Question

Consider the formation of p-nitrophenol from p-nitrophenyl trimethyl acetate. The process is known as enzymatic hydrolysis and it occurs in the presence of the enzyme elastase. Along with the formation of p-nitrophenol, trimethyl acetic acid is also formed which is an undesired byproduct. p-nitrophenol is an important intermediate in the manufacture of several pharmaceuticals. Your role as a Chemical Engineer is to maximize the production of p-nitrophenol. The reactions can be denoted as:
E+S → P+ES R1
ES+PE+A R2
where e denotes the enzyme elastase, denotes the substrate p-nitrophenyl trimethyl acetate, es denotes enzyme-substrate intermediate and A denotes the trimethyl acetic acid. The rate of the reactions 1 and 2 are given by:
kg Cs KM + C r2 = kxCp
where Cs and Cp denote the concentrations of the substrate and the product, k, and ky are the rate constants given by 0.015 s' and 0.0026 s. Ky is the Michaelis - Menten constant and is given by 5.53 mol/m!. All the reactants and products are in the liquid phase. The initial concentrations of S and E are 0.5 mol/m3 and 0.001 mol/m..Consider the above reaction to occur in a batch reactor for 15 minutes.
a. Plot the concentration profiles of S, P and A as a function of time in a single figure.
b. Plot the selectivity of P with respect to 5 as a function of time b.

Answer 1

Solution :

cs=zeros(9001);

ca=zeros(9001);

cp=zeros(9001);

psi=zeros(9001);

t=[0:0.1:900];

cs(1)=0.5;

ce(1)=0.001;

cp(1)=0;

ca(1)=0;

psi(1)=0;

for i=1:1:9000

cs(i+1)=cs(i)-0.1*((0.015*cs(i))/(5.53+cs(i)));

cp(i+1)=cp(i)+0.1*((0.015*cs(i))/(5.53+cs(i))-0.0026*cp(i));

ca(i+1)=ca(i)+0.1*0.0026*cp(i);

psi(i+1)=((cp(i+1)-cp(i)))/((cs(i)-cs(i+1)));

end

plot(t,cs,t,cp,t,ca);

plot(t,psi);