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Chapter 5 - Gasses
Gasses take the shape/volume of container, compress, mix with other gasses, exert pressure on surroundings
Measuring Pressure
Units: kiloPascals (kPa), torr (mmHg), atmospheres (atm) = 101.3kPa = 760 torr
Barometer
Manometer
Air pressure pushes down on a pool of liquid
Air pressure and gas pressure push down on liquid inside a U
mercury, forcing it to rise up a narrow tube.
shaped tube. The difference in height of the liquid on both
1mm of Hg=1 torr.
sides of the U indicates the pressure of the gas.
Gas Laws
Boyle’s Law
 A gas’s volume at constant
temperature is inversely
related to its pressure.
 As pressure increases,
volume decreases
 PV=k
 P1V1=P2V2
Charles’s Law
 A gas’s volume at constant
pressure is directly related
to its Kelvin temperature.
 As temperature increases,
volume increases.
 VT=b
 V1T1=V2T2
Gay-Lussac’s Law
 A gas’s pressure at constant
volume is directly
proportional to its Kelvin
temperature.
 As temperature rises,
pressure rises.
 P1/T1=P2/T2
Combined Gas Law
P1V1/T1=P2V2/T2
Ideal Gas Law
PV=nRT
 P=atm OR kPa
 V=Liters
 n=moles
 R=0.0821 L atm/mol K OR
8.31 L kPa/mol K
 T=Kelvins
Molar Mass from PV=nRT
 n=g/MM
 PV=(g/MM)RT
 MM=gRT/PV
Graham’s Law
 Effusion = gas particles
escaping through a small
opening
 The rate of effusion of a gas
is inversely proportional to
the square root of its molar
mass.
 (rateA)(sqrt(MMA)) =
(rateB)(sqrt(MMB))
 (rateA)/(rateB) =
(sqrt(MMB))/(sqrt(MMB))
 Typically, A is denoted the
gas with the smaller molar
mass
Van Der Waals Equation
[P+a(n/v)2] (V-nb) = nRT
 Real gas particles have
attractions so their actual
pressure is less than that of
ideal gasses. Add an
additional value to account
for this error.
 Real gas particles occupy
volume, and reduce the
volume available in the
container. Subtract a value
to correct this error.
 Real gasses approach ideal
behavior at high
temperature and low
pressure.
Avogadro’s Law
 The volume of a gas at
constant temperature and
pressure is directly
proportional to moles.
 V1/n1=V2/n2
Dalton’s Law
 The total pressure of gasses
in a container is the sum of
the pressures that the
gasses would exert alone
 Ptotal=P1 + P2 + P3...
 Ptotal=(ntotalRT)/V
 X1=n1/ntotal
 X1=P1/Ptotal
To find partial pressure
1. Find total pressure
2. Find mole fraction of each
gas
3. Multiply mole fraction of
each gas by total pressure.
Page 1 of 17
Kinetic Molecular Theory
1.
2.
3.
4.
Particles are in constant random motion and collide to cause pressure.
Volume of particles is negligible.
There are no attractive or repulsive forces between particles.
Average kinetic energy is directly proportional to Kelvin temperature.
Intuition behind Gas Laws using KMT
Boyle’s Law (P, V)
Charles’s Law (V, T)
As the volume of a container
As the temperature of a gas
decreases, the gas particles collide
increases, the energy of particles
with the walls more offten,
increases, forcing the container to
increasing pressure.
expand.
Gay-Lussac’s Law (P, T)
As the temperature of a gas
increases, the energy of the
particles increases, so they collide
with the walls more.
Chapter 6 - Thermochemistry
Vocabulary
Energy
Kinetic energy
Potential energy
Chemical energy
System
Surroundings
Exothermic reaction
Endothermic reaction
First law of thermodynamics
Temperature
Heat
the ability to do work or produce heat
the energy of motion
the energy of position/composition
the heat transfer that accompanies a chemical reaction
the part of the universe which we focus on
everything else in the universe
a reaction in which heat is given off by the system
a reaction in which heat absorbed by the system
energy in the universe is constant AKA...
law of conservation of energy: energy cannot be created or destroyed
a measure of the kinetic energy in a system based on the random motion of
particles
the transfer of energy between two objects due to difference in temperature.
Units
Joule (J)
= 𝐿 × 𝑘𝑃𝑎 =
𝑘𝑔 × 𝑚2
𝑠2
Calorie (cal)
quantity of heat that raises the temperature of 1 gram of water by 1°C; equal
to 4.184 joules
Kilocalorie (Cal) (kcal)
1000 cal
Kilojoule (kJ)
1000 kJ
Chemical Energy
In chemical reactions, energy is either lost or gained. This change is based on the potential energy of the
reactants and the products. In exothermic reactions, the potential energy of the reactants exceeds that of the
products, so energy is lost as heat is given off. In endothermic reactions, the potential energy of the products
exceeds that of the reactants, so energy is gained and heat is absorbed.
Page 2 of 17
Internal energy (E) is the sum of all the kinetic energy and potential energy in a system. ∆E is the change in
internal energy due to change in heat or work.
∆E=q+w
q=heat and w=work
-q
+q
-w
-w
system is exothermic and loses heat
system is endothermic and gains heat
work is done by system which loses energy
work is done on system which gains energy
Chemical Work
Involves the expansion of gasses, not the pushing of boxes up hills
P=F/A and F=PA
W=FD
W=(PA)D
W=P(AD)
-W=P(V)
W(J)=-P(kPa) ∆V(L)
Because I’m Bad at Exothermic/Endothermic
Exothermic
∆H is negative
Heat is released
∆H is written on the reactant side of the equation
Heat is evolved
Heat exits the system and enters the surroundings
Work is done by the system (gas expands)
Endothermic
∆H is positive
Heat is absorbed, required
∆H is written on the reactant side of the equation
Heat is involved
Heat enters the system and exits the surroundings
Work is done on the system (gas is compressed)
Enthalpy
Enthalpy (H) = E + PV
H = E + PV ∆E = qp + PV
∆H = ∆E + PV qp = ∆E + PV
H = qp
At constant pressure (qp) the change in enthalpy of the system is equal to the heat of the reaction.
Thermochemical Equations
Exothermic Reaction (heat produced, given off)
Endothermic Reaction (heat absorbed)
CH4 + 2O2  CO2 + 2H2O (∆H = -890kJ)
CH4 + 2O2  CO2 + 2H2O + 890kJ
H2 + 2C + N  2HCN (∆H = +135kJ)
H2 + 2C + N + 135kJ  2HCN
*Negative signs are never used in thermochemical equations.
Calculations with Thermochemical Equations
Determine the amount of heat absorbed by 32.5 g of carbon in H2 + 2C + N + 135kJ  2HCN.
32.5𝑔𝐶 ×
𝑚𝑜𝑙
135𝑘𝐽
×
= 183𝑘𝐽
12.01𝑔𝐶 2𝑚𝑜𝑙 𝐶
Page 3 of 17
Calorimetry
This is the science of measuring heat. The heat change of a chemical reaction is measured with a calorimeter. A
coffee-cup calorimeter consists of 2 cups inside each other, filled with water. The heat absorbed by the water, as
indicated by its change in temperature, is equal to the heat released by the heated metal or reaction.
Heat Capacity
ℎ𝑒𝑎𝑡 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑
This is a property of a substance described as𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒.
𝐽
𝐽
𝐽
𝐽
Specific Heat Capacity: 𝑔℃ or 𝑔 𝐾
Molar Heat Capacity: 𝑚𝑜𝑙℃ or 𝑚𝑜𝑙 𝐾
∆𝐻
Specific Heat (C) = 𝑔∆𝑇  ∆𝐻 = 𝑚𝐶∆𝑇
Calculations with Metals
1. Use the equation ∆H=mC∆T to solve for unknown variables.
Calculate the temperature for mercury if 160g of the metal absorb 1500J of heat.
𝐶=
∆𝐻
1500
=
= 67℃
𝑚∆𝑇 160 × 0.14
Calculation with Chemical Reactions
1. Use the equation ∆H=mC∆T to solve for the change in enthalpy of the resulting solution (product). The
number of grams of solution is the same as the grams of reactant because mass is constant.
2. Multiply the ∆H of water by -1 to find the ∆H of the reaction.
3. Divide the ∆H of the reaction by the number of moles of the limiting reagent. The limiting reagent limits
the extent of the reaction.
Calculate the ∆H for the reaction of 25.0mL of 1.0M HCl with 25.0mL of 1.0M NaOH, both at 25°C, if the
temperature of the solution rises to 32°C. The density of the water is 1g/mL.
We know that there are 50g of reactant, so there must be 50g of solution.
𝐽
∆𝐻𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = (50𝑔) (4.18
) (7℃) = 1500𝐽
𝑔℃
∆𝐻𝑟𝑥𝑛 = −1.5𝑘𝐽
Neither of the reactants is the limiting reagent, so we can find the moles of either reactant. There are 0.025
moles of HCl, so we divide -1.5kJ/.025 = 60kJ/mol.
Hess’s Law
This law states that: from a set of reactants to a set of products, the enthalpy change is the same whether the
reaction occurs in one step of in a series of steps. We can add two reactions, and then add the corresponding
enthalpy changes. These 2 rules that must be followed:
1. If you reverse the direction of a chemical equation you must change the sign of ∆H.
2. If you multiply each coefficient by a number, you must multiply ∆H by that number.
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2 B(s) + 3 H2(g) → B2H6(g)
Use the following information:
2 B(s) + 3/2 O2(g) → B2O3(s)
B2H6(g) + 3 O2(g) → B2O3(s) + 3 H2O(g)
H2(g) + ½ O2(g) → H2O(l)
H2O(l) → H2O(g)
2 B(s) + 3/2 O2(g)
3 H2(g) + 3 ½ O2(g)
B2O3(s) + 3 H2O(g)
3 H2O(l)
2 B(s) + 3 H2(g)





B2O3(s)
3 H2O(l)
B2H6(g) + 3 O2(g)
3 H2O(g)
B2H6(g)
∆H = –1273 kJ
∆H = –2035 kJ
∆H = –286 kJ
∆H = +44 kJ
H = –1273 kJ
H = (–286 kJ) 3
H = +2035 kJ
H = (+44 kJ) 3
ΔH = + 36 kJ
Standard Enthalpy of Formation
1. Write the chemical equation.
2. Look up the enthalpy of formation of each compound in a reference table.
3. Add all the values together using PRODUCTS minus REACTANTS!
Chapter 7 - Atomic Structure and Periodicity
Electromagnetic Radiation
Radiant energy that exhibits wavelike behavior and travels at the speed of light
As energy increases, frequency increases and wavelength decreases
Waves
Wavelength Distance between two consecutive peaks or troughs; lambda
Frequency
Waves per second passing a given point; nu
Meters (m)
Hertz (s-1)
Page 5 of 17
Electromagnetic spectrum
Particle Wave Duality
In the early 1700’s Issac Newton propsed that that light is a particle emitted from objects. Though he had not
experiment or evidence, people believed him because he was Issac Newton. Then in 1801, Thomas Young
performed the double slit experiment and observed constructive and destructive interference patterns, and
determined that light is a wave. In 1900: Max Planck observed effects of blackbody radiation, and named the
smallest particle of energy a “quantum” and said that energy can only be gained or lost in whole number
amounts of quanta. His formula for calculating the energy of one quantum was E=hν, where h is planck’s
constant (6.626 x 10-34). In 1905, Albert Einstein determined that light is also quantized particle, and named a
quantum of light a “photon”. He used the formula E=mc2 to calculate the mass of a photon, implying that energy
has mass. In 1923, Louis deBroglie determined that matter can act as a wave, and used a combined the above
formulas to produce λ=h/mν. deBroglie wavelengths can be calculated with this formula. We can conclude that
energy is a form of matter, which exhibits both wave and particle properties.
Hydrogen Spectrum and Bhor Model of the Atom
When light is projected through a prism, a continuous spectrum is observed, meaning all colors of the
electromagenetic spectrum are seen. However, when light from a single element is viewed, a line spectrum is
observed, meaning that only particular wavelengths are visible. This is a result of the fact that energy is
quantized. According to the Bhor Model of the atom, electrons are located in different “levels”, and move
between the levels (from ground state to excited states) by absorbing and emitting light, but this energy is
quantized so only wavelengths associated with those particular amounts of energy can be seen in a line
spectrum. The wavelengths of the light associated with electron transitions can be derived from the formula
𝑍2
𝐸 = −2.178 × 10−18 (𝑛2
𝑓𝑖𝑛𝑎𝑙
𝑍2
− 𝑛2
). Then, λ=hc/E can be used. Bhor’s equation only works for hydrogen, so Z,
𝑖𝑛𝑖𝑡𝑖𝑎𝑙
the nuclear charge, is always equal to 1. N is the integer corresponding to the energy level of the electron.
Quantum Mechanical Model of the Atom
A man named Shrodinger, treating electrons as waves, solved an equation to get wave functions and predict the
location of electrons. The wave functions were orbitals, regions with 90% probability of holding an electron.
Orbitals are characterized by 4 quantum numbers: principal, angular momentum, magnetic, and spin.
Page 6 of 17
1. The value principal quantum number (n) corresponds to the size and energy of the orbital, which is
determined by the element’s period on the periodic table. The value of n can be any integer greater
than 1.
2. The value of the angular momentum quantum number (l) corresponds to the shape of the orbital.
Although this quantum number is represented by a numeral, it is represented by a letter (s, p, d, f) on
the periodic table. The value of l can be any integer greater than 0 but less than n. (l cannot equal n)
3. The value of the magnetic quantum number(ml) corresponds to the orientation of the orbitals and the
number of orbitals. The value of ml can be any integer between –l and +l.
4. The value of the spin quantum number(ms) corresponds to the direction of the electron’s spin. Each
orbital can contain two electrons, one of each spin direction. The value of ms can be either ½ or -½.
The Pauli Exclusion Principle states that in an atom, each electron has a set four unique quantum numbers,
stated in the form (n, l, ml, ms)
The first two quantum numbers can be combined to form the names of specific orbitals: 1s, 2s, 2p etc.
Electron Configuration
According to the Aufbau Principle, each occupied orbital is listed in order of increasing energy. Energy levels are
not necessarily listed in order due to shielding, where electrons closer to the nucleus block out electrons in
outer orbitals. Instead, the periodic table is followed to obtain:
1s2, 2s2, 2p2, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6, 6s2, 4f14, 5d10, 6p6
The superscripts for any electron configuration should add up to the number of electrons in the atom.
Hund’s Rule: electron spins are aligned in the lowest energy configuration. Don’t pair them unless necessary.
Shorthand electron configuration is abbreviated using the previous noble gas in brackets to obtain:
Titanium: [Ar] 4s2, 3d2
Periodic Trends
Atomic Radius
This is defined as half the distance between two identical nuclei. The metallic radius half the distance between
two metal atoms, and the covalent radius is half the distance between two atoms in a diatomic molecule.
As we move down a group, atomic radius increases. This is because more “levels” are added. As we move across
a period, atomic radius decreases because protons being added pull electrons closer to them. Even though
electrons are added, the protons added have a greater impact on atomic radius.
Ionization Energy
This is defined as the energy needed to remove and electron from an atom in the gas phase. It takes a
considerably higher amount of energy to remove non-valence electrons.
As we move down in a group, ionization energy decreases. This is because valence electrons become farther and
farther away from the nucleus and its bind to the atom. As we move across a period, ionization energy increases
because electrons pulled more tightly by increasing numbers of protons and are closer to the nucleus.
Page 7 of 17
Electron Affinity
This is defined as the energy change when atoms gain electrons. Energy is given off so values are negative.
As we move down a group, less energy is given off because larger atoms are can more easily gain electrons. As
we move across a group, more energy is given off because smaller atoms can more easily gain electrons.
Adding an electron to Francium takes a lot of energy.
Electronegativity
This is defined as the ability of an atom to attract electrons in a bond.
As we move down a group, electronegativity decreases because the nucleus is farther from other atoms, and
cannot pull electrons toward it as strongly. As we move across a period, electronegativity increases because
additional protons can effectively attract electrons.
Heisenburg Uncertainty Principle
We cannot precisely know both the position and momentum of an electron at a given time.
History of the Periodic Table
Dobereiner (1817)
Newlands (1864)
Mendeleev (1872)
Moseley (1914)
triads
octaves
elements!)
by atomic number
Chapters 8 and 9 – Bonding
Types of Chemical Bonds
Electrostatic attraction between closely packed, oppositely charged metal and nonmetal ions, form ionic bonds.
The energy of ionic bonds can be calculated using Coulomb’s Law, where Q = the charge of each ion, and r =
distance between ions (nm): (Negative answer = attraction, positive answer = repulsion)
E = 2.31x10-19 J nm (Q1Q2/r)
For example, the energy of a NaCl bond, given r = 0.276nm, would be: 2.31x10-19 (1*-1/0.276) = -8.37x10-19J. The
negative answer indicates an attractive force, and that a bond forms if the system can lower its energy.
In covalent bonds two nuclei of nonmetals share one or more electrons. The forces between atoms include p+
to p+ and e- to e- repulsions and p+ to e- attractions. Atoms adjust their distance to achieve bond lengths that
minimize the sum of these energies.
Atomic bonds range from covalent to polar covalent to ionic. Differences in electro-negativity between two
atoms result in polar covalent bonds. Molecules with polar bonds that don’t directionally cancel each other out
have dipole moments and are polar molecules. For example the two polar bonds in CO2 point in opposite
directions so is no dipole moment present, and CO2 is not a polar molecule. On the other hand, H2O has a bent
shape, and the oxygen pulls electrons from both hydrogen atoms, so a dipole moment is present.
Page 8 of 17
Calculating Energy of Formation
There are two ways of doing this: standard energy of formation (∆𝐻𝑓° ), the change in energy when 1 mole of the
compound is formed from the elements in their standard states, and lattice energy, the change in energy when
separated gas ions combine to form an ionic compound.
M(s) + X2(g)  ½MX(s)
standard energy of formation (∆𝐻𝑓° ),
lattice energy
M+(g) + X-(g)  MX(s)
°
To Calculate (∆𝐻𝑓 ) for LiF using a lattice energy diagram: Li(s) + ½ F2(g) → LiF(s)
1
2
3
4
5
Li(s) → Li(g)
Li(g) → Li+(g) + e½ F2(g) → F(g)
F(g) + e– → F–(g)
Li+(g)+ F–(g)→ LiF(s)
∆H = 161 kJ/mol
∆H = 520. kJ/mol
∆H=(154 kJ)/2 =77 kJ
∆H = –328 kJ/mol
∆H=–1047kJ/mol
sublimation
ionization energy (IE1)
bond dissociation (F–F)
electron affinity
lattice energy
Covalent Bonds
Single bonds between two electrons are the lowest in energy, and have the longest bond length. Double bonds,
sharing 4 electrons, have intermediate bond energies and bond lengths. Triple bonds, sharing 6 electrons,
require the most energy, and have the shortest bond lengths.
Enthalpy Calculations
∆H= Σ E of bonds broken – Σ E of bonds formed
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Bonds Broken: 4 C–H, 2 O=O | Bonds Formed: 2 C=O, 4 O–H
4(C – H) +2(O=O) – 2(C=O) – 4(O – H)
4(413) + 2 (495) – 2(799) – 4(467) = -824kJ
Lewis Dot Structures
Lewis Dot Structures are used to show the arrangement of valence electrons, their quantity being equal to the
group number of the particular atom. According to the octet rule, 8 electrons are required to acquire a noble
gas configuration, filling the valence shell of an atom. Hydrogen, being an exception to this rule shares 2
electrons and makes only one bond. To draw a Lewis Dot Structure, follow these steps:
1. Find the total number of valence electrons
from all the elements in the compound.
2. Draw the central atom, allowing the most
symmetrical molecule
3. Connect each atom using a single bond.
4. Fill each atom’s valence shell with electron
pairs.
5. If there are not enough electrons to satisfy the
octet rule on each atom, make double or triple
bonds as needed.
CO2: Carbon = 4 electrons;
Oxygen = (6 electrons)x2=12. 16 total electrons
O C O
O–C–O
Count 6 for each oxygen, 4 for the carbon, and 2 for
each bond.
The above arrangement requires 20 electrons, but the
atoms only had 16 to start. Remove an electron pair
from each of an oxygen and a carbon, and create a
double bond. Repeat for the other CO bond.
Page 9 of 17
6. Remember that halogens NEVER form double
or triple bonds
Freshman Year Method
1. List each atom in the molecule, count the
valence electrons and bonds (shared
electrons) using the periodic table and the
octet rule. Make a chart for valence and
shared electrons. Account for extra electrons
due to charge. Valence + Shared =8 (octet rule)
O=C=O
SO32-:
S
O
O
O
2-
Valence
6
6
6
6
2
26
Shared
2
2
2
2
-2
6
=8
=8
=8
=8
=8
6/2=3 bonds
2. Choose an arrangement of atoms (carbon in
center, hydrogen or halogens on outside). And
add the calculated number of bonds; single
bonds first, then multiple bonds if necessary
3. Add electron pairs to complete all octets.
4. Check your work by comparing the number of
electrons drawn (including 2 per bond) to the
calculated number of valence electrons.
6+2+6+6+3(2) = 26
Exceptions to the Octet Rule
LESS than an octet
Be, B, Al can form electron
deficient molecules. Be = 2
bonds; B, Al =3 bonds.
The atoms in AlCl3 have 24
electrons total, so they form a
structure where Al only has 6.
MORE than an octet:
Nonmetals in periods 3 – 6 can
have 8, 10, or 12 electrons in
their valence shells. SF4 has 34
electrons, and forms a structure
with sulfur having 10 electrons
in its valence shell.
Molecular Shapes: VSEPR Model
The VSEPR (Valence Shell Electron Pair Repulsion) Model is used to determine the shapes of molecules based on
atom positions by minimizing electron repulsions. An occupied area can be an electron pair, a single bond, a
double bond, or a triple bond.
Occ. areas
2
3
3
4
4
4
5
5
5
5
Bonds/lone pairs
2 bonds
3 bonds
2 bonds, 1 long pair
4 bonds
3 bonds, 1 lone pair
2 bonds, 2 lone pairs
5 bonds
4 bonds, 1 lone pair
3 bonds, 2 lone pairs
2 bonds, 3 lone pairs
Example
BeCl2
BH3
SO2
CH4
NH3
H2O
PF5
SF4
ClI3
I3-
Shape name
Linear
Trigonal planar
Bent
Tetrahedral
Trigonal pyramidal
Bent
Trigonal bipyramidal
See-saw
T shaped
Linear
Bond angles
180°
120°
120° (<120°)
109°
107°
105°
90°, 120°
90°, 120° (<90, <120)
90° (<90°)
180°
Page 10 of 17
6
6
6
6 bonds
5 bonds, 1 lone pair
4 bonds, 2 lone pairs
SF6
BrF5
XeF4
Octahedral
Square pyramidal
Square planar
90°
90° (<90°)
90°
Polarity of Molecules
Molecules are polar if dipole moments are present. Shapes are not determining factors for polarity, for example,
bent molecules are not always polar. H2O is polar but O3 is not. In general, symmetrical molecules are nonpolar.
Resonance Structures
Resonance occurs when there are multiple Lewis structures for a molecule, meaning pairs of electrons can move
around stationary atoms. Sometimes all resonance forms will be equally stable, but often one is favorable over
the others. Formal charge is used to evaluate different resonance structures. The Lewis structure that has the
most formal charges closest to zero is the most stable.
Formal charge = (group number) – (number of electrons) – (number of bonds)
Bonding Theory: Hybridization
This is a theory based on the mixing of atomic orbitals to from new “hybrid” orbitals for bonding.





4 equivalent orbitals: s and p valence orbitals mix to form 4 sp3 orbitals, each composed of 1 part s and 3
parts p orbitals. (4 lobes) all sigma bonds.
3 equivalent orbitals: s and p valence orbitals mix to form 3 sp2 orbitals, each composed of 1 part s and 2
parts p, along with 1 unhybridized p orbital. (5 lobes) 3 sigma and 1 pi bond (involves double bond)
2 equivalent orbitals: s and p valence orbitals mix to form 2 sp orbitals, each composed of 1 par s and 2
parts p, along with 2 unhybridized p orbitals. (6 lobes) 2 sigma and 2 pi bonds
5 equivalent orbitals: one s, three p, and one d valence orbital mix to form 5 sp3d orbitals.
6 equivalent orbitals: one s, three p and two d valence orbitals mix to form 6 sp3d2 orbitals.
Only elements with d orbitals may have more than an octet.
To determine the type of hybrid orbital, first count the number of occupied areas. Then combine one s orbital
with as many p orbitals as needed, and add one or two d orbitals if necessary. Any unused p orbitals remain
unhybridized.
Bonding Theory: Molecular Orbitals
Each molecular orbital may hold 2 electrons, and represents the probability of finding an electron, calculated
from Schrödinger equation. A molecular orbital diagram shows atomic orbitals overlapping to produce MOs.
(1) Draw lines to represent atomic valence orbitals (s, p, etc) for both atoms in the bond.
(2) Fill the atomic orbitals with arrows to represent the number of electrons in each orbital.
(3) Draw lines to represent molecular orbitals below and above the middle of each atom.
a. The lines below and above represent bonding and antibonding, due to conservation of energy.
(4) Add the number of electrons in the atomic orbitals combined, and fill in the molecule orbitals with that
number of arrows.
Page 11 of 17
Bond Order
(# 𝑏𝑜𝑛𝑑𝑖𝑛𝑔 𝑒 − ) − (# 𝑎𝑛𝑡𝑖𝑏𝑜𝑛𝑑𝑖𝑛𝑔 𝑒 − )
2
(2) − (0)
(2) − (2)
= 1 (𝑠𝑖𝑛𝑔𝑙𝑒 𝑏𝑜𝑛𝑑)
𝐻𝑒2
= 0 (𝑛𝑜 𝑏𝑜𝑛𝑑)
2
2
Paramagnetic compounds have unpaired electrons and are attracted to magnetic fields. O2 is paramagnetic.
𝐻2 =
Diamagnetic compounds do not have unpaired electrons (all electrons are paired) and are repelled from
magnetic fields. Nitrogen is diamagnetic.
Page 12 of 17
Chapter 13 – Chemical Equilibrium
Equilibrium Constant
In reactions such as 2 SO2(g) + O2(g) ⇄ 2 SO3(g), 2 reactions occur simultaneously. Equilibrium is reached when the
rate of forward reaction is equal to the rate of reverse reaction. At this point, the concentrations of reactants
and products are constant (but not necessarily equal).
The Law of Mass Action, formulated by Cato Maximillian Guldberg and Peter Waage, states that:
for a reaction: jA + kB ⇄ lC + mD, the equilibrium expression(K) is:



([𝐶]𝑙 )([𝐷]𝑚 )
([𝐴]𝑗 )([𝐵]𝑘 )
= 𝐾, where brackets = mol/L (M)
When K > 1, there are more products at equilibrium, and it is said that equilibrium lies to the right.
When K = 1, [reactants] = [products] at equilibrium.
When K < 1 there are more reactants at equilibrium, and it is said that equilibrium lies to the left
Pressure Equilibria
K may also be written in terms of pressure. For N2(g) + 3 H2(g) ⇄ 2 NH3(g):
([𝑁𝐻3 ]2 )
= 𝐾𝑐
([𝑁2 ])([𝐻2 ]3 )
2
𝑃𝑁𝐻3
3 = 𝐾𝑝
𝑃𝑁2 𝑃𝐻2
Using PV=nRT, we can say that
P = (n/V)RT → P = (mol/L) RT
Derivation of Kp from Kc
To write an equilibrium expression for Kp, we would write:
2
𝑃𝑁𝐻3
3 = 𝐾𝑝
𝑃𝑁2 𝑃𝐻2
Because P = (n/V)RT, we can replace each “PA” with “[A]”:
([𝐶]𝑙 )(𝑅𝑇)([𝐷]𝑚 )(𝑅𝑇)
= 𝐾𝑝
([𝐴]𝑗 )(𝑅𝑇)([𝐵]𝑘 )(𝑅𝑇)
Factoring out each (RT), we have:
([𝐶]𝑙 )([𝐷]𝑚 )
× (𝑅𝑇)(𝑙+𝑚)−(𝑗+𝑘) = 𝐾𝑝
([𝐴]𝑗 )([𝐵]𝑘 )
We can reduce this to:
𝐾𝑐 (𝑅𝑇)∆𝑛 = 𝐾𝑝
Heterogeneous Equilibria



Homogeneous Equilibria: all reactants/products in one phase
Heterogeneous Equilibria: reactants/products in more than one phase
Solids and liquids are not included in Keq
Reaction Quotient
Comparing the reaction quotient, Q, of a certain given state, (not necessarily equilibrium) to the equilibrium
constant, K, can help predict which way the reaction will shift. Q can be calculated using the formula:
([𝐶]𝑙 )([𝐷]𝑚 )
𝑄=
([𝐴]𝑗 )([𝐵]𝑘 )
If Q is larger than K, there is too much product (as the numerator is too large) and the reaction will shift to the
left, trying to generate more reactant. If Q is smaller than K, there is too much reactant, and the reaction will
shift to the right, trying to generate more product. If Q=K then the reaction is at equilibrium.
Page 13 of 17
ICE Charts
ICE Charts can be used to calculate the states of each component when a reaction is at equilibrium, using the
initial states of each component and the change each component undergoes.
Given initial and equilibrium states; asked to solve for K
PCl5 is put in a flask at 0.50 atm and 523 K. At equilibrium, the pressure of PCl5 is 0.16 atm. Calculate Kp.
Initial
Change
Equilibrium
PCl5(g)
0.50
-0.34
0.16
⇄
PCl3(g)
0
+0.34
0.34
+
Cl2(g)
0
+0.34
0.34
𝐾𝑝 =
(0.34)(0.34)
= 0.72
(0.16)
Given initial state and K; asked to solve for equilibrium state
Calculate equilibrium concentrations if 1.00 mol of each component is mixed in a 1.00 L flask (Kc = 5.10)
Initial
Change
Equilibrium
CO(g)
+
H2O(g)
CO2(g)
+
H2(g)
⇄
1.00
1.0e 0
1.00
1.00
Q = 1.00 <5 .10 = K, so the reaction shifts right
-x
-x
+x
+x
1.00 – x
1.00 - x
1.00+x
1.00 – x
𝐾𝑝 =
(1.00 − 𝑥)(1.00 − 𝑥)
= 5.10
(1.00 + 𝑥)(1.00 + 𝑥)
Take the square root of both sides
and solve for x
Steps to Solve Equilibrium Calculations
1. Write a balanced equation
2. Write the equilibrium expression
3. Write the initial concentrations (ICE chart)
4. Calculate Q and determine the direction of shift
5. Define the change and list the equilibrium concentrations or pressures (ICE chart)
6. Substitute into expression and solve for K
7. Check answers to make sure they give the correct K value (?)
Shortcut (to avoid solving quadratic equations)
The value of x in the “–x” maybe be considered negligible and ignored in calculating the equilibrium constant if:


The equilibrium constant, K, is less than 10-3
The calculated X is less than 5% of the initial concentration (or pressure)
Given that K=4.66 x 10-3
I
C
E
N2O4(g)
0.800
(-x)
0.800 (–x)
⇄
2 NO2(g)
0
+2x
2x
(2𝑥)2
= 4.66 × 103
0.800
x = 0.0305M
LeChatelier’s Principle
If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to alleviate that stress.
There are 3 ways to apply stress: change the amount (or concentration) of one component, change the pressure
of the atmosphere by changing the volume of the container, or change the temperature of the environment.
Page 14 of 17
Change in Amount (Concentration)
 If more reactant is added, the reaction adjusts by creating more product, shifting the equilibrium right.
 If more product is added, the reaction adjusts by creating more reactant, shifting the equilibrium left.
 If a component is added, the equilibrium shifts to reduce the concentration of that component.
Change in Pressure
 Adding an inert (noble) gas does not affect on the position of the equilibrium (even with more particles)
 If the size of the container is decreased, the equilibrium will shift to favor the side with fewer moles.
Change in Temperature
 In an exothermic reaction, heat is considered a product. If heat is added, the reaction will shift left.
 In an endothermic reaction, heat is considered a reactant. If heat is added, the reaction will shift right.
Chapters 14 and 15 – Acids and Bases
Types of Acids
Arrhenius Acids produce protons (H+ ions) in water, while Bronsted-Lowry Acids donate protons in water.
Arrhenius Bases produce OH- ions in water, while Bronsted-Lowry Bases accept protons in water. An acid
donates a proton to a base, forming a conjugate base/acid. This type of equation can be generalized as such:
HA(aq)+H2O(l) ⇄ A-(aq) + H3O+(aq)
Acid Strength
Ka, or the acid dissociation constant, describes the extent to which an acid will dissociate in water. A higher Ka
denotes an acid that dissociates more completely. Ka is written similarly to an equilibrium expression, omitting
the liquid water. Also, though H+ only exists in water as H3O+, it can be reduced to just H+ for convenience.
Strong acids dissociate completely in water, leaving essentially no “acid”. For this, the equilibrium lies far to the
right, and the reaction is assumed to reach completion, so the Ka value needs not to be calculated. The six strong
acids: HCl, HBr, HI, HNO3, H2SO4, HClO4, exhibit these properties, and yield inactive conjugate bases.
Weak acids do not dissociate completely in water, so the equilibrium lies far to the left. They have small Ka
values (ranging from 10-2 to 10-30), and yield weak conjugate bases. All acids, except the 6 strong acids, are weak.
Autoionization of Water
When water reacts with an acid, it accepts protons, but when it reacts with a base, it donates protons. For this,
water is considered amphoteric, meaning it can behave both like an acid and like a base. This can be written as:
H2O(l) + H2O(l) ⇄ H3O+(aq) + OH-(aq)
Ka = Kw = [H3O+][OH–] = 1.0 x 10-14 (always!)
Because, experimentally, at 298K, [H3O+] = [OH–] = 1.0 x 10-7 M
Polyprotic Acids
Polyprotic acids yield more than one (often 2-3) proton in water. For example, H2SO4 yields 2 protons in water.
Page 15 of 17
pH Scale
pH is a measure of acidity, equal to –log[H+]. (pOH can be calculated using –log[OH-]). Thus, pH + pOH = 14, just
as [H3O+][OH–] = 1.0 x 10-14
All of the following four items ([H3O+], [OH–], pH, pOH) can be calculated given any one of them.
pH of Acid Solutions
In solutions of strong acids, the molarity of the acid is equal to the molarity of the H+ ion. (100% dissociation)
In solutions of weak acids, the molarity of the H+ ion must be calculated using an ICE chart:
What is the pH of 1.0 M HNO2, given that Ka = 4.0x10–
4
?
+
HNO2(aq) ⇄
H (aq) +
NO2–(aq)
1.0
0
0
1.0 – x
x
x
x = [H+] = 0.020 M, so pH = 1.70
X is fairly easy to calculate:
 coefficients are always 1 for acid/base chem
 Ka is small enough for x to be negligible, so a
quadratic equation is not needed
Bases
Strong bases, (group 1 or 2 metal hydroxides), completely dissociate in water. 1.0 M NaOH 1.0M OH-, pH = 14
Weak bases commonly have a nitrogen atom attached to three groups (with one lone pair of electrons).
Examples: pyridine, aniline, methylamine, dimethylamine, trimethylamine
The dissociation of bases is measured by Kb, the base dissociation constant. To calculate the pH of a weak base,
use an ICE chart. A weak base will react with water to accept an H+, and leave an OH-.
CH3NH2 + H2O ⇄ CH3NH3 + OH-
Polyprotic Acids
Polyprotic acids yield more than on proton (H+) in water. For example, 3 protons will dissociate from H3PO4.
These protons each dissociate separately and each have their own Ka values. Ka1>Ka2>Ka3, meaning the first
proton to dissociate has the highest Ka, and each successive proton has a smaller Ka. Solving these problems
requires using one successive ICE chart for each proton that dissociates. It is also important to keep track of
initial concentrations in each ICE chart.
Take H3PO4, for example. In the first step, the H3PO4 completely dissociates into H+ and H2PO4-. In the second
step, H2PO4- dissociates into H+ and HPO42-. The initial concentrations of H+ AND H2PO4 are equal to its final
concentration found from the first ICE chart, the initial concentration of HPO42- is 0. In addition, the value of“x”
in the ICE chart is considered negligible, indicating two things: pH can be based off of the final H+ concentration
solved in the first ICE chart, and that x=Ka2. In the third step, the initial concentrations are again transferred
from the final concentrations found in the second ICE chart.
Percent Dissociation
Percent dissociation is defined as the amount of a weak acid or base that is dissociated at equilibrium. (100x
amount dissociated/initial concentration) this basically X/I in an ICE chart.
Page 16 of 17
Hydrolysis of Salts
A salt is an ionic compound formed from an acid and a base. When a salt is placed in water one of the resulting
ions may react with water to alter the pH. The ions found in salts may come from (1) a strong acid and a strong
base, (2) a strong acid and a weak base or (3) a weak acid and a strong base.
(1) Ions that come from strong acid/strong base salts such as (Na+, Cl-) are “useless conjugate bases/acids”
that don’t react with water. Placing a strong acid/strong base salt into solution does not change the pH.
(2) Ions that come from weak acid/strong base (NaF) salts yield neutral conjugate acids (Na+) and weak
conjugate bases (F-) that react with water, taking H+ and forming OH-. This raises the pH of the solution.
a. NaF  Na+ + Fb. F- + H2O ⇄ HF + OHc. [HF][OH]/[F-]=Kb
i. Ka*Kb=Kw  Kb=Kw/Ka
d. [HF][OH]/[F-]=Kw/Ka
e. Solve for [OH-], find pOH, find pH
(3) Ions that come from strong acid/weak bases (NH4Cl) salts yield useless neutral conjugate bases (Cl-) and
weak conjugate acids (NH4+) that react with water, donating H+. This lowers the pH of the solution.
a. NH4Cl  NH4+ + Clb. NH4+ + H2O ⇄ NH3 + H3O+
c. [NH3][H+]/[NH4+]=Ka
d. Solve for [H+], find pH
Lewis Acids and Bases
Lewis acids accept electron pairs, while Lewis bases donate electron pairs. Things that have space for a lone
pairs (Be, B, Al) are often Lewis acids, while things that have lone pairs (NH3) are often Lewis bases.
Effect of Structure on Acidity
[to be continued…]
Page 17 of 17