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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) Assuming the collision factor is the same for both situations, estimate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

Short Answer

Expert verified
The balanced equation for the reaction catalyzed by urease is: \( \mathrm{NH}_2\mathrm{CONH}_2 + H_2O \xrightarrow{urease} CO_2 + 2NH_3 \). The difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction is approximately 51.4 kJ/mol.

Step by step solution

01

Write the balanced equation for the reaction catalyzed by urease.

The enzyme urease catalyzes the reaction of urea with water to produce carbon dioxide and ammonia. The balanced equation for this reaction is: \( \mathrm{NH}_2\mathrm{CONH}_2 + H_2O \xrightarrow{urease} CO_2 + 2NH_3 \)
02

Calculate the difference in activation energies for the uncatalyzed and enzyme-catalyzed reactions.

We can use the Arrhenius equation to estimate the difference in activation energies for the uncatalyzed and enzyme-catalyzed reactions. The Arrhenius equation is: \( k = A \times e^{\frac{-E_A}{RT}} \) where \(k\) is the rate constant, \(A\) is the Arrhenius constant (also known as the collision factor), \(E_A\) is the activation energy, \(R\) is the gas constant (\(\approx 8.314 \text{ J mol}^{-1} \text{ K}^{-1}\)), and \(T\) is the temperature in Kelvin. We are given that the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \text{ s}^{-1}\) without the enzyme at \(100^{\circ} \text{C}\), and with a rate constant of \(3.4 \times 10^{4} \text{ s}^{-1}\) in the presence of the enzyme at \(21^{\circ} \text{C}\). Since the problem states that the collision factor is the same for both situations, we specifically can write: \( \frac{k_{uncatalyzed}}{k_{catalyzed}} = \frac{A \times e^{\frac{-E_A^{uncatalyzed}}{RT_{uncatalyzed}}}}{A \times e^{\frac{-E_A^{catalyzed}}{RT_{catalyzed}}}} \) Where: - \(k_{uncatalyzed} = 4.15 \times 10^{-5} \text{ s}^{-1}\) - \(k_{catalyzed} = 3.4 \times 10^{4} \text{ s}^{-1}\) - \(T_{uncatalyzed} = 373.15 \text{ K}\) (converting from Celsius to Kelvin) - \(T_{catalyzed} = 294.15 \text{ K }\) (converting from Celsius to Kelvin) Now we need to find the difference in activation energies, \(\Delta E_A = E_A^{uncatalyzed} - E_A^{catalyzed}\), using the given information. Since the collision factors (A) in the numerator and denominator are the same, they cancel out: \( \frac{k_{uncatalyzed}}{k_{catalyzed}} = \frac{e^{\frac{-E_A^{uncatalyzed}}{RT_{uncatalyzed}}}}{e^{\frac{-E_A^{catalyzed}}{RT_{catalyzed}}}} \) Taking the natural logarithm of both sides: \( \ln{\frac{k_{uncatalyzed}}{k_{catalyzed}}} = \frac{-E_A^{uncatalyzed}}{RT_{uncatalyzed}} + \frac{E_A^{catalyzed}}{RT_{catalyzed}} \) Rearranging the equation for \(\Delta E_A\): \( \Delta E_A = \left( \frac{-RT_{catalyzed}}{RT_{uncatalyzed}} + 1 \right)R \ln{\frac{k_{uncatalyzed}}{k_{catalyzed}}} \) Plugging in the given values and solving for \(\Delta E_A\): \( \Delta E_A = \left( \frac{-294.15}{373.15} + 1 \right)(8.314) \ln{\frac{4.15 \times 10^{-5}}{3.4 \times 10^{4}}} \) \( \Delta E_A \approx 51.4 \text{ kJ mol}^{-1} \) The difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction is approximately 51.4 kJ/mol.

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Key Concepts

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

Arrhenius Equation
The Arrhenius equation is foundational to understanding the effects of temperature on the rate of a chemical reaction. It delineates how the rate constant (k) of a reaction is related to the temperature and the energy barrier that must be overcome for reactants to transform into products - known as the activation energy (EA). The equation is expressed as

\[ k = A \times e^{\frac{-E_A}{RT}} \]

where A is the pre-exponential factor or frequency factor, which is a measure of the likelihood that collisions between reactant molecules will be successful in producing a reaction. The exponential factor, e^{\frac{-E_A}{RT}}, represents the fraction of molecules that have sufficient energy to overcome the activation energy barrier at a given temperature. R is the universal gas constant, and T is the absolute temperature in Kelvin. Through this equation, we can predict how reaction rates will change with varying temperatures. Moreover, it helps in comparing the effects of different catalysts, as seen in enzyme-catalyzed reactions, by evaluating the changes in activation energy.
Activation Energy
Activation energy (EA) is a critical concept in chemical kinetics. It is the energy required to initiate a chemical reaction; essentially, it's the proverbial 'hump' that reactants must overcome to be converted into products. This energy barrier ensures that molecules need a certain threshold energy to react when they collide. The lower the activation energy, the faster the reaction, because a greater proportion of the reactant molecules can overcome this barrier at a given temperature. Enzymes, being potent biological catalysts, often work by significantly reducing the activation energy of a reaction, allowing the reaction to proceed much faster under the same conditions—as illustrated by the marked difference in rate constants for the enzyme-catalyzed and uncatalyzed reactions in our exercise.
Rate Constant
The rate constant (k) of a reaction is the proportionality constant in the rate equation that relates the reaction rate to the concentrations of reactants. For a first-order reaction such as the urea hydrolysis in the exercise, the rate is directly proportional to the concentration of one reactant, and the rate constant provides a measure of how quickly the reaction occurs. As the Arrhenius equation implies, the rate constant is temperature-dependent and also varies inversely with activation energy. Knowing the rate constants at different temperatures (which must be in Kelvin for computations), one can use the Arrhenius equation to reveal insights about the mechanism of the reaction and the comparative effectiveness of catalysts, including enzymes.
Chemical Kinetics
Chemical kinetics is the study of reaction rates and the factors affecting these rates. It involves the exploration of how different conditions—such as temperature, concentration, and the presence of a catalyst—influence the speed of chemical reactions. Understanding kinetics is crucial for designing chemical processes, preserving food, developing pharmaceuticals, and explaining natural phenomena. In the context of our exercise, principles of chemical kinetics are used to compare the rates of uncatalyzed and enzyme-catalyzed reactions, highlighting the remarkable ability of enzymes to accelerate reactions, a cornerstone of biochemical processes in living organisms.

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Most popular questions from this chapter

(a) Two reactions have identical values for \(E_{a} .\) Does this ensure that they will have the same rate constant if run at the same temperature? Explain. (b) Two similar reactions have the same rate constant at \(25^{\circ} \mathrm{C}\), but at \(35^{\circ} \mathrm{C}\) one of the reactions has a larger rate constant than the other. Account for these observations.

(a) If you were going to build a system to check the effectiveness of automobile catalytic converters on cars, what substances would you want to look for in the car exhaust? (b) Automobile catalytic converters have to work at high temperatures, as hot exhaust gases stream through them. In what ways could this be an advantage? In what ways a disadvantage? (c) Why is the rate of flow of exhaust gases over a catalytic converter important?

A reaction \(A+B \longrightarrow C\) obeys the following rate law: Rate \(=k[\mathrm{~B}]^{2}\). (a) If \([\mathrm{A}]\) is doubled, how will the rate change? Will the rate constant change? Explain. (b) What are the reaction orders for \(\mathrm{A}\) and \(\mathrm{B}\) ? What is the overall reaction order? (c) What are the units of the rate constant?

(a) Define the following symbols that are encountered in rate equations for the generic reaction \(\mathrm{A} \longrightarrow \mathrm{B}:[\mathrm{A}]_{0}, t_{1 / 2}[\mathrm{~A}]_{t}, k\) (b) What quantity, when graphed versus time, will yield a straight line for a first-order reaction? (c) How can you calculate the rate constant for a first- order reaction from the graph you made in part (b)?

The gas-phase reaction \(\mathrm{Cl}(g)+\mathrm{HBr}(g) \longrightarrow \mathrm{HCl}(g)+\mathrm{Br}(g)\) has an overall enthalpy change of \(-66 \mathrm{~kJ}\). The activation energy for the reaction is \(7 \mathrm{~kJ}\). (a) Sketch the energy profile for the reaction, and label \(E_{a}\) and \(\Delta E\). (b) What is the activation energy for the reverse reaction?

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