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Chemistry 2710 Review Problem for the March 8 Lecture: Solution

Apply the equilibrium approximation to the first step. The result is tex2html_wrap_inline35 . The rate of the reaction is

displaymath37

The overall reaction is tex2html_wrap_inline39 . If this were an elementary reaction, the rate would be proportional to the product of the hydrogen and iodine concentrations. This is exactly what we get with the approximate rate law, so we can't tell that this reaction isn't elementary from the rate law. There are at least two ways to show experimentally that this reaction isn't elementary, one of which is based on kinetics:

  1. When we say that the first step is fast, we mean that it is fast compared to the rate of the second step under normal experimental conditions. However, since these rates depend on the concentrations, we can imagine reducing the tex2html_wrap_inline41 concentration and increasing the tex2html_wrap_inline43 concentration so that this is no longer the case. If we can make the second step reasonably fast, then the equilibrium approximation is no longer appropriate. Rather, we should use the steady-state approximation to deal with the highly reactive radical intermediate. The result of this approximation is

    displaymath45

    (Try to derive this equation for yourself.) This is one of the rare cases where, in a relatively simple mechanism, the functional form of the SSA is different from that of the EA. The dependence of the rate on the concentration of tex2html_wrap_inline43 is quite different from what we would find in an elementary reaction. It only remains to know whether the experimental conditions required to bring out this relationship are feasible.

  2. Free radicals can be observed directly by electron spin resonance (sometimes called electron paramagnetic resonance). The observation of iodine radicals would of course strongly suggest a non-elementary reaction.


Marc Roussel
Wed Mar 8 11:05:00 MST 2000