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:

All the data was performed on polymath educational. If you have the educational version of plymath copy paste the following code to get the required values of moles of glycol formed. You can also get it for free from the polymath education free website which is free for 15 days.

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