT )

∆(PV ) = ∆ngasRT

∆ngas = moles of gaseous products – moles of gaseous reactants

Therefore,∆Hrxn = ∆Urxn +∆ngasRT

Also, for solids and liquids: ∆H ≈ ∆U

Heats of ReactionOther Methods to Evaluate ∆H1. Heat of Formation-�Heats�of�formation�at�standard�state�(25ºC�and�1�atm),�ΔHºf can be found in tables.-�The�ΔHºf of pure elements at 25ºC and 1 atm are ZERO. Examples: C(s,�graphite), Br2(l), Hg(l), Cl2(g), Al(s)

For Gases: Standard state is at 1.00 atm * proper state must be indicated For Solids: Standard state is at 1.00 M

Therefore,∆Hrxn = Σ(n∆H◦

f , products)–Σ(n∆H◦f , reactants)

Example: Calculate�the�ΔHrxn upon the combustion of 1 mole of glucose, C6H12O6.� Given:�� ΔHºf (CO2)�� =�-393.5�kJ/mol� � ΔHºf (H2O)�� =�-285.6�kJ/mol� � ΔHºf�(glucose)�� =�-1274��kJ/mol

C6H12O6 + 6O2 → 2CO2 + 6H2O

∆Hrxn = Σ(n∆H◦f , products)–Σ(n∆H◦

f , reactants)

=

[(6mol)

(−285.6kJ

mol

)+ (6mol)

(−393.5kJ

mol

)]–

[(1mol)

(−1274kJ

mol

)+ (6mol)

(0kJ

mol

)]

= −2800.6kJ

Example: Methanol (CH3OH)�is�often�used�as�a�fuel�in�high�performance�engines�in�race�cars. Using the data above and below, calculate the enthalpy of combustion per gram of


methanol and compare that with the enthalpy of combustion per gram of octane (C8H18). Given: �ΔHºf�(methanol)��=�-239�kJ/mol� � ΔHºf�(octane)�� =�-269�kJ/mol

2CH3OH + 3O2 → 2CO2 + 4H2O


f , reactants)

=

[(4mol)

(−285.6kJ

mol

)+ (2mol)

(−393.5kJ

mol

)]–

[(2mol)

(−239kJ

mol

)+ (3mol)

(0kJ

mol

)]

= −1451.4kJ

2C8H18 + 25O2 → 16CO2 + 18H2O


f , reactants)

=

[(18mol)

(−285.6kJ

mol

)+ (16mol)

(−393.5kJ

mol

)]–

[(2mol)

(−269kJ

mol

)+ (25mol)

(0kJ

mol

)]

= −10898.8kJ

2. Hess’ Law- Since enthalpy is a state function it follows that in going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.-�Evaluation�of�ΔH�when�ΔHºf data is not available-�The�ΔH�of�such�reactions�can�be�evaluated�using�several�steps�but�with�the�net�reaction�� equal to the desired reaction.

DESIRED RXN: N2�(g)�� +�2O2�(g) ⇒ 2NO2�(g)�� ΔH1�� =�68�kJ N2�(g)�� +�O2�(g) ⇒ 2NO (g)�� ΔH2�� =�180�kJ 2NO (g)� +�O2�(g)� ⇒ 2NO2�(g)�� ΔH3�� =�-112�kJ NET: N2�(g)�� +�2O2�(g) ⇒ 2NO2�(g)�� ΔH2�+�ΔH3�=�68�kJ

Note:1.�If�the�reaction�is�reversed,�the�sign�of�ΔH�is�also�reversed.2.�The�magnitude�of�ΔH�is�directly�proportional�to�the�quantities�of�reactants�and�products�in��the�reaction.�If�the�coefficients�in�a�balanced�reaction�are�multiplied�by�an�integer,�the�value��of�ΔH�is�multiplied�by�the�same�integer.

Example: Diborane (B2H6)�is�a�highly�reactive�boron�hydride,�which�was�once�considered�as�a�possible�rocket�fuel�for�the�U.S�space�program.�Calculate�ΔH�for�the�synthesis�of�diborane�from its elements.

2B(s) + 3H2(g) → B2H6(g)

Using the following data: Reaction ∆H 2B�(s)�� +�3/2O2�(g)�� ⇒ B2O3�(s) -1273 kJ B2H6�(g)�� +�3O2�(g)�� ⇒ B2O3�(s)�+�3H2O�(g) -2035 kJ H2�(g)�� +�1/2O2�(g)�� ⇒ H2O (l) -286 kJ H2O (l) ⇒ H2O�(g) 44 kJ 3. Evaluation of Lattice Enthalpy, ∆Hlattice (Born-Haber Cycle)- the energy needed to separate an ionic lattice into gaseous ions or the energy released when component ions form ionic compounds. Since, this cannot be determined exactly by experiment, but can be envisioned as the formation of an ionic compound occurring in a series of steps.

Reaction ∆H Li (s)�� ⇒ Li�(g)� � � ΔHsub�� =�161�kJ Li�(g)� � ⇒ Li+ (g)�+�e� � 1st�Ionization�Energy� =�520�kJ 1/2F2�(g)� � ⇒ F�(g)� � � ΔHdiss� � � =�77�kJ


F�(g)�+�e�� ⇒ F- (g)� � � Electron�Affinity�� =�-328�kJ Li (s)�+�1/2F2�(g) ⇒ LiF (s)�� ΔHºf� � � =�-617�kJWhat is the lattice enthalpy of LiF?

4. ∆Hºf from Bond Enthalpies

Example: H2(g) + F2(g) → 2HF(g)

∆Hrxn = Σ(n∆H, bonds formed)–Σ(n∆H, bonds broken)

Therefore, ∆H = BEH−H +BEF−F –2BEH−F

Calculate�the�ΔHof of HF. Given that BE of H-H is 432 kJ, F-F is 154 kJ and H-F is 565 kJ.

The Second Law of Thermodynamics- basis for predicting if a reaction will be spontaneous or non-spontaneous.- spontaneous if a process will cause an increase in entropy of the universe- The entropy of the universe is increasing.

Spontaneous - occurs without outside intervention - it occurs and liberates energy at the same timeNon-spontaneous - requires energy to carry out

Measure of spontaneity: S or entropy.

Entropy- Degree of disorder/randomness in a system-�difficult�to�measure�and�quantify- not all reactions with ↑ΔS�are�spontaneous

Gibb’s Free Energy (∆G)- expendable energy of the system-�better�measure�because�it�takes�into�account�ΔH�and�ΔS

At constant pressure:

∆G = ∆H − T∆S (The AGAHTAS equation)

Therefore, ∆G < 0:�(-)��Spontaneous�Process�at�any�T ∆G > 0:�(+)�Non-spontaneous�Process�at�any�T ∆G = 0: System at Equilibrium

Cases Common in Chemical Reactions

ΔH T ΔS = ∆G

- any + -

+ any - +

Cases Common in Physical Changes

ΔH T ΔS = ∆G

- low - -

- high - +

+ high + -

+ low + +


In physical transformations:

∆S =qrevTrev

∆S =∆Htransition

Ttransition

Example:�Melting:�� ΔHfusion�=�qrev Tmelt�=�Trev

Where, qr is the heat released or absorbed while T is temperature in Kelvin.

Let us see why water freezes spontaneously at -10ºC while not at 10ºC.Given:� ΔHº�� =�6.03�x�103�J/mol� � ΔSº�=�(6.03�x�103�J/mol)/273�K Ttransition��=�273�K�� ΔSº�=�22.1�J/mol�K

T (ºC) T (K) ∆Hº (J/mol) ∆Sº (J/mol K) T∆S (J/mol) ∆Gº (J/mol)

-10 263 6.03 x 103 22.1 5.81 x 103 2.2 x 102

0 273 6.03 x 103 22.1 6.03 x 103 0

10 283 6.03 x 103 22.1 6.25 x 103 -2.2 x102��Note:�Both�ΔS�and�ΔG�are�STATE FUNCTIONS!

Therefore,∆S = Σ(n∆S, products)–Σ(n∆S, reactants)

∆G = Σ(n∆G, products)–Σ(n∆G, reactants)

Example: Consider the reaction 2SO2�(g)�+�O2�(g) ⇒ 2SO3�(g)

The reaction was carried out at 25ºC and 1 atm. Calculate�the�ΔH,�ΔS�and�ΔG�using�the�following�data:

Substance ∆Hf (kJ/mol) ∆S (J/mol K)� SO2�(g)�� -297� � � 248� SO3�(g)�� -396� � � 257� O2�(g)� � � 0� � � 205

Is the process spontaneous or non-spontaneous?

Example:

1. Given: ∆Hf = 1440 cal/mol ; Tf = 0◦C

a) at− 5◦C

∆G = ∆H − T∆S

= ∆H − T∆Hf

Tf

= 1440 cal/mol − 268K1440 cal/mol

273K= 27.64 cal/mol

∆S =1440 cal/mol

273K= 5.27 cal/molK

= 5.27 eu

b) at 0◦C

∆G = ∆H − T∆S

= 1440 cal/mol − 273K5.27cal

molK= 0

c) at 5◦C

∆G = ∆H − T∆S

= 1440 cal/mol − 278K5.27cal

molK= −25.06 cal/mol

Using Standard Entropy of Formation (Sfº)Example: Given the data below:

6CO2 + 6H2O → C6H12O6 + 6O2

ΔH�=�2806�kJ/mol


ΔS�(CO2�)�=�214�J/mol�K�� ΔS�(glucose)�=�212�J/mol�KΔS�(H2O)�=�70�J/mol�K� � � ΔS�(O2)�=�205�J/mol�K

Is photosynthesis a spontaneous at 25ºC?

∆G = ∆H − T∆S

= ∆H − T [Σ(n∆S, products)–Σ(n∆S, reactants)]

= ∆H − T

[(1mol)

(212J

molK

)+ (6mol)

(205J

molK

)]–

[(6mol)

(214J

molK

)+ (6mol)

(70J

molK

)]

= 2806kJ − 298K

(−262J

K

)

= 2806kJ − 298K

(−0.262kJ

K

)

= 2884.08kJ (not spontaneous at 25oC)

Third Law of Thermodynamics- At absolute zero, or 0 K, -273oC, the entropy of a pure crystalline solid is zero.

S0K = 0

*With this, we are able to evaluate the absolute values of entropy. That is, if we start heating a�pure�crystalline�solid,�from�T�=�0�K�to�T�we�get:

∆S = Sf − Si

∆S = Sf − S0K

∆S = Sf

(No�delta�(ΔS)�when�reported)

Chemical Kinetics- area concerned with the speed or rates of reaction- how fast or slow a reaction occurs

Rate- measured change of concentration of reactants/products per unit time- you can use any reactant/product to state rate Example: A ⇒ B

reactant: rate =−∆[A]

∆t(rate of disappearance of A)

product: rate =∆[B]

∆t(rate of appearance of B)

The Rate of Reaction: A decreases with time B increases with time

Also dependent on Stoichiometry

Example: 2A ⇒ B reactant: rate = −1

2

∆[A]

∆t product: rate =

∆[B]

∆taA+ bB → cC + dD

Reactant A: rate = −1

a

∆[A]

∆t Product C: rate =

1

c

∆[C]

∆t

Reactant B: rate = −1

b

∆[B]

∆t Product D: rate =

1

d

∆[D]

∆t


Theories on Reaction RatesCollision Theory -�rate�is�proportional�to�(number�of�collisions)/(time)�[frequency�of�collision]- ↑frequency, ↑rate- effective collision is needed

Effective Collision1.�colliding�molecules:�energy�≥�minimum�energy�requirement� (activation energy (Ea)�or�energy�barrier) activation energy: the minimum amount of energy required to initiate a chemical reaction2. proper orientation is needed

Reaction Profile-�plot�of�the�potential�energy�vs.�time�of�a�reaction�(progress)

Conclusions from Reaction Profiles:- the one with a lower Ea will occur faster ↓Ea: faster ↑Ea: slower- you can determine the energy of the reactants and products- you can determine if the reaction is exothermic/endothermic- if the reaction is reversible, you can compute for the Ea of both directions- you can also determine the number of steps in the reaction - single step: concerted� -�no.�of�peaks�=�no.�of�steps

Proper Orientations- if the molecules are not properly oriented, they will not form the products

Transition State Theory- the reactants will form a transition site before forming the products- unstable arrangement of atoms- reactants in a short-lived, high-energy, intermediate state

� Initial:� A�+�B-B�� Transition:�A--B--B� � Final:�A-B�+�B

- exact structure cannot be determined- the rate of the reaction is dependent on the energy required to form the transition state ↓energy requirement: faster ↑energy requirement: slower-�peaks�in�reaction�profile:�transition�state

- multistep reactions have many transition states

Factors Affecting Rates of Reactions1. Nature of Reactants - some reactions are inherently fast/slow - dependent on the Ea of the reaction� -�liquid�vs.�gas�(different�phases) - Ea

- position in the reactivity series

2. Surface Area - ↑SA, ↑ rate - due to the high number of molecules exposed to collision

3. Temperature - ↑T, ↑KE, ↑mobility, ↑frequency of collision, ↑rate - not dependent if the reaction is exothermic/endothermic - Arrhenius Equation - relates rate with temperature� -�k:�specific�rate�constant


4. Concentration of Reactants - Law of Mass Action ↑conc'n, ↑no. of molecules - ↑conc'n, ↑frequency of collision, ↑rate��-�exception:�Zero�Order�(concentration�does�not�affect�rate)

Rate Law Expression� � � � � � � k� �=�specific�rate�constant rate = k[A]x[B]y[C]z� [A]� �=�molar�concentration� � � � � � � x/y/z� �=�orders�of�reaction�

order of reaction - magnitude of effect of change in concentration zero order: [A]0�=�1

Determination of Order-�always�experimentally�determined�and�not�always�=�coefficient�in�balanced�equation

Method of Initial Rates- given a set of experimental data comparing different rates of reaction- choose one reactant to vary with the other reactants being constant

- get the general ratio: rateArateB

=

([A]A[A]B

)x

- solve for x to get the order-�overall�order�=�∑(n�orders)-�with�all�orders�determined,�you�can�find�the�value�of�k.

5. Presence of Catalyst - a substance that increases the rate of the reaction without itself being consumed Types of Catalysis: 1. Homogenous - reactants and catalysts are dispersed in the same phase�� intermediate:�valleys�in�the�reaction�profile

2. Heterogenous - reactants and catalysts are not of the same phase Example: catalytic hydrogenation

- the only factor that affects activation energy - catalysts lower the activation energy - catalysts change the reaction into a multi-step reaction with lower Ea

Chemical EquilibriumMolecular EquilibriumDefinitionsKinetic Definition:�rate�of�forward�reaction�=�rate�of�backward�reaction

In a reversible reaction: aA(g) + bB(g) � cC(g) + dD(g)

Rate Laws: Kf [A]a[B]b = Kb[C]c[D]d In equilibrium, we assume order is � � � � � � � equal�to�coefficient

Then, Keq = KC =Kf

Kb=

[C]c[D]d

[A]a[B]b KP =

PCcPD

d

PAaPB

b

Thermodynamic Definition:�ΔG�=�0

∆G = ∆G◦ +RTlnQ Q = reaction quotient


Q =

[C]c[D]d

[A]a[B]b

at equilibrium:

0 = ∆G◦ +RTlnKeq

lnKeq =−∆G◦

RT

Keq = e−∆G◦

RT

Types of Equilibrium:1. Homogenous Equilibrium - reactants and products are of the same phase

Example: 2HI(g) � H2(g) + I2(g) KC =[H2][I2]

[HI]2

2. Heterogenous Equilibrium - reacyants and products are present in different phases Example: CaCO3(g) � CaO(s) + CO2(g) KC = [CO2] ; KP = PCO2

KC does not include concentration terms of pure solids/liquids

AgCl(s) � Ag+(aq) + Cl−(aq) KSP = [Ag+][Cl−] KSP = solubility product constant

KSP is for partially soluble salts

CH3COOH +H2O � H3O

+ + CH3COO−

Ka =

[H3O+][CH3COO−]

[CH3COOH]

- any mixture with acetic acid is always in equilibrium - Ka�=�acid�dissociation�constant� Kb�=�base�dissociation�constant

Factors Affecting EquilibriumLe Chatelier's Principle- when a stress factor is applied to a system in equilibrium, the system will shift to relieve the stress�(relieve�the�balance)� factors:�changes�in�concentration,�P,�V,�T,�addition�of�catalyst

2A(g) +B(g) � P(g) + heat

Stresses:Addition of A: shift to the RIGHT Removal of A: shift to the LEFT Addition of B: shift to the RIGHT Removal of B: shift to the LEFT Addition of P: shift to the LEFT Removal of P: shift to the RIGHT

↑P (resulting in ↓V)�shift�ot�the�RIGHT� � ↓P (resulting in ↑V)�shift�to�the�LEFT�↑P,�shift�to�the�lesser�number�of�moles)� � ↓P,�(shift�to�the�higher�number�of�moles)

↑T: shift to the LEFT ↓T shift to the RIGHT↑T favors endothermic processes ↓T favors exothermic processes

catalyst: no shift in equilibrium, they affect both the forward and backward reactions

Example:Fe3+︸︷︷︸orange

+CNS−︸︷︷︸colorless

� FeCNS2+︸︷︷︸blood red

Stresses:Addition of Fe(NO3)3: darker (↑[Fe3+])Addition of KSCN: darker (↑[CNS-])Addition of AgNO3: Ag+�+�SCN- ⇒�AgSCN�(white�precipitate)�LEFT�(↓[CNS-])Addition of NaF: lighter (forms FeF6

3-)�(↓[Fe3+])Addition of KCl: lighter (forms FeCl6

3- (↓[Fe3+])�or�by�dilution)� or no shift: KCl does not react with the species

CH3COOH +H2O � H3O+ + CH3COO−


Stresses:Addition of NaCH3COO: NaCH3COO ⇒ Na+�+�CH3COO- LEFT(common-ion�effect:�presence�of�a�common�ion�suppresses�the�ionization�of�an�electrolyte)Addition of HCl: HCl ⇒ H+�+�Cl-�LEFT�(common-ion�effect)Addition of NaOH: NaOH ⇒ Na+�+�OH- (neutralizes H3O

+)�RIGHTAddition of NaCl: no shift

CaCO3(g) � CaO(s) + CO2(g)

Stresses:Addition of CO2: LEFTAddition of CaO: no shift (solid, is not involved in Keq)Addition�of�Inert�Gas�(constant�V):�↑PTOTAL

Finding KC from equilibrium concentrations

1. Given: CO(g) + Cl2(g) � COCl2(g)

Equilibrium Concentrations: CO : 1.2× 10−4M Cl2 : 0.054M COCl2 : 0.14M

KC =[COCl2]

[CO][Cl2]=

(0.14M)

(1.2× 10−4M)(0.054M)= 216.05

2. Given: H2(g) + I2(g) � 2HI(g)

Initial Concentrations: H2 : 1× 10−3M I2 : 2× 10−3M

Equilibrium Concentrations: HI : 1.87× 10−3M

ICE tableH2 I2 HI

I 1× 10−3 M 2× 10−3 M 0 MC* -x M -x M +2x ME 1× 10−3 − x M 2× 10−3 − x M 2x M

*C-row follows stoichiometric coefficient

[HI] = 1.87× 10−3M = 2x x = 9.35× 10−4M

[H2] = 1× 10−3 − 9.35× 10−4M = 6.5× 10−5M

[I2] = 2× 10−3 − 9.35× 10−4M = 1.065× 10−3M

KC =[HI]2

[H2][I2]=

(1.87× 10−3M)2

(6.5× 10−5M)(1.065× 10−3M)= 50.51

Finding equilibrium concentrations from KC

3. Given: H2(g) + I2(g) � 2HI(g)

Initial Concentrations: H2 : 1mol I2 : 2mol in 1.0L

KC = 50.5

H2 I2 HII 1 M 2 M 0 MC -x M -x M +2x ME 1-x M 2-x M 2x M

KC =[HI]2

[H2][I2]

50.5 =(2x)2

(1− x)(2− x)

46.5x2 − 151.5x+ 101 = 0

x = 2.92 x = 0.93

[HI] = 2× 0.93 = 1.86M

[H2] = 1− 0.93M = 0.07M

[I2] = 2− 0.93M = 1.07M


4. Given: COCl2(g) � CO(g) + Cl2(g)

Initial Concentrations: COCl2 : 3.50× 10−3M CO : 1.11× 10−5M Cl2 : 3.25× 10−6M

KC = 1.19× 10−10

Is the system in equilibrium?

QC =[CO][Cl2]

[COCl2]=

(1.11× 10−5M)(3.25× 10−6M)

(3.50× 10−3M)= 1.19× 10−8

Note:

If KC > QC , there are more reactants than in equilibrium (forward shift)

If KC = QC , there is equilibrium

If KC < QC , there are more products than in equilibrium (backward shift)

KC < QC , therefore, there is no equilibrium

COCl2(g) CO(g) Cl2(g)I 3.50× 10−3 M 1.11× 10−5 M 3.25× 10−6 MC +x M -x M -x ME 3.50× 10−3 + x M 1.11× 10−5 − x M 3.25× 10−6 − x M

KC =[CO][Cl2]

[COCl2]

1.19× 10−10 =(1.11× 10−5M)(3.25× 10−6M)

(3.50× 10−3M)

x = 1.12× 10−5 x = 3.15× 10−6

[COCl2] = 3.50× 10−3M

[CO] = 7.95× 10−6M

[Cl2] = 1.00× 10−7M

Ionic Equilibrium- between weak acids and weak bases - any solution of a weak acid or a weak base is in equilibrium

Example 1:

CH3COOH +H2O � H3O+ + CH3COO−

Ka =

[H3O+][CH3COO−]

[CH3COOH]

Ka�=�acid�dissociation�constant

Ka for CH3COOH�=�1.8x10-5

- it is a small value, which means that the concentration of the products is much less than the concentration of the reactants

Example 2:

NH3 +H2O � NH4+ +OH−

Kb =[NH3OH][OH−]

[NH3]

Kb�=�base�dissociation�constantKb for NH3�=�1.8x10

-5

Ka from % ionization

1. Given: HA+H2O � H3O+ +A−

0.010 M HA; 2.0% ionization

HA H3O+ A−

I 0.01 M 0 M 0 MC -(0.02×0.01) M +(0.02×0.01) M +(0.02×0.01) ME 9.8× 10−3 M 2× 10−4 M 2× 10−4 M

Ka =[H3O

+][A−]

[HA]=

(2× 10−4M)(2× 10−4M)

(9.8× 10−3M)= 4.08× 10−6


1. Given: HA+H2O � H3O+ +A−

0.010 M HA; 2.0% ionization

HA H3O+ A−

I 0.01 M 0 M 0 MC -(0.02×0.01) M +(0.02×0.01) M +(0.02×0.01) ME 9.8× 10−3 M 2× 10−4 M 2× 10−4 M

Ka =[H3O

+][A−]

[HA]=

(2× 10−4M)(2× 10−4M)

(9.8× 10−3M)= 4.08× 10−6

% ionization from Ka

2. Given: CH3COOH +H2O � H3O+ + CH3COO−

0.010 M CH3COOH; Ka = 1.8× 10−5

CH3COOH H3O+ CH3COO−

I 0.01 M 0 M 0 MC -x M +x M +x ME 0.01-x M x M x M

Ka =[H3O

+][CH3COO−]

[CH3COOH]

1.8× 10−5 =(x)(x)

0.01− x

Rounding off: IfC

Keq≥ 100, then you can round off

Therefore:

1.8× 10−5 =x2

0.01

x =√(1.8× 10−5)(0.01)

= 4.24× 10−4 = [H3O+]

%ionization =ionized form

unionized form× 100 =

4.24× 10−4

0.01× 100 = 4.24%

Note: kay sir Quiming lang yata pwede yung round-off thing

pH = −log[H3O+] pOH = −log[OH−]

autoionization of H2O: H2O + H2O � H3O+ + OH−

at 25◦C, [H3O+] = [OH−] = 1.0× 10−7M

KW = equilibrium constant for H2O

= [H3O+][OH−]

= (1.0× 10−7M)(1.0× 10−7M)

= 1.0× 10−14

−logKW = −log[H3O+] +−log[OH−]

14 = pH + pOH

1. pH of strong acids and bases: HCl → H+ + Cl− full ionization

Given: 0.1 M of HCl

pH = −log[0.1M ] = 1


2. pH of weak acids and bases

Given: HCN +H2O � H3O+ + CN−

0.2 M HCN; Ka = 4.9× 10−10

HCN H3O+ CN−

I 0.2 M 0 M 0 MC -x M +x M +x ME 0.2-x M x M x M

Ka =[H3O

+][CN−]

[HCN ]

4.9× 10−10 =(x)(x)

0.2− x

0.2

4.9× 10−10≥ 100, therefore you can round off

4.9× 10−10 =x2

0.2

x =√(4.9× 10−10)(0.2)

= 9.90× 10−6 = [H3O+]

pH = −log[H3O+]

= −log(9.90× 10−6M)

= 5.00

Note: if roundable: pH = -log√Ka × C

3. Salts: formed by an acid plus a base a. Strong Acid + Strong Base = Neutral Salt Example: HCl +NaOH → NaCl +H2O Na+ is a metal Cl-�is�a�very�weak�base�(conjugate�of�a�very�strong�acid)

b. Weak Acid + Strong Base = Basic Salt Example: CH3COOH +NaOH → NaCH3COO +H2O Na+ is a metal CH3COO- : CH3COO− +H2O � CH3COOH +OH−

c. Strong Acid + Weak Base = Acidic Salt Example: HCl +NH4OH → NH4Cl +H2O NH4

+ : NH4+ +H2O � NH3 +H3O

+ Cl-�is�a�very�weak�base�(conjugate�of�a�very�strong�acid)

d. Weak Acid + Weak Base = Acidic/Basic/Neutral - it is dependent on the Ka of the acid and the Kb of the salt Ka�=�Kb, Neutral Salt

Example: CH3COOH +NH4OH → NH4CH3COO +H2O

Ka of CH3COOH = Kb of NH4OH Ka�>�Kb, Acidic Salt Ka�<�Kb, Basic Salt

chemistry 14 outline guide

Documents

ha h2

1s 2s 2p

0m

1s 2s

ch3 cooh

valence electrons needed

heat energy required

2s