Question

Ammonia gas enters the reactor of a nitric acid plant mixed with 30% more dry air than is required for the complete conversion of the ammonia to nitric oxide and water vapor. If the gases enter the reactor at 75 ◦ C (167 ◦ F), if conversion is 80%, if no side reactions occur, and if the reactor operates adiabatically, what is the temperature of the gases leaving the reactor? Assume ideal gases.

Answer 1

4NH3+ 5O2-à4NO2+6H2O

Delta total = [(4*91.3 kJ/mol) + (6*-241.8 kJ/mol.) – {5*0 kJ/mol.) + (4*-45.9 kJ/mol.}
=-902 Kj/mole

Since 4 moles of NH3 react standard heat of reaction = 4*-902*1000=3608000 joules

Material balance : 4 moles of NH3 entering, Oxygen required= 5moles. Since air contains 21%Oxygen moles of air required= 5/0.21=23.81

Air supplied = 30%, air supplied =23.81*1.3=30.95 moles

Conversion is 80%.

Moles of NH3 converted = 4*0.8= 3.2 Oxygen consumed = 5*0.8= 4, moles of NO2 formed =3.2 and water= 3.2*1.5= 4.8 moles (4moles of NH3 produces 6 moles of water).

Oxygen remaining = 30.95*0.21-4=2.5 moles

N2 supplied = 30.95*0.79=24.45 moles

Products contain : NH3= 4-3.2= 0.8 moles, Oxygen =2.5, moles NO= 3.2 moles, H2O= 4.8 moles and Nitrogen = 24.45 moles

Since the reactor is operated under adiabatic conditions. Standard heat of reaction will rise the temperature of products if reference temperature is taken as 75 deg.c

Q= heat of reaction = Enthalpy of products-+Standard heat of reaction- enthalpy of reactants

At reference temperature is 75, enthalpy of reactants = 0

Since the reactor is operated under adiabatic conditions. Standard heat of reaction will rise the temperature of products if reference temperature is taken as 75 deg.c. let the final temperature be T.

Products:

Products	Moles	Specific heat	Enthalpy =
NH3	0.8	35.49 joules/mole.K	0.8*35.49*(T-75)=28.4(T-75)
Oxygen	2.5	29.39	2.5*29.39*(T-75)= 73.475*(T-75)
NO2	3.2	36.97	3.2*36.97*(T-75)=118.3*(T-75)
Water	4.8	33.57	4.8*33.57*(T-75)=161.136*(T-75)
N2	24.45	29.12	24.45*29.11*(T-75)= 712*(T-75)
Total			1093.31*(T-75)

1093.31*(T-75)=3608000

T= 3300.07+75= 3375.07 deg.c