Question

At body temperature \(\left(37^{\circ} \mathrm{C}\right),\) the rate constant of an enzyme-catalyzed decomposition is \(2.3 \times 10^{14}\) times that of the uncatalyzed reaction. If the frequency factor, \(A,\) is the same for both processes, by how much does the enzyme lower the \(E_{\mathrm{a}}\) ?

Short answer

The enzyme lowers the activation energy by 85.29 kJ/mol.

Step-by-step solution

- Understand the Arrhenius Equation

The Arrhenius equation relates the rate constant, k, to the activation energy, Ea, by the formula: \ \[ k = A e^{-\frac{E_a}{RT}} \] \ where: \ - k is the rate constant, \ - A is the frequency factor, \ - Ea is the activation energy, \ - R is the gas constant (8.314 J/mol·K), \ - T is the temperature in Kelvin.

- Set up the Ratio of Rate Constants

Given that the rate constant of the enzyme-catalyzed reaction is \(2.3 \times 10^{14}\) times that of the uncatalyzed reaction, set up the ratio: \ \[ \frac{k_{\text{catalyzed}}}{k_{\text{uncatalyzed}}} = 2.3 \times 10^{14} \]

- Express Each Rate Constant Using the Arrhenius Equation

The rate constants for the catalyzed and uncatalyzed reactions can be written as: \ \[ k_{\text{catalyzed}} = A e^{-\frac{E_{a, \text{catalyzed}}}{RT}} \] \ and \ \[ k_{\text{uncatalyzed}} = A e^{-\frac{E_{a, \text{uncatalyzed}}}{RT}} \] \ Since frequency factor \(A\) is the same for both processes, we can focus on the exponential part.

- Set Up the Equation from the Ratio

Using the given ratio: \ \[ \frac{A e^{-\frac{E_{a, \text{catalyzed}}}{RT}}}{A e^{-\frac{E_{a, \text{uncatalyzed}}}{RT}}} = 2.3 \times 10^{14} \] \ This simplifies to: \ \[ e^{-\frac{E_{a, \text{catalyzed}}}{RT} + \frac{E_{a, \text{uncatalyzed}}}{RT}} = 2.3 \times 10^{14} \]

- Natural Logarithm to Simplify the Equation

Take the natural logarithm of both sides to simplify: \ \[ -\frac{E_{a, \text{catalyzed}}}{RT} + \frac{E_{a, \text{uncatalyzed}}}{RT} = \text{ln}(2.3 \times 10^{14}) \] \ Simplify further: \ \[ -\frac{E_{a, \text{catalyzed}} - E_{a, \text{uncatalyzed}}}{RT} = \text{ln}(2.3 \times 10^{14}) \] \

- Solve for the Difference in Activation Energies

Rearrange to solve for the difference in activation energies: \ \[ E_{a, \text{uncatalyzed}} - E_{a, \text{catalyzed}} = RT \times \text{ln}(2.3 \times 10^{14}) \] \ Given: \ \[ R = 8.314 \text{J/mol·K} \] \ \[ T = 37^\text{{\textdegree}} \text{C} = 310 \text{K} \] \ Now, calculate the difference: \ \[ E_{a, \text{uncatalyzed}} - E_{a, \text{catalyzed}} = 8.314 \times 310 \times \text{ln}(2.3 \times 10^{14}) \] \ \[ E_{a, \text{uncatalyzed}} - E_{a, \text{catalyzed}} = 8.314 \times 310 \times 33.14 \] \ \[ E_{a, \text{uncatalyzed}} - E_{a, \text{catalyzed}} = 85286.126 \text{ J/mol} \] \ Convert to kJ/mol: \ \[ E_{a, \text{uncatalyzed}} - E_{a, \text{catalyzed}} = 85.29 \text{ kJ/mol} \]

- Conclusion

The enzyme lowers the activation energy by 85.29 kJ/mol.

Key concepts

activation energy

Activation energy is a crucial concept in chemical reactions. It represents the minimum energy required for reactants to undergo a transformation into products. This energy barrier must be overcome for a reaction to proceed. Imagine activation energy as a hill that reactants need to climb over. The higher the hill, the harder it is for the reaction to happen.

When discussing enzyme catalysis, activation energy becomes especially important. Enzymes work by lowering this energy barrier, allowing reactions to occur more easily and quickly. Without enzymes, many biological reactions would simply take too long to sustain life.

In the context of the Arrhenius equation, activation energy (\(E_a\)) is the exponent in the exponential function. Lowering the activation energy means that the exponential term increases, thereby raising the rate constant (k), which signifies a faster reaction rate.

Consider the example provided: The enzyme decreases the activation energy considerably, causing the rate constant of the enzyme-catalyzed reaction to be substantially higher than that of the uncatalyzed reaction.

enzyme catalysis

Enzyme catalysis is the process by which enzymes accelerate chemical reactions. Enzymes are specialized proteins that serve as biological catalysts. They speed up reactions by providing an alternative reaction pathway with a lower activation energy.

The key properties of enzyme catalysis include:

• **Specificity**: Enzymes are highly specific to their substrates (the molecules they act upon). This specificity is due to the unique active site of the enzyme, which precisely fits the substrate.
• **Efficiency**: Enzymes can increase reaction rates by a factor of millions, making biochemical reactions feasible at the mild conditions of temperature and pH that are typical in biological systems.
• **Control**: Enzyme activity can be regulated by various factors such as inhibitors, activators, and changes in environmental conditions (pH, temperature).

In enzyme catalysis, once the substrate binds to the active site of the enzyme, the enzyme-substrate complex forms. This complex then goes through a series of steps to convert the substrate into product, releasing it and allowing the enzyme to act again on new substrate molecules.

In our example, the enzyme significantly lowers the activation energy required for the reaction, making the reaction rate approximately 2.3 x 10^14 times faster compared to the uncatalyzed reaction.

reaction rate

The rate of a chemical reaction refers to how quickly reactants are converted to products. Reaction rates can vary greatly depending on several factors, including temperature, concentration of reactants, and the presence of a catalyst. In general, a higher reaction rate means that the reaction proceeds faster.

One way to quantitatively understand reaction rates is using the Arrhenius equation:
\( k = A e^{-\frac{E_a}{RT}} \)
This equation shows how the rate constant (\(k\)) of a reaction depends on the activation energy (\(E_a\)), the gas constant (\(R\)), and the temperature (\(T\)). From it, we see that increasing the temperature or decreasing the activation energy will result in a higher rate constant, and thus a faster reaction.

In enzyme catalysis, the rate constant for the catalyzed reaction is significantly higher than the uncatalyzed reaction due to the lower activation energy. As a result, enzyme-catalyzed reactions happen at a much faster rate. For example, in the problem provided, the presence of an enzyme causes the reaction rate to increase by a factor of 2.3 x 10^14, demonstrating the powerful effect enzymes have on reaction rates.