Chemistry 116 Fall 2024 Lecture 17 PDF

10/7/24 Chemistry 116 Fall 2024 Lecture 17 Focus Area 4A : Thermochemistry Calorimetry 10...

10/7/24 Chemistry 116 Fall 2024 Lecture 17 Focus Area 4A : Thermochemistry Calorimetry 100 µm Glaucophane Na(Na,K,Ca) Paulingite 2(Mg,Fe) (Al,Si) 3Al23Si 2O242:17 H2O 8O22(OH) Half Dome, Yosemite 1 Seven elements exist at diatomic molecules at standard conditions 1 18 IA VIIIA 1A 8A 1 Periodic Table of the Elements 2 H Hydrogen 2 IIA 13 IIIA 14 IVA 15 VA 16 VIA 17 VIIA He Helium 1.008 2A 3A 4A 5A 6A 7A 4.003 3 4 5 6 7 8 9 10 Li Lithium Be Beryllium B Boron C Carbon N Nitrogen O Oxygen F Fluorine Ne Neon 6.941 9.012 10.811 12.011 14.007 15.999 18.998 20.180 11 12 13 14 15 16 17 18 Na Sodium Mg Magnesium 3 IIIB 4 IVB 5 VB 6 VIB 7 VIIB 8 9 VIII 10 11 IB 12 IIB Al Aluminum Si Silicon P Phosphorus S Sulfur Cl Chlorine Ar Argon 22.990 24.305 3B 4B 5B 6B 7B 8 1B 2B 26.982 28.086 30.974 32.066 35.453 39.948 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Potassium Ca Calcium Sc Scandium Ti Titanium V Vanadium Cr Chromium Mn Manganese FeIron Co Cobalt Ni Nickel Cu Copper Zn Zinc Ga Gallium Ge Germanium As Arsenic Se Selenium Br Bromine Kr Krypton 39.098 40.078 44.956 47.867 50.942 51.996 54.938 55.845 58.933 58.693 63.546 65.38 69.723 72.631 74.922 78.971 79.904 83.798 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Rubidium Sr Strontium Y Yttrium Zr Zirconium Nb Niobium Mo Molybdenum Tc Technetium Ru Ruthenium Rh Rhodium Pd Palladium Ag Silver Cd Cadmium In Indium SnTin Sb Antimony Te Tellurium I Iodine Xe Xenon 85.468 87.62 88.906 91.224 92.906 95.95 98.907 101.07 102.906 106.42 107.868 112.414 114.818 118.711 121.760 127.6 126.904 131.294 55 56 57-71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Cesium Ba Barium Hf Hafnium Ta Tantalum W Tungsten Re Rhenium Os Osmium Ir Iridium Pt Platinum Au Gold Hg Mercury Tl Thallium Pb Lead Bi Bismuth Po Polonium At Astatine Rn Radon 132.905 137.328 178.49 180.948 183.84 186.207 190.23 192.217 195.085 196.967 200.592 204.383 207.2 208.980 [208.982] 209.987 222.018 87 88 89-103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 Fr Francium Ra Radium Rf Rutherfordium Db Dubnium Sg Seaborgium Bh Bohrium Hs Hassium Mt Ds Rg Meitnerium Darmstadtium Roentgenium Copernicium Cn Uut Ununtrium Fl Flerovium Uup Lv Uus Uuo Ununpentium Livermorium Ununseptium Ununoctium 223.020 226.025 unknown unknown unknown unknown 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 Lanthanide Series La Lanthanum Ce Cerium Pr Nd Praseodymium Neodymium Pm Sm Promethium Samarium Eu Europium Gd Gadolinium Tb Terbium Dy Dysprosium Ho Holmium Er Erbium Tm Thulium Yb Ytterbium Lu Lutetium 138.905 140.116 140.908 144.243 144.913 150.36 151.964 157.25 158.925 162.500 164.930 167.259 168.934 173.055 174.967 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Actinide Series Ac Actinium Th Thorium Pa Protactinium U Uranium Np Neptunium Pu Plutonium Am Cm Americium Curium Bk Berkelium Cf Californium Es Einsteinium Fm Fermium Md Mendelevium No Nobelium Lr Lawrencium 227.028 232.038 231.036 238.029 237.048 244.064 243.061 247.070 247.070 251.080 257.095 258.1 259.101 © 2014 Todd Helmenstine sciencenotes.org 2 1  10/7/24 The specific heat is an intensive property When the heat capacity is divided by the mass, we get an intensive property: 18,140 J Substance Symbol (state) Specific Heat (J/g °C) cFe = = 0.45 J/ g oC helium He(g) 5.193 (800 g)(50 oC) water H2O(l) 4.184 ethanol C2H6O(l) 2.376 90,700 J ice H2O(s) 2.093 (at −10 °C) cFe = = 0.45 J/ g oC water vapor H2O(g) 1.864 (4000 g)(50 oC) nitrogen N2(g) 1.040 air 1.007 oxygen O2(g) 0.918 q aluminum Al(s) 0.897 c = carbon dioxide CO2(g) 0.853 (m)(DT) argon Ar(g) 0.522 iron Fe(s) 0.449 copper Cu(s) 0.385 q = m c DT lead Pb(s) 0.130 gold Au(s) 0.129 J o silicon Si(s) 0.712 == g C g oC Why do gases and liquids have much higher specific heats than metals? 3 Example Problems 1. If a beaker containing 800.0 g of water is heated from 21 oC to 85 oC, what quantity of heat has been added to the water? q = m c DT = (800.0 g) (4.184 J / g oC) (85 oC – 21 oC) = 214,220 J = 210,000 J = 210 kJ 2. Suppose a piece of an unknown metal with a mass of 217 g absorbs 1430 J of heat, and its temperature rises from 24.5 oC to 39.1 oC. What is the specific heat of the metal? q c = (m)(DT) 1430 J = = 0.451 J / g oC (217 g)(39.1 oC – 24.5 oC) 4 2  10/7/24 Calorimetry is a technique for the experimental measurement of heat changes heat released == q is “negative” heat absorbed == q is “positive” 5 A simple calorimeter can be highly effective! A 248 g piece of an unknown metal is heated to 314.0 oC and is dropped into a calorimeter containing 390 g of water, in which the water is initially at 22.6 oC. The final temperature is 38.7 oC. What is the specific heat of the metal? qlost = – qgained qlost, metal = – qgained, water m c DT (metal) = – [m c DT (water)] (248 g) (c) (38.7 oC – 314.0 oC) = – [(390 g) (4.184 J/g oC) (38.7 oC – 22.6 oC)] – [(390 g) (4.184 J/g oC) (16.1 oC)] c = (248 g) (– 275.3 oC) c = 0.39 J / g oC 6 3  10/7/24 Measurement of heat changes in chemical reactions Example : Heat of neutralization Experiment: 50.0 mL of 1.00 M HCl(aq) reacts with 50.0 mL of 1.00 M NaOH(aq). If the initial temperature of both solutions is 22.0 oC, and the final temperature is 28.9 oC, what is the heat of neutralization of HCl(aq) in J/mole? Assume that the specific heat of the solution is equivalent to the specific heat of water. The density of the solution is 0.9987 g / mL. HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) q = m c DT 0.9987 g solution 100.0 mL solution x o o = (99.87 g solution) (4.184 J / g C) (6.9 C) 1 mL solution = 2883 J = 2900 J = 99.87 g solution moles HCl = 50.0 mL 10 –3 L 1.00 mole HCl x x = 0.0500 mole HCl 1 mL 1L 2900 J Temperature of surroundings increased, so qsolution = = 58,000 J/mole = 58 kJ/mole 0.0500 mole HCl reaction is exothermic, and q = - 58 kJ 7 Heat of combution can be measured in a “bomb” calorimeter qrxn = – (qwater + qbomb) = – [ (m c DT)water + (C DT)bomb] 8 4  10/7/24 Example Problem Suppose that 0.963 g of liquid benzene (C6H6) undergoes combustion in a bomb calorimeter with C = 784 J/oC, and 925 mL of water. The temperature of the calorimeter and water increases by 8.39 oC as a result of the reaction. What is the heat of combusion of benzene in kJ/mole? C6H6(l) + O2(g) CO2(g) + H2O(l) C6H6(l) + O2(g) 6 CO2(g) + 3 H2O(l) C6H6(l) + 15/2 O2(g) 6 CO2(g) + 3 H2O(l) 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(l) qrxn = – (qwater + qbomb) = – [ (m c DT)water + (C DT)bomb] = – [(925 g) (4.184 J / g oC) (8.39 oC) + (784 J/oC) (8.39 oC)] = – (32,500 J + 6,580 J) = – (32,500 + 6,600 J) = –39,100 J = – 39.1 kJ 9 Benzene combustion, continued. 0.963 g C6H6 1 mole C6H6 x = 0.0123 mole C6H6 78.12 g C6H6 – 39.1 kJ = – 3180 kJ / mole !!! 0.0123 mole C6H6 10 5  10/9/24 Chemistry 116 Fall 2024 Lecture 18 Focus Area 4A : Thermochemistry Calorimetry The Enthalpy Function Thermochemical Equations Standard Enthalpy of Formation Orthoclase KAlSi3O8 Calcite, CaCO3 1 The halogens span three states of matter at standard conditions vaporization Fluorine Bromine sublimation Chlorine Iodine 2 1  10/9/24 Heat of combustion can be measured in a “bomb” calorimeter q lost = – q gained qrxn = – (qwater + qbomb) = – [ (m c DT)water + (C DT)bomb] 3 Example Problem Suppose that 0.963 g of liquid benzene (C6H6) undergoes combustion in a bomb calorimeter with C = 784 J/oC, and 925 mL of water. The temperature of the calorimeter and water increases by 8.39 oC as a result of the reaction. What is the heat of combusion of benzene in kJ/mole? C6H6(l) + O2(g) CO2(g) + H2O(l) C6H6(l) + O2(g) 6 CO2(g) + 3 H2O(l) C6H6(l) + 15/2 O2(g) 6 CO2(g) + 3 H2O(l) 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(l) q lost = – q gained qrxn = – (qwater + qbomb) = – [ (m c DT)water + (C DT)bomb] = – [(925 g) (4.184 J / g oC) (8.39 oC) + (784 J/oC) (8.39 oC)] = – (32,500 J + 6,580 J) = – (32,500 + 6,580 ) J = –39,100 J = – 39.1 kJ 4 2  10/9/24 Benzene combustion, continued. 0.963 g C6H6 1 mole C6H6 x = 0.0123 mole C6H6 78.12 g C6H6 – 39.1 kJ = – 3180 kJ / mole !!! 0.0123 mole C6H6 5 The conservation of energy is stated by the First Law of Thermodynamics DE = q + w == The change in INTERNAL ENERGY of a chemical system is the sum of the work done and the heat exchanged. Some conventions : Heat added TO the system == “positive” q Heat removed FROM the system == “negative” q Work done ON the system == ”positive” w Work done BY the system == “negative” w DE Key Point : The internal energy E is a STATE FUNCTION – the path of the change in internal energy does not matter !! 6 3  10/9/24 The Enthalpy is a state function describing "heat content" H = E + PV For chemical reactions at constant pressure, DP = 0, DH = DE + D(PV) = DE + PDV + VDP so VDP = 0 ! PDV == “Pressure – volume work” Compression : change in volume is negative (DV < 0) If the work done is positive when the work done on the system (w is positive) volume change is negative, then w = – PDV So, w = – PDV For systems at constant pressure DH = DE + PDV = q + w – w Because PDV = – w = q First Law of Thermodynamics: DE = q + w The heat of reaction at constant pressure is equal to the enthalpy change ! = q – PDV 7 Thermochemical equations relate enthalpy changes to chemical reactions H2(g) + ½ O2(g) H2O(l) DH = – 285.8 kJ (exothermic) H2(g) + ½ O2(g) H2O(g) DH = – 241.8 kJ (exothermic) The physical state of the reactants and products will affect the value of the enthalpy change !! C(s) + 2H2(g) + ½ O2(g) CH3OH(l) DH = – 239.2 kJ 2 C(s) + 4H2(g) + O2(g) 2 CH3OH(l) DH = – 478.4 kJ Enthalpy of reaction is an extensive property ! Standard enthalpy of formation == enthalpy of reaction for the formation of ONE MOLE of a substance from the elements in their STANDARD STATES H2(g) + ½ O2(g) H2O(l) DH f o = – 285.8 kJ (exothermic) A special type of notation for a specific type of enthalpy change ! 8 4  10/12/24 Chemistry 116 Fall 2024 Lecture 19 Focus Area 4A : Thermochemistry Calculation of Enthalpy of Reaction Orthoclase Almandine Garnet Fe2Al3(SiO4)3 Eclogite KAlSi3O8 Calcite, CaCO3 1 The Enthalpy is a state function describing "heat content" H = E + PV For chemical reactions at constant pressure, DP = 0, DH = DE + D(PV) = DE + PDV + VDP so VDP = 0 ! PDV == “Pressure – volume work” Compression : change in volume is negative (DV < 0) If the work done is positive when the work done on the system (w is positive) volume change is negative, then w = – PDV So, w = – PDV For systems at constant pressure DH = DE + PDV = q + w – w Because PDV = – w = q First Law of Thermodynamics: DE = q + w The heat of reaction at constant pressure is equal to the enthalpy change ! = q – PDV 2 1  10/12/24 Thermochemical equations relate enthalpy changes to chemical reactions H2(g) + ½ O2(g) H2O(l) DH = – 285.8 kJ (exothermic) H2(g) + ½ O2(g) H2O(g) DH = – 241.8 kJ (exothermic) The physical state of the reactants and products will affect the value of the enthalpy change !! C(s) + 2H2(g) + ½ O2(g) CH3OH(l) DH = – 239.2 kJ 2 C(s) + 4H2(g) + O2(g) 2 CH3OH(l) DH = – 478.4 kJ Enthalpy of reaction is an extensive property ! Standard enthalpy of formation == enthalpy of reaction for the formation of ONE MOLE of a substance from the elements in their STANDARD STATES H2(g) + ½ O2(g) H2O(l) DH f o = – 285.8 kJ (exothermic) A special type of notation for a specific type of enthalpy change ! 3 Thermochemical properties are tabulated by substance Example Problem : What is the chemical equation corresponding to the formation of liquid carbon tetrachloride from the elements in their standard states? C(s) + 2 Cl2(g) CCl4(l) DH = – 128.2 kJ DHo f = – 128.2 kJ / mole 4 2  10/12/24 Example Problem : Enthalpy Calculation When 1.34 g of zinc metal reacts with 60.0 mL of 0.750 M hydrochloric acid, 3.14 kJ of heat is produced and hydrogen gas is released. What is the enthalpy change for the reaction? a. Balanced chemical equation b. Enthalpy change in kJ/mole of Zn Zn(s) + HCl(aq) ZnCl2(aq) + H2(g) Zn(s) + 2 HCl(aq) ZnCl2(aq) + H2(g) 1 mole Zn 1.34 g Zn x = 0.0205 mole Zn 65.38 g Zn 0.0600 L HCl sol'n x 0.750 mole HCl = 0.0450 mole HCl | 0.0450 mole / 2 mole = 0.0225, so Zn is limiting 1 L HCl sol'n – 3.14 kJ DH = = – 153 kJ / mole Zn 0.0205 mole Zn 5 Hess’s Law gives us a method for calculating enthalpies Suppose we want to know the standard enthalpy of formation for FeCl3(s): Fe(s) + 3/2 Cl2(g) FeCl3(s) Fe(s) + Cl2(g) FeCl2(s) DHo = – 341.8 kJ FeCl2(s) + ½ Cl2(g) FeCl3(s) DHo = – 57.7 kJ Fe(s) + 3/2 Cl2(g) FeCl3(s) DHo = – 399.5 kJ 6 3  10/12/24 A bit more challenging example... Let's use experimental measurements of enthalpies of reaction to calculate the standard enthalpy of formation for AlCl3(s)... Al(s) + 3/2 Cl2(g) AlCl3(s) DHof = ? Experimental Measurements HCl(g) HCl(aq) DHo = – 75 kJ ½ H2(g) + ½ Cl2(g) HCl(g) DHo = – 93 kJ AlCl3(aq) AlCl3(s) DHo = + 323 kJ Al(s) + 3 HCl(aq) AlCl3(aq) + 3/2 H2(g) DHo = – 525 kJ 7 Putting the pieces together... Al(s) + 3/2 Cl2(g) AlCl3(s) 1. Al(s) + 3 HCl(aq) AlCl3(aq) + 3/2 H2(g) DHo = – 525 kJ 2. 3 HCl(g) 3 HCl(aq) DHo = – 225 kJ 3. 3/2 H2(g) + 3/2 Cl2(g) 3 HCl(g) DHo = – 279 kJ 4. AlCl3(aq) AlCl3(s) DHo = + 323 kJ Al(s) + 3/2 Cl2(g) AlCl3(s) DHo = – 706 kJ DHo f (AlCl3 , s) = -706 kJ / mole 8 4  10/12/24 Hess's Law can be extended to a more general form DHo f Elements in their standard states at 25 oC are reference states. CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) DHo = S DHof (products) – S DHof (reactants) DHo = [DHo f (CO2) + 2 DHo f (H2O)] – [2 DHo f (O2) + DHo f (CH4)] DHo = [(1 mole CO2)(-393.5 kJ/mole) + (2 mole H2O)(-285.8 kJ/mole)] – [2 mole O2)(0 kJ/mole)] + (1 mole CH4)(-74.6 kJ/mole)] 9 Be sure to keep track of your negative signs ! ! DHo = [ (1 mole CO2) (-393.5 kJ/mole) + (2 mole H2O) (-285.8 kJ/mole) ] – [ (2 mole O2) (0 kJ/mole) + (1 mole CH4) (-74.6 kJ/mole) ] DHo = [ (-395.5 kJ) + (-571.6 kJ) ] – [ ( 0 kJ) + (-74.6 kJ) ] DHo = [ - 967.1 kJ ] – [ - 74.6 kJ ] = - 892.5 kJ 10 5  10/12/24 Practice Problem Let's calculate the standard enthalpy change DHo for the reaction of solid lithium hydroxide with carbon dioxide gas to form solid lithium carbonate and liquid water. LiOH(s) + CO2(g) Li2CO3(s) + H2O(l) 2 LiOH(s) + CO2(g) Li2CO3(s) + H2O(l) DHo = S DHo f (P) –S DHo f (R) DHo = [DHo f (Li2CO3) + DHo f (H2O)] – [2 DHo f (LiOH) + DHo f (CO2)] DHof 11 Practice Problem, continued. DHo = [DHo f (Li2CO3) + DHo f (H2O)] – [2 DHo f (LiOH) + DHo f (CO2)] DHo = [ (1 mole Li2CO3) (-1216.0 kJ/mole) + (1 mole H2O) (-285.8 kJ/mole) ] – [ (2 mole LiOH) (-487.5 kJ/mole) + (1 mole CO2) (-393.5 kJ/mole) ] DHo = [ (-1216.0 kJ) + (-285.8 kJ) ] – [ (-975.0 kJ) + (-393.5 kJ) ] DHo = [ -1501.8 kJ ] – [ -1368.5 kJ ] = - 133.3 kJ 12 6  10/14/24 Chemistry 116 Fall 2024 Lecture 20 Lecture Topics Recap Orthoclase Eclogite KAlSi3O8 Calcite, CaCO3 1 1. Reaction Chemistry Chemical Reactivity Types Balance Chemical Equations Metathesis Reactions – Predict reactivity through solubility rules Reaction Stoichiometry Acid-Base Reactions – Predict reaction products by neutralization Limiting Reactants Oxidation-Reduction Reactions – Analysis through change in oxidation states 2. Volume and Density Conversions Convert between volume units Convert density between kg/m3 and g/cm3 3. Combustion Analysis Determination of empirical molar mass from masses of CO2 and H2O or CO2, H2O and NO2 Titration analysis to determine molar mass, and then true molecular formula 2 1  10/14/24 4. Thermochemistry Kinetic and potential energy First Law of Thermodynamics : Definitions of heat and work Enthalpy function : Equivalence of q and DH for reactions at constant pressure Thermochemical equations Definition of the Standard Enthalpy of Formation Hess’s Law Enthalpy of reaction DHo from tabulated values of DHfo 3 How do we determine uncertainty for a quotient or a product ? Suppose we want to determine the uncertainty in the density of a substance from laboratory measurements. In this case we would have a quotient. m = 7.629 ± 0.002 g d = 4.04 g/cm3 v = 1.89 ± 0.02 cm3 In order to propagate the error in this measurement, we must determine the relative uncertainties first. The reason is that we are DIVIDING measurements to get a different type of measurement, so we cannot simply add the squares of the absolute uncertainties as we did for a sum or difference. In this case, we must calculate the relative uncertainties in each measurement first, and then carry out the calculation of the uncertainty as before: e1 = 0.002 / 7.629 x 100 % = 0.02 % e = [ 0.02 % 2 + 1 % 2 ] ½ = 1.0004 ½ = 1% e2 = 0.02 / 1.89 x 100 % = 1 % So, in terms of the relative uncertainty : d = 4.04 g/cm3 ± 1 % The absolute uncertainty is then 0.01 x 4.04 g/cm3 = 0.04 g/cm3, so in terms of absolute uncertainty, d = 4.04 ± 0.04 g/cm3 4 2 Chemistry 116 Fall 2024 Lecture 21 Focus Area 1 : Atoms 1A : Electromagnetic Radiation 1B : Quantum Theory and the Bohr Model Calcite, CaCO3 Granite Half Dome, Yosemite Quantum theory starts with the visible emission spectrum of atomic hydrogen ! Balmer Series (visible) H2(g) 2H(g) 121.6 nm Key Observation : Only specific energies of Lyman Series light were observed to be emitted from the (ultraviolet) hydrogen atom !!  Light is composed of oscillating electric and magnetic fields Electromagnetic Spectrum James Clerk Maxwell visible range == 425 – 750 nm University College, London Rumford Medal, 1860 All electromagnetic radiation travels with the same speed in a vacuum Wavelength == meters / wave == l Frequency second == n == waves / == meters waves speed of light = c = l n == x wave second == meters / second = 3.00 x 10 m/s 8 Interactions between waves can produces interference patterns Nodal points in 1 dimension Nodal surfaces in 2 dimensions Practice Problems 1. What is the frequency of green light with l = 514 nm ? 514 nm = 514 x 10 –9 m c = l n c 3.00 x 10 m/s 8 = n = = 5.84 x 10 14 /s l 514 x 10 –9 m 2. What region of the electromagnetic spectrum would correspond to a frequency of 1.80 x 10 18 /s ? c = l n c 3.00 x 10 m/s 8 = l = = 1.67 x 10 – 10 m = 1.67 Å (recall that 1 Å = 10 – 10 m) n 1.80 x 10 18 /s 1 nm 1.67 x 10 – 10 m x = 0.167 nm 10 –9 m Max Planck showed that light is emitted in discrete "packets" classical physics prediction Max Planck Universität München Nobel Prize in Physics, 1918 For any temperature, there must be a specific energy at which light is emitted, which is directly proportional to the frequency: E=hn h = 6.626 x 10 – 34 J s == Planck's constant Five spectral series are observed for the hydrogen atom Light energy is emitted from the atom in an organized, predictable pattern ! The electron in the hydrogen atom must be losing energy, and this energy is released as electromagnetic radiation. Electromagnetic radiation is released as PHOTONS, which have both wave and particle properties. The electron must only be allowed to have a specific set of energies within the atom. Therefore, these energy states in the hydrogen atom are QUANTIZED. The energy of the light emitted by the atom must be equal to the difference between different energy states. The Bohr model of the hydrogen atom explains the emission spectra ! Energy levels must be more closely spaced with increasing energy !! Neils Bohr University of Copenhagen Nobel Prize in Physics, 1922 ℎ𝑐 !"# 1 1 ∆𝐸 = ℎ𝜐 = = 2.179 × 10 J %− % 𝜆 𝑛$ 𝑛& Practice Problems 1. Calculate the wavelength of light emitted when an electron in the hydrogen atom undergoes a transition from n = 3 to n = 1. To what region of the electromagnetic spectrum does this wavelength correspond? 2. A line in the Paschen series of the hydrogen emission spectrum has l = 1280 nm. To what region of the electromagnetic spectrum does this wavelength correspond? What are the initial and final states of the electron involved in this transition? Sure, but what about helium? H: 2-body problem He: 3-body problem  10/18/24 Chemistry 116 Fall 2024 Lecture 22 Focus Area 1: Atoms Quantum Mechanics The Wavefunction Atomic Orbitals S22- ions Pyrite, FeS2 Ammonite Calcite, CaCO3 Trilobite Half Dome, Yosemite 1 Practice Problems 1. Calculate the wavelength of light emitted when an electron in the hydrogen atom undergoes a transition from n = 3 to n = 1. To what region of the electromagnetic spectrum does this wavelength correspond? 2 1  10/18/24 2. A line in the Paschen series of the hydrogen emission spectrum has l = 1280 nm. To what region of the electromagnetic spectrum does this wavelength correspond? What are the initial and final states of the electron involved in this transition? 3 Sure, but what about helium? H: 2-body problem He: 3-body problem 4 2  10/18/24 If waves can behave like matter, can matter behave like a wave? Louis de Broglie University of Paris Nobel Prize in Physics, 1929 h l = mv Electron at 1.00 x 107 m/s l = 7.27 x 10 – 11 m = 72.7 pm = 0.0727 nm Baseball at 85.0 mph = 40.0 m/s l = 4 x 10 – 34 m ~ 4 x 10 – 22 pm ! Softball at 40.0 mph = 17.9 m/s l = 2 x 10 – 34 m ~ 2 x 10 – 22 pm ! 5 l How can we completely describe the behavior of an electron? Quantum (Wave) Mechanics – the laws of physics for very small objects with speeds close to the speed of light What we need is... Erwin Schrödinger Friedrich Wilhelm University Berlin A mathematical function, like Newton's laws but Nobel Prize in Physics, 1933 that will describe a particle emitting a wave... the WAVEFUNCTION To predict motion motion, energies of the electron To take into account complex potential energies For the mathematics to reduce to classical mechanics as objects get larger and speeds get lower Wolfgang Pauli Federal Technical Institute of Technology Zurich Nobel Prize in Physics, 1945 6 3  10/18/24 Quantum Mechanics starts with the Schrödinger Equation ℎ! 𝜕! 𝜕! 𝜕! "Eigenvalue Problem" == Solving for the value of Y will − + + + 𝑉(𝑥, 𝑦, 𝑧! Ψ = 𝐸 Ψ give back the wavefunction Y along with the energy 8𝜋 !𝑚 𝜕𝑥 ! 𝜕𝑦 ! 𝜕𝑧 ! "OPERATOR" calculates kinetic and potential energies Operator Algebra Many solutions – describe the "stationary states" of the hydrogen f(x) = x + 2 O = ( )2 atom O [ f(x) ] = (x+2)2 = x2 + 4x + 4 Must specify the potential energy expression f(x) = x2 O=Ö O [ f(x) ] = Ö x2 = ± x 7 The Quantum Numbers index the solutions to the Schrödinger Equation Key Point : Solutions to the Schrödinger equation tell us what types of motion the electron is ALLOWED to have ! First Solution : PRINCIPAL quantum number, n == the "energy level" quantum number Allowed values : n = 1, 2, 3... For 1 - electron atoms or ions (H, He+, Li2+), energy of Bohr model is reproduced EXACTLY !! n=7 397 nm BUT... this is as far as the Bohr model would go ! Balmer Series 8 4  10/18/24 Angular momentum describes the TYPE of motion the electron is allowed to have Second Solution : Angular momentum quantum number, Allowed values : = 0, 1, 2, 3... n-1 The number of values of that are allowed depends on the energy level AND is equal to the energy level number The specific value of defines a type of ORBITAL – a region of space in which the electron is ALLOWED to move 2p If = 1, the electron is in a p-type orbital n=2 = 0, 1 only : If = 0, the electron is in an s-type orbital 2s 1s n=1 = 0 only : If = 0, the electron is in an s-type orbital 9 Angular momentum, continued. If = 2, the electron is in a d-type orbital 3d 3p n=3 3s = 0, 1, 2 only 3d 2p Key Points: n=2 2s Atomic orbitals are mathematical functions (also called wavefunctions !) that describe the motion an electron is allowed to have based on its energy. The atomic orbitals along with their energies are solutions to the Schrödinger Equation. n=1 1s 10 5  10/21/24 Chemistry 116 Fall 2024 Lecture 23 Focus Area 1: Atoms Quantum Mechanics The Wavefunction Atomic Orbitals Electron Spin 1m Zebra Rock East Kimberly, Australia Earth and Space Chemistry 2023, 7, 2042-2049. Calcite, CaCO3 Half Dome, Yosemite 1 How can we completely describe the behavior of an electron? Quantum (Wave) Mechanics – the laws of physics for very small objects with speeds close to the speed of light What we need is... Erwin Schrödinger Friedrich Wilhelm University Berlin A mathematical function, like Newton's laws but Nobel Prize in Physics, 1933 that will describe a particle emitting a wave... Method : Differential Equations (difficult) the WAVEFUNCTION To predict motion motion, energies of the electron To take into account complex potential energies For the mathematics to reduce to classical mechanics (Newton’s Laws) as objects get larger Wolfgang Pauli and speeds get lower Federal Technical Institute of Technology Zurich Nobel Prize in Physics, 1945 Method : Matrix Mechanics (extremely difficult) 2 1  10/21/24 Quantum Mechanics starts with the Schrödinger Equation ℎ! 𝜕! 𝜕! 𝜕! "Eigenvalue Problem" == Solving for the value of Y will − + + + 𝑉(𝑥, 𝑦, 𝑧! Ψ = 𝐸 Ψ give back the wavefunction Y along with the energy 8𝜋 !𝑚 𝜕𝑥 ! 𝜕𝑦 ! 𝜕𝑧 ! Operator Algebra "OPERATOR" calculates kinetic and potential energies f(x) = x + 2 O = ( )2 O [ f(x) ] = (x+2)2 = x2 + 4x + 4 Many solutions – describe the "stationary states" of the hydrogen atom f(x) = x2 O=Ö Must specify the potential energy expression O [ f(x) ] = Ö x2 = ± x Key Points: We propose a wavefunction based on all the physics of We iterate on the wavefunction until the Ψ we put in the proton-electron interaction is the same as the Ψ we get out (convergence) We operate on the wavefunction with the energy operator In so doing, we generate many solutions to the (“Hamiltonian” operator) Schrödinger Equation (many Ψ , many E ) 3 Angular momentum describes the TYPE of motion the electron is allowed to have Second Solution : Angular momentum quantum number,