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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
| 3Prof. Noel S. Quiming
4 | Chemistry 14: Fundamentals of General Chemistry I
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
| 5Prof. Noel S. Quiming
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:↓↑
Differentiating�Electron:�(2,�1,�+1,�-½)� � Valence�Electrons:�6
25Mn: 1s22s22p63s23p64s23d5
Ground state: 1s:↓↑ 2s:↓↑ 2p: ↓↑ ↓↑ ↓↑ 3s: ↓↑ 3p: ↓↑↓↑↓↑4s: ↓↑3d: ↓_ ↓_ ↓_ ↓_ ↓_�Differentiating�Electron:�(3,�2,�+2,�-½)� � Valence�Electrons:�2
51Sb: 1s22s22p63s23p64s23d104p65s24d105p3�=�[Kr]5s24d105p3
6 | Chemistry 14: Fundamentals of General Chemistry I
Differentiating�Electron:�(5,�1,�+1,�-½)� � Valence�Electrons:�5
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
| 7Prof. Noel S. Quiming
- 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
8 | Chemistry 14: Fundamentals of General Chemistry I
- 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
| 9Prof. Noel S. Quiming
#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-
10 | Chemistry 14: Fundamentals of General Chemistry I
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
| 11Prof. Noel S. Quiming
- 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
12 | Chemistry 14: Fundamentals of General Chemistry I
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: ↓_ ↓_ ↓_
| 13Prof. Noel S. Quiming
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: ↓_ ↓_ __ __ __
14 | Chemistry 14: Fundamentals of General Chemistry I
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: ↓_ ↓_ ↓_
| 15Prof. Noel S. Quiming
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
16 | Chemistry 14: Fundamentals of General Chemistry I
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
| 17Prof. Noel S. Quiming
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
18 | Chemistry 14: Fundamentals of General Chemistry I
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
| 19Prof. Noel S. Quiming
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)
20 | Chemistry 14: Fundamentals of General Chemistry I
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
| 21Prof. Noel S. Quiming
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%
22 | Chemistry 14: Fundamentals of General Chemistry I
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
| 23Prof. Noel S. Quiming
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.
24 | Chemistry 14: Fundamentals of General Chemistry I
-�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
| 25Prof. Noel S. Quiming
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)
26 | Chemistry 14: Fundamentals of General Chemistry I
-�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)
| 27Prof. Noel S. Quiming
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
28 | Chemistry 14: Fundamentals of General Chemistry I
� -�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
| 29Prof. Noel S. Quiming
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
30 | Chemistry 14: Fundamentals of General Chemistry I
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
| 31Prof. Noel S. Quiming
- 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
32 | Chemistry 14: Fundamentals of General Chemistry I
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
| 33Prof. Noel S. Quiming
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
34 | Chemistry 14: Fundamentals of General Chemistry I
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)
| 35Prof. Noel S. Quiming
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
36 | Chemistry 14: Fundamentals of General Chemistry I
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.
| 37Prof. Noel S. Quiming
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
38 | Chemistry 14: Fundamentals of General Chemistry I
� 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.
| 39Prof. Noel S. Quiming
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
40 | Chemistry 14: Fundamentals of General Chemistry I
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
∆Hrxn = Σ(n∆H◦f , products)–Σ(n∆H◦
f , reactants)
=
[(4mol)
(−285.6kJ
mol
)+ (2mol)
(−393.5kJ
mol
)]–
[(2mol)
(−239kJ
mol
)+ (3mol)
(0kJ
mol
)]
= −1451.4kJ
2C8H18 + 25O2 → 16CO2 + 18H2O
∆Hrxn = Σ(n∆H◦f , products)–Σ(n∆H◦
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
| 41Prof. Noel S. Quiming
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 + +
42 | Chemistry 14: Fundamentals of General Chemistry I
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
| 43Prof. Noel S. Quiming
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
44 | Chemistry 14: Fundamentals of General Chemistry I
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
| 45Prof. Noel S. Quiming
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
46 | Chemistry 14: Fundamentals of General Chemistry I
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−
| 47Prof. Noel S. Quiming
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
48 | Chemistry 14: Fundamentals of General Chemistry I
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
| 49Prof. Noel S. Quiming
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
50 | Chemistry 14: Fundamentals of General Chemistry I
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
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