Question

In your lab you are studying a certain biochemical bacterial metabolic pathway (shown below). You have initial data that suggests that if you could inhibit this pathway you would have the "next penicillin", but first you need to determine the activation energy for the rate determining step of this process. enzyme + substrate → product 1 + product 2 It is known that under the conditions you are performing your experiment the rate law for the reaction is: rate = k[enzyme][substrate]2. You set up an experiment at 319 K where [enzyme] = 0.00475 M and [substrate] = 1.69 M. A plot of ln[enzyme] versus time (seconds) gives a straight line relationship with a slope of -0.005387. Previous studies have shown that a plot of ln(k) at a series of different temperatures gives a y-intercept of 0.62647. What is the activation energy for this reaction?

Answer 1

From the data

first plot of ln[enzyme] vs time is a frist-order reaction kinetics plot

The slope of the straight like is rate constant -k

So, k = 0.005387 s-1

Another plot of lnk vs 1/T gave y-intercept = 0.62647

y-intercept = lnA = 0.62647

A = 1.703 is pre-exponential factor

So activation energy Ea would be,

lnk = lnA - Ea/RT

T = 319 K

Feeding other values from above,

Arrhenius equation,

lnk = -Ea/RT + lnA

ln(0.005387) = 0.62647 - Ea/8.314 x 319

Ea = 15.516 kJ/mol