The production of ethylene glycol from ethylene chlorohydrin and sodium bicarbon

Question

The production of ethylene glycol from ethylene chlorohydrin and sodium bicarbonate: CH_2OHCH_2CI + NaHCO_3 rightarrow (CH_2OH)_2 + NaCI + CO_2 is carried out in a semibatch reactor, A 1.5 molar solution of ethylene chlorohydrin is fed at a rate of 0.1 mole/min to 1500 L of a 0.75 molar solution of sodium bicarbonate. The reaction is elementary and carried out isothermally at 30 degree C where the specific reaction rate (rate constant) is 5.1 L/mol/h. Higher temperature produce unwanted side reaction products. The reactor can hold a maximum 2500 L of liquid. Assume constant density and CO_2 leaves the reactor as fast as it is formed. Plot the conversion, reaction rate, concentration of reactants and products and number of moles of glycol formed as a function of time.

Explanation / Answer

DIFERENTIAL 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]

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.