The \(k_{\text {cat }}\) for alkaline phosphatase-catalyzed hydrolysis of methylphosphate is approximately \(14 / \sec\) at \(\mathrm{pH} 8\) and \(25^{\circ} \mathrm{C}\). The rate constant for the uncatalyzed hydrolysis of methylphosphate under the same conditions is approximately \(10^{-15} /\) sec. What is the difference in the free energies of activation of these two reactions?

Short Answer

Expert verified
To obtain the answer, one must first calculate the activation energy for both the catalyzed and uncatalyzed reactions using the Arrhenius equation. Then, the difference between these calculated activation energies is computed to find the difference in the free energies of activation.

Step by step solution

01

Understanding the Arrhenius equation

The Arrhenius equation is \(k = Ae^{-Ea/RT}\), where \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the gas constant (8.314 J/mol·K) and \(T\) is the temperature in Kelvin. In order to compare the activation energy of the two given reactions, we'll first need to isolate \(E_a\), which gives us \(E_a = -ln(k/A) * RT\). However, because we are after the difference in activation energy between the two reactions, we can ignore the \(A\) pre-exponential factor, as it will cancel out in the differencing. So, we'll be primarily using the simplified equation \(Ea = -ln(k) * RT\).
02

Calculating activation energy for the catalyzed reaction

First, we'll calculate the \(Ea_{catalyzed}\) using the information about the catalyzed reaction. Thus, by using \(Ea = -ln(k) * RT\) and the values \(k_{catalyzed} = 14/1\sec\) and \(T = 25^{\circ}C = 298.15K\) (remember to convert from Celsius to Kelvin), we get \(Ea_{catalyzed}= -ln(14)*8.314*298.15 J/mol\).
03

Calculating activation energy for the uncatalyzed reaction

Similarly, we can calculate \(Ea_{uncatalyzed}\) using \(k_{uncatalyzed}= 10^{-15} /1\sec\) and the same temperature, resulting in \(Ea_{uncatalyzed} = -ln(10^{-15})*8.314*298.15 J/mol\).
04

Calculating the difference in activation energies

The next step is to calculate the difference in activation energies \(ΔEa = Ea_{uncatalyzed} - Ea_{catalyzed}\). This will provide the difference in the free energies of activation of the reactions.

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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 plays a crucial role in understanding how temperature affects the rate of a chemical reaction. It is mathematically expressed as \( k = Ae^{-Ea/RT} \), where \( k \) is the rate constant of the reaction, \( A \) is the pre-exponential factor (also known as the frequency factor), Ea is the activation energy necessary for the reaction to occur, R is the universal gas constant (8.314 J/mol·K), and T is the temperature in Kelvin. To understand how a catalyzed and an uncatalyzed reaction differ, we can compare their rate constants via the Arrhenius equation. By isolating the activation energy, we can gain insights into how much energy is required to initiate each type of reaction.
Activation energy is a vital concept when explaining the effect of catalysts; it shows how a catalyst lowers the activation energy, thereby increasing the reaction rate. By comparing the differences in activation energy between reactions, as seen in the exercise solution, it becomes clear why catalyzed reactions proceed at a much faster rate than their uncatalyzed counterparts.
Activation Energy
Activation energy, represented by the symbol \( E_a \) in scientific equations, is the minimum amount of energy required to start a chemical reaction. It is the energy barrier that reactants must overcome to be converted into products. A higher activation energy means that fewer molecules have the necessary energy to undergo the reaction at a given temperature, leading to a slower reaction rate. It is this energy that is significantly lowered by the presence of a catalyst in a catalyzed reaction, explaining the drastic differences in reaction rates when comparing catalyzed and uncatalyzed reactions. The concept of activation energy is key to comprehending the inner workings of enzyme kinetics and the impact catalysts have on making reactions more efficient.
Catalyzed Reaction
A catalyzed reaction is one that is accelerated by the addition of a catalyst, a substance that increases the rate of the reaction without undergoing any permanent chemical change itself. In biological systems, enzymes are natural catalysts that facilitate various biochemical reactions necessary for life. In the given exercise, alkaline phosphatase accelerates the hydrolysis of methylphosphate. Enzymes like alkaline phosphatase operate by providing an alternative reaction pathway with a lower activation energy compared to the uncatalyzed pathway, hence significantly enhancing the reaction rate. This concept is crucial as it explains the mechanism through which living organisms achieve rapid and highly regulated metabolic processes.
Uncatalyzed Reaction
In contrast to a catalyzed reaction, an uncatalyzed reaction proceeds without the assistance of a catalyst. It relies solely on the kinetic energy of the reactants and environmental conditions, such as temperature and pressure, to reach the activation energy threshold. As a result, uncatalyzed reactions typically occur at a much slower rate because the probability of reactant molecules colliding with sufficient energy to surpass the activation energy barrier is lower. The vast difference in rate constants of the catalyzed and uncatalyzed hydrolysis of methylphosphate highlights the significant role enzymes play in increasing reaction rates, which the exercise solution illustrates by calculating the immense difference in activation energies.

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Most popular questions from this chapter

As noted on page \(423,\) a true transition state can bind to an enzyme active site with a \(K_{\mathrm{T}}\) as low as \(7 \times 10^{-26} M .\) This is a remarkable number, with interesting consequences. Consider a hypothetical solution of an enzyme in equilibrium with a ligand that binds with a \(K_{\mathrm{D}}\) of \(10^{-27} M .\) If the concentration of free enzyme, \([\mathrm{E}],\) is equal to the concentration of the enzyme-ligand complex, [EL], what would \([\mathrm{L}],\) the concentration of free ligand, be? Calculate the volume of solution that would hold one molecule of free ligand at this concentration.

Another consequence of tight binding (problem 9 ) is the free energy change for the binding process. Calculate \(\Delta G^{\circ}\) ' for an equilibrium with a \(K_{\mathrm{D}}\) of \(10^{-27} M .\) Compare this value to the free energies of the noncovalent and covalent bonds with which you are familiar. What are the implications of this number, in terms of the nature of the binding of a transition state to an enzyme active site?

Tosyl-L-phenylalanine chloromethyl ketone (TPCK) specifically inhibits chymotrypsin by covalently labeling His \(^{57}\). Tosyl-L-phenylalanine chloromethyl ketone (TPCK) a. Propose a mechanism for the inactivation reaction, indicating the structure of the product(s). b. State why this inhibitor is specific for chymotrypsin. c. Propose a reagent based on the structure of TPCK that might be an effective inhibitor of trypsin.

In which of the following environmental conditions would digestive enzyme Y be unable to bring its substrate(s) to the transition state? a. At any temperature below optimum b. At any pH where the rate of reaction is not maximum c. At any pH lower than 5.5 d. At any temperature higher than \(37^{\circ} \mathrm{C}\)

At \(35^{\circ} \mathrm{C},\) the rate of the reaction catalyzed by enzyme \(\mathrm{A}\) begins to level off. Which hypothesis best explains this observation? a. The temperature is too far below optimum. b. The enzyme has become saturated with substrate. c. Both \(A\) and \(B\). d. Neither A nor B.

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