Question $4\mathrm{.}16$ of the textbook.

$4\mathrm{.}16.$ Consider the following chain$-$reaction mechanism for the high$-$temperature formation of

nitric oxide, i$.$e$.,$ the Zeldovich mechanism:

O $+$ N$2->$ NO $+$ N $(2$a$)$

N $+$ O$2->$ NO $+$ O $(2$b$)$

The reaction rates of Eqs. $(2$a$)$ and $(2$b$)$ are k$1$f and k$2$f $,$ respectively.

A$.$ Write out expressions for d$[$NO$]/$dt and d$[$N$]/$dt$.$

B$.$ Assuming N atoms exist in steady state and that the concentrations of O$,$ O$2,$ and N$2$

are at their equilibrium values for a specified temperature and composition, simplify your

expression obtained above for d$[$NO$]/$dt for the case of negligible reverse reactions. $($Answer:

d$[$NO$]/$dt $=2$k$1$f $[$O$]$eq$[$N$2]$eq$)$

C$.$ Write out the expression for the steady$-$state N$-$atom concentration used in part B$.$

D$.$ For the conditions given below and using the assumptions of part B$,$ how long does it

take to form $50$ ppm $($mole fraction$\backslash$times $106)$ of NO$?$ T $=2100$ K$,\backslash$rho $=0\mathrm{.}167$ kg$/$m$3,$ M W $=$

$28\mathrm{.}778$ kg$/$kmole$,\backslash$chi O$,$eq $=7\mathrm{.}6\backslash$times $105$ mole fraction, $\backslash$chi O$2,$eq $=3\mathrm{.}025\backslash$times $103$ mole fraction,

$\backslash$chi N$2,$eq $=0\mathrm{.}726$ mole fraction, and k$1$f $=1\mathrm{.}82\backslash$times $1014$ exp $38370/$T with units of cm$3/$gmol$-$s$.$

E$.$ Calculate the value of the reverse reaction rate coefficient for the first reaction, i$.$e$.,$ O $+$ N$2$ NO $+$ N

for a temperature of $2100$ K$.$

F$.$ For your computations in part D$,$ how good is the assumption that reverse reactions are

negligible? Be quantitative.

G$.$ For the conditions of part D$,$ determine numerical values for $[$N$]$ and $\backslash$chi N$.$ Note: k$2$f $=$

$1\mathrm{.}8\backslash$times $1010$T exp $4680/$T with units of cm$3/$gmol$-$s$.$

Hint: The answer to part $($D$)$ is: about $37$ msec. The answer to part $($G$)$ is: about $9\mathrm{.}4\backslash$times $103$

ppm$.$