Question

The rate constant of a reaction at 500 K and 700 K are 0.02 s$^{-1}$ and 0.2 s$^{-1}$ respectively. The activation energy of the reaction (in kJ mol$^{-1}$) is ($R = 8.3 \, \text{J K}^{-1} \text{mol}^{-1}$)

• A. 66.90
• B. 33.45
• C. 22.30
• D. 44.45

Hint:

The Arrhenius equation helps calculate activation energy using rate constants at two different temperatures.

Answer 1

The Correct Option is B

Using the Arrhenius equation: \(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\), where \(k_1 = 0.02 \, \text{s}^{-1}\) at \(T_1 = 500 \, \text{K}\), \(k_2 = 0.2 \, \text{s}^{-1}\) at \(T_2 = 700 \, \text{K}\), and \(R = 0.0083 \, \text{kJ K}^{-1} \text{mol}^{-1}\). First, calculate: \[ \ln\left(\frac{0.2}{0.02}\right) = \ln(10) \approx 2.303 \] \[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{500} - \frac{1}{700} = \frac{1}{1750} \] Substitute: \[ 2.303 = \frac{E_a}{0.0083} \times \frac{1}{1750} \] \[ E_a = 2.303 \times 1750 \times 0.0083 \approx 33.45 \, \text{kJ mol}^{-1} \] This matches option 2. Thus, the correct answer is 33.45.