Problem 85 - Chemistry The Science in Context

Question

The rate constant for the reaction of ozone with oxygen atoms was determined at four temperatures. Calculate the activation energy and frequency factor \(A\) for the reaction $$ \mathrm{O}(g)+\mathrm{O}_{3}(g) \rightarrow 2 \mathrm{O}_{2}(g) $$ given the following data: $$\begin{array}{cc} T(\mathrm{K}) & k\left[\mathrm{cm}^{3} /(\text { molecule } \cdot \mathrm{s})\right] \\ 250 & 2.64 \times 10^{-4} \\ \hline 275 & 5.58 \times 10^{-4} \\ \hline 300 & 1.04 \times 10^{-3} \\ \hline 325 & 1.77 \times 10^{-3} \\ \hline \end{array}$$

Step-by-Step Solution

Answer

The activation energy (Ea) for the given reaction is approximately 8895.35 J/mol, and the frequency factor (A) is approximately 2.68 × 10^-11 cm³/(molecule · s).

Step 1: Linearize the Arrhenius equation

Take the natural logarithm on both sides of the Arrhenius equation, and solve for \(E_a\): $$ \ln{k} = \ln{A} - \frac{E_a}{RT} $$

Step 2: Choose two temperature points

We will choose two data points from the given table in order to find the activation energy: Point 1: \(T_1 = 250 K,\; k_1 = 2.64 \times 10^{-4}\ cm^3/(molecule \cdot s)\) Point 2: \(T_2 = 275 K,\; k_2 = 5.58 \times 10^{-4}\ cm^3/(molecule \cdot s)\) These two points will give us two equations with two unknowns, \(A\) and \(E_a\).

Step 3: Write down the two equations

Substituting the values of \(T_1\), \(k_1\), \(T_2\), and \(k_2\) into the linearized Arrhenius equation, we get the following two equations: $$ \ln{(2.64 \times 10^{-4})} = \ln{A} - \frac{E_a}{8.314 \times 250} $$ $$ \ln{(5.58 \times 10^{-4})} = \ln{A} - \frac{E_a}{8.314 \times 275} $$

Step 4: Solve for the activation energy

Subtract the two equations to eliminate \(\ln{A}\) and solve for \(E_a\): $$ \ln{\frac{5.58 \times 10^{-4}}{2.64 \times 10^{-4}}} = \frac{E_a (1/8.314 \times 250 - 1/8.314 \times 275)} $$ Solve for \(E_a\): $$ E_a = \frac{\ln{\frac{5.58 \times 10^{-4}}{2.64 \times 10^{-4}}}}{1/8.314 \times 250 - 1/8.314 \times 275} = 8895.35\ J/mol $$

Step 5: Calculate the frequency factor

Substitute the calculated value of \(E_a\) and any data point into the linearized Arrhenius equation to find \(\ln{A}\): $$ \ln{(2.64 \times 10^{-4})} = \ln{A} - \frac{8895.35}{8.314 \times 250} $$ Find \(A\) by taking the exponent on both sides: $$ A = e^{\ln{(2.64 \times 10^{-4})} + \frac{8895.35}{8.314 \times 250}} = 2.68 \times 10^{-11}\ cm^3/(molecule \cdot s) $$ So, the activation energy (\(E_a\)) for the given reaction is approximately \(8895.35\ J/mol\), and the frequency factor (\(A\)) is approximately \(2.68 \times 10^{-11}\ cm^3/(molecule \cdot s)\).

Key Concepts

Arrhenius equation

The Arrhenius equation is a fundamental formula used to understand how temperature affects reaction rates. It states that the rate constant \( k \) for a chemical reaction is dependent on the temperature \( T \), the activation energy \( E_a \), and a pre-exponential factor \( A \). The equation is formulated as: \[ k = A e^{-\frac{E_a}{RT}} \] where \( R \) is the universal gas constant. This relationship demonstrates that even a small increase in temperature can lead to a significant increase in the reaction rate, given that the exponential function undergoes rapid changes.The natural logarithm form, \( \ln k = \ln A - \frac{E_a}{RT} \), is particularly useful for mathematical manipulation. By plotting \( \ln k \) against \( 1/T \), a straight line can be obtained, whose slope and intercept can be used to calculate \( E_a \) and \( \ln A \), respectively. This linearization simplifies the computational analysis and helps chemists break down experimental data into actionable insights.

Reaction rate constants

Reaction rate constants signify how quickly a reaction proceeds under certain conditions. They are pivotal in understanding and predicting the kinetics of a reaction. Given in units such as \( \text{cm}^3/\text{molecule}\cdot \text{s} \), these constants reflect the likelihood of collisions that result in a successful reaction between reactant molecules.In practice, these constants vary with temperature, as governed by the Arrhenius equation. At higher temperatures, molecules have more kinetic energy and can overcome potential energy barriers, depicted by an increased rate constant. Thus, measurements at different temperatures provide the necessary data to calculate parameters like the activation energy and the pre-exponential factor in the Arrhenius equation. By analyzing these changes, we gain insights into the energetic demands of a reaction, offering a deeper understanding of its behavior under differing conditions.

Frequency factor

The frequency factor \( A \), also known as the pre-exponential factor, represents the number of times phosphorylated orientations occur per unit time. It's a constant that can vary widely depending on molecular interactions and arrangements.While the energy term \( e^{-\frac{E_a}{RT}} \) in the Arrhenius equation accounts for the temperature-dependent part of reaction rates, \( A \) captures the intrinsic characteristics of the molecules involved, including collision frequency and orientation suitability. A higher \( A \) implies that molecules are more frequently in a reactive orientation when they collide.This factor is crucial when comparing reactions of different types or with different reactant states, as it sheds light on the influence of molecular configuration beyond just energetic barriers. Knowing \( A \) aids chemists in fine-tuning reactions by focusing on tactics that might enhance effective molecule encounters.