The production of ethylene glycol from ethylene chlorohydrin and sodium bicarbonate ↑ is carried out in a semibatch reactor. A 1.5-molar solution of ethylene chlorohydrin is fed at a rate of 0.1 mole/minute to 1500 dm3 of a 0.75-molar solution of sodium bicarbonate. The reaction is elementary and carried out isothermally at 30°C where the specific reaction rate is 5.1 dm3/mol/h. Higher temperatures produce unwanted side reactions. The reactor can hold a maximum of 2500 dm3 of liquid. Assume constant density.

See explanation.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

For which the differential equations in terms of the time variation are:

And the implicit equations:

Thus, on the attached pictures you will find the conversion, concentrations, rate and volume profiles as well as the number of moles of ethylene glycol considering a maximum volume of 2500 dm³.

This is the code I used on matlab:

-

t=0;

na(1,1)=0;

nb(1,1)=1125;

nc(1,1)=0;

nd(1,1)=0;

V(1,1)=1500;

k=length(na);

C_A(1,1)=0;

C_B(1,1)=0.75;

x(1,1)=0;

t(1,1)=0;

r_A(k,1)=0;

while V(k,1)<2500

   na(k+1,1)=na(k,1)+0.01*(6+r_A(k,1)*V(k,1));

   nb(k+1,1)=nb(k,1)+0.01*(r_A(k,1)*V(k,1));

   nc(k+1,1)=nc(k,1)+0.01*(-r_A(k,1)*V(k,1));

   nd(k+1,1)=nd(k,1)+0.01*(-r_A(k,1)*V(k,1));

   V(k+1,1)=V(k,1)+0.01*(4-(-r_A(k,1))*V(k,1)*44/1000);

   r_A(k+1,1)=-5.1*C_A(k,1)*C_B(k,1);

   C_A(k+1,1)=na(k+1,1)/V(k+1,1);

   C_B(k+1,1)=nb(k+1,1)/V(k+1,1);

   C_C(k+1,1)=nc(k+1,1)/V(k+1,1);

   C_D(k+1,1)=nd(k+1,1)/V(k+1,1);

   x(k+1,1)=(nb(1,1)-nb(k+1,1))/nb(1,1);

   t(k+1,1)=t(k,1)+0.01;

   k=k+1;

end

-

Best regards.

answer: a chemical reaction

explanation:

im not wrong