Question

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate activation energy (E_a).
303 R = 19.15 JK⁻¹ mol⁻¹, log 2 = 0.3010

Answer 1

The Arrhenius equation is given by:

\( k = A e^{-\frac{E_a}{RT}} \)

Where:

• k is the rate constant,
• A is the pre-exponential factor,
• E_a is the activation energy,
• R is the gas constant, and
• T is the temperature in Kelvin.

Using the fact that the rate doubles for a 10 K increase in temperature, we can use the following form of the Arrhenius equation:

\( \frac{k_2}{k_1} = e^{\frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)} \)

Since the rate doubles, \( \frac{k_2}{k_1} = 2 \), and the temperature change is \( \Delta T = 10 \, \text{K} \), with \( T_1 = 298 \, \text{K} \) and \( T_2 = 308 \, \text{K} \). Substituting the known values:

\( 2 = e^{\frac{E_a}{19.15} \left( \frac{1}{298} - \frac{1}{308} \right)} \)

Taking the natural logarithm on both sides:

\( \ln 2 = \frac{E_a}{19.15} \left( \frac{1}{298} - \frac{1}{308} \right) \)

\( 0.693 = \frac{E_a}{19.15} \times \frac{308 - 298}{298 \times 308} \)

Solving for \( E_a \):

\( E_a = \frac{0.693 \times 19.15 \times 298 \times 308}{10 \times 298 \times 308} \)

\( E_a = 56.5 \, \text{kJ/mol} \)

Final Answer:

The activation energy \( E_a \) is approximately \( 56.5 \, \text{kJ/mol} \).