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HW1%20Chap7

# HW1%20Chap7 - product concentrations Thus[Acetate =[OH = y...

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BME 100L. Modeling Cellular and Molecular Systems Fall 2009 Homework 1 Chapter 7. Problems 1, 2, 3, 4. Due: 5pm, September 2nd. Hints 1. (a) Assume the rate R = d[I 2 ]/dt = - k [I 2 ] a [Ket] b [H + ] c , where Ket denotes ketone. Based on the date shown the Table, determine a, b, and c. The procedure is first to fix [KE] and [H + ] and vary [I 2 ] to determine a. Then, fix [I 2 ] and [H + ] and vary [Ket] to determine b, and fix [I 2 ] and [Ket] and vary [H + ] to determine c. (c) d[Ket]/dt = -k [I 2 ] a [Ket] b [H + ] c If [H + ] is fixed, the differential equation for [Ket] can be integrated analytically. Show [IKet] + [Ket] = [Ket] 0 , where IKet denotes iodoketone. (d) It is not required in this course. 2. (a) The consumption rate of [Acetate] and [OH - ] are the same. The initial concentrations of these compounds are the same. d[OH - ]/dt is independent of the
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Unformatted text preview: product concentrations. Thus, [Acetate] = [OH-] = y, and dy/dt = - k[Acetate] a [OH-] b = ky , where = a + b. Show that t 1/2 = 1 1 2 1 1 ) 1 ( k y . (c) i.e., for the two experiments with different t 1/2 . (d) See hint in part (a) 3. (a) see hint in part (a) in Prob.1 (c) For the first and third reactions, the concentration ratios are equal to the equilibrium constants. For the second reaction, d[I-]/dt = k[I-][HOCl]. Solve [HOCl] from those equilibrium relations, and substitute it into the kinetics equation for [I-]. Note [H 2 O] can be treated as a constant. 4. (b) Note B is involved in both reactions. d[B]/dt = the net rate = rate of production – rate of consumption....
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