chm115_lecture14a - 15-1Chemistry 115Lecture 14...

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Unformatted text preview: 15-1Chemistry 115Lecture 14 OutlineChapter 5Hesss LawStandard Heats of ReactionEXAM II: Thursday, March 108:00pm, Elliott Hall of MusicChapters 5, 6, 7, 8 (Lewis dots)Recitation: Review Chapters 5, 6, 7, Lewis dots14-2H =-890 kJ mol-1CH4(g) + 2O2(g) CO2(g) + 2H2O(l)CO2(g) + 2H2O(l) CH4(g) + 2O2(g)H =890 kJ mol-1Heat inNote: H reverses sign with a reverse in reaction direction14-3Concept QuestionsConsider the following dissociation reaction:NH4NO3(s)+ H2O NH4+(aq)+ NO3-(aq)+ H2O H = 25.7 kJ/mol1) Is the reaction exothermic, or endothermic?A) ExothermicB) EndothermicC) Neither2) If we start with 240 g of ammonium nitrate, how much heat is transferred?A) 240 kJB) 123 kJC) 77 kJD) Insufficient Info14-4We have the following information (given):A: CO(g)+ O2(g)CO2(g)HA= -283.0 kJB: N2(g)+ O2(g)2NO(g)HB= 180.6 kJNow lets look at a more complicated example:EXAMPLE2: What is H for the following rxn?CO(g)+ NO(g)CO2(g)+ N2(g) H = ?QUESTION: How do we get Hrxn?ANSWER:We manipulate Eqns A and B to get proper rxn, and keep track of changes in Hs.To get - B => reverse rxn B and multplying by In this case we want: Rxn A - Rxn B 14-5HACO(g)+ O2 (g)CO2(g)-283.0 kJ- BNO(g)N2(g)+ O2(g)-180.6 kJ2TotalCO + NO + O2CO2+ N2+ O2-373.3 kJ Cancel common factorsCO(g)+ NO(g)CO2(g)+ N2(g) Hrxn= -373.3 kJSo lets write out Rxn A Rxn BDont forget to multiply H by 1/2Result:14-6Why does this work?...
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This note was uploaded on 04/09/2011 for the course CHM 115 taught by Professor Towns during the Spring '08 term at Purdue University-West Lafayette.

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chm115_lecture14a - 15-1Chemistry 115Lecture 14...

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