In your lab you are studying a certain biochemical bacterial metabolic pathway (
ID: 1061353 • Letter: I
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]^2[substrate]^2.
You set up an experiment at 325 K where [enzyme] = 0.00287 M and [substrate] = 1.57 M. A plot of 1/[enzyme] versus time (seconds) gives a straight line relationship with a slope of 0.001174. Previous studies have shown that a plot of ln(k) at a series of different temperatures gives a y-intercept of 0.60377. What is the activation energy for this reaction?
Explanation / Answer
1/At = 1/Ao + kt
Second order with respect to enzyme
1/Enzyme vs time
The slope of this line is rate constant
k = 0.001174
ln K vs 1/T
y - intercept is ln A = 0.60377
The mainequation is
ln K = -Ea /RT + ln A
ln 0.001174 = -Ea / 8.314*325 + 0.60377
Ea = 19.863 KJ