P4-230 The production of ethylene glycol from ethylene chlorohydrin and sodium bicarbonate CH2OHCH2Cl+ NaHCO3 (CH2OH)2+ NaCI + CO2 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.

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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answer7,570,82356 cubic centimeters;