Chem 1061 1st Edition Lecture 16 Outline of Last Lecture II. Determining ∆H of a ReactionIII. Heat lost to both water and Calorimetera. Bomb Calorimeterb. CombustionIV. Stoichiometry of Thermochemical EquationsV. Balancing CoefficientsVI. Thermochemical EquivalenceOutline of Current Lecture VII. Hess’s Law: Adding Reactions and EnthalpiesVIII. Standard Enthalpies of a Reactiona. Standard Conditionb. Heat of FormationCurrent LectureHess’s Law : Adding Reactions and EnthalpiesFind ∆H of a “missing” reaction if you know the others.∆H only depends on initial and final states (doesn’t depend on path taken to get there)Target Equation:NH3(g) + HCL(g) NH4Cl(s) ∆H =?Given:1) ½N2(g) + 3/2H2(g) NH3(g) ∆H=45.9kJ2) ½H2(g) + ½ Cl(g) HCl(g) ∆H=-92.3kJ3) ½ N2(g) + 2H2(g) + ½ Cl2(g) NH4Cl ∆H=-314.4kJFlip 1, so NH3 is a reactant (also flip ∆H sign)NH3(g)1/2N2(g) + 3/2 H2(g) ∆H=45.9Flip 2, so HCl is a reactant (also flip ∆H sign)HCl(g)1/2H2(g) + ½Cl2(g) ∆92.3kJThese 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.MAKE SURE EQUATIONS MATCH.Reactants: NH3(g) + HCl(g) + 1/2N2(g) + 2H2(g) + ½Cl(g) Products: ½ N2(g) + 3/2H2(g) + ½ H2(g) + 1/2Cl2(g) + NH4Cl(s)THEY MATCH!Add up 1-3 so they balance/cancel out to the targetDo the same manipulations to ∆H. ∆H overall = ∆H1 +∆H2 + ∆H345.9kJ + 92.3 kJ + (-314.4) kJ = -176.2 kJStandard Enthalpies of a ReactionStandard Condition (state)Gas= 1 atm + ideal behavioraqueous solutions = 1 Mpure substances = must be in stable form at 1 atm and 25ºCSymbol used to show variable has been measured with all substances in their standard forms (states)∆H(rxn)ºStandard Formation1 mol of a compound forms from elements in their standard form.∆H(f)º of MgO(s)Mg(s) + 1/2O2(g) MgO(s)0 0 -601.8kJKnowelements are defined at 0kJcompounds differ from
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