Questions: A particular reaction has a rate constant of 0.101 M^-1 s^-1 at 25°C and a rate constant of 0.332 M^-1 s^-1 at 45°C. Calculate the activation energy (Ea) for the reaction in kJ / mol. Enter your answer with no units.

Solution Steps

The Arrhenius equation relates the rate constant \( k \) to the temperature \( T \) and activation energy \( E_a \) as follows:

\[ k = A e^{-\frac{E_a}{RT}} \]

where:

• \( k \) is the rate constant,
• \( A \) is the pre-exponential factor,
• \( E_a \) is the activation energy,
• \( R \) is the universal gas constant (\(8.314 \, \text{J/mol K}\)),
• \( T \) is the temperature in Kelvin.

For two different temperatures, the Arrhenius equation can be rearranged to:

\[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

where:

• \( k_1 = 0.101 \, \text{M}^{-1} \text{s}^{-1} \) at \( T_1 = 25^\circ \text{C} = 298 \, \text{K} \),
• \( k_2 = 0.332 \, \text{M}^{-1} \text{s}^{-1} \) at \( T_2 = 45^\circ \text{C} = 318 \, \text{K} \).

Calculate the natural logarithm of the ratio of the rate constants:

\[ \ln\left(\frac{0.332}{0.101}\right) = \ln(3.2871) \approx 1.1892 \]

Calculate the difference in reciprocal temperatures:

\[ \frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{318} - \frac{1}{298} \approx -0.0002105 \, \text{K}^{-1} \]

Rearrange the equation to solve for \( E_a \):

\[ E_a = -\frac{R \cdot \ln\left(\frac{k_2}{k_1}\right)}{\left(\frac{1}{T_2} - \frac{1}{T_1}\right)} \]

Substitute the known values:

\[ E_a = -\frac{8.314 \cdot 1.1892}{-0.0002105} \approx 46900 \, \text{J/mol} = 46.9 \, \text{kJ/mol} \]

Final Answer