# quiz11s11 - CHEMISTRY 444.11 QUIZ 1 Spring 2011(11S Feb 17...

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Problem 1 (8 points) Starting from the continuous probability distribution for one-dimensional diffusion, show how one would determine the average squared displacement in a model describing isotropic diffusion in two dimensions. The 1-dimensional continuous probability distribution is given by: P ( x , t ) dx = 1 4 π Dt x 2 / 4 Dt e dx For two-dimensional isotropic diffusion, with uncorrelated probabilities, we write: P ( x , y , t ) dx dy = 1 4 π Dt x 2 / 4 Dt e dx Λি Νয় Μ৏ Ξ৯ Πਏ Ο৿ 1 4 π Dt y 2 / 4 Dt e dy Λি Νয় Μ৏ Ξ৯ Πਏ Ο৿ P ( x , y , t ) dx dy = 1 4 π Dt x 2 + y 2 ( ) / 4 Dt e dx dy Transforming to plane polar coordinates, r 2 = x 2 + y 2 : P ( x , y , t ) dx dy = 1 4 π Dt r 2 ( ) / 4 Dt e dx dy P ( r , φ , t ) dr d φ = 1 4 π Dt r 2 ( ) / 4 Dt e r dr d φ Integrating out the angular part P ( r , t ) dr = 1 4 π Dt r 2 ( ) / 4 Dt e r dr d φ 0 2 π = 1 4 π Dt r 2 ( ) / 4 Dt e r dr Λি Νয় Μ৏ Ξ৯ Πਏ Ο৿ d φ 0 2 π P ( r , t ) dr = 1 2 Dt r 2 ( ) / 4 Dt e r dr The average squared displacement is: r 2 = r 2 0 P ( r , t ) dr = r 2 1 2 Dt r 2 ( ) / 4 Dt e r dr Λি Νয় Μ৏ Ξ৯ Πਏ Ο৿ 0 = 1 2 Dt r 2 ( ) / 4 Dt r 3 e dr Λি Νয় Μ৏ Ξ৯ Πਏ Ο৿ 0 Using Table 2.3, the integral is evaluated as: r 2 = 4 Dt CHEMISTRY 444.11 Spring, 2011 (11S) QUIZ 1 Feb. 17, 2011 NAME: Score ______/20

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Problem 2 (2 Points) Assuming an ideal gas model for Neon (Ne) and Krypton (Kr) and isotropic, two-dimensional diffusion, what is the ratio: r 2 Kr r 2 Ne Solution: r 2 Kr r 2 Ne = D Kr D Ne = σ Ne 2 m Ne σ Kr 2 m Kr = d Ne 2 m Ne d Kr 2 m Kr = d Ne d Kr m Ne m Kr = 0.2382 nm 0.3362 nm 20.1797 g 83.798 g = 0.7085 ( ) 0.490727 ( ) = 0.348
Problem 1 (10 points) Boundary centrifugation is performed on a solution of cytochrome c (M=13,400 gr/mol) in water at T=20° C ( ρ =0.998 gr cm 3 , η =1.002 cP) using a centrifuge operating at 40000 revolutions per minute (rpm). The following data are for the position of the boundary layer as a function of time: Time (hours) X b (cm) 0 4.00 2.5 4.11 5.2 4.23 12.3 4.57 19.1 4.91 What is the sedimentation coefficient of cytochrome c under these conditions? Show all work and give your answer in the appropriate units Solution: Plot the data as ln(x b /x b,t=0 ) versus time, t. The slope is ϖ 2 s . We fit the last 4 data points as a straight line with the parameters a and b defining the line y = at + b with a being the slope. Results of Fitting (using gnuplot’s Marquardt- Levenburg algorithm CHEM444 Section 11/81 Spring 2011 (11S) QUIZ 2 Feb. 24, 2011 NAME: Score ______/10

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The sedimentation coefficient is then ϖ 2 s = 0.0107 hr 1 s = 0.0107 hr 1 ϖ 2 1 hr 3600sec Λি Νয় Μ৏ Ξ৯ Πਏ Ο৿ = 2.97x10 6 sec 1 ϖ 2 = 2.97x10 6 sec 1 40000 rev min Λি Νয় Μ৏ Ξ৯ Πਏ Ο৿ 2 2 π radians rev Λি Νয় Μ৏ Ξ৯ Πਏ Ο৿ 2 1 min 60 sec Λি Νয় Μ৏ Ξ৯ Πਏ Ο৿ 2 = 1.7x10 13 sec Problem 2 (5 points) For the following data giving reactant concentration versus time for the following general chemical transformation: A products Time (sec) [A] (M) 1 0.12 2 0.074 3 0.044 4 0.027 5 0.016 6 0.009 8 0.0036 10 0.0013 determine the order of the reaction (i.e., is this first or second order reaction). Show work and use any means (i.e., graphical or other) to arrive at your answer. Please give a rationale for your conclusion
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