Question

For a reaction having three steps, the overall rate constant is \( K = \frac{k_1 k_3}{k_2} \). The values of \( E_{a1} \), \( E_{a2} \), and \( E_{a3} \) (activation energies stepwise) are 40, 50, and 60 kJ mol\(^{-1}\) respectively. 
Then the overall \( E_a \) (activation energy) of the reaction is 
\(\underline{\hspace{3cm}}\).

• A. 30 kJ mol\(^{-1}\)
• B. 40 kJ mol\(^{-1}\)
• C. 50 kJ mol\(^{-1}\)
• D. 60 kJ mol\(^{-1}\)

Hint:

A quick shortcut for such problems is to remember the relationship between the overall activation energy and the individual activation energies based on the rate constant expression. If \(K = \frac{k_1 k_3}{k_2}\), then the overall activation energy is \(E_a = E_{a1} + E_{a3} - E_{a2}\). Essentially, activation energies from the numerator are added, and those from the denominator are subtracted.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For complex reactions, the overall rate constant is often a composite of rate constants from multiple elementary steps. The overall activation energy is derived from the Arrhenius equation.
Step 2: Key Formula or Approach:
If an overall rate constant is given as a product or quotient of elementary rate constants ($K = \frac{k_{1}^{n}k_{3}^{m}}{k_{2}^{p}}$), the overall activation energy relates algebraically: $E_{a} = nE_{a1} + mE_{a3} - pE_{a2}$.
Step 3: Detailed Explanation:
Given the composite rate constant expression: $K = \frac{k_{1}k_{3}}{k_{2}}$.
Taking the natural logarithm ($ln$) on both sides:
\[ \ln K = \ln k_{1} + \ln k_{3} - \ln k_{2} \]
According to the Arrhenius equation, differentiating $ln k$ with respect to temperature ($T$) yields: $\frac{d(\ln k)}{dT} = \frac{E_{a}}{RT^{2}}$.
Applying this derivative to our equation:
\[ \frac{E_{a (\text{overall})}}{RT^{2}} = \frac{E_{a1}}{RT^{2}} + \frac{E_{a3}}{RT^{2}} - \frac{E_{a2}}{RT^{2}} \]
Multiplying entirely by $RT^{2}$:
\[ E_{a} = E_{a1} + E_{a3} - E_{a2} \]
Plugging in the provided numerical values:
\[ E_{a} = 40 + 60 - 50 = 100 - 50 = 50 \text{ kJ mol}^{-1}. \]
Step 4: Final Answer:
The overall activation energy is 50 kJ mol$^{-1}$.