Question

A certain reaction is 50 complete in 20 minutes at 300 K and the same reaction is 50 complete in 5 minutes at 350 K. Calculate the activation energy if it is a first order reaction.
Given: \[ R = 8.314 \, \text{J K}^{-1} \, \text{mol}^{-1}, \quad \log 4 = 0.602 \]

Hint:

The Arrhenius equation relates the rate constant of a reaction to the activation energy, allowing us to calculate activation energy when rate constants are known at different temperatures.

Answer 1

1. Calculation of Activation Energy (Ea) for the First-Order Reaction:

Given Data:
- Reaction is 50% complete in 20 minutes at 300 K and 50% complete in 5 minutes at 350 K.
- \( R = 8.314 \, \text{J K}^{-1} \, \text{mol}^{-1} \)
- \( \log 4 = 0.602 \)

Step 1: Use the Arrhenius Equation:
The Arrhenius equation relates the rate constant \(k\) to the temperature \(T\) and activation energy \(E_a\):
\[ k = A e^{-\frac{E_a}{RT}} \] where \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. We can use the two given temperatures to find the activation energy.

Step 2: Use the Integrated Rate Law for a First-Order Reaction:
For a first-order reaction, the rate constant \( k \) is related to the time for 50% completion by:
\[ k = \frac{0.693}{t_{1/2}} \] where \( t_{1/2} \) is the time for the reaction to reach 50% completion. From the data: - At 300 K, \( t_{1/2} = 20 \, \text{minutes} \) - At 350 K, \( t_{1/2} = 5 \, \text{minutes} \)

Step 3: Calculate the Rate Constants:
For 300 K, the rate constant \( k_1 \) is:
\[ k_1 = \frac{0.693}{20} = 0.03465 \, \text{min}^{-1} \] For 350 K, the rate constant \( k_2 \) is:
\[ k_2 = \frac{0.693}{5} = 0.1386 \, \text{min}^{-1} \]

Step 4: Use the Arrhenius Equation to Find \( E_a \):
Now, use the Arrhenius equation in its logarithmic form to calculate the activation energy:

\[ \log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] Substitute the values into the equation:
\[ \log \left( \frac{0.1386}{0.03465} \right) = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{300} - \frac{1}{350} \right) \] \[ \log (4) = \frac{E_a}{19.028} \left( \frac{50}{15000} \right) \] \[ 0.602 = \frac{E_a}{19.028} \times 0.00333 \] \[ 0.602 = \frac{E_a}{5700.34} \] \[ E_a = 0.602 \times 5700.34 = 3431.8 \, \text{J/mol} \]

Final Answer:
The activation energy \( E_a \) of the reaction is approximately \( 3432 \, \text{J/mol} \).