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At \(25^{\circ} \mathrm{C}\), an enzyme accelerates a reaction by a factor of \(10^{5}\) over the uncatalyzed reaction in water. If the effect of the enzyme is solely to reduce the energy of the transition state, by what amount does it reduce the energy of the transition state \(\left(E_{A}\right)\) ?

Short Answer

Expert verified
The enzyme reduces the transition state energy by approximately 28.5 kJ/mol.

Step by step solution

01

Understanding the Arrhenius Equation

The Arrhenius equation relates the rate constant of a reaction to the temperature and activation energy: \[ k = A \cdot e^{-E_A/(RT)} \]where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_A \) is the activation energy, \( R \) is the gas constant \( (8.314 \, \text{J/mol K}) \), and \( T \) is the temperature in Kelvin.
02

Calculate Temperature in Kelvin

Convert the given temperature from Celsius to Kelvin: \[ T = 25 + 273.15 = 298.15 \, \text{K} \]
03

Setting Up the Ratio of Rate Constants

Given that the enzyme accelerates the reaction by a factor of \( 10^5 \), the catalyzed and uncatalyzed rate constants are related by: \[ \frac{k_{cat}}{k_{uncat}} = 10^5 \]
04

Express the Ratio Using Arrhenius Equation

Substitute the Arrhenius equation into the ratio equation:\[ \frac{A \cdot e^{-E_{A(cat)}/(RT)}}{A \cdot e^{-E_{A(uncat)}/(RT)}} = 10^5 \]After simplification:\[ e^{-(E_{A(cat)} - E_{A(uncat)})/(RT)} = 10^5 \]
05

Solve for the Activation Energy Difference

Take the natural logarithm of both sides to solve for the difference in activation energies:\[ -(E_{A(cat)} - E_{A(uncat)}) / (RT) = \ln(10^5) \]\[ E_{A(cat)} - E_{A(uncat)} = -R \cdot T \cdot \ln(10^5) \]
06

Calculate the Energy Reduction

Substitute \( R = 8.314 \; \text{J/mol K} \), \( T = 298.15 \; \text{K} \), and \( \ln(10^5) = 5 \cdot \ln(10) \approx 11.51\) \[ E_{A(cat)} - E_{A(uncat)} = -8.314 \cdot 298.15 \cdot 11.51 \]\[ E_{A(cat)} - E_{A(uncat)} \approx -28,491.4 \; \text{J/mol} \]

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Key Concepts

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

Activation Energy
Activation energy is the minimum amount of energy required to initiate a chemical reaction. It plays a significant role in determining the speed or rate of the reaction. In enzyme kinetics, enzymes act as catalysts that significantly lower the activation energy, thereby speeding up reactions. A lower activation energy means that more molecules have enough energy to reach the transition state at a given temperature.
Here are some key aspects to understand about activation energy:
  • It represents a barrier that reactants must overcome to be converted into products.
  • The units are typically given in joules per mole (J/mol).
  • Activation energy is specific to each reaction and varies with conditions like temperature and pressure.
To calculate changes in activation energy due to enzyme action, we apply formulas like the Arrhenius equation, which relates temperature, rate constants, and activation energy.
Arrhenius Equation
The Arrhenius equation is a fundamental formula used to describe how the rate constant, which measures the speed of a reaction, changes with temperature and activation energy. This equation is expressed as:\[ k = A \cdot e^{-E_A/(RT)} \]where:
  • **k** is the rate constant.
  • **A** is the pre-exponential factor, representing the frequency of collisions with correct orientation.
  • **E_A** is the activation energy.
  • **R** is the universal gas constant (8.314 J/mol K).
  • **T** is the temperature in Kelvin.
The Arrhenius equation shows that higher temperatures or lower activation energies result in a larger value for the rate constant, meaning the reaction proceeds faster. When enzymes lower the activation energy, the rate constant increases significantly, emphasizing their role in biochemistry as powerful catalysts.
Transition State Theory
Transition state theory is an important concept in understanding how reactions occur. It suggests that during a reaction, reactants move through a high-energy state, known as the transition state, before becoming products. The activation energy is the energy difference between the reactants and this transition state.
The main ideas of transition state theory include:
  • It depicts how chemical reactions develop, emphasizing the transition state's pivotal role.
  • In enzyme-catalyzed reactions, enzymes stabilize the transition state, reducing the activation energy needed.
  • According to this theory, the transition state is at the top of the energy barrier separating reactants and products.
This theory helps explain why enzymes are so effective in facilitating reactions. They provide an alternative pathway with a lower energy barrier, thereby accelerating the reaction speed. Transition state theory, combined with the Arrhenius equation, offers an insightful framework for analyzing reaction kinetics in both catalyzed and uncatalyzed conditions.

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Most popular questions from this chapter

Given the following three data tables of substrate concentrations and initial velocities for enzymes that obey Michaelis-Menten kinetics, estimate \(K_{\mathrm{M}}\) for each enzyme in molar units. a. \begin{tabular}{|c|c|} \hline\([\mathbf{S}](\mathrm{mM})\) & \(v_{0}\left(\mathrm{mM} \cdot \mathrm{sec}^{-1}\right)\) \\ \hline 1 & \(266.7\) \\ \hline 3 & \(553.8\) \\ \hline 5 & \(705.9\) \\ \hline 50 & \(1121.5\) \\ \hline 500 & \(1191.7\) \\ \hline 5000 & \(1199.2\) \\ \hline \end{tabular} b. \begin{tabular}{|c|l|} \hline IS] (nM) & \(v_{0}\left(\mathrm{mM}^{2} \cdot \mathrm{min}^{-1}\right)\) \\\ \hline 4 & \(123.5\) \\ \hline 5 & \(137.4\) \\ \hline 6 & \(148.5\) \\ \hline 10 & \(177.3\) \\ \hline 100 & \(240.2\) \\ \hline 1000 & \(249.0\) \\ \hline \end{tabular} C. \begin{tabular}{|c|l|} \hline\([\mathrm{S}](\mathrm{mM})\) & \(\nu_{0}(\) Mohour \(-1)\) \\ \hline 1 & \(0.00\) \\ \hline 10 & \(0.01\) \\ \hline 100 & \(0.07\) \\ \hline 200 & \(0.10\) \\ \hline 1000 & \(0.17\) \\ \hline 5000 & \(0.19\) \\ \hline \end{tabular}

Catalytic antibodies are generally less efficient than natural enzymes that catalyze the same reactions. True/False

The table below lists initial velocities measured for an enzymatic reaction at different substrate concentrations in the presence and absence of an inhibitor. The enzyme concentration is identical in both reactions. a. Graph a Lineweaver-Burk plot for each set of data. b. What are the values of \(V_{\max }\) and \(K_{M}\) for each experiment? c. What is the inhibition mechanism ? d. If the concentration of inhibitor is \(100 \mathrm{nM}\), what is the value of \(K_{1}\) ?

Why is triose phosphate isomerase considered to be an example of a "perfect enzyme"?

An enzyme inhibitor is observed to alter the \(K_{\mathrm{M}}\) but not the \(V_{\max }\) of a reaction. This inhibitor is most likely: a. A noncompetitive inhibitor. b. A competitive inhibitor. c. An allosteric inhibitor. d. A substrate-dependent noncompetitive inhibitor. e. A covalent inhibitor.

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