Arrhenius Equation: Activation Energy from Two Rate Constants

Concept Overview

This question tests the understanding and application of the Arrhenius equation, which quantitatively relates the rate constant of a chemical reaction to temperature and activation energy. Specifically, it focuses on the two-point form of the Arrhenius equation, allowing us to determine the activation energy ($E_a$) if we know the rate constants at two different temperatures. This is a fundamental concept in chemical kinetics for understanding reaction mechanisms and predicting reaction rates.

Worked Solution

Step 1: Recall the Arrhenius equation. The Arrhenius equation describes the temperature dependence of reaction rates: $k = A e^{-E_a/RT}$ Here, $k$ 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 absolute temperature.

Step 2: Express the Arrhenius equation for two different temperatures. Let $k_1$ be the rate constant at temperature $T_1$, and $k_2$ be the rate constant at temperature $T_2$. We can write the Arrhenius equation for each case: $k_1 = A e^{-E_a/RT_1}$ $k_2 = A e^{-E_a/RT_2}$ The pre-exponential factor $A$ and the activation energy $E_a$ are assumed to be constant over the temperature range considered.

Step 3: Divide the second equation by the first equation. Dividing the expression for $k_2$ by the expression for $k_1$ eliminates the pre-exponential factor $A$: $\frac{k_2}{k_1} = \frac{A e^{-E_a/RT_2}}{A e^{-E_a/RT_1}}$ This simplifies to: $\frac{k_2}{k_1} = e^{-E_a/RT_2 + E_a/RT_1}$ $\frac{k_2}{k_1} = e^{\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)}$

Step 4: Take the natural logarithm of both sides. To isolate the terms involving activation energy and temperature, we take the natural logarithm (ln) of both sides of the equation: $\ln\left(\frac{k_2}{k_1}\right) = \ln\left(e^{\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)}\right)$ Using the property $\ln(e^x) = x$, we get: $\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ This is the two-point form of the Arrhenius equation.

Step 5: Rearrange the equation to solve for activation energy ($E_a$). To calculate $E_a$, we can rearrange the equation: $E_a = R \frac{\ln\left(\frac{k_2}{k_1}\right)}{\left(\frac{1}{T_1} - \frac{1}{T_2}\right)}$ Alternatively, it can be written as: $E_a = R \frac{\ln\left(\frac{k_2}{k_1}\right)}{\left(\frac{T_2 - T_1}{T_1 T_2}\right)}$ $E_a = \frac{R \cdot T_1 T_2}{T_2 - T_1} \ln\left(\frac{k_2}{k_1}\right)$ In this formula, $k_1$ and $k_2$ are the rate constants at absolute temperatures $T_1$ and $T_2$, respectively. The ideal gas constant $R$ is typically $8.314 \, \text{J/mol} \cdot \text{K}$. Ensure that temperatures are in Kelvin.

Key Takeaways:

• The Arrhenius equation relates reaction rate constants to temperature and activation energy.
• The two-point form of the Arrhenius equation, $\ln(k_2/k_1) = (E_a/R)(1/T_1 - 1/T_2)$, allows calculation of $E_a$ from rate constants at two temperatures.
• Always use absolute temperatures (Kelvin) and ensure consistent units for $R$ and $E_a$.

Answer: $E_a = R \frac{\ln\left(\frac{k_2}{k_1}\right)}{\left(\frac{1}{T_1} - \frac{1}{T_2}\right)}$