Problem 84 - Chemistry: An Atoms-Focused Approach

Question

Does reducing the activation energy of a reaction by \(\frac{1}{2}\) increase its rate constant by a factor of \(2 ?\)

Step-by-Step Solution

Answer

Explain. No, reducing the activation energy by half does not always increase the rate constant by a factor of 2. The relationship between the rate constant and activation energy can be analyzed using the Arrhenius equation. When calculating the ratio of the new rate constant to the old rate constant, we get \(e^{\frac{E_a}{2RT}}\), which is not equal to 2 for all values of \(E_a\), \(R\), and \(T\). The actual increase in the rate constant depends on the specific values of the activation energy and temperature.

Step 1: Understand the Arrhenius equation

The Arrhenius equation relates the rate constant, activation energy, and temperature of a reaction. The equation is given by: \[k = A \cdot e^{-\frac{E_a}{RT}}\] where \(k\) is the rate constant, \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the ideal gas constant, and \(T\) is the temperature.

Step 2: Derive the equation for the new rate constant

Let the original activation energy be represented as \(E_a\). When we reduce the activation energy by half, the new value will be represented as \(\frac{E_a}{2}\). We can rewrite the Arrhenius equation for the new rate constant, \(k'\), by substituting the new activation energy in the equation: \[k' = A \cdot e^{-\frac{E_a / 2}{RT}}\]

Step 3: Find the ratio between the new and old rate constants

To see whether the rate constant increases by a factor of 2 or not, we will take the ratio of the new rate constant to the old rate constant: \[\frac{k'}{k} = \frac{A \cdot e^{-\frac{E_a / 2}{RT}}}{A \cdot e^{-\frac{E_a}{RT}}}\]

Step 4: Simplify the equation

We can cancel out the pre-exponential factor \(A\) from the numerator and denominator, which simplifies the equation to: \[\frac{k'}{k} = \frac{e^{-\frac{E_a / 2}{RT}}}{e^{-\frac{E_a}{RT}}}\] Now, let's apply the exponent property: \(\frac{e^a}{e^b} = e^{a - b}\). Therefore, the equation becomes: \[\frac{k'}{k} = e^{-\frac{E_a / 2}{RT} + \frac{E_a}{RT}}\]

Step 5: Check if the ratio is equal to 2

If the ratio \(\frac{k'}{k}\) is equal to 2, the rate constant will indeed increase by a factor of 2. Otherwise, it won't. Let's evaluate the expression: \[\frac{k'}{k} = e^{-\frac{E_a / 2}{RT} + \frac{E_a}{RT}} = e^{\frac{E_a}{2RT}}\] As \(e^{\frac{E_a}{2RT}}\) isn't equal to 2 for all values of \(E_a\), \(R\), and \(T\), the rate constant will not always increase by a factor of 2 when the activation energy is reduced by half. The actual increase in the rate constant depends on the values of the activation energy and temperature.

Key Concepts

Arrhenius Equation

The Arrhenius Equation is a fundamental formula that helps us understand the relationship between the rate of a chemical reaction and factors like temperature and activation energy. It is expressed mathematically as:\[k = A \cdot e^{-\frac{E_a}{RT}}\]Here,

• \( k \) stands for the rate constant, an essential parameter that dictates the speed of the reaction.
• \( A \) is the pre-exponential factor or frequency factor, representing the number of times reactant molecules face the right orientation to react.
• \( E_a \) is the activation energy, the minimum energy that must be overcome for a reaction to occur.
• \( R \) is the ideal gas constant.
• \( T \) is the temperature in Kelvin.

This equation signifies that as temperature increases, or as the activation energy decreases, the rate constant \( k \) will increase.
Hence, it shows the sensitivity of reaction rates to the aforementioned factors. Understanding the Arrhenius Equation is key to controlling industrial chemical reactions and predicting how reactions proceed in atmospheric conditions.

Activation Energy

Activation Energy \( E_a \) is a crucial concept in chemical kinetics. It defines the minimum energy that reacting species must possess for a reaction to take place. Think of activation energy as a kind of barrier that reactants must cross in order to transform into products.

• The higher the activation energy, the slower the reaction, since fewer molecules have enough energy to get over the barrier initially.
• Lowering the activation energy, such as through a catalyst or other interventions, can significantly boost the rate of a reaction.

In the exercise, we considered what happens when the activation energy is halved.
Reduction in activation energy increases the rate constant but not by precisely doubling it. The relationship isn't linear but rather exponential based on the Arrhenius Equation since the expression \(e^{\frac{E_a}{2RT}}\) governs the outcome when changes are made to \( E_a \).
This makes understanding the specifics of activation energy crucial when assessing how reaction rates change in various situations.

Rate Constant

The Rate Constant \( k \) is critical in determining the speed of a reaction. It's a numerical value that tells us just how fast reactants are getting converted into products in a given reaction. The interplay of the rate constant with activation energy and temperature eventually determines the reaction's speed.

• In the Arrhenius equation, an increase in the temperature or a decrease in activation energy will lead to an increase in \( k \).
• The actual value of \( k \) can vary widely depending on the specific reaction and conditions.
• The units of the rate constant vary with the order of the reaction, e.g., \( \text{mol}^{1-n}\cdot \text{L}^{n-1}\cdot \text{s}^{-1}\) for a reaction of order \( n \).

In terms of the exercise discussed, halving the activation energy results in significant changes to the rate constant,
but exactly how much it changes depends on the specifics like the original \( E_a \) value and the temperature involved.
Using the formula \(k' = A \cdot e^{-\frac{E_a / 2}{RT}}\), we can analyze these changes—but it's clear that they don't simply double due to the precise exponential relationship described in the Arrhenius Equation.