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Hess’s Law 1. (a) Heptane, C7H16, DHof = –44.9 kcal/mol C7H16 + 11 O2 → 7 CO2 + 8 H2O ΔH = [7(–94.0) + 8(–68.3)] – [–44.9 + 11(0.0)] = –1160. kcal/mol (b) Octane, C8H18, DHof = –49.8 kcal/mol 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O ΔH = [16(–94.0) + 18(–68.3)] – [2(–44.9) + 25(0.0)] = –2634 kcal/2 mol = –1317 kcal/mol (c) 2,2,3-Trimethylpentane, C8H18, DHof = –52.6 kcal/mol 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O ΔH = [16(–94.0) + 18(–68.3)] – [2(–52.6) + 25(0.0)] = –2634 kcal/2 mol = –1317 kcal/mol 2. 250 g 1 mol 1320 kcal = 2880 kcal/mol 1 1 4 . 1 8 8 g 1 m o l 3. In order to compare the fuels, one must find calories per gram for each fuel: Heptane = 1160 cal/mol = 11.6 cal/g Octane = 1320 cal/mol = 11.6 cal/g 2,2,3-Trimethylpentane = 1320 cal/mol = 11.6 cal/g 114.19 g/mol To three significant digits, all are equivalent in the number of calories per gram of fuel burned. 4. First, find cal/g for the burning of the ethanol: 2 C2H5OH + 7 O2 → 4 CO2 + 6 H2O ΔH = [4(–94.0) + 6(–68.3)] – [2(–56.2) + 7(0.0)] = –673 cal/2 mol = –337 cal/mol Ethanol: 336.7 cal/mol = 7.308 cal/g 46.07 g/mol For 10%, 0.1(7.308 cal/g) = 0.7308 cal/g 2,2,3-Trimethylpentane: 1317 cal/mol = 11.53 cal/g For 90%, 0.9(11.53 cal/g) = 10.38 cal/g Adding the two together gives 11.11 cal/g, which is less than pure 2,2,3-trimethylpentane. Solutions 100.20 g/mol 114.19 g/mol 114.19 g/mol Pushing the Envelope: A NASA Guide to Engines 81 TEACHERS SECTIONPDF Image | NASA Guide to Engines
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