Lecture 39 Thermochemistry 2 Read HW BLB 5 6 5 7 8 8 BLB 5 63 67a b 69 73 75 83 85 8 65a 67a c 92a 18 72a b 74 Sup 5 1 7 8 11 13 Know Hess s Law heats of formation enthalpy of reactions estimating reaction enthalpy from bond energies Check out the grad u lator on the Chem110 website FINAL SKILL CHECK TEST DEADLINE MONDAY APRIL 27 Need help Get help TAs in CRC 211 Whitmore and SI hours on Chem 110 website my office hours Mon 12 30 2 Tues 10 30 12 in 324 Chem Bldg or 326 Chem F in al Exa m MONDAY May 4 12 20 pm Concept Final review session with Sheets Thursday 4 30 6pm in 108 Forum Please work through the Concept Exam on lecture note page before the review session bring it with you along with any questions you may have This review is meant to complement to the review sessions that your TAs will be holding in which they will go over the practice exams Sheets Page 1 Lecture 39 Enthalpy review enthalpy is an stuff you have property how much H for a reaction is reverse reaction in magnitude in sign to H for the H for a reaction depends on the of the reactants and products gas liquid solid H for a sum of steps is the same as H for the overall process Hess s law Why is this true H is a state function path independent history independent depends only on current states T P etc Sheets Page 2 Lecture 39 Hess s law H for an overall reaction is equal to the sum of the individual steps a consequence of H being a state function Sheets Page 3 Lecture 39 Example Given the following information 2 SO3 g H 196 kJ 2 S s 3 O2 g 2 SO3 g 790 kJ 2 SO2 g O2 g What is Hrxn for S s O2 g SO2 g Sheets Page 4 Lecture 39 Heat of formation enthalpy of formation Hf heat released or absorbed when combine to form a compound combine elements compound Hf standard enthalpy of formation Hf H to form 1 mole of the compound from its elements which are all in their standard states standard state most stable form of element gas liquid solid at standard temperature and pressure T 25 C 298 K P 1 atm by definition an element in its standard state has Hf 0 Hf of O2 g S s C graphite 0 Hf of C diamond 0 NOTE Hf values found in BLB Appendix C Sheets Page 5 Lecture 39 Heat of formation cont Examples C graphite 2H2 g 1 2 O2 g 1 atm 25 C CH3OH l Hf 238 6 kJ reactants elements in std state NOTE phase state is important l s g these reactants might not have whole numbers for coefficients Why Because we are making 1 mole of the product 1 atm Br2 l Br2 g H f 30 71 kJ 25 C Sheets Page 6 Lecture 39 Example For which of the following reactions at 25 C 1 atm does H rxn H f H2 g F2 g 2HF g NO g 1 2 O2 g NO2 g 2C graphite 3H2 g 1 2 O2 g C2H5OH l Pb s Cl2 g PbCl2 s S s O3 g SO3 g Sheets Page 7 Lecture 39 Standard enthalpy of a reaction H rxn heat of reaction Hrxn when all reactants and products are in their standard states obtain H rxn from H f H rxn n Hf products m Hf reactants n m are stoichiometric coefficients of each individual product and reactant respectively NOTE this is an application of Hess s Law Sheets Page 8 Lecture 39 H rxn example calculate H rxn C2H5OH l 3 O2 g 2 CO2 g 3 H2O l C2H5OH l Hf kJ mol 277 7 O2 g 0 CO2 g 393 5 H2O l 285 8 substance o Hrxn Sheets n H products m H reactants o f o f Page 9 Lecture 39 Estimating Hrxn from bond dissociation energies Hrxn n D broken m D formed energy is released H is when bonds form n m bonds helps us understand origins of Hrxn Sheets Page 10 Lecture 39 Example 2 H2 O2 2 H2O Hrxn draw Lewis structures of reactants and products Bond H H O H O2 D kJ mol 436 463 495 reactants broken bond D Sheets products formed bond D Page 11 Lecture 39 Example The following reaction is used to produce methanol CO g 2 H2 g CH3OH Hrxn 128 kJ H C O 2 H H H C O H H Use the following info to calculate the O bond enthalpy approximate C Bond C H C O H H C O O H D kJ mol 413 799 436 358 463 reactants broken bond D In BLB Table 8 4 D C Sheets products formed bond D O 1072 kJ mol Page 12 Lecture 39 Example What is H for the following reaction Is this an exothermic or endothermic reaction Fe2O3 s 3 H2O l 2 Fe OH 3 s H f kJ mol Fe2O3 s 824 2 H2O l 285 8 H2O g 241 8 Fe OH 3 s 823 0 Sheets Page 13 Lecture 39 Before next class Read HW BLB 15 1 15 3 BLB 15 13 14 21 Sup 15 1 4 Know concept of equilibrium equilibrium constant Keq Work through the concept final on lecture notes website Good luck studying for finals Please start now Answers p 4 Hrxn 297 kJ p 9 Hrxn 1366 7 kJ p 11 Hrxn 485 kJ p 12 1060 kJ mol p 13 Hrxn 35 6 kJ endothermic Sheets Page 14 Lecture 39
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