The elementary gas-phase reaction (ch3)3cooc(ch3)3→c2h6+2ch3coch3 is carried out isothermally in a flow reactor with no pressure
drop. the specific reaction rate at 50 ºc is 10-4 min-1 and the activation energy is 85 kj/mol. pure di-tert-butyl peroxide enters the reactor at 10 atm and 127 ºc and a molar flow rate of 2.5 mol/min. a.calculate the reactor volume and space time to achieve 90% conversion in a cstr.
b.assume that the reaction is reversible with kc = 0.025 mol2/dm6. calculate the equilibrium conversion. redo part
a.to achieve a conversion that 90% of the equilibrium conversion. polymath or a similar numerical solver is very useful for solving the set of equations.
When: (CH3)3COOH(CH3)3⇒ C2H6 + 2CH3CoCH3 A B 2C SO first for A: When feed rate to reactor = FA(o), and Effluent rate from reactor= FA=Fa(o)(1-x)
Second for B: When feed rate to reactor = 0, and Effluent rate from reactor FB= FA(o)X
Third for C: When feed rate tp reactor = 0, and the Effluent rate from reactor FC= 2FA(o)X
So, the total feed rate to reactor FT(o) = FA(o) and the total Effluent rate from reactor FT = FA(o)(1+2X) For the CSTR: V= FA(o) X / γa where -γ = KCA We can get the total concentration at any point CT and the entrance CT(o) from this equation: CT= FT/Q= P/ZRT and CT(o) = FT(o)/ Q(o) = P(o)/ Z(o)RT(o) We can neglect the changes in Z (the compressibility factor) So: Q = Q(o) FT/FT(o) P(o)/P T/T(o) Because there is no pressure drop and for the isothermal system we can assume P & T are constants: ∴ Q = Q(o) FT/ FT(o) ∴ CA = FA / Q = (FA(o)(1-X)) / (Q(o)(1+2X)) = CA(o)((1-X)/(1+2X)) -----> eq 1 When K (the reaction rate constant) at 127C° can be evaluated with R (the gas constant) 8.314 J/mole.K° = 0.08205 L.atm/mole.°K
When, K2=K1(exp)[E/R ( (1/T1) - (1/T2)] when T1 by kelvin = 50+273= 323 K° and T2= 127+273 = 400 K° and activition energy = 85 KJ/mol = 85000 J/mol ∴ K2 =10^-4 (exp) [85000/8314 ((1/323) - (1/400)] = 0.0443 min^-1 We can get the concentration of (A) at 10 atm and 127C° (400K°) by:
∴ The reactor volume of the CSTR is: V = FA(o)X / KCA---> eq 2 by substitute eq 1 in eq 2 ∴ V=[ FA(o)x * (1+2x)] / [ KCA(o) * (1-X) ] = [ 2.5 * 0.9 * (1+1.8)] / [0.0443 * 0.305* 0.1] = 4662.7 L - The equilibrium conversion: according to the reaction: A⇔B + 2C and we have: -γ = KfCA - Kb CB( C(C)^2) = 0 and KC = Kf / Kb = (CB* CC^2)/ CA= 0.025(as it is agiven)
when we have CA(o) (the initial concentration)= 0.305 mol/L, So we have this eq: [(XCA(o)) (2XCA(o))^2] / [(1+2X)^2 CA(o) (1-X) ] = 0.025 By the matlab function solve ∴ X= X eq = 0.512 for 90% of the equilibrium conversion ∴ X= 0.9 X eq = 0.4608
