Actually Chapman proposed a set of 6 reactions, shown below:
|(1)||O + O -- > O2|
|(2)||O + O2 -- > O3|
|(3)||O + O3 -- > 2 O2|
|(4)||O3 -- > O + O2|
|(5)||2 O3 -- > 3 O2|
|(6)||O2 -- > O + O|
rate1 = k1[O]2 .1) Write expressions for the rates of change of the concentrations of O, O2, and O3. 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 [O3] 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 [O2] 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 [O3] and [O2] separately, but it is possible to solve for the ratio [O3]/[O2] in terms of rate constants only. Do so.
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