Question

The rate constant of a reaction at 500 K and 700 K are \(0.02\,\text{s}^{-1}\) and \(0.07\,\text{s}^{-1}\) respectively. Calculate the value of activation energy (\(E_a\)).

Hint:

Higher activation energy means greater temperature sensitivity of the rate constant. Always convert \(E_a\) into kJ mol\(^{-1}\) for final answers.

Answer 1

Step 1: Use the Arrhenius equation: \[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \]
Step 2: Substitute the given values: \[ k_1 = 0.02\,\text{s}^{-1}, \quad k_2 = 0.07\,\text{s}^{-1} \] \[ T_1 = 500\,\text{K}, \quad T_2 = 700\,\text{K} \]
Step 3: Calculate each term: \[ \ln\left(\frac{0.07}{0.02}\right) = \ln(3.5) = 1.253 \] \[ \frac{1}{500} - \frac{1}{700} = \frac{200}{350000} = 0.0005714 \]
Step 4: Substitute values into the equation: \[ 1.253 = \frac{E_a}{8.314} \times 0.0005714 \]
Step 5: Solve for \(E_a\): \[ E_a = \frac{1.253 \times 8.314}{0.0005714} \] \[ E_a \approx 1.82 \times 10^{4}\,\text{J mol}^{-1} \] \[ E_a \approx 18.2\,\text{kJ mol}^{-1} \]