Answer:
The answer is in the picture attached
Explanation:
A six-membered aromatic ring is one of the most common structures in natural compounds.
The aldehyde group in position one consist in one carbon double-bonded to oxygen and bonded to hydrogen.
The alkoxy group in position three consist in a oxygen bonded to an alkyl group (The simplest alkyl group is methyl -CH₃-).
There is, in position four, an hydroxyl group -OH. bonded to this six-membered aromatic ring.
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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
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 =
Let us assume that collision factor (A) is same for both the reactions.
Hence,
=
= 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.