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Practice of Kinetics (Comprehensive Chemical Kinetics, Volume 1)

Practice of Kinetics (Comprehensive Chemical Kinetics, Volume 1)

Practice of Kinetics (Comprehensive Chemical Kinetics, Volume 1)

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2 CONSISTENCY WITH EQNS. OF TYPE -d[A]/dt = k[A]"[BIb 373Combining eqns. (65), (66) and (67), we obtainwhich, irrespective <strong>of</strong> the values <strong>of</strong> a and b, shows thatorwi, 0 as ail 3 0 (c+ve)w,+O as a[,+ -c (c-ve)In other words, the experimental determinations made towards the end <strong>of</strong> areaction contribute very little to the least squares estimate <strong>of</strong> the rate coefficient.Eqn. (68) is <strong>of</strong> little use in computation, however, unless we can make some statementabout a(ai,). There are two clearcut situations which we can discuss. Thefirst <strong>of</strong> these situations is that the value <strong>of</strong> o(a,) does not depend onjand so withinany one run, o(ai,) is a constant. In other words, were it possible to determinea large number <strong>of</strong> values <strong>of</strong> each all, the width <strong>of</strong> the normal distribution would bethe same in each case. Since it is only relative weights which are important incomputation, we can omit such constant factors from the weights [as was done ineqn. (65)] and writewhen a(.[,) is independent <strong>of</strong>j. The alternative situation which permits <strong>of</strong> discussionis that which arises when o(aij) is proportional to aij so that a(ai,)/ai, remainsconstant throughout the run; this situation is described by saying that the relativeerror in ai, is constant. In this caseThe choice between these two alternatives can only be made by considering theexperimental technique employed to determine the concentration <strong>of</strong> A.A simple example <strong>of</strong> common occurrence will illustrate the difference betweenthe two weighting procedures. Consider a simple first order disappearance <strong>of</strong>A whose rate equation isordadt--Ika in differential form,-In a = k(t - to) in integral form.so that f(a) = In a.References p. 407

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