$N{O}_x$ from lightning is a very efficient source for tropospheric $O_3$ because it is emitted in the upper troposphere where the lifetime of NOx is long. Assume the loss of NOx is by $N{O}_2$ conversion to $HN{O}_3.$

a$.$ Show that the lifetime of NOx is proportional to:

$(1+\frac{[NO]}{[N{O}_2]})$

b$.$ The $N\frac{O}{N}{O}_2$ conversion ratio is largely controlled by the null cycle:

$NO+O_{3}N{O}_2+O_2$

$N{O}_2+h$

$u{O}_{2}NO+O_3$

Derive from this mechanism an expression for the $N\frac{O}{N}{O}_2$ concentration ratio as a function of $k_1,k_2$ and $O_3.$ Calculate the value of this ratio in the upper troposphere $(10$km altitude, $($:$T=220K_{80}pp{b}_3\}$ and in surface air $,($:$30ppb{O}_3\}$ using $k_1=2{10}^{-12}$exp$(-\frac{1400}{T})c{m}^3$ molecules ${?}^{-1}{s}^{-1},k_2=1{10}^{-2}{s}^{-1}$ and scale height $H=7\mathrm{.}4km.$ You should find that the $N\frac{O}{N}{O}_2$ concentration ratio is much higher in the upper troposphere than in surface air, implying by a much longer lifetime for NOx. $($Note$,$ to solve this problem you will need the pressure at the two altitudes. You will see later in the course that the pressure with altitude $(h)$ equals $p=p_0$exp$(-\frac{h}{H}),$ where po is the pressure at $h=0km$ and is $101$kPa or $1$atm. You can use the Universal gas constant is $R=82\mathrm{.}1$atm $($:$c\frac{m^3}{K}\frac{?}{\text{m}\text{o}\text{l}\text{e}}\}.$