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) H2S3) 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 itExamples: 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 DIAGRAMS1) 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 = mcDT2) 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 A2BRate = 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 LAWFirst 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 REACTANTSZero 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: +XX = 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 = 56.9J1) 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 ANSWERCreate an ICE table but only for Ag+ and SO4

2-

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 *

0.1Solve 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