Chem 1061 1st Edition Lecture 15 Outline of Last Lecture I. Units of EnergyII. EnthalpyIII. Exothermic and Endothermic ProcessesIV. CalorimetryV. Specific Heat CapacityVI. Molar Heat CapacityVII. Calorimetry Outline of Current Lecture II. Determining ∆H of a ReactionIII. Heat lost to both water and Calorimetera. Bomb Calorimeterb. CombustionIV. Stoichiometry of Thermochemical EquationsV. Balancing CoefficientsVI. Thermochemical EquivalenceCurrent LectureDetermining ∆H of a ReactionCoffee Cup Calorimeter calculationT(initial) = 22.4 degree CT(final) = 23.0 degree CV = 85.0 g H2OReaction = flash bulb (Zr)Equation: Zr(s) + O2(g) ZrO2(s) ∆H = ?Zr = 0.0223 gO2 = In excessq(lost) + q(gain) = 0q(rxn) + q(H2O) = 0∆H = q(rxn)/mol ZrO2These 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.0.0223 g Zr x (1mol Zr / 91.22g Zr) x (1mol ZrO2/ 1mol Zr) = 0.000244 mol ZrO2q(rxn) + c(H2O) x m(H2O) x ∆T = 0q(rxn) + (4.184 J/gºC) x (85.0 g) x (23.0ºC – 22.4ºC) = 0q(rxn) = -213.38 Jq(rxn)/mol ZrO2 = (-213.38J / 0.000244 mol) = -8.75 x 10^5 J/mol = -875 kJHeat lost to both water and calorimeterCalorimeter is constant = c(cal) = J/K = J/ºCNo mass component because you cannot have one half of a calorimeter or one and onehalf of one. q(gain) = q(H2O) + q(cal)q(cal) = c(cal) x ∆TBomb calorimeter: way to do constant volume calorimetryused for combustion reactionsStoichiometry of Thermochemical Equations Reaction + ∆H on appropriate side∆H = + endothermic (heat in, heat is reactant)∆H = - exothermic (heat out, heat is product)1) Reverse reaction, reverse the sign on ∆HH2O (l) H2 (g) + 1/2 O2 (g) ∆H= 286 K/JH2 (g) + ½ O2 (g) H2O (l) ∆H = -286 K/J2) Change magnitude of coefficient, then the magnitude of ∆H is changed PROPORTIONALLY Balancing Coefficients ∆H specified for particular amount (often 1 mol, ∆H(f)º)use fractional coefficients as necessaryC(s) + 1/2 O2(g) CO(g) ∆H=-110.5 kJThermochemical Equations-110.5 is thermochemically equivalent to:1 mol C(s)½ mol O2(g)1 mol CO (g)Ex: 2P(s) + 2Br2(l) 2PBr3(g) ∆H=-243kJWhat is the enthalpy when 2.63g P(s) reacts with excess Br2(l)?2.63 g P(s) x (1 mol P(s) / 30.97 g P(s)) x (-243 kJ/ 2 mol P(s)) = -10.3
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