exp7 - Experiment 7, page 1 Version of August 6, 2009...

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Experiment 7, page 1 Version of August 6, 2009 Experiment 446.7 KINETICS OF THE REDUCTION OF METHYLENE BLUE BY ASCORBIC ACID Theory Understanding a chemical reaction requires one to determine the mechanism of reaction, predict it by some plausible set of elementary steps, determine the rate constants involved, and predict the magnitudes of the various parameters, if possible. Mechanisms can be quite involved, as for example in the catalyzed decomposition of ozone or the formation of HCl from H 2 and Cl 2 , even though the overall reaction may appear simple. Consider the simplest single-step mechanisms. For example, one is the transformation of a reactant A to product, which is of first order at all times: oducts Pr A (7.1) For such a first-order reaction, the rate equation is: A A C k dt dC v 1 , (7.2) where C A is the reactant concentration and k 1 is the first-order rate constant . The solution of this equation gives the manner in which the concentration of reactant changes with time: t k C t C A A 1 ) 0 ( ) ( ln . (7.3) This is the linearized form of the rate equation. A plot of the logarithm of the concentration versus t is a straight line, with a slope of – k 1 . From the slope of such a plot, one extracts the rate constant. Some reactions do not go to completion as assumed in the previous paragraph. In the following two-step mechanism, A is converted to B , and B is converted to A , both by first-order processes. A B B A k k   1 1 (7.4) The equation for the disappearance of A by this mechanism is: B A A C k C k dt dC 1 1 (7.5) The equation for the rate of appearance of B is identical in form. The solution of this equation is:  t k k C C C t C A A A A 1 1 ) ( ) 0 ( ) ( ) ( ln , (7.6)
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Experiment 7, page 2 Version of August 6, 2009 where the concentration at any time t is given by C A ( t ). In particular, after a very long time (indicated by ), the reaction may produce a state in which both A and B are present. For this mechanism, the linearized form of the equation is also logarithmic, however the argument of the logarithm is a difference of concentrations, so one has to form a function of concentrations to determine the rate constants, but these are available from the measured concentrations of A . The situation is shown in Figure 7.1. This mechanism illustrates the approach to dynamic equilibrium. It appears from equation (7.6) that the only information from such a graph is the sum of the rate constants. However, at dynamic equilibrium (i.e. at very long times), the rates of these two processes are equal, which allows one to determine the ratio of the rate constants. 1
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This note was uploaded on 02/02/2012 for the course CHEM 446 taught by Professor Staff during the Fall '11 term at University of Delaware.

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exp7 - Experiment 7, page 1 Version of August 6, 2009...

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