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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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This note was uploaded on 01/17/2010 for the course BME 100 taught by Professor Yuan during the Fall '07 term at Duke.
- Fall '07