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Chapter 14: Problem 86

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

Expert verified
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# (褋) 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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Urease Catalysis
Urease is an enzyme that serves as a biological catalyst to enhance the rate of the reaction between urea and water. This reaction involves breaking down urea into carbon dioxide and ammonia.
Without the presence of the urease enzyme, this reaction occurs very slowly, as evident by the low reaction rate constant at elevated temperatures like 100掳C. However, when urease is involved, it significantly increases the reaction rate even at much lower temperatures, such as 21掳C, by providing an alternative pathway that requires less energy.
  • The reaction with urease: \[ NH_2CONH_2 + H_2O \xrightarrow{Urease} CO_2 + 2NH_3 \]
  • Urease acts on the substrate urea, drastically reducing the time it takes to reach equilibrium compared to the uncatalyzed reaction.
Enzymes like urease are critical in many biological processes because they allow reactions to occur rapidly and efficiently under mild conditions, unlike industrial processes that might require higher temperatures.
Activation Energy
Activation energy is the minimum energy required to start a chemical reaction. It is often a barrier that must be overcome for the reactants to transform into products.
Without catalysts, reactions like urea hydrolysis have higher activation energy, which explains the slow rate of reaction at 100掳C. With the urease enzyme, the activation energy is significantly lowered, making it much easier for the reaction to proceed.
  • The enzyme stabilizes the transition state, thereby reducing the energy required to transform urea into products.
  • The difference in activation energies for catalyzed versus uncatalyzed reactions is a key reason enzymes are so effective.
In this case, even though the reactions at 21掳C and 100掳C are at different temperatures, the enzyme's role in lowering activation energy explains why a lower overall temperature can still yield a quicker reaction when urease is present.
Reaction Rate Constant
A reaction rate constant is a numerical value that indicates how fast a reaction proceeds. It can vary significantly with the use of catalysts like enzymes.
In the previously mentioned reaction, without urease, the rate constant at 100掳C is quite low: \( 4.15 \times 10^{-5} \text{s}^{-1} \). However, when urease is added at 21掳C, the rate constant becomes much higher: \( 3.4 \times 10^4 \text{s}^{-1} \).
  • The increase in rate constant signifies how enzymes speed up reactions effectively.
  • Higher rate constants in enzymatic reactions indicate more production of products in a given time.
Overall, the difference in rate constants with and without an enzyme demonstrates the enzyme's efficiency in lowering the necessary energy to achieve a faster reaction.
Temperature Effects on Rate
Temperature can greatly influence the rate of chemical reactions because it affects the particles' kinetic energy.
Generally, an increase in temperature results in faster reactions due to increased collision frequency and energy between molecules. However, in this enzymatic reaction with urease, the reaction is faster at a lower temperature compared to the non-catalyzed version at a higher temperature because urease lowers the activation energy pathway.
  • Without urease, you'd typically expect reactions at 100掳C to proceed faster than those at 21掳C.
  • However, with the enzyme, even a low temperature is sufficient for fast reactions since the activation energy is lower.
This shows how enzymes allow reactions to proceed quickly without the need for high temperatures, which is crucial in maintaining cellular processes and enzyme stability.

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Most popular questions from this chapter

You have studied the gas-phase oxidation of HBr by \(\mathrm{O}_{2}\) : $$4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Br}_{2}(g)$$ You find the reaction to be first order with respect to HBr and first order with respect to \(\mathrm{O}_{2}\) . You propose the following mechanism: $$ \begin{array}{c}{\operatorname{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{HOOBr}(g)} \\ {\operatorname{HOOBr}(g)+\operatorname{HBr}(g) \longrightarrow 2 \mathrm{HOBr}(g)} \\\ {\operatorname{HOBr}(g)+\operatorname{HBr}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g)}\end{array}$$ (a) Confirm that the elementary reactions add to give the overall reaction. (b) Based on the experimentally determined rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?

Consider the following reaction: $$2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in \(\mathrm{NO}\) . Write the rate law. (b) If the rate constant for this reaction at 1000 \(\mathrm{K}\) is \(6.0 \times 10^{4} M^{-2} \mathrm{s}^{-1}\) what is the reaction rate when \([\mathrm{NO}]=0.035 M\) and \(\left[\mathrm{H}_{2}\right]=0.015 M ?\) (c) What is the reaction rate at 1000 \(\mathrm{K}\) when the concentration of \(\mathrm{NO}\) is increased to 0.10 \(\mathrm{M}\)while the concentration of \(\mathrm{H}_{2}\) is 0.010\(M ?\) (d) What is the reaction rate at 1000 \(\mathrm{K}\) if \([\mathrm{NO}]\) is decreased to 0.010 \(\mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]\) is increased to 0.030 \(\mathrm{M}\) ?

(a) What factors determine whether a collision between two molecules will lead to a chemical reaction? (b) Does the rate constant for a reaction generally increase or decrease with an increase in reaction temperature? (c) Which factor is most sensitive to changes in temperature-the frequency of collisions, the orientation factor, or the fraction of molecules with energy greater than the activation energy?

The activation energy of an uncatalyzed reaction is 95 \(\mathrm{kJ} / \mathrm{mol} .\) The addition of a catalyst lowers the activation energy to 55 \(\mathrm{kJ} / \mathrm{mol}\) . Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C},\) (b) \(125^{\circ} \mathrm{C} ?\)

Based on their activation energies and energy changes and assuming that all collision factors are the same, rank the following reactions from slowest to fastest.\( \begin{aligned} \text { (a) } E_{a} &=45 \mathrm{kJ} / \mathrm{mol} ; \Delta E=-25 \mathrm{kJ} / \mathrm{mol} \\ \text { (b) } E_{a} &=35 \mathrm{kJ} / \mathrm{mol} ; \Delta E=-10 \mathrm{kJ} / \mathrm{mol} \\ \text { (c) } E_{a} &=55 \mathrm{kJ} / \mathrm{mol} ; \Delta E=10 \mathrm{kJ} / \mathrm{mol} \end{aligned}\)
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