... burnt into KJ/mol. This was done using the following formula: Amount of energy given off [Energy] ___________________________________ x number of grams per mol Amount Burned (g) I was given the following bond values as below: Bonds Value of Bonds (KJ/Mol) C-H 412 O=O 496 C=O 743 O-H 463 C-O 360 C-C 348 Results Ethanol C2H5OH + 3O2 2CO2 + 3H2O H H O=O O H-O-H | | || H - C - C - O - H O=O C=O H-O-H | | O H H O=O || H-O-H C=O Theoretical H-C x5 = (412x5) O=O x3 O=C x4 H-O x 6 C-C x1 = 348 (496 x 3) + O-C x1 = 360 ------------- (743 x4) (463 x6) O-H x1 = 463 1488 ---------- ----------- -------- 2972 2778 3231 4719 5750 4719 - 5750 = -1031 Practical Start Temperature End Temperature Start Mass End Mass Mass Used 27 47 167.4 165.3 2.1 25 54 213.36 210.92 2.44 50g x 200 x 4.2 92000 ------------------- x 46 = --------- = 92 2.1 1000 50g x 290 x 4.2 114811.4754 -------------------- x 46 = ----------------- = 114.8114754 2.44 1000 Butanol C4H9OH + 6O2 4CO2 + 5H2O H H H H O=O O=C=O O-H-O | | | | O=O O=C=O O-H-O H - C - C - C - C - C - O - H + O=O O=C=O + O-H-O | | | | O=O O=C=O O-H-O H H H H O=O ...

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