Final_10_Key

# Final_10_Key - Chem 111(Chemical Kinetics Kahn 2010 Final...

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1 Chem 111 (Chemical Kinetics), Kahn 2010 Final Exam Your Name: Q1 (30) Q2 (30) Q3 (24) Q4 (15) Q5 (35) Q6 (20) Q7 (20) TH (26) Total (200) 1. a) You are given a familiar-looking printout of Mathematica file that illustrates analysis of multi-wavelength kinetics of an enzymatic reaction. Unlike typical examples or answer keys, this printout is sparse in comments (similar to some homeworks by our students). However, it shows all the relevant Mathematica commands and results. Write a one-page explanation about this analysis. Summarize the key findings about the reaction under study. Highlight any clever approaches that were taken to learn as much as possible from this data. (20 pts) The analysis starts with plotting 3D graph of noisy absorbance data vs. and time. This reveals the disappearance of reactant ( max ≈ 300 nm) and appearance of product ( max ≈ 500 nm). The action spectra show no isosbestic point suggesting the presence of at least one intermediate (with max ≈ 400 nm) Singular value analysis leads to three significant singular values, three non-random basis spectra, and three non-random basis profiles, confirming that three species contribute to absorbance. Considering the time-dependence of absorbance, it is reasonable to suggest that we have a consecutive A B C process. Individual double-exponential fits to basis profiles are challenging due to a large number of unknowns. However, C0 can be determined for each fit based on the value at completion of the reaction. One of the basis profiles yields reasonably well determined rate coefficients (0.97 and 0.53), which are further con- firmed via global fit. Note that we do not know which step each of these rate coefficients corresponds to. Armed with the knowledge that we have only three species, we reconstruct the absorbance matrix and display it as a 3D graph. This time the formation and disappearance of intermediate is clearly visible, lending support to the A B C model. Because we were not given the pure spectra of species, we have to extract that information out from the absorbance data. The key idea here is that at time zero, we only have A, and at the end, we only have C. The mixing was quick enough and the data was collected long enough, so we can get spectra of pure A and C from data.

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2 We note that the intermediate reaches maximal concentration around 1.8 seconds. Because we think that we know the rate constants from SVD analysis, we can calculate how much A is left after 1.8 seconds, and how much C has formed by 1.8 seconds. Subtracting the contributions of A and C from observed (SVD-reconstructed) data, we obtain the spectrum of B. This was a clever approach! Armed with spectra of A, B, and C we can calculate the real concentration profiles for A, B, and C via matrix algebra. Global fit to analytical rate expressions describing the consecutive A B C process yields rate constants with small errors.
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Final_10_Key - Chem 111(Chemical Kinetics Kahn 2010 Final...

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