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 $4.15 \times 10^{-5} s^{-1}$ at $100^{o}C$. In the presence of the enzyme in water, the reaction proceeds with a rate constant of $3.4 \times 10^{4} s^{-1}$ 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.

Explanation:

The reaction equation is as follows.

       Urea + Water $\rightarrow CO_{2} + NH_{3}$

Hence, it is given that,

without enzyme: Rate = $4.15 \times 10^{-5} s^{-1}$ at $100^{o}C$

with enzyme: Rate = $3.4 \times 10^{4} s^{-1}$ at $21^{o}C$

                      Rate = $3.4 \times 10^{4} s^{-1}$ 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 = $\frac{-E_{a}}{RT} + ln A$

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

Hence,   $\Delta E = RT ln (\frac{K_{with enzyme}}{K_{without enzyme}})$

                         = $8.314 \times J/K mol \times (100 + 273) K \times ln (\frac{3.4 \times 10^{4}}{4.15 \times 10^{-5}})$

                         = 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.