with explanations please! The enzyme urease catalyzes the reaction of urea (NH_2

Question

with explanations please!

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 degree 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 degree C. Part A Write out the balanced equation for the reaction catalyzed by urease. Express your answer as a chemical equation. Identify all of the phases in your answer Part B If the rate of the catalyzed reaction were the same at 100 0Cas it is at 21 degree C, what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? Express your answer using two significant figures.

Explanation / Answer

Part A :

NH2CONH2 + H2O ---> 2NH3 + CO2

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Part B

For the uncatalysed reaction : ln k1 = -Eun/R * (1/294) + lnA

For catalysed reaction : ln k2 = -Ecat/R *(1/294) + ln A

ln (k1) -ln(k2) = -Eun/R * (1/294) + lnA+ Ecat/R *(1/294) - ln A

or, ln(k1/k2) = 1/R*294 (-Eun+Ecat)

or, ln(4.15*10^-5/3.4*10^4) = 1/8.314 J/K/mol * 294 (-Eun+Ecat)

or, Ecat-Euncat = -50166.97 J = -50.167 kJ

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