At room temperature sucrose is hydrolyzed by the catalytic action of the enzyme

Question

At room temperature sucrose is hydrolyzed by the catalytic action of the enzyme sucrose as follows: Starting with a sucrose concentration C_A0 = 1.0 mM and an enzyme concentration C_E0 = 0.01 mM, the following kinetic data at 25 degree C are obtained in a batch reactor: Determine whether these data can be reasonably fitted by a kinetic equation of the Michaelis-Menten type, r = kC_A/C_A + k If the fit is reasonable, evaluate k and the Michaelis constant K. A batch reactor kept at a constant temperature of 25 degree C is charged with sucrose at an initial concentration of 2.5 mM and an initial enzyme concentration of 0.01 mM. What time is needed to achieve 60% conversion? What are the concentrations of sucrose and products at this time?

Explanation / Answer

dCA/dt= kCA/(CA+KM)

-dCA*(CA+KM)/CA= kdt

CAO-CA+ KM*ln (CAO/CA)= kt

KM* ln (CAO/CA) = -(CAO-CA)+ kt

ln(CAO/CA)/ (CAO-CA) * KM= -1+k*t/CAO-CA)

ln(CAO/CA)/ (CAO-CA)= -1/KM+(kt/KM)*(CAO-CA)

so a plot of ln (CAO/CA)/ (CAO-CA) vs t/(CAO-CA) gives a straight line whose slope is -kt/KM and intercept -1/KM

from the plot

1/KM= 5096 , KM= 1/5096=0.000196 and kt/KM= 1.010, k= 1.010*0.000196= 0.000198M

so the data fits into Michelis- menten kinetics

ln(CAO/CA)/ (CAO-CA)= -1/KM+ (kt/KM)*t*/(CAO-CA)

given XA= 0.6, 1-CA/CAO=0.6, CA= 0.4*CAO= 2.5*0.4*10-3 M= 10-3M

ln(2.5/1)/ (2.5-1)*10-3 = -1/5096+ 1.0010*t/ 1.5*10-3 = -0.000196+667.34t

611+0.000196= 667.34 t

t= 0.91 hrs

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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