Question

The rate constant is doubled when temperature increases from 27°C to 37°C. Activation energy in kJ is:

• A. 54 kJ
• B. 100 kJ
• C. 50 kJ
• D. 20 kJ

Hint:

The Arrhenius equation allows you to calculate the activation energy of a reaction from the change in rate constants at different temperatures.

Answer 1

The Correct Option is A

Use the Arrhenius equation to calculate the activation energy \( E_a \): \[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] Substituting the given values, we find \( E_a = 54 \, \text{kJ/mol} \).
Final Answer: \[ \boxed{54 \, \text{kJ/mol}} \]