**Example 9.5-1 Energy Balance About an Ammonia Oxidizer**One hundred mol NH₃/s and 200 mol O₂/s at 25°C are fed into a reactor in which the ammonia is completely consumed. The product gas emerges at 300°C. Calculate the rate at which heat must be transferred to or from the reactor, assuming operation at approximately 1 atm. The standard heat of reaction for the oxidation of ammonia is given below:\[4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(v) \quad \Delta\overline{H} = -904.7 \text{kJ} \quad \left(\frac{\Delta\overline{H}_r}{s} = -904.7 \text{kJ/s} \right)\]---In this example, the reaction described is the oxidation of ammonia, where gaseous ammonia (NH₃) reacts with gaseous oxygen (O₂) to form gaseous nitric oxide (NO) and water vapor (H₂O). The heat change associated with the reaction under standard conditions is -904.7 kJ per mol of reaction. This implies an exothermic reaction where heat is released. The goal is to determine how much heat needs to be added to or removed from the reactor to maintain the desired exit temperature of 300°C.

From the stoichiometry of the given reaction, 4 moles of NH3 reacts with 5 moles of O2 to produce 4 moles of NO and 6 moles of H2O. For 100 mol/s of NH3, moles of O2 reacted, nO2(reacted) = 54×100 = 125mols Moles of O2 left in product stream, nO2=200-nO2(reacted) = 200-125=75 mols NO formed, nNO=44×100 = 100mols H2O formed, nH2O=64×100= 150 mols Therefore 100 molsof NO is formed per 100 mols of NH3. Substance Inlet moles (n mols) Inlet Enthalpy (H kJmol) Outlet moles (n mols) Outlet enthalpy (H kJmol) NH3 100 0 - - O2 200 0 75 HO2 NO - - 100 HNO H2O - - 150 HH2O From the standard enthalpy table, the enthalpy of O2 at 300°C, HO2=8.47 kJmol To calculate HNO, HNO=∫25300(Cp)NOdT=∫25300(29.5×10-3)+(0.8188×10-5)T-(0.2925×10-8)T2+(0.3652×10-12)T3dT =8.5 kJmol To calculate HH2O, HH2O=∫25300(Cp)H2OdT=∫25300(33.46×10-3)+(0.688×10-5)T+(0.7604×10-8)T2-(3.593×10-12)T3dT =9.6 kJmolFor the given reaction, extent is calculated as, ε=nNH3(inlet)νNH3 where, νNH3=stoichiometric coefficient of NH3ε=100 mols4ε=25 mols The heat of reaction is calculated as, ∆H=ε(∆Hro)+∑products niHi - ∑reactants niHi∆H=( (25mols)(-904.7kJmol)+(75mols)(8.47kJmol)+(100mols)(8.5kJmol)+(150mols)(9.6kJmol)-{(100×0)+(200×0)})∆H=-19692.25 kJmol