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# HW3 - Homework 3 Feb 8 2007 1 a By using the dilute...

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Homework 3 Feb. 8, 2007

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1. a. By using the dilute solution standard states, K = [G-1-P][ADP] [G][ATP] = 770 × 10 - 4 10 - 3 = 77 . 0 where G-1-P and G refer to glycerol-1-phosphate and glycerol, respectively. Δ G 0 298 = - RT ln K = - 8 . 314 × 298 × ln 77 = - 10 . 76 kJ mol - 1 . b. 1 : G + ATP G-1-P + ADP , Δ G 0 1 = - 10 . 76 2 : ADP + D i G-1-P + H 2 O, Δ G 0 2 = 31 . 10 1 + 2 : G + P i G-1-P + H 2 O. Hence, Δ G 0 298 = Δ G 0 1 + Δ G 0 2 = - 10 . 76 + 31 . 10 = 20 . 2 kJ mol - 1 K = e - Δ G 0 298 RT = 2 . 9 × 10 - 4 . 2. a. Δ G = Δ G 0 + RT ln [ADP][P i ] [ ATP ] = - 31 . 0 + 8 . 314 × 10 - 3 × 298 × ln 0 . 5 × 10 - 3 × 2 . 5 × 10 - 3 1 . 25 × 10 - 3 = - 48 . 1kJ mol - 1 . b. At constant temperature and pressure, w max = w rev = Δ G = - 48 . 1 kJ mol - 1 . 1
c. Let PC and C be phosphocreatine and creatine, respectively. 1 : PC + H 2 O C + P i , Δ G 0 1 = - 43 . 1 2 : P i + ADP ATP + H 2 O , Δ G 0 2 = 31 . 0 1 + 2 : PC + ADP C + ATP , Δ G 0 3 = Δ G 0 1 + Δ G 0 2 Δ G 0 3 = - 43 . 1 + 31 . 0 = - 12 . 1kJ mol - 1 K = e - Δ G 0 3 RT = 132 . 3 a. Let G-6-P and G be glucose-6-P and glucose, respectively. 1 : G-6-P + ADP ATP + G , Δ G 0 1 = 16 . 7 2 : ATP + H 2 O ADP + P i , Δ G 0 2 = - 31 . 0

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HW3 - Homework 3 Feb 8 2007 1 a By using the dilute...

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