Question

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

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

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): \(8.314 J / (mol \cdot K)\)

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.

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, \((Ea_2 - Ea_1)\): \[(Ea_2 - Ea_1) = -RT \ln{\frac{k_2}{k_1}}\]

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: \[(Ea_2 - Ea_1) \approx -2.57 \times 10^{4} J/mol\]

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.

Key concepts

Arrhenius Equation

The Arrhenius equation is a crucial formula in chemistry that explains how reaction rates vary with temperature. It is expressed as follows:
\[k = Ae^{\frac{-Ea}{RT}}\]
In this equation, \(k\) is the rate constant, \(A\) represents the pre-exponential factor (or collision factor), \(Ea\) denotes the activation energy, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin. The equation firmly establishes the relationship between the rate constant and temperature, showing how an increase in temperature generally leads to a faster reaction.

• Rate Constant \(k\): Shows how fast a reaction progresses.
• Pre-exponential Factor \(A\): Reflects the frequency of collisions with correct orientation.
• Activation Energy \(Ea\): The minimum energy necessary for the reaction to occur.
• Gas Constant \(R\): Usually taken as \(8.314 \text{ J/mol⋅K}\).
• Temperature \(T\): Must be in Kelvin for accurate calculations.

The beauty of the Arrhenius equation lies in its ability to help predict reaction kinetics by considering the influence of temperature and activation energy. By rearranging the equation, one can solve for different variables, as was done in the original exercise to find the difference in activation energies between enzyme-catalyzed and uncatalyzed reactions.

Enzyme Catalysis

Enzyme catalysis is a process where enzymes speed up chemical reactions. It plays a critical role in biological systems, ensuring that necessary reactions occur swiftly enough to sustain life. Enzymes achieve this by reducing the activation energy required for a reaction to proceed, making it easier for the reaction to take place.
Several factors contribute to enzyme catalysis:

• Specificity: Enzymes are highly specific, meaning they usually catalyze one specific type of reaction or react with a particular substrate.
• Active Site: This is the region of the enzyme where substrate molecules bind, and reactions occur. The unique shape and chemical environment of the active site facilitate the conversion of substrates into products.
• Lowering Activation Energy: By stabilizing the transition state, enzymes effectively lower the energy hurdle needed for the reaction, enhancing the reaction rate.

In the original exercise, the enzyme allowed for a significant increase in the reaction rate by lowering the activation energy from that of the uncatalyzed reaction. The rate constant increased from \(6.50 \times 10^{-4} \mathrm{min}^{-1}\) to \(1.67 \times 10^{4} \mathrm{min}^{-1}\), illustrating enzyme efficiency.

Rate Constant

In chemical kinetics, the rate constant is a numerical value that helps define the rate at which a reaction occurs. It appears in the rate equation for the reaction and changes with factors like temperature and the presence of catalysts (such as enzymes).
The rate constant is crucial for understanding reaction kinetics:

• The rate constant, \(k\), varies with temperature and is specific to each reaction.
• The higher the rate constant, the faster the reaction at a given temperature.
• The presence of catalysts, like enzymes, can significantly increase \(k\) by lowering the activation energy for a reaction.

In our case, enzyme catalysis significantly increased the rate constant of the reaction, allowing the enzyme-catalyzed process to proceed much faster than the uncatalyzed one. By comparing the rate constants of catalyzed versus uncatalyzed reactions, one can understand how much influence an enzyme has on the reaction pathway and speed.