P$9-6($Tribology$)$ Why you change your motor oil? One of the major reasons for engine$-$oil degradation is

the oxidation of the motor oil. To retard the degradation process, most oils contain an antioxidant $[$see

Ind. Eng. Chem. $26,902(1987)].$ Without an inhibitor to oxidation present, the suggested mechanism

at low temperatures is

$I_2{k}_{0}2I*$

$I*+RH{k}_{i}R*+HI$

$R*+O_2{k}_{p1}R{O}_2$

$R{O}_2+RH{k}_{p2}\frac{R}{O}OH+R*$

$2R{O}_2{k}_t$ inactive

where $I_2$ is an initiator and $RH$ is the hydrocarbon in the oil.

When an antioxidant is added to retard degradation at low temperatures, the following additional ter$-$

mination steps occur:

$R{O}_2*+AH{k}_{41}\frac{R}{O}OH+A*$

$A*+R{O}_2{k}_{A2}$ inactive

$($a$)$ Derive a rate law for the degradation of the motor oil in the absence of an antioxidant at low tem$-$

peratures.

$($b$)$ Derive a rate law for the rate of degradation of the motor oil in the presence of an antioxidant for

low temperatures.

$($c$)$ How would your answer to part $($a$)$ change if the radicals $I*$ were produced at a constant rate in

the engine and then found their way into the oil?

$($d$)$ Sketch a reaction pathway diagram for both high and low temperatures, with and without antioxidant.

$($e$)$ See the open$-$ended problem G$.2$ in Appendix G and on the CRE Web site for more on this

problem.