The given question is incomplete. The complete question is as follows.
The enzyme urease catalyzes the reaction of urea, (
), with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of
at
. In the presence of the enzyme in water, the reaction proceeds with a rate constant of
at
.
If the rate of the catalyzed reaction were the same at
as it is at
, what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions?
Express your answer using two significant figures.
Explanation:
The reaction equation is as follows.
Urea + Water ![\rightarrow CO_{2} + NH_{3}](https://tex.z-dn.net/?f=%5Crightarrow%20CO_%7B2%7D%20%2B%20NH_%7B3%7D)
Hence, it is given that,
without enzyme: Rate =
at
with enzyme: Rate =
at
Rate =
at
It is known that,
ln \frac{K_{2}}{K_{1}} = \frac{-E_{a}}{R}{\frac{1}{T_{2}} - \frac{1}{T_{1}}][/tex]
and, ln K = ![\frac{-E_{a}}{RT} + ln A](https://tex.z-dn.net/?f=%5Cfrac%7B-E_%7Ba%7D%7D%7BRT%7D%20%2B%20ln%20A)
Let us assume that collision factor (A) is same for both the reactions.
Hence, ![\Delta E = RT ln (\frac{K_{with enzyme}}{K_{without enzyme}})](https://tex.z-dn.net/?f=%5CDelta%20E%20%3D%20RT%20ln%20%28%5Cfrac%7BK_%7Bwith%20enzyme%7D%7D%7BK_%7Bwithout%20enzyme%7D%7D%29)
= ![8.314 \times J/K mol \times (100 + 273) K \times ln (\frac{3.4 \times 10^{4}}{4.15 \times 10^{-5}})](https://tex.z-dn.net/?f=8.314%20%5Ctimes%20J%2FK%20mol%20%5Ctimes%20%28100%20%2B%20273%29%20K%20%5Ctimes%20ln%20%28%5Cfrac%7B3.4%20%5Ctimes%2010%5E%7B4%7D%7D%7B4.15%20%5Ctimes%2010%5E%7B-5%7D%7D%29)
= 63672.8 J/mol
= 63.67 kJ/mol (as 1 kJ = 1000 J)
Thus, we can conclude that the difference in the activation energy between the catalyzed and uncatalyzed reactions is 63.67 kJ/mol.