CHEM 101 1nd Edition Lecture 21Outline of Last Lecture I. First Law of ThermodynamicsII. Enthalpy Outline of Current Lecture I. EnthalpyII. CalorimetryCurrent LectureI. Enthalpya. Consider the formation of wateri. H2(g) + ½ O2(g)  H2O(g) + 241.8 kJii. Exothermic reaction  Heat is a "product" and ΔH = -241.8 kJb. Examplei. The standard enthalpy of formation for AlCl3 (s) is the enthalpy change for what reaction?1. Al(s) + 3/5 Cl2(g)  AlCl3(s)a. Only 1 mol of AlCl3 is producedb. Reactants must be in elemental formi. Cl, N, O, F, Cl, Br, and I are diatomic when in elemental formc. Equation must be balanced.c. Using Enthalpyi. Making liquid H2O from H2 + O2 involves 2 exothermic steps1. First step ΔHfinal H2O2. Second Step H2O(g)  H2O(l) ΔHii. Example of Hess's Lawa. H2(g) + ½ O2(g)  H2O(g) -242 kJb. H2O(g)  H2O(l) -44 kJc. H2(g) + ½ O2(g)  HaO(l) -286 kJ2. Forming H2O can occur in a single step or in a two-step. ΔHtotal is the same no matter which path is taken.iii. Example1. Acetic acid is made by the reaction CH3OH(l) + CO(g) --> 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.CH3CO2H(l). Determine the enthalpy change for this reaction from the enthalpies of the three reactions below. a. 2 CH3OH(l) + 3 O2(g)  2 CO2(g) + 2 H2O(l) H1b. CH2CO2H(l) + 2 O2  2 CO2(O + 2 H2)(l) H2c. 2 CO(g) + O2(g)  2 CO2(g) H3i. Find just 1 Methanol ii. Find just 1 CO2iii. Find just 1 Acetic Acid, on reactant side1. Reverse reaction, and add opposite enthalpy changed. ½ ΔrH1 + ½ Δr3 - ΔrH2i. Put everything in it's place and simplifyii. If it simplifies down, which it does, you have the correct equationd. Using Standard Enthalpy Valuesi. Example 11. Consider the Following Combustion of ethanea. C2H6(g) + 7/2 O2(g)  2 CO2(g) + 3 H2O(l)2. If 365 kJ of heat is released (negative), how many grams of C2H6 were combusted? ΔH = -1561 kJ/mol for C2H6 combustiona. Plan: heat real eased  mol C2H6  g C2H6b. -365 kJ(1 mol / -1561 kJ)(30 g / 1 mol C2H6) = 7.01 g C2H6ii. Example 21. What is the standard enthalpy change if 2.20 g of chromium is oxidized to C2O3(s)?a. First write the reaction:i. 2 Cr(s) + 3/2 O2  Cr2O3(s)b. Next figure out how many moles of product can be formedi. g Cr  mole Cr  mol Cr2O3  kJc. 2.20 g Cr(1 mol Cr / 52 g)(1 mol Cr2O3 / 2 mol Cr)(-1139.7 kJ / 1 mol Cr2O3) = - 24.1 kJ i. The kJ come from tables in the book, however you will be given them just know how to use themii. Negative because you real ease 24.1 kJII. Calorimetrya. Calorimetry can be used to measure energy change for a chemical reactionb. "Coffee Cup Calorimeter"i. Done at content pressure, deterring ΔHii. Assume no energy transfer to the surroundingsiii. The reaction is carried out in solutioniv. Exothermic reactions will result in increased temperaturev. Endothermic reaction will result in decreased temperaturevi. With the mass and specific heat capacity of the solution (usually water)and the temperature change can determine ΔHc. Examplei. Suppose you place 0.500 g of magnesium chips into 100 mL of 1.00 M HClin a coffee cup calorimeter. Find qp in J and ΔHrxn in kJ/mol Mg. 1. Reaction: Mg(s) + 2HCl(aq) --> MgCl2(aq) + H2(g)2. Data: a. Temperature increases from 22.21 to 24.46 degrees Cb. Specific heat capacity of solution = 4.20 J/g degree Cc. Density of Solution = 1.00 g/mL3. Calculate amount of Mga. 0.50g(1 mol / 24.3 g) = .0206 mol4. Calculate qsolution = C x m x ΔTa. q = (4.20 J / g C)(100.0 g HCl + .500 g Mg)(24.46 - 22.21 C) = 945 J5. Calculate qra. qr + qsolution = 0b. qr = -945 J6. Calculate ΔH per mol of Mga. ΔH = (-945 J / .0206 mol Mg) = 45,872 J/mol or 45.9 kJ/molreleased or -45.9