Question

Consider two reactions having the same pre-exponential factor (A) occurring at the same temperature (T). \[ A \xrightarrow{E_{a1}} B \] \[ C \xrightarrow{E_{a2}} D \] \[ E_{a1} = 5E_{a2} \] Find out the correct expression.

• A. \( \frac{k_1}{k_2} = e^{\frac{E_{a2}}{RT}} \)
• B. \( \frac{k_1}{k_2} = e^{\frac{E_{a1}}{RT}} \)
• C. \( \frac{k_1}{k_2} = e^{\frac{4E_{a1}}{RT}} \)
• D. \( \frac{k_1}{k_2} = e^{\frac{E_{a2}}{SRT}} \)

Hint:

The ratio of rate constants can be found using the Arrhenius equation by plugging in the activation energies and temperature.

Answer 1

The Correct Option is C

Step 1: Apply the Arrhenius equation.
The Arrhenius equation is: \[ k = Ae^{-\frac{E_a}{RT}}, \] where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature. Step 2: Use the given information.
The ratio of the rate constants \( \frac{k_1}{k_2} \) can be written as: \[ \frac{k_1}{k_2} = e^{\frac{E_{a1} - E_{a2}}{RT}}. \] Given that \( E_{a1} = 5E_{a2} \), substitute this into the equation: \[ \frac{k_1}{k_2} = e^{\frac{4E_{a1}}{RT}}. \] Final Answer: \[ \boxed{\frac{k_1}{k_2} = e^{\frac{4E_{a1}}{RT}}}. \]