Chem 16 Solutions

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108. Two common sugars, glucose or C6H12O6, and sucrose or C12H22O11, are both carbohydrate4s. Their standard enthalpies of formation are: Glucose = -1273 kJ/mol and Sucrose = -2221 kJ/mol. Using this data: a) Calculate the molar enthalpy of combustion to CO2 (g) and H2O (l). b) Calculate the enthalpy of combustion per gram of each sugar. C6H12O6 + 6O2 --&gt; 6CO2 + 6H2O (-383.5 x 6 + -285.8 x 6) - (-1273) = -2742.8kJ C12H22O11 + 12O2 --&gt; 12CO2 + 11H2O (-383.5x12 + -285.8x11) - (-2221) = -5524.8kJ -2742.8kJ / 180g/mole = -15.24kJ/g -5524.8kJ / 342g/mole = -16.14kJ/g 107. Three common hydrocarbons that contain four carbons are listed here, along wiht their standard enthalpies of formation: Formula Delta H in kJ per mol 1,3-Butadiene C4H6(g) 111.9 1-Butene C4H8(g) 1.2 n-Butane C4H10(g) -124.7 a. For each of these substances, calculate the molar enthalpy of combustion to CO2(g) and H2O(l). b. Calculate the fuel value in kJ/g for each of these compounds. c. FOr each hydrocarbon determine the percentage of hydrogen by mass. d. By comparing your answers for part (b) and (c), propose a relationship between hydrogen content and fuel value in hydrocarbons. a) for a we first need to find a balanced equation for when the hydrocarbons combust to form CO2 and H20. Then we plug in the deltaHf values and plug these into the equation. a) C4H6 + 11/2O2 ==&gt; 4CO2 + 3H2O Delta Hrxn = [4DeltaHf(CO2)+3DeltaHf(H2O)] - [DeltaHf(C4H6) + 11/2DeltaHf(O2)] = [4(-393.5kJ) + 3(-285.83kJ)] - [111.9kJ + 11/2(0kJ)] = -2543.39kJ C4H8 + 6O2 ==&gt; 4CO2 + 4H2O Delta H rxn = [4DeltaHf(CO2) + 4DeltaHf(H2O)] - [DeltaHf(C4H8) + 6DeltaHf(O2)] = [4(-393.5kJ) + 4(-285.83kJ)] - [1.2kJ +6(0kJ)] = -2718.52kJ C4H10 +13/2O2 ==&gt; 4CO2 +5H2O DeltaHrxn = [4DeltaHf(CO2) + 5DeltaHf(H2O)] - [DeltaHf(C4H10) + 13/2DeltaHf(O2)] = [4(-393.5kJ) + 5(-285.83kJ)] - [-124.7kJ +