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F09_230_L9pre[1] - CHEM 230 F09 Lecture 9 Chapter 6...

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CHEM 230 F09 Lecture 9 Chapter 6: Thermodynamics First Law Session ID: 230 Homework 2 due Wed 9/30 at 10pm
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a) C12H22O11(s) 12 C(s) + 11 H2O(g) + heat b) heat + 2KClO3(s) 2 KCl(s) + 3O2(g) b) C(s) + O2(g) CO2(g) + heat H2SO4 C Self-Lighting Candle Frozen Beaker Heat + Ba(OH)2° 8H2O(s) + 2NH4SCN(s) Ba(SCN)2(s) + 2NH3(g) + 10 H2O(l) http://www.youtube.com/watch? v=vgh76gPSg3M
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Measuring Enthalpy Heat exchange: Temperature change q = mC T qa= -qb Phase change Hvap, Hfus Measure with calorimeter Reactions Themochemical equations (adding, reversing, stoichiometry) Hess’s Law Heats of formation Bond enthalpy
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Calorimetry Calorimeter: device used to measure the heat absorbed or evolved during a physical or chemical change Constant volume: measures U = qv Constant pressure: measures H = qp = heat of reaction
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Problem A calorimeter with heat capacity equivalent to 0.739 grams of water is used to measure the heat of combustion of 0.303 g of sucrose (C12H22O11 MW = 342 g/mol). The temperature increase was found to be 5.00oC. Calculate the would be released by this reaction C12H22O11 (s) + O2(g) g CO2(g) + H2O(g) qrxn = -qCal = -mCH2O T = -0.739g (4.184 J )(+5oC) goC = -15.46 J Cs,H2O = 4.184 J/goC
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Concept Question Is the energy that we just calculated, the energy associated with the combustion reaction as written? A calorimeter with heat capacity equivalent to 0.739 grams of water is used to measure the heat of combustion of 0.303 g of sucrose (C12H22O11 MW = 342 g/mol). The temperature increase was found to be 5.00oC. Calculate the would be released by this reaction C12H22O11 (s) + O2(g) L CO2(g) + H2O(g) Hrxn= - 15.46 J? Yes or No
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Thermochemical Equations C(s, graphite) + O2(g)→ CO2(g) ΔH = -393.509 kJ 2C(s, graphite) + 2O2(g)→ 2CO2(g) ΔH = -787.018 kJ (*2) 4C(s, graphite) + 4O2(g)→ 4CO2(g) ΔH = -1574.04 kJ (*4) CO2(g) → C(s, graphite) + O2(g) ΔH = 393.509 kJ (reversed) Which are exothermic, which endothermic?
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Assuming gasoline and diesel engines consume the same number of moles of fuel, and lose the same amount of
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