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Chapter 14: Problem 85

A certain enzyme catalyzes a biochemical reaction. In water, without the enzyme, the reaction proceeds with a rate constant of $6.50 \times 10^{-4} \mathrm{~min}^{-1}\( at \)37^{\circ} \mathrm{C} .$ In the presence of the enzyme in water, the reaction proceeds with a rate constant of $1.67 \times 10^{4} \mathrm{~min}^{-1}\( at \)37^{\circ} \mathrm{C}$. Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

Short Answer

Expert verified
The difference in activation energies for the uncatalyzed and enzyme-catalyzed reaction is approximately \(2.57 \times 10^{4} J/mol\). This negative value indicates that the activation energy for the enzyme-catalyzed reaction is lower than the activation energy for the uncatalyzed reaction, which is a typical behavior of enzymes to speed up reactions.

Step by step solution

01

Write down the known values

First, let's list the known values: - Rate constant for the uncatalyzed reaction (k1): \(6.50 \times 10^{-4} \mathrm{min}^{-1}\) - Rate constant for the enzyme-catalyzed reaction (k2): \(1.67 \times 10^{4} \mathrm{min}^{-1}\) - Temperature (T): \(37^{\circ} \mathrm{C}\), or (T in Kelvin): \(37 + 273.15 = 310.15 K\) - Gas constant (R): \(8.314 J / (mol \cdot K)\)
02

Write down the Arrhenius equation for both cases

The Arrhenius equation is given by: \[k = Ae^{\frac{-Ea}{RT}}\] where k is the rate constant, A is the collision factor (also known as the pre-exponential factor), Ea is the activation energy, R is the gas constant, and T is the temperature. For the uncatalyzed reaction: \[k_1 = Ae^{\frac{-Ea_1}{RT}}\] For the enzyme-catalyzed reaction: \[k_2 = Ae^{\frac{-Ea_2}{RT}}\] The problem states that the collision factor (A) is the same in both cases, so we can use the same variable A for both equations.
03

Divide the two Arrhenius equations and solve for the difference in activation energies

Now, let's divide the enzyme-catalyzed reaction's Arrhenius equation by the uncatalyzed reaction's Arrhenius equation: \[\frac{k_2}{k_1} = \frac{Ae^{\frac{-Ea_2}{RT}}}{Ae^{\frac{-Ea_1}{RT}}}\] The collision factor (A) cancels out: \[\frac{k_2}{k_1} = e^{\frac{-(Ea_2 - Ea_1)}{RT}}\] Rearrange the equation to find the difference in activation energies, \((Ea_2 - Ea_1)\): \[(Ea_2 - Ea_1) = -RT \ln{\frac{k_2}{k_1}}\]
04

Plug in the known values and perform the calculation

Substitute the given values (k1, k2, R, and T) into the equation: \[(Ea_2 - Ea_1) = -8.314 J/mol \cdot K \cdot 310.15 K \cdot \ln{\frac{1.67 \times 10^{4}}{6.50 \times 10^{-4}}}\] Now perform the calculation: \[(Ea_2 - Ea_1) \approx -2.57 \times 10^{4} J/mol\]
05

Interpret the result

The difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction is approximately \(2.57 \times 10^{4} J/mol\). Since the value is negative, this indicates that the activation energy for the enzyme-catalyzed reaction is lower than the activation energy for the uncatalyzed reaction. Enzymes typically lower the activation energy of a reaction, which allows the reaction to proceed more quickly.

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Most popular questions from this chapter

Many primary amines, \(\mathrm{RNH}_{2}\), where \(\mathrm{R}\) is a carboncontaining fragment such as $\mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2},$ and so on, undergo reactions where the transition state is tetrahedral. (a) Draw a hybrid orbital picture to visualize the bonding at the nitrogen in a primary amine (just use a C atom for "R"). (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate?

(a) Most commercial heterogeneous catalysts are extremely finely divided solid materials. Why is particle size important? (b) What role does adsorption play in the action of a heterogeneous catalyst?

Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) or silica \(\left(\mathrm{SiO}_{2}\right) .(\mathbf{a})\) Why is this an effective way of utilizing the catalyst material compared to having powdered metals? (b) How does the surface area affect the rate of reaction?

(a) Develop an equation for the half-life of a zero-order reaction. (b) Does the half-life of a zero-order reaction increase, decrease, or remain the same as the reaction proceeds?

For the elementary process $\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)$ the activation energy \(\left(E_{a}\right)\) and overall \(\Delta E\) are $154 \mathrm{~kJ} / \mathrm{mol}\( and \)136 \mathrm{~kJ} / \mathrm{mol}$, respectively. (a) Sketch the energy profile for this reaction, and label \(E_{a}\) and \(\Delta E\). (b) What is the activation energy for the reverse reaction?
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