CHE132S2005Ex01Key

# CHE132S2005Ex01Key - T so T = 42.3 x 10 6(4.18x400000 =...

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CHE132 Answer Key Midterm 1 on 2/16/05 Question Form 0 Form 1 Form 2 Form 3 1 B A E D 2 A E D C 3 D C B A 4 B A E D 5 C B A E 6 C B A E 7 B A E D 8 C B A E 9 D C B A 10 A E D C 11 E D C B 12 B A E D 13 B A E D 14 E D C B 15 B A E D 16 A E D C 17 E D C B 18 A E D C Question 19. A 400 kg bear consumes 2.7 kg of glucose. Given that the specific heat capacity of the bear (C bear ) is 4.18 J g -1 K -1 , calculate the increase in temperature of the bear assuming that all the energy from the combustion of glucose is used to raise the bear’s body temperature. H combustion for glucose = -2825.8 kJ mol -1 q reaction = n H combustion = 2700/180.2 x -2825.8= -42.3 x 10 3 kJ q reaction = -q bear q bear = 42.3 kJ = nC
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Unformatted text preview: T so T = 42.3 x 10 6 /(4.18x400000) = 25.3 K Correct equation 2 points, correct q reaction 2 points Question 20 Calculate G° reaction for the following reaction at 350 K given that S° reaction = -574.9 J K-1 and given H f ° for each reactant and product. 2NH 3 (g) + O 2 (g) NH 4 NO 3 (s) + H 2 O(l) H f ° kJ mol-1 : NH 3 (g) = -45.9, O 2 (g) = 0, NH 4 NO 3 (s) = -365.6, H 2 O(l) = -285.83 H° reaction = (-365.6+-285.83) – (2x-45.9) = -651.43 - -91.8 = -559.63 G° reaction = H° reaction- T S° reaction = -559.63 x 10 3 – (-350x574.9) = -358.4 kJ mol-1 Correct H° reaction 2 points, correct equation 2 points....
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## This note was uploaded on 07/15/2009 for the course CHE 131 taught by Professor Kerber during the Fall '08 term at SUNY Stony Brook.

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