Question

Calculate activation energy for a reaction if it's rate doubles when temperature is raised from $20^\circ\text{C}$ to $35^\circ\text{C} \left( \text{R} = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \right)$}

• A. $17.336 \text{ kJ}$
• B. $26.900 \text{ kJ}$
• C. $34.673 \text{ kJ}$
• D. $44.236 \text{ kJ}$

Hint:

Shortcut: Rate doubles $\Rightarrow \ln 2 = 0.693$

Answer 1

The Correct Option is C

Concept: Arrhenius equation: \[ \ln \frac{k_2}{k_1} = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

Step 1: Convert temperatures. \[ T_1 = 20^\circ\text{C} = 293 \text{ K}, \quad T_2 = 35^\circ\text{C} = 308 \text{ K} \]

Step 2: Given: \[ \frac{k_2}{k_1} = 2 \Rightarrow \ln 2 = 0.693 \]

Step 3: Substitute values. \[ 0.693 = \frac{E_a}{8.314} \left(\frac{1}{293} - \frac{1}{308}\right) \]

Step 4: Calculate temperature term. \[ \frac{1}{293} - \frac{1}{308} = \frac{308 - 293}{293 \times 308} = \frac{15}{90244} \approx 1.662 \times 10^{-4} \]

Step 5: Solve for $E_a$. \[ 0.693 = \frac{E_a}{8.314} \times 1.662 \times 10^{-4} \] \[ E_a = \frac{0.693 \times 8.314}{1.662 \times 10^{-4}} \approx 34673 \text{ J mol}^{-1} \] \[ E_a \approx 34.673 \text{ kJ mol}^{-1} \]

Step 6: Conclusion.
Thus, activation energy = $34.673 \text{ kJ}$. Final Answer: Option (C)