The enzyme urease catalyzes the reaction of urea, (NH2CONH2), with water to prod

Question

The enzyme urease catalyzes the reaction of urea, (NH2CONH2), 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 ×105s1 at 100 C. In the presence of the enzyme in water, the reaction proceeds with a rate constant of 3.4 ×104s1 at 21 C.

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?

Explanation / Answer

we have to use the equation E = RTln( k/k ), remembering that the collision factor is the same for both conditions.

therefroe

In the presence of the enzyme in water

E = 8.314 JK¹mol¹ (21 +273) K ln(3.4x10^4 s¹ )
= 25418 Jmol¹
= 25.42 kJmol¹

In water, without the enzyme,

E = 8.314 JK¹mol¹ (100 +273) K ln(4.15x10^-5 s¹ )
=-31320 Jmol¹
=-31.32 kJmol¹

so

the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction so simply find the difference = 31.32 kj + 25.418 kj = 56.738 kj

EaEac = between 50 - 64 kJ.

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