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Unformatted text preview: Answer Key Post‐lab Quiz Kinetics Glucose Lab Chemistry 223 Fall, 2010 1. You quenched the reaction after 25 minutes. By that time, the rate of reaction was about half of what it was initially. Suppose you were able to quench the reaction with an uncertainty of ± 10 s. a) what is the relative standard deviation in the reaction time? (1) 25 minutes *60 s min‐1 = 1500 s. RSD = 10 s /1500 s = 1/150 (or 0.0067 or 0.7%) b) would you expect the relative standard deviation in concentration due to the uncertainty in quench time to be GREATER THAN, EQUAL TO, or LESS THAN the uncertainty in reaction duration (while you may do a propagation of error calculation to figure this out, qualitative reasoning is a lot faster and entirely adequate) (1) c) Explain your reasoning in b) (1) Because the rate is slower than at the beginning of the reaction, the rate of product formation is slower than it was initially. Thus, if the signal changes slowly, the effect of time uncertainties is less than when the signal changes rapidly. At equilibrium, the signal is time‐independent, so the closer one is to equilibrium, the less RSD is due to uncertainties in timing. 2. Which would have a greater influence on the accuracy of the working curve, a 10% change in the activity of GLUCOSE OXIDASE or a 10% change in the activity of HORSERADISH PEROXIDASE? (1) Explain your reasoning. (1) The rate of diimine formation is proportional to the rate of glucose consumption, which is related to the activity of glucose oxidase. As long as the horseradish peroxidase is active enough to rapidly consume all the H2O2 produced by glucose/glucose oxidase, it doesn't much matter what the rate constant is. ...
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This note was uploaded on 01/06/2011 for the course CHEM 223 taught by Professor Scheeline during the Fall '08 term at University of Illinois at Urbana–Champaign.
- Fall '08