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

Correct Answer: 100

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 \( 100 \, \text{kJ/mol} \).

Answer 2

Step 1: Given data.
The overall reaction consists of first-order elementary steps:
\[ \text{A + B} \xrightarrow[\text{Step 1}]{k_1, E_{a1}=300} \text{C} \xrightarrow[\text{Step 2}]{k_2, E_{a2}=200} \text{P} \] and the overall rate constant is expressed as:
\[ k = \frac{k_1 k_2}{k_3} \] where \( k_3 \) corresponds to a side or reverse step with activation energy \( E_{a3} \).
Given that the overall activation energy \( E_a = 400 \, \text{kJ mol}^{-1} \).

Step 2: Relation between activation energies.
For a reaction where: \[ k = \frac{k_1 k_2}{k_3}, \] taking logarithms and differentiating with respect to \( \frac{1}{T} \) (using Arrhenius law \( k = A e^{-E_a/RT} \)), we get:
\[ E_a = E_{a1} + E_{a2} - E_{a3} \]
Step 3: Substitute the given values.
\[ 400 = 300 + 200 - E_{a3} \] \[ 400 = 500 - E_{a3} \] \[ E_{a3} = 100 \, \text{kJ mol}^{-1} \]
Step 4: Final Answer.
\[ \boxed{E_{a3} = 100 \, \text{kJ mol}^{-1}} \]