Question

Consider the following transformation involving first-order elementary reaction in each step at constant temperature as shown below: $\text{A} + \text{B} \xrightarrow{\text{Step 1}} \text{C} \xrightarrow{\text{Step 2}} \text{P}$ Some details of the above reaction are listed below:If the overall rate constant of the above transformation ($k$) is given as $k = \frac{k_1 k_2}{k_3}$ and the overall activation energy ($E_a$) is $400 \, \text{kJ mol}^{-1}$, then the value of $E_{a3}$ is ______ $\text{kJ mol}^{-1}$ (nearest integer).

Answer 1

Given Information:

Overall rate constant: $K = \frac{k_1 k_2}{k_3}$
Overall activation energy: $E_a = 400 \, \text{kJ/mol}$
Activation energies for each step:
$E_{a1} = 300 \, \text{kJ/mol}, \, E_{a2} = 200 \, \text{kJ/mol}, \, E_{a3} = ?$

Using the Arrhenius Equation:

The overall rate constant $K$ and overall activation energy $E_a$ can be determined by combining the individual rate constants and activation energies as follows:

$K = \frac{k_1 k_2}{k_3}$

According to the Arrhenius equation, we can write: $\ln K = \ln \left(\frac{k_1 k_2}{k_3}\right) = \ln k_1 + \ln k_2 - \ln k_3$ The corresponding activation energy $E_a$ for $K$ is: $E_a = E_{a1} + E_{a2} - E_{a3}$

Substituting the Given Values:

$400 = 300 + 200 - E_{a3}$

Solving for $E_{a3}$:

$E_{a3} = 500 - 400 = 100 \, \text{kJ/mol}$

Conclusion:

The value of $E_{a3}$ is $100 \, \text{kJ/mol}$.