What is Rate Law of the reaction below?$$\ce{4 HBr + O2 -> 2 H2O + 2 Br2}$$ How do we approach finding the rate law of reaction? Can this be found out only through experiments?
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Preamble
For a reaction with no intermediate steps, you could derive the rate law from the reaction equation. For a generic equation $$\ce{$a$A + $b$B -> C}$$ the rate is given by $$r = k\cdot c^{a}(\ce{A})\cdot c^{b}(\ce{B}).$$ Rigorously, this is only possible for elementary reactions. A prominent example for a strictly unimolecular reaction is radioactive decay. Also quite commonly known is the photolytic or thermal cleavage of hydrogen or bromine: \begin{align} \ce{H2 &->[\Delta T] 2H. }\\ \ce{Br2 &->[h\cdot\nu] 2Br. } \end{align} An example for a bimolecular reaction would be $$\ce{2NO2 -> NO3 + NO}.$$
Reaction mechanism of the combustion of hydrogen bromide
This seems to be a reaction with a fairly complicated mechanism, involving multiple species, most of them radical. I would assume, that it is somewhat similar to the combustion of hydrogen, that I have described here.
A quick search on some forums will turn out a reaction mechanism, which I think is wrong. But I will show it here anyways, since it gives you a first clue, about how complicated this reaction might be. (Note: These are supposed to be gas phase reactions.) \begin{align}\ce{ HBr + O2 &-> HOOBr\\ HOOBr + HBr &-> 2 HOBr\\ HOBr + HBr &-> H2O + Br\\ }\end{align} The following mechanism is taken from Lilian G. S. Shum and Sidney W. Benson Int. J. Chem. Kin., 2004, 15 (4), 341-380. I think this is a much more suitable and credible approach to the problem. (Note: Most of the reactions are in equilibrium.) \begin{align}\ce{ HBr + O2 &~<=> HO2 + Br\\ HO2 + HBr &~<=> H2O2 + Br\\ H2O2 + 2HBr &~->T[wall] 2H2O + Br2\\ Br + Br + M &~<=> Br2 + M\\\hline 4HBr + O2 &~<=> 2H2O + 2Br2\\ }\end{align}
Unfortunately I was unable to determine what $\ce{M}$ refers to in this case. It might be any inerat gas used as a solvent in the reaction or the wall.
Rate law of the combustion of hydrogen bromide
As a consequence of the mechanism you cannot find the rate law from theory, but only from experiments. It is important to realise, that most of the reactions are in equilibrium and the whole system is dependent on the concentrations of each of the involved species. Therefore the kinetics of this reaction may be expressed as a complicated system of coupled differential equations.
However, Shum and Benson conclude, that the first step will be the rate-determining and the rate expression is given through $$\frac{\mathrm{d}[\ce{H2O}]}{\mathrm{d}t} = 2k_g[\ce{HBr}][\ce{O2}].$$ Therefore it can formally be regarded as a bimolecular reaction.
answered Jan 20 '15 at 3:48
