Question

Nitric oxide (NO) is a gas that decomposes in air, therefore it is usually made when it is needed immediately. The most common use of nitric oxide is as fertilizer. In order to produce nitric oxide, 25 kmol/h of ammonia (NH3) and a stream of oxygen (02), both at 25°C and 1 atm, are fed into a continuous reactor packed with catalyst in which ammonia is completely consumed. The product gas emerges at 226 °C and 1 atm. Ammonia is burned to form nitric oxide in the following reaction: NH3(g) + 1.250(9) NO(g) + 1.5H,0(1) Based on the above information, answer the following questions. (a) Construct a flowchart and label the streams for the process described. [5 marks] (b) Determine the oxygen feed rate (kmol/h) that would correspond to 60 % excess 02. [4 marks] (c) Determine the rate of heat (kW) at which heat must be transferred to or from the reactor. [18 marks] (d) After 2 years, the percentage conversion of ammonia has decreased from 100 % to 70% due to the degradation of the catalyst in the continuous reactor. Propose a solution to increase the percentage conversion of ammonia without changing the catalyst or the operation pressure/temperature. [2 marks]

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Answer 1

[ 12 25 Pelato NH₃ = 25 lemol (g) Reactor [T=226C, P = 1 atm I NO(g) unreacted O₂ [g O2 (g) [T = 25C, P-latm H₂O(e) [T=226°C, Pzlatm] 12NH2 + 50 (g) (g) 9 Η NO (g) + 3H2O les From stoichometry we can say u moly of WH₂ completely reacts with smoked 25 kmol of NH₃ require = Sexes knol of les kool of 02. to exeet = 60 oxygen bed rate 125 +60% of 125 % of 125 1 kmol 125 + 3x43)= sokma oxygen feed rate - so kmal

29.3763 mol-k 31.104 I mol-k 30.51I mol-c 58.514 J molt mo Fast Total Enthalpy rate in Property Table: component specific Heat capacity T=25c T= 226C [cp, 25 [CP, 226 NH3(g) 37 J mol-k O2 (g) No' (g) H₂oce) Heat of reaction A Hoon ngo4 kJ Q since, such properties and heat of reactions are not mentioned in the question! so I am using above values obtained from google Assumption Heat of reaction does not vary with temprature. Analysis Abblying the energy balance across the reactor, COH) rate out wher, o= Heat removal rate from the reactor our rate of energy seland cline toren

To calculate specific enthalky of stream, I am taking de afa enthalpy Table reference state specific specific component enthalpy enthalpy at at inlet outlet NH3 Cp, 25 x 25 =37x25 =925 J/mole p, 25x25 229.376x25 € 73414 Il mole O2 CP, 22 x 226 = 31104x226 Famil NO (p, 226*226 H₂O = 30.518226 Ilmes) CP 26 x 226 -58.slux 226 (3224.164 I mel Molar Flow rate at outh Stoichiometric table - Molar Flow component ratemoles at Inlet NH3 25kmol h Roboch sokmo h u mol NO 135 kmol h. 75 но- h

[Total Enthalby rate in = (Enthalpy rate ] NH3 +Enthalpy sate Jo. 25 kmol X 9253 mole 734.4 I male [sokman 61623 59845 kJ hr 16.623 kJ sec mo Total Enthalpy rate Jout = Enthal by rutales + Centrally + Enthalfy rate] H₂O NO - MI 75 kmol x 31.64 x 2265 25kmol x 30.51 226 h mol mol + [ * 13224-144] seco et 851866.6 kJ 153.29 kJ hr Now substituting there values in eah , AH+[16.623-153.29] = 0 Х Algoux35 904 6J X 25 kmol mol H DH = 904 ILJ x 25 x 10 mol mol ch An = 6277 78 KI sec '1h 36.okee

From @ and ② (6277,78 k of [16:223 - 153:29] Ic] Sec secJ Q = 6146.7 les See az 6140.7 kw This much rate of heat muut be transferseed from the reactor. Atgan – ive It means that reaction is exothermic in nature. If we want to increase the percentage Conversion of Ammonia, then we have to increase the Temprature of reactor. Increase in Temprature will result into increase in rate constant which will further result into increase in rate of reaction. And this increase in rate of reaction will resu into increase in conversion,