CHEM 101 1nd Edition Lecture 22Outline of Last Lecture I. EnthalpyII. CalorimetryOutline of Current Lecture I. Measuring Heat of CombustionII. Measuring Heat of ReactionsIII. Using Standard Enthalpy ValuesCurrent LectureI. Measuring Heat of Combustiona. Constant Volume “Bomb” Calorimeteri. Used to evaluate heats of combustion of fuels and the caloric value of foods1. Determine ΔU a. Since volume does not change there is no energy transfer as work2. The “bomb” is placed in water-filled container with well-insulated walls3. “Bomb” is filled with pure oxygen4. Sample is ignited5. Burn combustible sample6. Measure heat evolved in reactiona. qr + qbomb + qwater = 0ii. Some heat from the reaction warms the water1. qwater = (specific heat)(water mass)( ΔT)iii. Some heat fro the reaction warms the “bomb”1. qbomb = (heat capacity, J/K)( Δ)iv. Total Heat Evolved = qtotal = qwater + qbombII. Measuring Heat of Reactionsa. Calculate heat of combustion of octane (unbalanced)i. C8H18 + O2 CO2 + H2O1. Burn 1.00g of octane2. Temp rises from 25.00 to 33.20 degrees C3. Calorimeter contains 1200 g waterThese 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.4. Heat capacity of bomb – 837 J/Kii. Step 11. Calculate heat transferred from reaction to water2. q = (4.134 J/gK)(1200 g)(8.20 K) = 41,170 Jiii. Step 21. Calculate heat transferred from reaction to bomb2. q = (bomb heat capacity)( ΔT)3. q = (837J/K)(8.20 K) = 6820 Jiv. Step 31. Total heat evolved2. 41,170 J + 6860 J = 48,030 J3. Heat of combustion of 1.00 g of octane = -48.0 kJIII. Using Standard Enthalpy Valuesa. In general, when ALL enthalpies of formation are known, ΔHrxn = Σ ΔHf (products) – Σ ΔHf (reactants)i. Δ always = final – initialb. If the ΔH is -1561 kJ/mol C2H6, how much heat is released if 20.0 g of the CO2(g) isproduced?i. C2H6(g) + O2(g) CO2(g) + H2O(l)ii. Step 11. Balance the reaction2. 1 C2H6(g) + 7/2 O2(g) 2 CO2(g) + 3 H2O(l)a. It’s okay to have fractional coefficients for thermodynamic problemsb. Since ΔH is -1561 kJ/mol C2H6, when one mole of C2H6(g) is consumed, 1561 kJ of heat is producediii. Step 21. Plan:a. g CO2 mol CO2 mol C2H6 heat2. 20.0 g CO2(1 mol CO2 / 44.0 g CO2)(1 mol C2H6 / 2 mol CO2)(1561 kJreleased / 1 mol C2H6) = 355 kJ heat releasediv. If 365 kJ of heat is released, how many grams of C2H6 were combusted?1. Plan:a. heat released mol C2H6 g C2H62. 365 kJ(1 mol C2H6 / 1561 kJ)(30.1 g C2H6 / 1 mol C2H6) = 7.04 g combustedc. Is the reaction spontaneous?i. ΔG = ΔH – TΔS1. ΔG < 0 a. The process is spontaneous in the direction written under standard conditions2. ΔG = 0a. The process is at equilibrium under standard conditions3. ΔG > 0a. The process in not spontaneous in the direction written under standard
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