Question

The rate constant is doubled when temperature increases from \(27^{\circ} \mathrm{C}\) to \(37^{\circ} \mathrm{C}\). What is activation energy in kJ ?

• A. 21.32
• B. 34.12
• C. 53.60
• D. 43.54

Hint:

Apply the Arrhenius equation between two temperatures after converting to kelvin.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The Arrhenius equation for two temperatures is \(\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{T_2 - T_1}{T_1 T_2}\right)\).

Step 2: Data
\(T_1 = 300\text{ K}\), \(T_2 = 310\text{ K}\), \(k_2/k_1 = 2\), \(R = 8.314\text{ J K}^{-1}\text{mol}^{-1}\).

Step 3: Calculation
\(E_a = \dfrac{2.303 \times 8.314 \times \log 2 \times 300 \times 310}{10}\)
\(E_a = \dfrac{2.303 \times 8.314 \times 0.3010 \times 93000}{10} \approx 53600\text{ J mol}^{-1}\)

Step 4: Convert
\(E_a \approx 53.6\text{ kJ mol}^{-1}\).

Final Answer:
The activation energy is about 53.6 kJ. \[ \boxed{\text{(C) }53.60} \]