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BS,MBA, PHD
Adelphi University/Devry
Apr-2000 - Mar-2005
HOD ,Professor
Adelphi University
Sep-2007 - Apr-2017
A researcher is studying the degradation kinetics of a drug Yin the presence of substance Xthat follows the rate law RATE = -k[Y]m[X]n. Drug Y has a lmax at 515 nm, and no other species in the reaction absorbs at this wavelength, so the kinetics can be followed with an absorbance spectrometer set to this wavelength. It has been experimentally determined that ε = 4.99 X104 cm-1M-1 for drug Y at 515 nm. The optical path length l for all the absorbance measurements was 1.00 cm.
Prior work in which [X] >> [Y] was maintained, found that m = 1. To determine n, the rate order with respect to the molar concentration of X, the researcher prepared two dilutions from a 1.00 M solution of X, one to give a 0.75 M solution in X, and another 0.50 M in X. The condition [X] >> [Y] held for these dilutions.The following data from three separate trials (runs), one for each concentration of X, was obtained with [Y] initially the same for each run and the time interval set to 5 seconds (rather the 1 second as is in our labs):
1.(4p)Now using Excel, transform the remainingA entries(in the range of 1.0 to 0.1) into[Y]values (convert the absorbance into the molar concentration of Y).Perform a graphical analysis on each run to find kobs for each of the three trials (the three different concentrations of X). Present below in tabular form kobs for each [X].
2.(6p)Plot the kobs vs. [X] recorded in the above table as ln(kobs) vs. ln[X]using Excel. Paste-in this plot below along with the linear trend-line equation and R2 value (from Excel). State the rate order n and the overall rate constant kfor this system.Show all work, explaining clearly how you derived the values of n and k.
We can tell from question 1 that this is first order, because In of Concentration(M) VS. Time gives the best linear graph.
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