P4-230 The production of ethylene glycol from ethylene chlorohydrin and sodium bicarbonate CH2OHCH2Cl+ NaHCO3 (CH2OH)2+ NaCI + C
O2 is carried out in a semibatch reactor. A 1.5 molar solution of ethylene chloro-hydrin is fed at a rate 0.1 mole/minute to 1500 dm^3 of a 0.75 molar solution of sodium bicarbonate. The reaction is elementary and carried out isother-mally at 30�C where the specific reaction rate is 5.1 dm^3/mol/h. Higher temperatures produce unwanted side reactions. The reactor can hold a maximum of 2500 dm^3 of liquid. Assume constant density. (a) Plot the conversion, reaction rate, concentration of reactants and products, and number of moles of glycol formed as a function of time.
1 answer:
Answer:
DIFFERENTIAL EQUATIONS
d(Ca)/d(t) = - K*Ca*Cb + (Fao/V)*(1-(Ca/1.5)) # Concentration of A
d(Cb)/d(t) = - K*Ca*Cb - (Fao*Cb)/(V*1.5) # Concentration of B
d(Cc)/d(t) = K*Ca*Cb - (Fao*Cc)/(V*1.5) # Concentration of C
d(Cd)/d(t) = K*Ca*Cb - (Fao*Cd)/(V*1.5) # Concentration of D
d(Ce)/d(t) = K*Ca*Cb - (Fao*Ce)/(V*1.5) # Concentration of E
# AUXILIARY EQUATIONS
K = 5.1 # Reaction Rate Constant, [dm3/mol.h]
r = K*Ca*Cb # Reaction Rate, [mol/dm3.h]
V = Vo + (Fao/1.5)*t # Volume, [dm3]
X = 1 - (Cb*V)/(0.75*Vo) # Conversion
Nc = 0.75*Vo*X # Moles of C formed
# INITIAL CONDITIONS
Ca(0) = 0
Cb(0) = 0.75
Cc(0) = 0
Cd(0) = 0
Ce(0) = 0
Fao = 6 # Feed of A, [mol/h]
Vo = 1500 # Initial Volume, [dm3]
t(0) = 0 # Initial Time, [h]
t(f) = 250 # End Time, [h]
Explanation:
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