The rates of most reactions double when their temperature is raised from 298 K to 308 K. Calculate their activation energy. [R = 8.314 JK^-1 mol^-1]

52.905 kJ mol^-1

The correct answer is option 1. \(52.905\text{k J mol}^{-1}\).

To determine the activation energy (\(E_a\)) using the temperature dependence of reaction rates, we can apply the Arrhenius equation and its logarithmic form. The given problem states that the reaction rate doubles when the temperature is increased from 298 K to 308 K. We can use the following relationship derived from the Arrhenius equation:

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

Given:

\( k_2 = 2k_1 \)

\( T_1 = 298 \, \text{K} \)

\( T_2 = 308 \, \text{K} \)

\( R = 8.314 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1} \)

Let us plug in these values:

\(\ln 2 = \frac{E_a}{8.314} \left( \frac{1}{298} - \frac{1}{308} \right)\)

First, calculate the temperature difference in the denominators:

\(\frac{1}{298} - \frac{1}{308} = \frac{308 - 298}{298 \times 308} = \frac{10}{298 \times 308}\)

\(\frac{10}{298 \times 308} = \frac{10}{91784} \approx 1.089 \times 10^{-4}\)

Now, use \(\ln 2 \approx 0.693\):

\(0.693 = \frac{E_a}{8.314} \times 1.089 \times 10^{-4}\)

Solving for \(E_a\):

\(E_a = \frac{0.693 \times 8.314}{1.089 \times 10^{-4}}\)

\(E_a \approx \frac{5.76222}{1.089 \times 10^{-4}} = 52905 \, \text{J/mol} = 52.905 \, \text{kJ/mol}\)

So, the activation energy (\(E_a\)) is 52.905 kJ mol\(^{-1}\)