Actually Chapman proposed a set of 6 reactions, shown below:
(1) | O + O -- > O_{2} |
(2) | O + O_{2} -- > O_{3} |
(3) | O + O_{3} -- > 2 O_{2} |
(4) | O_{3} -- > O + O_{2} |
(5) | 2 O_{3} -- > 3 O_{2} |
(6) | O_{2} -- > O + O |
rate_{1} = k_{1}[O]^{2} .1) Write expressions for the rates of change of the concentrations of O, O_{2}, and O_{3}. For example, noting that oxygen atoms are consumed in steps 2 and 3 and produced in steps 4 and 6. So the rate of change of oxygen atom concentration, d[O]/dt, is the rate of step 4 plus twice that of step 6 minus the rates of steps 2 and 3. Express these rates of change in equations.
2) Chapman examined the equilibrium state of this model, that is, a state in which the rates of change of all species are zero. Take the expressions obtained in exercise 1, set them equal to zero, and solve for [O] in terms of the rate constants. (Hint: take two of the rate equations and solve for [O_{3}] in terms of the rate constants and the other concentrations. Set these two expressions equal to each other, and solve for [O]. You should find that [O_{2}] drops out, and you can solve for [O] in terms of rate constants only.)
3) It turns out not to be possible to solve for [O_{3}] and [O_{2}] separately, but it is possible to solve for the ratio [O_{3}]/[O_{2}] in terms of rate constants only. Do so.
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