Chem 167 Final Review

1. Chem 167 Final Review Part 2

2. Resonance Structures • Compound that cannot be represented by only one Lewis structure. Determine resonance structures: 1) Ozone, O3 2)CO32- 3) Benzene, C6H6

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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.

11. 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

12. Draw Band Diagrams 1) Aluminum 2) P-doped Si 3) N2

13. 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

14. 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

15. 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)

16. 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

17. 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.

18. 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

19. Calorimetry Answer • DE = qv = CvDT = 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

20. 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.

21. 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  5.643kJ Q = (150g/18g/mol)(6.01kJ/mol) = 50.08kJ • Water Q = (150g)(4.184J/gC)(25C) = 15690J  15.69kJ • Answer: 71.4kJ

22. 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.

23. 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

24. Bond Dissociation Answer • Bonds broken - bonds formed • (H-H + Cl-Cl) – (2*H-Cl) • (435kJ + 243kJ) – (2*431kJ) • Answer: -184kJ

25. 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 

26. Hess’s Law Answer • Leave equations 1 and 2, flip equation 3 • CH4O, CH2O, N2, and 3H2 cancel out • Answer: -37 + -46 + 65 = -18kJ

27. 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.

28. 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

29. 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]tand x-axis: t the slope = -k • Integrated rate laws are in y=mx + b format

30. 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

31. 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.

32. 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 + HClk1 [H2] [ICl] fast reaction: HI + IClI2 + HCl k2  [HI] [ICl]

33. 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

34. 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 H2(g) + I2(g) Start with 0.5M HI at equilibrium 0.0534M I2

35. 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: 0.0534M • Kc = [I2][H2]/[HI]2 • Kc = 0.018

36. 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)

37. 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

38. 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

39. Le Chatelier’s PrincipleN2O4(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.

40. 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 [SO42-] = 0.1M Ksp = 1.5 * 10 -5

41. 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

42. 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

43. Common Ion Effect Answers • A) decrease • B) increase • C) decrease • D) increase

44. 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