CHM 1220 1st Edition Lecture 1 Outline of Last Lecture (Recap of CHM 1210 topics)1. Periodicity/Periodic Trends Ionization energy, electron affinity, electronegativityCharge of Cations/anions based upon group2. Naming & Polyatomic Ions (reference chpt. 4 pg 141)Recognize names and structure interchangeably3. Stoichiometric Calculations (reference chpt. 7 pg 266)Find/Use mole ratios from balanced chemical equationsLimiting reagent, percent yield, molar mass determination4. Intermolecular interactions (reference chpt. 6 pg 238)Hydrogen-bonding, dispersion forces, dipole-dipoleIonic vs. covalent bonding5. Oxidation-reduction chemistry (reference chpt. 8 pg 332)Determination of oxidation state of atomIdentify what is being oxidized/reduced and what the oxidizing/reducing agents are6. Electrolytes/Non-electrolytes (reference chpt. 8 pg 318)Identify strong/weak/non-electrolytesRecognize Brønsted-Lowry acids & bases7. Enthalpy (reference chpt. 9 pg 371)First Law of ThermodynamicsSystem, surroundings, universe and sign conventionsHess’s Law and other ways to calculate enthalpiesOutline of Current Lecture (Chapter 11 – Properties of Solutions):1. Enthalpy of solutions2. Lattice Energy3. Born-Haber Cyclea. Practice problem4. Calculation of U Current Lecture (Chapter 11 – Properties of Solutions):1. Enthalpy of Solution (pg. 465)i) Enthalpy of solutions occurs when solutes are dissolved in solvents.These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.∆Hsoln = ∆Hionic bonds + ∆Hdipole-dipole + ∆H ion-dipole2. Lattice Energy (U) (pg.466)i) Lattice energy forms its free ions in the gas phase and is the energy released from 1 mol of an ionic compound from its free ions in the gas phase. Crystal lattice is formedM+(g) + X-(g) → MX(s)*k = proportionality constant depending on lattice structure ∆Hsoln = ∆Hhydration + ∆Hionic bonds∆Hsoln = ∆Hhydration – U3. Born-Haber Cycle (pg.466)i) This is a series of steps to explain the formation of ionic solidsii) Steps with practice problemNa(s) + ½ Cl2(g) → NaCl(s) ∆Hf = -411.2 kJ(1) Sublimation of 1 mol Na(s) → Na(g)*This gives us ∆Hsub Hsub: Na(s) → Na(g)(2) Breaking bonds of ½ mol of Cl2(g)*This gives us ½ ∆HBE½Hbond energy: ½Cl2(g) → Cl(g) (3) Ionization of 1 mol Na(g) atoms*This gives us IE1 [ionization energy for the first element (Na)] HIE1: Na(g) → Na+(g) + e-(g)ExothermicEndothermicThis istheenergyrequired toremove ionsfromcrystallattice Key for Lattice Energy EquationQ1Charge of 1st ionQ2Charge of 2nd ionk* constantd Distance b/t 2 objects U=k⋅(Q1×Q2d)(4) Ionization of 1 mol Cl(g) atoms*This gives us EA1 HEA: Cl(g) + e-(g) → Cl-(g) (5) Formation of 1 mol NaCl(s) from ions(g)*Finally, this gives us U4. Calculation of U (lattice energy)i) Now that we have the information above in the red box, we can calculate the lattice energyii) Us the following equation to solve for UUNaCl = Hf - ½ HBE - HEA - Hsub - HIE1iii) Steps(1) Na+(g) + e-(g) → Na(g) -(HIE1) = -(495 kJ)(2) Cl-(g) → Cl(g) + e-(g) -(HEA) = -(-349 kJ)(3) Na(g) → Na(s) -(Hsub) = -(108 kJ)(4) Cl(g) → ½Cl2(g) -(½HBE) = -(½·242 kJ)(5) Na(s) + ½Cl2(g) → NaCl(s) Hf = -411 kJNa+(g) + Cl-(g) → NaCl(s)UNaCl = Hf - ½ HBE - HEA - Hsub - HIE1 = -786 kJU: Na+(g) + Cl-(g) →