Water$-$treatment processes that generate hydroxyl radicals are often

used to remove organic chemicals that contaminate groundwater. The

hydroxyl radical $(*OH)$ is a very reactive free$-$radical species that can be

thought of as a piece of a molecule. In an attempt to again become a

whole molecule, $*OH$ will attack almost any organic molecule, either

adding to a double bond or pulling off a hydrogen atom to become water.

Because of this property, the hydroxyl radical can be used to degrade

organic contaminants completely to carbon dioxide and water, providing

a remediation tool that leaves no harmful by$-$products.

In some applications, a small amount of toxic material like trictioro$-$

ethylene $($TCE$)$ must be destroyed in the presence of a much lager

quantity of "nontarget" material such as natural organic material $($Exer$-$

cise $9\mathrm{.}3),$ or$,$ in the case of industrial wastewater, some organic material

that is either environmentally benign or can subsequently be treated

more cheaply by biological methods. This material, usually characterized

instrumentally by the measurement of dissolved organic carbon $($DOC$),$

may compete for hydroxyl radical with the contaminant to the extent

that the economic feasibility of the process may depend on the outcome

of this competition.

Consider the removal of TCE. The rate equations for TCE and the

hydroxyl radical are

$\frac{d[TCE]}{dt}=-k_{oT}[*OH][TCE]$

and

$\frac{d[*OH]}{dt}=-k_{Or}[*OH][TCE]-k_{OD}[*OH][DOC]+S_{\text{O}\text{O}\text{H}}$

where it is assumed that the treatment produces the $*OH$ radical at a

constant rate $S_{-OH}.$ In this problem, DOC is assumed to be present in

sufficient excess that its concentration does not change significantly

during treatment. Thus, the experimentally determined DOC may not

change. If the rate constant of the resulting material also does not

change much, from the point of view of the effect of the material on

the system kinetics, there appears to be no change in the material. The

hydroxyl radical reacts essentially as fast as it is produced, so that

$d\frac{*OH}{d}t=0.$

$($a$)$ Show that the steady$-$state concentration of $*OH$ is given by

$[*OH]=\frac{S_{*OH}}{k_{OT}[TCE]+k_{OD}[DOC]}$

Substitution of this expression for $*OH$ into the TCE rate equation

yields

$\frac{d[TCE]}{dt}=-(\frac{K_{OT}[TCE]}{k_{OT}[TCE]+k_{OD}[DOC]})+S_{OH}$