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The enzyme carbonic anhydrase catalyzes the reaction \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HCO}_{3}{ }^{-}(a q)+\mathrm{H}^{+}(a q)\). In water, without the enzyme, the reaction proceeds with a rate constant of \(0.039 \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(1.0 \times 10^{6} \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). 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 energy between the uncatalyzed versus enzyme-catalyzed reaction is approximately 38,150.64 J/mol.

Step by step solution

01

Convert temperature to Kelvin

First, we need to convert the given temperature from Celsius to Kelvin. To do this, add 273.15 to the temperature in Celsius: 25°C + 273.15 = 298.15 K So, T = 298.15 K.
02

Write the Arrhenius equations for both the reactions

Apply the Arrhenius Equation to both the uncatalyzed and catalyzed reactions: For the uncatalyzed reaction: \(k_{uncatalyzed} = Ae^{\frac{-E_{a,uncatalyzed}}{RT}}\) For the enzyme-catalyzed reaction: \(k_{catalyzed} = Ae^{\frac{-E_{a,catalyzed}}{RT}}\)
03

Divide the catalyzed reaction by the uncatalyzed reaction

Since the collision factor is the same for both reactions, we can divide the enzyme-catalyzed reaction by the uncatalyzed reaction to eliminate the pre-exponential factor (A): \[\frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{Ae^{\frac{-E_{a,catalyzed}}{RT}}}{Ae^{\frac{-E_{a,uncatalyzed}}{RT}}}\] Simplify to get: \[\frac{k_{catalyzed}}{k_{uncatalyzed}} = e^{\frac{E_{a,uncatalyzed}-E_{a,catalyzed}}{RT}}\]
04

Solve for the difference in activation energy

Now, we have: \(e^{\frac{E_{a,uncatalyzed}-E_{a,catalyzed}}{RT}} = \frac{k_{catalyzed}}{k_{uncatalyzed}}\) Plug in the given values of rate constants and temperature: \(e^{\frac{E_{a,uncatalyzed}-E_{a,catalyzed}}{8.314 * 298.15}} = \frac{1.0 \times 10^6}{0.039}\) Take the natural logarithm of both sides: \(\frac{E_{a,uncatalyzed}-E_{a,catalyzed}}{8.314 * 298.15} = ln(\frac{1.0 \times 10^6}{0.039})\) Now, compute the expression on the right side: \(ln(\frac{1.0 \times 10^6}{0.039}) \approx 15.39\) Multiply both sides by 8.314 * 298.15: \(E_{a,uncatalyzed}-E_{a,catalyzed} \approx 15.39 * 8.314 * 298.15\) Calculate the difference in activation energy: \(E_{a,uncatalyzed} - E_{a,catalyzed} \approx 38150.64 \, J/mol\) So, the difference in activation energy between the uncatalyzed versus enzyme-catalyzed reaction is approximately 38,150.64 J/mol.

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Key Concepts

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

Carbonic Anhydrase
Carbonic anhydrase is an enzyme that plays a crucial role in speeding up the conversion of carbon dioxide and water into bicarbonate and protons. This reaction may seem simple, but it is vital for many physiological processes in humans and other organisms. Without carbonic anhydrase, the conversion rate of carbon dioxide to bicarbonate would be extremely slow, which could negatively impact processes such as respiration and pH balance.
One fascinating aspect of carbonic anhydrase is its efficiency. It can increase the rate of the reaction by over a million times compared to when the reaction occurs spontaneously without any catalysis. This rate enhancement demonstrates why enzymes like carbonic anhydrase are essential for life. They ensure that reactions occur at rates sufficient to support biological functions.
Enzyme Catalysis
Enzyme catalysis refers to the process of increasing the rate of a chemical reaction by the action of an enzyme. Enzymes, like carbonic anhydrase, are proteins that have specific structures allowing them to bind substrates and convert them into products more efficiently.
Key factors that contribute to enzyme catalysis include:
  • Enzyme specificity: Enzymes are highly specific to their substrates, which ensures that the correct substrate is transformed into the product.
  • Active sites: The particular region on an enzyme where the substrate binds and the reaction occurs.
  • Lowering activation energy: Enzymes reduce the activation energy required for a reaction, allowing the reaction to occur more quickly at a given temperature.

In the case of carbonic anhydrase, the enzyme drastically reduces the activation energy required to convert carbon dioxide and water into bicarbonate, facilitating swift and crucial physiological processes.
Arrhenius Equation
The Arrhenius equation is pivotal in understanding reaction kinetics. It illustrates how the rate of a reaction depends on temperature and activation energy. The equation is expressed as:\(k = Ae^{-E_a/RT}\),where
  • \(k\) is the rate constant,
  • \(A\) is the pre-exponential factor, which relates to the frequency of collisions and orientations,
  • \(E_a\) is the activation energy,
  • \(R\) is the universal gas constant, and
  • \(T\) is the temperature in Kelvin.

This equation shows that even a small decrease in activation energy, such as what happens in enzyme-catalyzed reactions, significantly boosts the reaction rate.
For instance, carbonic anhydrase catalysis embodies the reduction in activation energy, enhancing the reaction rate from 0.039 \(s^{-1}\) in an uncatalyzed state to an astonishing \(1.0 \times 10^6 \) \(s^{-1}\). This demonstrates the profound impact of enzymes on reaction kinetics and biological efficiency.

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Most popular questions from this chapter

The accompanying graph shows plots of \(\ln k\) versus \(1 / T\) for two different reactions. The plots have been extrapolated to the \(y\)-intercepts. Which reaction (red or blue) has (a) the larger value for \(E_{a}\) and (b) the larger value for the frequency factor, \(A\) ? [Section 14.5]

The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) is second order in \(\mathrm{NO}\) and first order in \(\mathrm{O}_{2}\). When \([\mathrm{NO}]=0.040 \mathrm{M}\), and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M}\), the observed rate of disappearance of \(\mathrm{NO}\) is \(9.3 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate of disappearance of \(\mathrm{O}_{2}\) at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of \(1.8\) ?

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?

Platinum nanoparticles of diameter \(\sim 2 \mathrm{~nm}\) are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of \(3.924 \AA\). (a) Estimate how many platinum atoms would fit into a \(2.0-\mathrm{nm}\) sphere; the volume of a sphere is \((4 / 3) \pi r^{3}\). Recall that \(1 \dot{A}=1 \times 10^{-10} \mathrm{~m}\) and \(1 \mathrm{~nm}=1 \times 10^{-9} \mathrm{~m}\). (b) Estimate how many platinum atoms are on the surface of a \(2.0\)-nm Pt sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one Pt atom can be estimated from its atomic diameter of \(2.8 \mathrm{~A}\). (c) Using your results from (a) and (b), calculate the percentage of \(\mathrm{Pt}\) atoms that are on the surface of a \(2.0-\mathrm{nm}\) nanoparticle. (d) Repeat these calculations for a \(5.0\)-nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\). At \(23^{\circ} \mathrm{C}\) and in \(0.5 \mathrm{M} \mathrm{HCl}\), the following data were obtained for the disappearance of sucrose: $$ \begin{array}{cc} \hline \text { Time (min) } & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right](\mathrm{M})} \\ \hline 0 & 0.316 \\ 39 & 0.274 \\ 80 & 0.238 \\ 140 & 0.190 \\ 210 & 0.146 \\ \hline \end{array} $$

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