A. Draw the process flow diagram noting Temps, pressures, and phase changesB. Preform a degree of freedom analysis (Acetylene is produced from methane in the reaction:2CH4 (g) → C₂H2 (g) + 3 H₂ (g)An undesired side reaction is the decomposition of acetylene:C₂H₂ (g) 2 C(s) + H₂ (g)Methane is fed to the reactor at 1500°C at a rate of 10 mol CH4/s. The reaction is carried outisothermally at a constant pressure of 1atm. The fractional conversion of methane is 0.6 and theyield of acetylene is 0.417 mol C₂H₂/mol CH4 consumed. Determine the magnitude and directionof heat transferred for this reaction.

Acetylene is produced from the methane.Process flow chart is shown below: Degree of freedom analysis: Number of unknowns = 5 (n1, n2, n3, n4, Q) Number of independent equations = 2 (C and H balance) Number of energy balance equation = 1 Number of other relations = 2 (conversion of methane and yield of ethylene) Therefore, DOF = 5-2-1-2 DOF = 0 The reactions are as follows: 2 CH4 g → C2H2 g + 3 H2 g ............ (1) C2H2 g → 2 C s + H2 g ............... (2) Fractional conversion of methane = 0.6 Methane reacted = 6 mol/s Therefore, n1 = 4 mol/s (10-6) yield of C2H2 = 0.417 Methane consumed = 6 mol/s Therefore, n2 = 2.5 mol/s Apply C atom balance 1×10 =1×4 + 2×2.5 + n4n4 = 1 mol Cs s Apply H atom balance 4×10 = 4×4 + 2×2.5 + 2×n3n3 = 9.5 mol H2 gs Enthalpy is calculated at a reference temperature. Reference temperature = 25 0C Heat capacity of each species is as follows: CH4 (g) = 0.079 kJ/mol.0C C2H2 (g) = 0.052 kJ/mol.0C H2 (g) = 0.031 kJ/mol.0C C(s) = 0.022 kJ/mol.0C Therefore, the enthalpy of products stream becomes, HCH4 = 4 mols×0.079 kJmol.°C×1500-25 °C= 466.1 kJsHC2H2 = 2.5 mols×0.052 kJmol.°C×1500-25 °C= 191.75 kJsHH2 = 9.5 mols×0.031 kJmol.°C×1500-25 °C= 434.3875 kJsHC = 1 mols×0.022 kJmol.°C×1500-25 °C= 32.45 kJsThus,∑ProductH = 1124.6875 kJs Enthalpy of feed stream is calculated as follows: ∑reactaantsH = 10 mols×0.079 kJmol.°C ×1500-25 °C∑reactaantsH= 1165.25 kJs Standard enthalpy change of reaction is calculated as follows: ∆Hr,10 = 1×227.48 kJmol + 3×0 - 2×-74.52 kJmol= 376.52 kJmol∆Hr,10 = 0 + 0 - 1×227.48 kJmol= -227.48 kJmolThus,∆Hr = 376.52-227.48= 149.04 kJmol CH4 reacted × 6 mol CH4 reacteds∆Hr= 894.24 kJsApply the energy balance ∑outH - ∑inH + ∆Hr = Q1124.6875 -1165.25 + 894.24 = QQ = 853.6775 kJs or kW As the sign of heat transfer is positive, it means heat is being transferred to the reactor.