Problem 37 - Chemistry and Chemical Reactivity

Question

Calculate the activation energy, \(E_{\alpha}\), for the reaction $$2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ from the observed rate constants: \(k\) at \(25^{\circ} \mathrm{C}=\) \(3.46 \times 10^{-5} s^{-1}\) and \(k\) at \(55^{\circ} \mathrm{C}=1.5 \times 10^{-3} \mathrm{s}^{-1}.\)

Step-by-Step Solution

Answer

The activation energy, \(E_{\alpha}\), is approximately 103.27 kJ/mol.

Step 1: Identify Known Values

You are given that the rate constant \( k \) at \( 25^{\circ}C \), which is \(298 \, K\), is \( 3.46 \times 10^{-5} \, s^{-1} \), and at \( 55^{\circ}C \), which is \(328 \, K\), is \( 1.5 \times 10^{-3} \, s^{-1} \).

Step 2: Use the Arrhenius Equation

The Arrhenius equation is \( k = A e^{-\frac{E_{\alpha}}{RT}} \), where \( A \) is the pre-exponential factor, \( E_{\alpha} \) is the activation energy, \( R \) is the gas constant (\(8.314 \, J/mol \, K\)), and \( T \) is the temperature in Kelvin.

Step 3: Apply the Arrhenius Equation to Two Temperatures

For two temperatures, the equation can be rearranged to: \[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_{\alpha}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]Let \( k_1 = 3.46 \times 10^{-5} \), \( k_2 = 1.5 \times 10^{-3} \), \( T_1 = 298 \, K \), and \( T_2 = 328 \, K \).

Step 4: Calculate the Natural Logarithm of Rate Constants Ratio

First, calculate the ratio \( \ln\left(\frac{1.5 \times 10^{-3}}{3.46 \times 10^{-5}}\right) \). This gives \( \ln(43.353) \). Evaluating gives \( \approx 3.770\).

Step 5: Calculate the Inverse Temperature Difference

Calculate \( \frac{1}{T_2} - \frac{1}{T_1} \):\[ \frac{1}{328} - \frac{1}{298} = -0.000304 \, K^{-1} \]

Step 6: Solve for Activation Energy \(E_{\alpha}\)

Rearrange the formula from Step 3 to solve for \( E_{\alpha} \):\[ E_{\alpha} = - \frac{R \cdot \ln\left(\frac{k_2}{k_1}\right)}{\frac{1}{T_2} - \frac{1}{T_1}} \]Substitute the known values:\[ E_{\alpha} = - \frac{8.314 \, J/mol \, K \cdot 3.770}{-0.000304 \, K^{-1}} \]Calculate the result to find \( E_{\alpha} \approx 103.27 \, kJ/mol \).

Step 7: Convert Activation Energy to kJ/mol

Since each mole is originally in \( J \), convert \( E_{\alpha} \) to \( kJ/mol \) by dividing by 1000:\[ E_{\alpha} = 103.27 \, kJ/mol \]

Key Concepts

Arrhenius Equation

The Arrhenius Equation is crucial for understanding how different factors affect the rate of chemical reactions. This equation is represented as \( k = A e^{-\frac{E_{\alpha}}{RT}} \). Here, \( k \) is the rate constant that measures how quickly a reaction proceeds. \( A \) is the pre-exponential factor, which is related to the frequency of collisions with the correct orientation.\The term \( e^{-\frac{E_{\alpha}}{RT}} \) involves the activation energy \( E_{\alpha} \), which is the minimum energy required for a reaction to occur. In this equation, \( R \) stands for the universal gas constant, and \( T \) is the temperature in Kelvin. Together, this mathematical relationship helps us predict how changes in temperature or activation energy will impact the reaction rate.

Rate Constants

Rate Constants, denoted by \( k \), are fundamental to determining the speed of chemical reactions. For every unique reaction, \( k \) can vary based on temperature and the presence of catalysts. In our case, with gases such as \( \text{N}_2\text{O}_5 \), the rate constant at \( 25^{\circ}C \) is \( 3.46 \times 10^{-5} \, s^{-1} \) and at \( 55^{\circ}C \) it becomes \( 1.5 \times 10^{-3} \, s^{-1} \). These values highlight how sensitive \( k \) is to temperature changes.\Calculating \( k \) is essential because it helps chemists and engineers predict how long reactions will take to reach a certain completion point. By evaluating these constants at different temperatures, we use the Arrhenius equation to determine the Activation Energy \( E_{\alpha} \).

Temperature Dependence

Understanding the Temperature Dependence of reactions is key to comprehending chemical kinetics. The Arrhenius equation dynamically demonstrates how even slight variations in temperature alter the reaction rate. At higher temperatures, molecules move faster, increasing the probability of successful collisions, resulting in a higher rate constant \( k \).\For example, when the temperature rises from \( 25^{\circ}C \) to \( 55^{\circ}C \), the rate constant grows significantly. This exemplifies that the reaction becomes much quicker, emphasizing the role temperature plays in reaction dynamics. By analyzing changes in \( k \) across temperatures, scientists can calculate the Activation Energy, which is a critical parameter in chemical processes.

Gas Constant

The Gas Constant, symbolized by \( R \), is a pivotal constant in physical chemistry. Its standard value is \( 8.314 \, J/mol \, K \). This constant bridges the concepts of energy and temperature in the context of gases. In the Arrhenius equation, \( R \) aids in expressing how changes in temperature influence the rate constant \( k \).\When engaging with formulas like the Arrhenius Equation, \( R \) helps translate the activation energy's influence into measurable units. Always ensuring \( T \) is in Kelvin is critical, as \( R \) is defined accordingly. Thus, \( R \) ensures a consistent understanding of thermal energy impacts across varied scientific calculations.