Question

For a reaction having three steps, the overall rate constant is \( K = \frac{k_1 k_2}{k_3} \). The values 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 ______

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

Hint:

For overall rate constants that are products or quotients of individual rate constants (e.g., $K = \frac{k_1 k_2}{k_3}$), the overall activation energy $E_a$ is found by summing the $E_a$ values for terms in the numerator and subtracting those in the denominator. That is, $E_a = \sum E_{a}(\text{numerator}) - \sum E_{a}(\text{denominator})$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to calculate the overall activation energy for a multi-step reaction, given the expression for its overall rate constant in terms of individual step rate constants and their respective activation energies.
Step 2: Key Formula or Approach:
The Arrhenius equation relates the rate constant \(k\) to the activation energy \(E_a\):
\[ k = A e^{-E_a/RT} \]
Taking the natural logarithm of both sides:
\[ \ln k = \ln A - \frac{E_a}{RT} \]
For an overall reaction, if its rate constant \(K\) is expressed as a product or quotient of individual step rate constants, the overall activation energy \(E_a\) can be found by summing and subtracting the individual activation energies.
Step 3: Detailed Explanation:
Given the overall rate constant expression:
\[ K = \frac{k_1 k_2}{k_3} \]
Take the natural logarithm of both sides:
\[ \ln K = \ln(k_1 k_2) - \ln k_3 \]
\[ \ln K = \ln k_1 + \ln k_2 - \ln k_3 \]
Now, substitute the Arrhenius equation (in logarithmic form) for each rate constant:
\[ \left(\ln A_{\text{overall}} - \frac{E_{a,\text{overall}}}{RT}\right) = \left(\ln A_1 - \frac{E_{a1}}{RT}\right) + \left(\ln A_2 - \frac{E_{a2}}{RT}\right) - \left(\ln A_3 - \frac{E_{a3}}{RT}\right) \]
Equating the terms containing the activation energies (which depend on \( -1/RT \)):
\[ -\frac{E_{a,\text{overall}}}{RT} = -\frac{E_{a1}}{RT} - \frac{E_{a2}}{RT} + \frac{E_{a3}}{RT} \]
Multiply by \(-RT\) to solve for the overall activation energy:
\[ E_{a,\text{overall}} = E_{a1} + E_{a2} - E_{a3} \]
Given values:
\( E_{a1} = 40 \text{ kJ mol}^{-1} \)
\( E_{a2} = 50 \text{ kJ mol}^{-1} \)
\( E_{a3} = 60 \text{ kJ mol}^{-1} \)
Substitute these values:
\[ E_{a,\text{overall}} = 40 \text{ kJ mol}^{-1} + 50 \text{ kJ mol}^{-1} - 60 \text{ kJ mol}^{-1} \]
\[ E_{a,\text{overall}} = 90 \text{ kJ mol}^{-1} - 60 \text{ kJ mol}^{-1} = 30 \text{ kJ mol}^{-1} \]
Step 4: Final Answer:
The overall activation energy of the reaction is 30 kJ mol\(^{-1}\).