2006 - answers

2006 - answers - Chem. 203 Oct. 12, 2006. Midterm 1:...

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Chem. 203 Oct. 12, 2006. Midterm 1: answers 1. Given H f for: 3C(g) + 6H(g) + O(g) → CH 3 -CO-CH 3 (g) at 298K where H f = E o + E thermal + nRT assume ideal gases and E o = - (2E C-C + 6E C-H - E C=0 – 0) bond energies additive then E C=O = - E o - 2E C-C - 6E C-H = - H f - E thermal - nRT- 2E C-C - 6E C-H = - H f - (3RT – 10x3/2RT) + 9RT - 2E C-C - 6E C-H neglecting vibr. contr. to E thermal = - H f - 21RT - 2E C-C - 6E C-H But H f = )) ( ( 3 3 g COCH CH H o f - )) ( ( 3 g C H o f - )) ( ( 6 g H H o f - )) ( ( g O H o f or = )) ( ( 3 3 g COCH CH H o f - 3x718.4 – 6x218 – 247.5 kJ mol -1 given for: CH 3 -CO-CH 3 (g) + 4O 2 (g) → 3CO 2 (g) + 3 ) ( 2 O H and H comb = -1814.4 kJ mol -1 then -1814.4 kJ mol -1 = )) ( ( 3 2 g CO H o f *+ )) ( ( 3 2 O H H o f - )) ( ( 3 3 g COCH CH H o f or )) ( ( 3 3 g COCH CH H o f = 1814.4 + )) ( ( 3 2 g CO H o f + )) ( ( 3 2 O H H o f kJ mol -1 = 1814.4 +3(-393.5) + 3(-285.8) = -223.5 kJ mol -1 Therefore: H f = - 223 - 3x718.4 – 6x218 – 247.5 = -3934.2 kJ mol -1 and E C=O = 3934.2 – 21RT – 2x348 – 6x412 = 714.17 kJ mol -1 * )) ( ( 2 O H H o f = )) ( ( 2 g H H o comb ; )) ( ( 2 g CO H o f = )) ( ( gr C H o comb Question 1. The answer could be provided in terms of the
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This note was uploaded on 07/24/2011 for the course CHEM 203 taught by Professor Galley during the Fall '09 term at McGill.

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2006 - answers - Chem. 203 Oct. 12, 2006. Midterm 1:...

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