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# HW5 - 7 a d[B = k1[A k2[B k3[B[C dt Assuming steady state...

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7. a. d [ B ] dt = k 1 [ A ] - k 2 [ B ] - k 3 [ B ][ C ] Assuming steady state of B , d [ B ] dt = 0 then k 1 [ A ] - k 2 [ B ] - k 3 [ B ][ C ] = 0 [ B ] = k 1 [ A ] k 2 + k 3 [ C ] d [ D ] dt = k 3 [ B ][ C ] = k 1 k 3 [ A ][ C ] k 2 + k 3 [ C ] b. By equilibrium assumption, K = [ AB ] [ A ][ B ] d [ D ] dt = k [ AB ][ C ] = kK [ A ][ B ][ C ] 8. a. Plot of 1 /c vs t is linear. Plot of ln( c ) vc t is nonlinear. 1

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b. Since the plot of 1 /c vs t is linear and we know that for second order reaction 1 c = kt + C where C is a constant. Hence, the date best fit second order kinetics. By linear regression, we obtain 1 c = 4 . 738 × 10 - 2 mM - 1 s - 1 c. Since the concentrations are changing rapidly, concentration distribution in the solution will be highly inhomogenous. Thus, it is necessary to mix the solution rapidly to ensure homogenous concentration. 10 a. dA dt = - k [ A ] Solving the ODE gives [ A ] = [ A 0 ] e - kt Given that [ A ] = 0 . 1[ A 0 ] at t = 1hr, 0 . 1[ A 0 ] = [ A 0 ] e - k × 1 k = 2 . 3 hr - 1 At t = 2hr, [ A ] = [ A 0 ] e - 2 . 3 × 2 = 0 . 01[ A 0 ] . b. d [ A ] dt = - k [ A ][ B ] Since A + B C and [ A 0 ] = [ B 0 ] at all time, 2
d [ A ] dt = k [ A ] 2 . Solving the ODE gives 1 [ A ] - 1 [ A 0 ] = kt.

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