CHEM 167 FINAL REVIEW Part 2
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CHEM 167 FINAL REVIEWPart 2
RESONANCE STRUCTURES
Compound that cannot be represented by only one Lewis structure.
Determine resonance structures:1) Ozone, O3
2) CO32-
3) Benzene, C6H6
RESONANCE ANSWERS
O3 has two resonance structures where the double bond switches between the 2 oxygens bonded to the central oxygen
CO32-
has three resonance structures where the double bond switches between the 3 oxygens bonded to the carbon, remember to write the structures in [ ] and include overall charge in upper right corner
C6H6 has two structures where double bonds
flip between carbons in an alternating pattern
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 orbitals2) There needs to be a hybrid orbital for each electron domain on the central atomExamples: What type of hybridization is present?1) O3
2) H2S
3) CO2
HYBRIDIZATION ANSWERS
O3 is sp2 because there are 3 electron domains on the central O
H2S is sp3 because there are 4 electron domains on S
CO2 is sp because there are 2 electron domains on C
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
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 top of valence band, dopant has less valence electrons than metal
N-type: on bottom of conduction band, dopant has more valence electrons than metal
Insulator: huge band gap, cannot be doped, conductivity nearly impossible
DRAW BAND DIAGRAMS
1) Aluminum2) P-doped Si3) N2
BAND DIAGRAMS ANSWERS
Aluminum is a metal/conductor so you would draw the conduction band directly on top of the valence band with no gap
Semiconductor in p-doped so you would draw the dopant on top of the valence band, because the semiconductor is Si you need the dopant to be from group 3 so that it will work (ex: B)
Nitrogen is an insulator and therefore will not conduct. Draw the conduction and valence bands apart with a large band gap
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 forcesExamples: 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 negative if the system is doing work. Work is positive 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 flow2 Types1) Constant pressure: coffee cupqcalorimeter = -qreaction qreaction = mcDT
2) Constant volume: bombqcalorimeter = 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
CALORIMETRY ANSWER
DE = qv = Cv DT = 10.17 * 5.67 = 57.66kJ
57.66 kJ is for the calorimeter, flip sign for qreaction
(-57.66kJ/1.435g) * (128.2g/1 mol) = -5,151 kJ/mol
DE units are kJ/mol so you needed to know the molar mass of the compound,128.2g/mol
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.
PHASE CHANGE ANSWER
You find out that you need to add up:Q=mcDT for ice, latent heat of ice, and Q=mcDT for water Ice:Q = (150g)(2.09J/gC)(18C) = 5643J 5.643kJQ = (150g/18g/mol)(6.01kJ/mol) = 50.08kJ WaterQ = (150g)(4.184J/gC)(25C) = 15690J 15.69kJ Answer: 71.4kJ
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 productsCalculated: bonds broken – bonds formedDH of bonds broken = positive because requires energyDH of bonds formed = subtracted because gives off energyExample: Calculate the bond dissociation energyH2 (g) + Cl2 (g) 2 HCl (g)
H—H:435kJ/mol, Cl—Cl:243kJ/mol, H—Cl:431kJ/mol
BOND DISSOCIATION ANSWER
Bonds broken - bonds formed (H-H + Cl-Cl) – (2*H-Cl) (435kJ + 243kJ) – (2*431kJ) Answer: -184kJ
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
HESS’S LAW ANSWER
Leave equations 1 and 2, flip equation 3 CH4O, CH2O, N2, and 3H2 cancel out Answer: -37 + -46 + 65 = -18kJ
DETERMINING RATE LAWS Instantaneous rate law: aA + bB cCRate = (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 0.4702 0.720 M 0.720 M 1.8803 0.360 M 0.180 M 0.117a) Determine the order with respect to each reactantb) Write the rate expression for the reaction.c) Find the value of the rate constant, k.
RATE LAW ANSWER
To determine the order of A use trials 1 and 3 because B is being held constant there. The change in A is by a factor of 2 and the rate change is by a factor of 4. This makes A second order
Use trials 1 and 2 to determine B. The change in B is by a factor of 4 and the rate changes by a factor of 4. This makes B first order.
Rate = K [A]2[B]
To find k pick a trial, k = 5.04
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 Zero 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/kSecond 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(K1/K2) = (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 stepslow reaction: H2+ IClHI + HCl k1 [H2] [ICl]
fast reaction: HI + IClI2 + HCl k2 [HI] [ICl]
REACTION MECHANISM ANSWER
To find the total reaction simply add up the reactants and products sides and take away compounds that appear on both sides.
Reaction : H2 + 2ICl I2 + 2HCl Intermediate is HI Rate determining step is the slow reaction
DYNAMIC EQUILIBRIUM
aA(g) + bB(g)cC(g) + dD(g)K = Kf/Kr Kc = [C]c[D]d/[A]a[B]b Kp = PC
cPDd/PA
aPBb
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 H2(g) + I2(g)
Start with 0.5M HI at equilibrium 0.0534M I2
DYNAMIC EQUILIBRIUM ANSWER
Create an ICE table Initial: HI: 0.5 M, I2: 0M, H2: 0M
Change: HI: -2X, I2: +X, H2: +X
X = 0.0534M Equilibrium: HI: 0.3932M, I2 and H2: 0.0534MKc = [I2][H2]/[HI]2
Kc = 0.018
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 right/products
If Q > K then reaction shifts to the left/reactants
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 product2) Change the volume or pressure changes moles of gaseous compounds3) Temperature change
exo: treat heat/energy as a productendo: treat heat/energy as a reactant
LE CHATELIER’S PRINCIPLEN2O4(G)2NO2(G) DH = 56.9J
1) NO2 is addeda. 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 increaseda. 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 initial [SO4
2-] = 0.1M Ksp = 1.5 * 10 -5
KSP ANSWER
Create an ICE table but only for Ag+ and SO42-
Initial: Ag+: 0M, SO42-: 0.1M
Change: Ag+: +2X, SO42-: +X
Equilibrium: Ag+: +2X, SO42-: 0.1M + X
Ksp = [Ag+]2 [SO42-] = [2X]2 [0.1+X] = 4X2 *
0.1
Solve for X 0.00612M, Ag+ = 2X = 0.012M
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 solubilitya) NaNO3 b) KBr c) CaCl2 d) Li2SO4
COMMON ION EFFECT ANSWERS
A) decrease B) increase C) decrease D) increase
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
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