Question

I need help in hurry, please

Sulfur dioxide (SO2) and hydrogen sulfide are toxic byproducts of many industrial processes. To prevent their release to the environment, they can be neutralized to form elemental sulfur through the Claus process, which follows the reaction: 2H2S + SO2 + 2H20 + 35 You have been tasked with neutralizing sulfur dioxide and hydrogen sulfide in a continuous reactor. The reagents arrive in stoichiometric proportion at 250°C and 100 kPa. The feed flowrate is 10 STP m/min. The reactor operates at 380°C and 250 kPa with 60% conversion of S02. Assume the entire process operates in the gas phase and all reactants and products are gases. Note: elemental sulfur exists as a solid in nature. The standard heat of formation of sulfur gas is +277 kJ/mol S. For all species you may use only the "a" term of the heat capacity to reduce the complexity of your calculations. The value of “a” for the heat capacity of gaseous sulfur is 27.46x103 kJ/mol°C. How much heat should be added or removed (specify which to maintain a temperature of 380°C in the reactor? Focus

Answer 1

component Assumption Gass behave as an ideal gas. 2 specific heat capacities of all the species are independent of tembrature, and pressure. 6 Heat of reaction Properties specific heat capacity of elemental sulphen=24.46 I specific heat capacity table, specific heat capacity (molo] H₂S (g) 34.6 SO2 (g) 39.87 33.552 Heat of formation table, component Heat of formation [, KI / moi] H₂S CG) -2016 SO2 (g) - 296.81 H₂O(g) - 241.82 specific heat capacity and heat of formation data is obtained from google. Because it was not mentioned in the question guol H₂O(g) Analysis y Reactor unreacted H2S Junneacted soz H₂O H₂S T=250c T=30°C 502 T=380 C = 16 STPM UP (W) Feed low rate m/min STP condition indicates, T=273k, P= Iban I Now using ideal gas equation to find the molar flow rate, i= PV (10spa) + (10m/min) RT 8314 34 ml-ce( 2731) i = 440,58 mol min

since H₂S and so are bed in stoichiometric proportion, mole fraction of Hasz mole braction of some molas flow rate ob HeS = ns = { x440-58 = 293.72 mol min me molar flow rate of soa = nso, 73x440.58 = 146.86 mol Stoichiometric table for 60% conversion of soa component molan Flow rate molar flow rate at inlet of reactor in product stream H₂S 293.72 293.72- 3.12 40.6x146.26€ 205 tony Soz 146.86 146.86(1-0.6) = 58.744 onin @ @ H2O 2x146.868016) = 176.232 3x(146-8506) = 264.348 2H₂S SO₂ 2 H₂O + 35 Heat ab sxn per mole of 502 = (2x (-241.82) +3 (277)]- (2x (-2016) + 1X(-296.21) = [2x (-241.82) +3x277]-[2xl-2016) +1x(-296-81)] seem of Heat of formation Sum of heat of formation of neactant = 685.37 KJ mol specific enthalby table with respect to oc reference temperature componeht Feed stream 2 Product stream [ Ilmol] To asoc of product Ahran 0 H₂S 34.682so 3416x380 SOZ 39.878380 39.87X250 ③ H2O 33.552x380 27,46 X 380

due torder Now applying the energy balance across the nearton, å + [total enthalpy] in - [Total Enthalby out later online where d= rate at which energy added to reactor (Total enthalpy] in Enxspecific enthalpy] Fred Stream 12931762 mol ) x (34.6 x 250]3+ 146.86 Mel x 33.27 +25] = [293.72 modelle xf457250) I]+[146-26 milho 39:274250) mei] mol min. mol - 4o045oS 105 I min = 4oolis kJ min [Total enthalpy out = ({ na specific enthalty product stream = [17488 346* 32.] + [58-74c4 x 39.27X120] + [176-232X 33.552x32.] + Ceasuri 7446073.285 I min - 7440.073 kJ min converted into product rate of energy consumption due torn= (HRK) & moles of so, = 685.57 ) f1462 X06 meta] = 60392.06 kJ min Now butting all the values in energy balance equation, 0 073 KI 160392.06 ks a = 636271635 kJ min This much heat must be added to the reactor to maintain its temprature at 385c.