Transcript Chem 167 Final Review - Iowa State University

CHEM 167 FINAL REVIEW
Part 2
RESONANCE STRUCTURES
Compound that cannot be represented by only
one Lewis structure.
Determine resonance structures:
1) Ozone, O3
2) CO323) Benzene, C6H6

SHAPES OF MOLECULES
Draw out the molecule in a Lewis dot structure
 Pay attention to the lone pairs that could be
present on the central atom.
 Lone pairs push bonds closer together and
farther away from the lone pair
 Pay attention to if you are being asked for the
electron configuration shape or molecular shape
An example of this would be NH3 just by looking at
domains it has a tetrahedral shape, but the lone
pair makes the molecular shape trigonal pyramidal

HYBRIDIZATION OF MOLECULES
Rules for the number of hybrids created:
1) The number of hybrids is equal to the number of
combined orbitals
2) There needs to be a hybrid orbital for each
electron domain on the central atom
Examples: What type of hybridization is present?
1) O3
2) H2S
3) CO2

OVERLAPPING OF ORBITALS
Bonds are formed by the overlapping of orbitals
also called constructive interference
 These bonds formed are sigma and pi bonds
 Sigma (s) bonds are formed by s-orbitals
overlapping and p-orbitals overlapping end-to-end
 Pi (p) bonds are formed by p-orbitals overlapping
on their sides and also contains a sigma bond
 A single bond is made of a sigma bond, a double
bond contains a sigma and pi bond, and a triple
bond contains 2 pi bonds and a sigma bond
 Sigma bonds exist in the middle and pi bonds exist
above and below the sigma bond

POLARITY OF MOLECULES
A molecule is polar if there is a partially positive
and partially negative area to it
Examples: CH3Cl, IF5, H2O
 A molecule is nonpolar if its charges are balanced
out and cancel
Examples: CO2, CH4
 Draw out these examples and see why they are
polar or nonpolar

PHASE DIAGRAMS
Demonstrate how a substance changes with
pressure and temperature
 Know how to read a general phase diagram
(know the sections and what the lines represent)
 Know the phase diagram for carbon, especially
the split between solid carbon where it is
diamond and graphite

CUBIC UNIT CELLS AND HCP
HCP: hexagonal close packing, has max
coordination number = 12; packing pattern:
ABAB
 Simple cubic : packing efficiency is 1 atom,
coordination number is 6 because it touches 6
other cells
 Body-centered cubic: packing efficiency is 2
atoms, coordination number is 8
 Face-centered cubic: packing efficiency is 4
atoms, coordination number is 12, which is the
maximum coordination number close packed.

BAND DIAGRAMS AND P-N JUNCTIONS
Bands are made up of infinite atoms. Conduction
band is made of anti-bonding orbitals. Valence
band is made of bonding orbitals.
 Metals (conductors): no band gap  conduction
 Semi-conductors: band gap, can be doped (p-type or
n-type) to decrease this gap and allow conduction
 P-type: on bottom of conduction band, dopant has
less valence electrons than metal
 N-type: on top of valance band, dopant has more
valence electrons than metal
 Insulator: huge band gap, cannot be doped,
conductivity nearly impossible

DRAW BAND DIAGRAMS
1) Aluminum
2) P-doped Si
3) N2
INTERMOLECULAR FORCES
Inside of a molecule not a bond
 London dispersion forces: present in all
molecules, due to electrostatic attractions
(random motion and temporary dipole)
 Polarizability: greater in larger molecules
because of more electrons and stronger dispersion
forces.
 Dipole-dipole: present in polar molecules, scales
with molecular polarity, stronger than dispersion
 Hydrogen bonding: between H and N,O, or F
only, reason for water’s high BP

VAPOR PRESSURE AND SURFACE TENSION
Vapor pressure: equilibrium between evaporation
and condensation, increases as temperature
increases, weaker intermolecular forces lead to
higher vapor pressure
 Surface tension: due to intermolecular forces
Examples: meniscus vs. water droplet
 Melting/Boiling point: low vapor pressure  high
MP and high BP. High surface tension high MP
and BP (from strong intermolecular forces)

POLYMERS, POLYMERIZATION, AND COPOLYMERS
Polymerization: ways of creating polymers, you
need to know two.
 Addition polymerization: initiation step (free
radical), propagation step (need C=C),
termination step (combine free radicals)
 Condensation polymerization: -OH of alcohol and
H combine to create H2O as byproduct
 Polymer types: isotactic, syndiotactic, atactic
 Copolymer types: alternating, block, graft
 Additives: plasticizers, pigments, fire retardants,
stabilizers

INTERNAL ENERGY AND P-V WORK
Made up of heat (q) and work (w), apply
magnitude of vectors in a diagram
 Heat: Exothermic is negative and heat/energy is
released from system to surroundings.
Endothermic is positive and heat/energy is
absorbed by system.
 Work: Work is positive if the system is doing
work. Work is negative if work is done on system
by surroundings.
 P-V work: If volume of products is greater than
reactants then work is done by system and is
positive. If volume is products is less than
reactants then work is negative.

CALORIMETRY
Calorimeter measures heat flow
2 Types
1) Constant pressure: coffee cup
qcalorimeter = -qreaction
qreaction = mcDT
2) Constant volume: bomb
qcalorimeter = Cv DT c = calorimeter constant
qcalorimeter = -DEreaction

Example: 1.435g C10H8 is combusted in a bomb
calorimeter what is DEreaction in kJ? Ti=20.28C and
Tf=25.95C Cv=10.17 kJ/C
PHASE CHANGES
Heating/cooling curve: areas of slope and latency
 Slope: q=mcDT; c is dependent on stage of matter
 Latency: where melting and vaporization occur
q=n* DHvap/fus
Example:
How much energy (in kJ) is required to melt 150.0 g of ice
from -18.00 C and bring the resulting liquid water up to
25.00 C? Specific heats: gas = 1.84 J/gC; liquid = 4.184
J/gC; solid = 2.09 J/gC. DHvap = 40.7 kJ/mol DHfus = 6.01
kJ/mol.

ENTHALPY/ENTROPY/GIBB’S FREE ENERGY
Enthalpy: measure of heat/energy. Positive =
endothermic. Negative = exothermic
 Entropy: measure of chaos or randomness of a
system
Both are calculated as Snproducts – Snreactants
 Gibb’s free energy: measure of spontaneity of a
reaction equal to DH – TDS. DG < 0 = spontaneous
 Know the table of how the sign on DH and DS will
give a spontaneous or nonspontaneous reaction or
if it is spontaneous only as certain temperatures.

BOND DISSOCIATION ENERGY
Standard enthalpy change in a reaction as reactants
turn to products
Calculated: bonds broken – bonds formed
DH of bonds broken = positive because requires
energy
DH of bonds formed = subtracted because gives off
energy
Example:
Calculate the bond dissociation energy
H2 (g) + Cl2 (g)  2 HCl (g)
H—H:435kJ/mol, Cl—Cl:243kJ/mol, H—Cl:431kJ/mol

HESS’S LAW
Way of finding the enthalpy of a reaction by
applying and manipulating known enthalpy values
of known reactions
Example: Find the ΔH for the reaction below:
N2H4(l) + H2(g)2NH3(g)

N2H4(l) + CH4O(l)CH2O(g) + N2(g) + 3H2(g) ΔH = -37 kJ
N2(g) + 3H2(g)2NH3(g)
ΔH = -46 kJ
CH4O(l)CH2O(g) +H 2(g)
ΔH = -65 kJ
DETERMINING RATE LAWS
Instantaneous rate law: aA + bB  cC
Rate = (1/c)(D[C]/Dt)=-(1/a)(D[A]/Dt)=-(1/b)(D[B]/Dt)
 Rate expression: 2A + B  A2B
Rate = K[A]x[B]y; x and y are the orders of the reactants, can only
be determined through experiment
 Overall order of a reaction is the sum of the orders of the
reactants.
Example:
For the reaction A + B AB , the following data were obtained.
Trial Initial [A]
Initial [B]
Initial Rate
1
0.720 M
0.180 M
0.470
2
0.720 M
0.720 M
1.880
3
0.360 M
0.180 M
0.117
a) Determine the order with respect to each reactant
b) Write the rate expression for the reaction.
c) Find the value of the rate constant, k.

INTEGRATED RATE LAW
First order: ln [A]t = -kt + ln [A]0, will produce a
straight line on a graph with y-axis: ln [A]t and
x-axis: t the slope=-k
 Second order: 1/ [A]t = kt + 1/ [A]0, will produce a
straight line on a graph with y-axis: 1/ [A]t and
x-axis: t the slope = k
 Third order: [A]t = kt + [A]0, will produce a
straight line on a graph with y-axis: [A]t and
x-axis: t the slope = k
 Integrated rate laws are in y=mx + b format

HALF-LIFE OF REACTANTS
Zero order: t1/2 = [A]0/2k
 First order: t1/2 = ln2/k = 0.693/k
 Second order: t1/2 = 1/k[A]0
Questions given for these will be extremely straight
forward and all you will need to do is insert values

ACTIVATION ENERGY AND ARRHENIUS EQUATION
K = A e ^ (-Ea/RT)
 Use this modified version to find Ea:
ln(K2/K1) = (Ea/R)(1/T2 – 1/T1)
 Again questions involving these equations will be
straight forward. Just makes sure to keep the K
and T values together that go in a pair.

REACTION MECHANISM
Mechanism is made up of elementary steps
 Rate determining step: one step will be the slowest
step and this is the rate determining step of the
reaction
 Molecularity: the molecularity of an elementary can
be determined by the number of different species that
make up the reactants.
Unimolecular>bimolecular>>termolecular
Example:
Write the total reaction, identify intermediates, and
pick out rate determining step
slow reaction: H2+ IClHI + HCl k1 [H2] [ICl]
fast reaction: HI + IClI2 + HCl k2 [HI] [ICl]

DYNAMIC EQUILIBRIUM
aA(g) + bB(g)cC(g) + dD(g)
K = Kf/Kr Kc = [C]c[D]d/[A]a[B]b Kp = PCcPDd/PAaPBb

K must be calculated at equilibrium and only for
compounds in the gaseous or aqueous state
 May need to construct an ICE table to calculate K
Example: Calculate Kc for the following reaction:
2HI 2H2(g) + 2I2(g)
Start with 0.5M HI at equilibrium 0.0534M I2

ACID IONIZATION CONSTANT
Calculated as [products]/[reactants]
 Summed acid dissociation reaction = multiplied
individual ionization constants
 Can also find an elementary step constant in the
total acid dissociation by dividing the quotient by
individual constant(s)

REACTION QUOTIENT (Q)
Is calculated the same way as an equilibrium
constant, but can be calculated with
concentrations taken at any point in the reaction
not just at equilibrium
 If Q < K then reaction shifts to the left/reactants
 If Q > K then reaction shifts to the right/products

LE CHATELIER’S PRINCIPLE
Provides ways that a system at equilibrium
moves/shifts to offset a stress or disturbance on
the system
Disturbances:
1) add/remove reactant or product
2) Change the volume or pressure  changes moles
of gaseous compounds
3) Temperature change
exo: treat heat/energy as a product
endo: treat heat/energy as a reactant

LE CHATELIER’S PRINCIPLE
N2O4(G)2NO2(G) DH = 56.9J
1) NO2 is added
a. Equilibrium will shift to consume N2O4 (g).
b. Equilibrium will shift to produce more NO2 (g).
c. Equilibrium will shift to consume the NO2 (g).
d. No effect on the equilibrium.
2) P is lowered by increasing V
a. Produce more N2O4 (g) to offset the pressure drop.
b. Shift to the right to produce more NO2 (g).
c . Shift to consume more NO2 (g).
d. No effect on the equilibrium.
3) Temperature is increased
a. Equilibrium will shift to the left.
b. Equilibrium will shift to the right.
c. Equilibrium will shift to produce more heat.
d. No effect on the equilibrium.
SOLUBILITY PRODUCT CONSTANT
Constant is calculated as an equilibrium
constant, use an ICE table
 Solid salts are not included, acids are included
 Can calculate pH from acid when Ka is given [H+]
 Change in the initial concentration (x) can be
treated as negligible when calculating constant
Example:
 Calculate M of Ag+ ions in solution of Ag2SO4
with equilibrium [SO42-] = 0.1M Ksp = 1.5 * 10 -5

COMMON ION EFFECT
If the same ion is added to a solution in which the
ion is already present it decreases the solubility of
the compound already present.
Example:
If NaCl is dissolved in solution determine which
compounds will increase or decrease NaCl solubility
a) NaNO3
b) KBr
c) CaCl2
d) Li2SO4

BRONSTED-LOWRY ACIDS AND BASES
Acid: a proton donor
 Base: a proton acceptor
 Conjugate acid: acid formed from base’s accepted
proton
 Conjugate base: base formed by acid donating
proton
