Exam2 Fall 2015_solutions

# Exam2 Fall 2015_solutions - Exam 2 Solutions 1 Solutions...

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Exam 2 Solutions 10/30/2015 1 Solutions 1. (a) Δ G (298 K) = - RT ln K = ( - 8 . 314 J K - 1 mol - 1 )(298 K) ln(2 . 2 × 10 6 ) = - 36 . 182 kJ mol - 1 Since Δ G is negative, this reaction is spontaneous at 298 K. (b) First we need to find Δ H . We can find this using Δ H and Δ H at 298 K . Δ H = Δ G + T Δ S = ( - 36182 J mol - 1 ) + (298 K)(23 . 5 J K - 1 mol - 1 ) = - 29 . 179 kJ mol - 1 Apply the Van’t Hoff equation ln K 2 K 1 = - Δ H R 1 T 2 - 1 T 1 K 2 = K 1 exp 29179J mol - 1 8 . 314J K - 1 mol - 1 1 310 K - 1 298 K = 1 . 39 × 10 6 (c) In the first condition Q = [ GAb ] [ G ][ Ab ] = (1 - 0 . 13) × 10 - 9 M (1 × 10 - 9 M )(0 . 13 × 10 - 9 M ) = 6 . 69 × 10 9 Since this number is larger than K at 310K, the reaction will go backwards . In the second condition Q = [ GAb ] [ G ][ Ab ] = (1 - 0 . 13) × 10 - 9 M (6 × 10 - 6 M )(0 . 13 × 10 - 9 M ) = 1 . 12 × 10 6 Since this number is smaller than K at 310K, the reaction will also go forwards .

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20.110/2.772 Exam 2 Solutions - Page 2 of 4 10/30/2015 (d) Since the Gleevec line and the drug B line intersect at 1 /T = 0 . 00322 K - 1 , the two drugs have the same equilibrium constant at body temperature and the same free energy of binding. Thus, they should be equally effective. Drug A is always
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