Ammonia is oxidized with air to form nitric oxide in the first step of the manufacture of nitric acid. Two reactions occur. In the first oxidation reaction, nitric oxide and water are formed. In the second oxidation reaction, nitrogen and water are formed.

The reactor is fed with gaseous ammonia and air. The ammonia feed rate is $250mo\frac{l}{m}in$ at a temperature of $25C$ and a pressure of $15$ bar whilst the air is fed at a temperature of $200C$ and a pressure of $12$ bar. Product leaves the reactor at $800C$ and $8$ bar at the following component flows: $180$molN$\frac{O}{m}in,350mol{H}_2\frac{0}{m}in,1432$mol$\frac{N_2}{m}in$ and $138$mol$\frac{O_2}{m}in.$

$($a$)$ Write down the reactions and determine the air supply to the reactor in moles and its composition in volume $\%$ if air is assumed to consist of nitrogen and oxygen.

$[7$ marks$]$

$($b$)$ Determine the heat transferred to or from the reactor to give these reactors the desired operating conditions. The answer should be given in SI unit.

To calculate the specific heat capacity, follow the convention of using average temperature. Assume ideal gas conditions and that the heat capacity is independent of pressure.

Draw a clearly labelled process flow diagram.

Data:

Heat of formation at $25C$ and $1$atm

$Hf(N{H}_3)=-46\mathrm{.}19k\frac{J}{m}ol$

$H_f(NO)=+90\mathrm{.}37k\frac{J}{m}ol$

$H_f(H_{2}O)=-241\mathrm{.}83k\frac{J}{m}ol$

$H_f(N_2)=0k\frac{J}{m}ol$

$H{f}_f(H_2)=0k\frac{J}{m}ol$

Heat capacity where temperature is in $C$

$N_2$:$29{10}^{-3}+0\mathrm{.}2199{10}^{-5}T+0\mathrm{.}5723{10}^{-8}{T}^{2}k\frac{J}{m}ol.C$

$O_2$:$29\mathrm{.}1{10}^{-3}+1\mathrm{.}158{10}^{-5}T-0\mathrm{.}6076{10}^{-8}{T}^{2}k\frac{J}{m}ol.C$

$N{H}_3$:$35\mathrm{.}15{10}^{-3}+2\mathrm{.}954{10}^{-5}T+0\mathrm{.}4421{10}^{-8}{T}^{2}k\frac{J}{m}ol.C$

$H_{2}O$:$33\mathrm{.}46{10}^{-3}+0\mathrm{.}688{10}^{-5}T+0\mathrm{.}7604{10}^{-8}{T}^{2}k\frac{J}{m}ol.C$

$NO$: $29\mathrm{.}5{10}^{-3}+0\mathrm{.}8188{10}^{-5}T-0\mathrm{.}2925{10}^{-8}{T}^2+0\mathrm{.}3652{10}^{-12}{T}^{3}k\frac{J}{m}ol.C$

$[18$ marks$]$