Answer:
See explanation.
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]CH_2OHCH_2Cl+NaHCO_3\rightarrow (CH_2OH)_2+NaCl+CO_2\\A+B\rightarrow C+D+E[/tex]
For which the differential equations in terms of the time variation are:
[tex]\frac{dn_A}{dt}= F_{A,0}+r_AV\\\frac{dn_B}{dt}=r_BV\\\frac{dn_C}{dt}=r_CV\\n_D=n_C\\\frac{dV}{dt}=v_0-\frac{r_CVM_{CO_2}}{\rho }[/tex]
And the implicit equations:
[tex]F_{A,0}=0.1mol/min*60min/1h=6.0mol/h\\v_0=(6.0mol/h)/(1.5mol/dm^3)=4dm^3/h\\rho=1000g/L[/tex]
[tex]r_A=-5.1C_AC_B\\r_B=r_A\\r_C=-r_A[/tex]
[tex]C_A=\frac{n_A}{V} \\C_B=\frac{n_B}{V}[/tex]
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:
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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
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Best regards.
