Problem 69 - Chemistry: An Atoms-Focused Approach

Question

The activation energy for the uncatalyzed decomposition of hydrogen peroxide at \(20^{\circ} \mathrm{C}\) is \(75.3 \mathrm{kJ} / \mathrm{mol} .\) In the presence of the enzyme catalase, the activation energy is reduced to \(29.3 \mathrm{kJ} / \mathrm{mol} .\) Use the following form of the Arrhenius equation, \(R T \ln \left(k_{1} / k_{2}\right)=E_{2,}-E_{2_{1}},\) to calculate how much larger the rate constant of the catalyzed reaction is.

Step-by-Step Solution

Answer

Answer: The rate constant of the catalyzed decomposition of hydrogen peroxide is approximately 51.5 times larger than the rate constant of the uncatalyzed decomposition at a temperature of 20°C.

Step 1: List down the given information

We are given the temperature \(T = 20^{\circ}\mathrm{C}\) and activation energies for both uncatalyzed and catalyzed reactions \(E_{1} = 75.3 \mathrm{kJ/mol}\) and \(E_{2} = 29.3 \mathrm{kJ/mol}\).

Step 2: Convert temperature to Kelvin

Before plugging the temperature into the Arrhenius equation, we need to convert it from Celsius to Kelvin. The temperature in Kelvin is given by \(T(K) = T(^\circ\mathrm{C}) + 273.15\). Thus, \(T(K) = 20 + 273.15 = 293.15\mathrm{K}\).

Step 3: Insert the given information into the Arrhenius equation

Now we can insert the values of the activation energies and the temperature into the Arrhenius equation: \(R T \ln \left(\frac{k_{1}}{k_{2}}\right) = E_{2} - E_{1}\) Since we are looking for how much larger the rate constant of the catalyzed reaction is, we need to find the ratio \(\frac{k_{2}}{k_{1}}\). Rearranging the equation, we get: \(\ln\left(\frac{k_{1}}{k_{2}}\right) = \frac{E_{2} - E_{1}}{RT}\) Taking the exponential of both sides, we find the ratio of the rate constants: \(\frac{k_{1}}{k_{2}} = e^\frac{E_{2} - E_{1}}{RT}\)

Step 4: Calculate the rate constant ratio

Now we can plug the values into the equation and find the ratio of the rate constants: \(\frac{k_{1}}{k_{2}} = e^\frac{29.3\,\text{kJ/mol} - 75.3\,\text{kJ/mol}}{8.314\,\text{J/mol}\cdot\text{K} \cdot 293.15\,\text{K}}\) First, we convert the activation energies from \(\mathrm{kJ/mol}\) to \(\mathrm{J/mol}\) by multiplying by \(1000\), so \(E_{2} = 29300 \, \mathrm{J/mol}\) and \(E_{1} = 75300\, \mathrm{J/mol}\): \(\frac{k_{1}}{k_{2}} = e^\frac{29300\,\text{J/mol} - 75300\,\text{J/mol}}{8.314\,\text{J/mol}\cdot\text{K} \cdot 293.15\,\text{K}}\) Now calculating the result: \(\frac{k_{1}}{k_{2}} = e^\frac{-46000\,\text{J/mol}}{2417068.1\,\text{J/mol}\cdot\text{K}}\) \(\frac{k_{1}}{k_{2}} = e^{-0.0190}\) Finally, we find the ratio of the rate constants for uncatalyzed and catalyzed reactions: \(\frac{k_{1}}{k_{2}} \approx 0.0194\) The rate constant of the catalyzed reaction is approximately \(1/0.0194 \approx 51.5\) times larger than the rate constant of the uncatalyzed reaction.

Key Concepts

Arrhenius Equation

The Arrhenius equation is a fundamental formula used in chemistry to describe how the rate constant of a reaction depends on temperature and activation energy. This equation helps understand the speed at which reactions occur.
The basic form of the Arrhenius equation is: \[ k = A \, e^{-\frac{E_a}{RT}} \]where:

• \(k\) is the rate constant, which tells how fast the reaction proceeds.
• \(A\) is the pre-exponential factor, a constant that represents the frequency of collisions between reactant molecules.
• \(E_a\) is the activation energy, the minimum energy required for the reaction to take place.
• \(R\) is the universal gas constant, with a value of 8.314 J/mol·K.
• \(T\) is the temperature in Kelvin.

This equation beautifully illustrates the relationship between activation energy and reaction speed:

• A lower activation energy \((E_a)\), means that the reaction can happen more easily, thus increasing the rate constant \((k)\).
• A higher temperature \((T)\) provides reactant molecules with more energy, increasing the likelihood that they will overcome the activation energy barrier.

This is why reactions usually speed up at higher temperatures.

Catalyzed Reaction

A catalyzed reaction involves the use of a catalyst to increase the reaction rate without undergoing any permanent chemical change itself. Catalysts work by lowering the activation energy that is needed for the reaction to occur.

In the example from the exercise, an enzyme called catalase acts as a catalyst in the decomposition of hydrogen peroxide. For the uncatalyzed reaction, the activation energy is quite high at 75.3 kJ/mol. However, with the catalase, this energy is reduced to only 29.3 kJ/mol. This substantial reduction in activation energy allows the reaction to proceed much more quickly.

Key characteristics of catalyzed reactions include:

• They do not alter the final equilibrium of the reaction, only the rate at which equilibrium is reached.
• Catalysts can participate in reactions without themselves being consumed, meaning they can be used repeatedly.
• Enzymes, like catalase, are biological catalysts that are particularly effective in speeding up reactions in living organisms.

Catalysts are incredibly important in both industrial processes and biological systems, where they often determine the efficiency and feasibility of the reactions.

Rate Constant

The rate constant in a chemical reaction is a crucial factor that determines the speed of the reaction. In the context of the Arrhenius equation, the rate constant \(k\) links the concentration of reactants to the rate at which they react.
The magnitude of the rate constant \(k\) can provide insights into the reaction:

• Higher rate constants indicate faster reactions, while lower rate constants signify slower reactions.
• The rate constant is sensitive to both the activation energy \((E_a)\) and the temperature \((T)\). A lower \(E_a\) or higher temperature increases \(k\), thus increasing the rate.

In the exercise example, when catalase is present, the significantly reduced activation energy results in a higher rate constant, making the catalyzed reaction approximately 51.5 times faster than the uncatalyzed one.
The rate constant's variability showcases its pivotal role in reaction dynamics, allowing chemists to control how quickly reactions occur.
Understanding and manipulating \(k\) helps in designing more efficient chemical processes and optimizing reaction conditions.