1. The elementary gas-phase reactionA + B + 2Cis carried out in a packed-bed reactor. The entering molar flow rates are FA-5 mol/s, F= 2 FAO,and Fx = 2FAO, with CAO-02 mol/dm³. The entering temperature is 325 K.Additional Information:CPA-CPR-CIC=20 cal/mol/K, k = 0.0002 dm/(kg-mol-s) @ 300KCPI = 20 cal/mol/K, a=0.00015 kg¹, E = 25 kcal/molAHR=-20 kcal/mol @298K, Ke-1000@ 305K(a) Write the mole balance, the rate law, Kc as a function of T, k as a function of T, and CA, CB,C, as a function of X, p, and T.(b) Write the rate law as a function of X, p, and T.(c) Show the equilibrium conversion isX₂ =3K-3KG2-2K (K-1)2(K-1)4and then plot Xe vs. T.(d) What are Σ, Cpi, ACP?(e) Write the energy balance for adiabatic operation.(f) Plot and then analyze Xe, X, p, and T versus W when the reaction is carried out adiabatically.Describe why the profiles look the way they do.

Write the reaction as follows: A +B→ 2C (a) Write the mole balance as follows: -rA = FA0dXdV Write the rate law for the reaction as follows: -rA = kCACB-CC2KcWrite the expression for the equilibrium constant as follows: Kc = Kc1T1exp∆HrxnoR1T1-1T The initial temperature, T1 = 305 K The initial equilibrium constant, Kc1 = 1000 The enthalpy of the reaction, ∆Hrxno = -20 kcal/mol Gas constant, R = 1.987 cal/molK Substitute the corresponding values in the above equation Kc = Kc1T1exp∆HrxnoR1T1-1TKc = 1000×exp-20 kcal/mol1.987 cal/mol1305 K-1TKc = 1000×exp-100651305 K-1TWrite the relation between rate constant and temperature as follows: k = k1expER1T1-1T The initial temperature, T1 = 300 K The initial rate constant, k1 = 0.0002 dm6kg·mol·s The activation energy, E = 25 kcal/mol Substitute the corresponding values in the above equation k = k1expER1T1-1Tk = 0.0002 dm6kg·mol·s×exp25 kcal/mol1.987 cal/molK1300 K-1Tk = 0.0002 dm6kg·mol·s×exp125821300 K-1TWrite the final concentrations of each component as follows: CA = CA01-XT0TPP0CA = CA01-XT0Ty CB = CB01-XT0TPP0aCB = CA0ΘB-XT0Ty CC = 2CA0XT0TPP0CC = 2CA0XT0Ty Where, ΘB = FB0FA0 = 2FA0FA0 = 2 Therefore, the required solutions are -rA = FA0dXdV, -rA = kCACB-CC2Kc, Kc = 1000×exp-100651305 K-1T, k = 0.0002 dm6kg·mol·s×exp125821300 K-1T, CA = CA01-XT0Ty, CB = CA0ΘB-XT0Ty, and CC = 2CA0XT0Ty.(b) Write the rate law in terms of X, P, and T as follows: -rA = kCACB-CC2Kc Substitute the corresponding values in the above equation. -rA = kCACB-CC2Kc-rA = kCA01-XT0TyCA02-XT0Ty-2CA0XT0Ty2Kc-rA = kCA0T0Ty21-X2-X-4X2Kc Therefore, the required soltuion is -rA = kCA0T0Ty21-X2-X-4X2Kc.(c) At equilibrium, -rA = 0 Substitute the value of -rA in the rate equation and calculate the equilibrium conversion as follows: -rA = kCA0T0Ty21-X2-X-4X2Kc0 = kCA0T0Ty21-Xe2-Xe-4Xe2Kc0 = 1-Xe2-Xe-4Xe2Kc4Xe2 = Kc2-Xe2-3Xe Kc-4Xe2-3KcXe+2Kc = 0The above equation is in the form of quadratic equation. The soltuion is Xe = --3Kc±-3Kc2-4Kc-42Kc2Kc-4Xe = 3Kc±3Kc2-8KcKc-42Kc-4Xe = 3Kc4-3Kc42-2KcKc4-12Kc4-1 Therefore, the required soltuion is Xe = 3Kc4-3Kc42-2KcKc4-12Kc4-1. Now draw the graph between Xe vs T as follows: From the above graph, we concluded that as the temperature increases the equilibrium conversion decreases.(d) Calculate the values of ∑ΘiCPi as follows: ∑ΘiCPi = ΘACPA+ΘBCPB= CPA+2CPB= 20 cal/molK+2 20 cal/molK= 60 cal/molK Calculate the value of ∆CP as follows: ∆CP = 2CPc-CPB-CPA= 220 cal/molK-20 cal/molK-20 cal/molK= 0 Therefore, the required solutions are ∑ΘiCPi = 60 cal/molK and ∆CP = 0.