Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is currently blended with gasoline as an automobile fuel. (a) Write a balanced equation for the combustion of liquid ethanol in air. (b) Calculate the standard enthalpy change for the reaction, assuming \(\mathrm{H}_{2} \mathrm{O}(g)\) as a product. (c) Calculate the heat produced per liter of ethanol by combustion of ethanol under constant pressure. Ethanol has a density of \(0.789 \mathrm{~g} / \mathrm{mL}\). (d) Calculate the mass of \(\mathrm{CO}_{2}\) produced per \(\mathrm{kJ}\) of heat emitted.

To write the balanced equation for the combustion of ethanol, we must remember that combustion occurs in the presence of oxygen (O2). The products formed will be carbon dioxide (CO2) and water (H2O). We'll write the unbalanced combustion equation of ethanol, C2H5OH, as: C2H5OH + O2 -> CO2 + H2O Now balance the equation by adjusting the coefficients. The balanced combustion equation of ethanol is: C2H5OH + 3O2 -> 2CO2 + 3H2O

To calculate the standard enthalpy change for the reaction, we can use the bond enthalpy values following the formula: ΔH = (sum of bond enthalpies of products) - (sum of bond enthalpies of reactants) For our combustion reaction, the bond enthalpies are given as: C-C: 347 kJ/mol C-H: 413 kJ/mol C-O: 358 kJ/mol O-H (ethanol): 463 kJ/mol O=O: 498 kJ/mol C=O: 799 kJ/mol O-H (water): 467 kJ/mol Ethanol molecule has 1 C-C bond, 5 C-H bonds, 1 C-O bond, and 1 O-H bond, whereas O2 molecule has 1 O=O bond. The products have 4 C=O bonds and 6 O-H bonds of water. Sum of bond enthalpies of reactants: 1(347) + 5(413) + 1(358) + 1(463) + 3(498) = 347 + 2065 + 358 + 463 + 1494 = 4727 kJ Sum of bond enthalpies of products: 4(799) + 6(467) = 3196 + 2802 = 5998 kJ ΔH = 5998 - 4727 = 1271 kJ/mol

To find the heat produced per liter of ethanol by its combustion under constant pressure, we first need to convert the density of ethanol to grams per liter (g/L): Density of ethanol = 0.789 g/mL Since we have 1000 mL in 1 liter, we multiply the density by 1000 to convert it to grams per liter: 0.789 g/mL * 1000 mL/L = 789 g/L Now, we need to find the heat produced per gram of ethanol. The molecular weight of ethanol (C2H5OH) is: (2 * 12.01) + (6 * 1.01) + (1 * 16.00) = 24.02 + 6.06 + 16.00 = 46.08 g/mol Heat produced per gram of ethanol = (1271 kJ/mol) / (46.08 g/mol) = 27.56 kJ/g Heat produced per liter of ethanol = (27.56 kJ/g) * (789 g/L) = 21746 kJ/L

To calculate the mass of CO2 produced per kJ of heat emitted, we need the stoichiometry of the balanced equation: C2H5OH + 3O2 -> 2CO2 + 3H2O 1 mol C2H5OH forms 2 mol CO2 From the balanced reaction, for each mole of ethanol, 2 moles of carbon dioxide are produced. Using the molar mass of CO2 (44.01 g/mol), we can find the mass of CO2 per mole of ethanol: mass of CO2 per mole of ethanol = 2 * 44.01 g/mol = 88.02 g/mol Next, we find the mass of CO2 produced per mol of ethanol: mass of CO2 produced per mol of ethanol = 88.02 g/mol Now, we know that the heat produced per mol of ethanol is 1271 kJ. So, we can find the mass of CO2 produced per kJ of heat emitted: mass of CO2 produced per kJ = (88.02 g/mol) / (1271 kJ/mol) = 0.0692 g/kJ