Question

"If the rate of the catalyzed reaction were the same at 100 ∘C as it is at 21 ∘C, what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? Express your answer using two significant figures."

Answer 1

The given question is incomplete. The complete question is as follows.

The enzyme urease catalyzes the reaction of urea, (
`NH_2CONH_2`), with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of
`4.15 * 10^(-5) s^(-1)` at
`100^(o)C`. In the presence of the enzyme in water, the reaction proceeds with a rate constant of
`3.4 * 10^(4) s^(-1)` at
`21^(o)C`.

If the rate of the catalyzed reaction were the same at
`100^(o)C` as it is at
`21^(o)C`, what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions?

Express your answer using two significant figures.

Step-by-step explanation:

The reaction equation is as follows.

Urea + Water
`\rightarrow CO_(2) + NH_(3)`

Hence, it is given that,

without enzyme: Rate =
`4.15 * 10^(-5) s^(-1)` at
`100^(o)C`

with enzyme: Rate =
`3.4 * 10^(4) s^(-1)` at
`21^(o)C`

Rate =
`3.4 * 10^(4) s^(-1)` at
`100^(o)C`

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 =
`(-E_(a))/(RT) + ln A`

Let us assume that collision factor (A) is same for both the reactions.

Hence,
`\Delta E = RT ln ((K_(with enzyme))/(K_(without enzyme)))`

=
`8.314 * J/K mol * (100 + 273) K * ln ((3.4 * 10^(4))/(4.15 * 10^(-5)))`

= 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.