CHEM 1211Exam # 3 Study Guide Lectures: 14-17Chapters: 5-7Lecture 14 (March 6)I. Hess’s law continued H°f1N2(g) + ½O2(G)  1N2O(g)1Na(s) + ½Cl2(g)  1NaCl(s)- Use standard statesEXOTHERMIC!H°rxn= Σn H°f products - Σn H°f reactants- Trying to get to low energy state1. H°C3H8= -104.7 KJ/molH°CO2= -393.5 KJ/molH°O2= 0.0 KJ/molH°H2O= -285.8 KJ/molCalculate the H°298 for this reaction:C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l)H°rxn: [(3)(-393.5) + (4)(-285.8)] – [(1)(-104.7)]H°rxn: -2,219 KJ/mol : EXOTHERMIC.II. Chapter 6: Atoms Electronic StructureIII. Color in Neon Lights, Fireworks, etc…Use: Barium, Copper, Strontium… molecular compounds (lightning bugs)A. Properties of Light (photon)- Osolating electric field - Wavelength: m, cm, nm, A- A = 1 x 10-10m or 1 x 10-8cm- Period: unit is seconds- Frequency: V = 1/PV= Frequency of Hertz (Hz)P= Period- Velocity: v= uƛƛ = wavelengthU= frequency in HzV= velocity in m/s or cm/sEX: VACUUM: 3.0 x 108 m/s or 3.0 x 1010 cm/sIV. Constant Frequency & Variable Wavelength- Frequency of photon is constantA. Photoelectric Effect: lights can strike the surface of some metals causing an e- to be ejected.Work fxn = Ф = hvo WAVE PARTICLE DUALITYB. Light & Energy: 1900 Max Planck (black body radiation)1. Energy is quantized2. Light has particle characterE = hv or E = hc / ƛ or E = nhvH= 6.626 x 10-34 J x SEphoton = Ʋ or 1/ƛLecture 15 (March 18)I. Electromagnetic RadiationA. Emission Spectrum: formed by electric current passing through a gas in a vacuumtube (at a very low pressure) which causes gas to emit lightEx: PrismLight  Prism  Detector- If we add lots of energy to a sample of elemental gas, we get an emission spectrum (elemental specific)1. In the hydrogen emission spectrum a band is observed at 4860A. What is the frequency and energy of this emission line?C = ƛv V = C / ƛ4860A 1 x 10 -10 m = 4.860 x 10-7m 1AV (visible: green light): 3.00 x 10 8 m/s = 6.17 x 1014 s-1 nHz 4.860 x 10-7mE= hvE= (6.626 x 10-34 J*s)(6.17 x 108 m/s) = 4.088 x 10-19 J / photon4.088 x 10 -19 J 6.02 x 10 23= 2.46 x 105 J / mol photon photon Mol photonE = 246 KJ / mol photonB. Absorption Spectrum: shine a bean of white light through gas- Indicates wavelengths of light that have been ABSORBEDDistanceC. Bohr Model of Atom: only works for hydrogen, explains atomic spectra- Energy of quantized state: - RhC / n2- N: integral numbers only ( 1 to infinity )- Radius of orbitals: n2 (0.0529nm)II. Predicting Emission Lines of HydrogenA. Rydberg Equation1 = RH 1 - 1ƛ n12 n22- RH : 1.097 x 107 m-1- RH: 1.097 x 10-2 nm-1- n refers to one of the infinite possible energy state of the electron in hydrogen- requirement: n1 < n2III. Electronic Transitions1. Move an electron from a lower energy level to a higher energy level, E= -RhC 1 - 1 n12 n222. Start a rotational mode of motion3. Begin a vibrations mode of motion4. Gain kinetic energy (translational motion) ex: waterA. Wave nature of electrons2πr = nƛDE Broglie said the wavelength of an object = h / mv2πr = nh / mv- Angular momentum = mvr – nh / 2πLecture 16 (March 20)I. Orbital Quantum Number- An atomic orbital is defined by three quantum numbersN: shell principleL: Angular Momentum Quantum NumberMl: Designates an orbital within a shell- Electrons arranged in shells and subshells made of orbitals- You cannot figure out Ml without L or L without N.1. Clicker Question One: Calculate Enthalpy H°f for H2S(g) given that;Enthalpy H°fO2: 0 KJ/molEnthalpy H°fSO2: -296.8 KJ/molEnthalpy H°fH2O: -285.8 KJ/molEnthalpy H°298: -1124 KJ/molAnswer: -20.6 KJ2. Clicker Question Two: Which is not a possible value for the principle quantum number?A. 4B. 2C. 1200D. 03. Clicker Question Three: Consider the third shell in Hydrogen. How many values ofthe angular momentum quantum number would be possible?A. 0B. 1C. 2D. 3E. 4A. The S Orbital- The 1st shell has n=1 (lowest energy) SO: l=0, ml=0- Since ml has a single value, the subshell has a single orbital - This subshell is labeled S and we call this orbital 1S: it has a spherical shape- State with the lowest energy state first when dealing with electronsB. The P Orbital- For n=2, l= 0 and 1- Two types of orbitals, 2 subshells- For l=0, ml= -1, 0, 1 (this is the p subshell with three orbitals- When l = 1, there is a planar node through the nucleus- Planar node: spot on orbital where the probability of find the electron is zero- Px = Py = Pz: Degenerate (have same energy)C. The D Orbital: third shell (2 planar nodes) - For l=2, ml= -2, -1, 0, 1, 2 (D subshell with five orbitals!)- Number of nodes = value of angular quantum number (l)- There are n2 orbitals in the nth shell.II. Spin Quantum Number- Last Quantum number is the spin quantum number, Ms- The spin quantum number values: Ms = +1/2 or -1/2- Quantum number tells us the spin and orientation of the magnetic field of the electron- Wolfgang Pauli (1925): Exclusion Principle- NO two electrons in an atom can have thesame set of four quantum numbers- Orbitals can only have two electrons max.- Doesn’t have to have two electrons though.1. Clicker Question Four: Which of these subshell types isn’t possible?A: 5s: n=5, l=0B: 4p: n=4, l=1C: 3d: n=3, l=2D: 2d: n=2, l=2E: 2p: n=2, l=1Lecture 17 (March 25)I. Using a Clicker for Spins- A: + ½ Spin- V: - ½ Spin- Separating Orbitals: AV_AVA. Clicker Question OneWhat quantum number would be used to determine the number of orbitals in a subshell?A. Angular Momentum Quantum NumberB. Magnetic Quantum NumberC. Principle Quantum NumberD. Spin Quantum NumberB. Clicker Question TwoThe angular momentum quantum number for a subshell is four. How many orbitals would we expect to find?A. 10B. 4C. 6D. 16E. 9C. Clicker Question ThreeWhat type of orbital is this?GD. Clicker Question FourWhen filling three degenerate orbitals with two electrons, you must first?A. Utilize Pauli Exclusion PrincipleB. Throw them in JailC. Understand DE Broglie’s PrincipleD. Apply Hund’s RuleII. Electron Configurations for HeliumZ=2Orbital Box Notation: 1s: Core electron2s: Valence/Outer Shell electronLithium:1s2 2s1 (Valence) [He] 2s1 (Condensed)Number of Valence Electrons: 1energy1s2s2px 2py 2pz1sIII. Hund’s Rule: Nitrogen: Z=71s2s2px2py2pz1s 2s 2pCore