Question

Calculate the activation energy of a reaction, whose rate constant doubles on raising the temperature from 300 K to 600 K.

• A. 3.45 kJ/mol
• B. 6.90 kJ/mol
• C. 9.68 kJ/mol
• D. 19.6 kJ/mol

Hint:

The Arrhenius equation helps determine activation energy when the rate constant changes with temperature.

Answer 1

The Correct Option is A

Step 1: Applying the Arrhenius Equation 
The Arrhenius equation relates the rate constant \( k \) to temperature \( T \): \[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] where: - \( k_2 = 2k_1 \) (since the rate constant doubles), - \( T_1 = 300 \) K, - \( T_2 = 600 \) K, - \( R = 8.314 \) J mol\(^{-1}\)K\(^{-1}\). 
Step 2: Substituting Values 
\[ \log \left(\frac{2k_1}{k_1} \right) = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{300} - \frac{1}{600} \right) \] \[ \log 2 = \frac{E_a}{2.303 \times 8.314} \times \left( \frac{2}{600} \right) \] 
Step 3: Solving for \( E_a \) 
Using \( \log 2 = 0.301 \), \[ 0.301 = \frac{E_a}{2.303 \times 8.314} \times \frac{1}{300} \] \[ E_a = \frac{0.301 \times 2.303 \times 8.314 \times 300}{1} \] \[ E_a \approx 3.45 { kJ/mol} \] 
Final Answer: The activation energy is 3.45 kJ/mol.