The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right),\) with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{s}^{-1}\) at \(100^{\circ} \mathrm{C} .\) In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{s}^{-1}\) at \(21^{\circ} \mathrm{C}\) . (a) Write out the balanced equation for the reaction catalyzed by urease. (b) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C},\) what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{Cas} \mathrm{com}-\) pared to that at \(21^{\circ} \mathrm{C} ?(\mathbf{d})\) On the basis of parts \((\mathrm{c})\) and \((\mathrm{d}),\) what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

Short Answer

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In the presence of the enzyme, the rate at \(21^{\circ}C\) is \(3.4 \times 10^{4} \mathrm{s}^{-1}\). If the rate is the same at \(100^{\circ}C\), we can use the Arrhenius equation to find the difference in activation energy: \(\Delta E_a = R \ln{\frac{k_\mathrm{uncat}}{k_\mathrm{cat}}} (T_\mathrm{cat} - T_\mathrm{uncat})\) \(\Delta E_a = (8.314 \mathrm{J K^{-1} mol^{-1}}) \ln{\frac{4.15 \times 10^{-5} \mathrm{s^{-1}}}{3.4 \times 10^{4} \mathrm{s^{-1}}}} (3.15 \times 10^2 \mathrm{K} - 2.94 \times 10^2 \mathrm{K})\) \(\Delta E_a \approx 9.7 \times 10^{4} \mathrm{J mol^{-1}}\) #tag_title# (с) Rate of Catalyzed Reaction at \(100^{\circ}C\) Compared to \(21^{\circ}C\) In actuality, the rate of the catalyzed reaction at \(100^{\circ}C\) would likely be higher than at \(21^{\circ}C\) due to the general increase in reaction rates with the increase in temperature. #tag_title# (d) Conclusion on Difference in Activation Energies The results from parts (b) and (c) imply that the difference in activation energies between the uncatalyzed and catalyzed reactions at these temperatures may be even significantly greater than \(9.7 \times 10^{4} \mathrm{J mol^{-1}}\). Thus, the enzyme urease greatly lowers the activation energy for the reaction between urea and water.

Step by step solution

01

(a) Balanced Equation for the Reactio +:+tag_content# The balanced equation for the reaction between urea and water catalyzed by urease enzyme is: \[NH_2CONH_2 + H_2O \xrightarrow{Urease} CO_2 + 2NH_3\]

(b) Difference in Activation Energy

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Most popular questions from this chapter

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