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Unformatted text preview: perlick (app582) HW10 quy (50970) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Calculate the standard enthalpy of formation of bicyclo[1.1.0]butane H H C C CH 2 H 2 C given the standard enthalpies of formation of 717 kJ mol 1 for C(g) and 218 kJ mol 1 for H(g) and the average bond enthalpies of 412 kJ mol 1 for C H and 348 kJ mol 1 for C C. 1. +312 kJ mol 1 2.- 124 kJ mol 1 3.- 36 kJ mol 1 correct 4.- 472 kJ mol 1 5. +175 kJ mol 1 Explanation: We can write an equation in which we com- pletely decompose bicyclo [1,1,0] butane: C 4 H 6 (bicyclobutane , g) 4 C(g) + 6 H(g) H rxn = 5 (BE C C ) + 6 (BE C H ) (The comment on the right comes from dissecting the structure given in the question and noting how many of each kind of bond is present). To find H for this reaction we can use Hess Law with formation enthalpies: H rxn = bracketleftBig 4 H f C(g) + 6 H f H(g) bracketrightBig- bracketleftBig 1 H f C 4 H 6 (g) bracketrightBig = [4 (717 kJ / mol) + 6 (218 kJ / mol)]- H f C 4 H 6 (g) = 4176 kJ / mol- H f C 4 H 6 (g) Now we use the comment on which bonds were broken: H rxn = 6 BE C C + 6 BE C H = 5 (348 kJ / mol) + 6 (412 kJ / mol) = 4212 kJ / mol We can set the two sides of the equation equal since they represent the same reaction: 4212 kJ / mol = 4176 kJ / mol- H f C 4 H 6 (g) H f C 4 H 6 (g) =- 36 kJ / mol ....
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- Fall '07