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: Relate the overall rate constant to the Arrhenius equation.
The Arrhenius equation relates the rate constant (k) to the activation energy (E\(_a\)): \[ k = A e^{-E_a / RT} \] We can apply this relationship to the overall rate constant K and the individual rate constants k\(_1\), k\(_2\), and k\(_3\). \[ K = A_{overall} e^{-E_{a,overall} / RT} \] \[ k_1 = A_1 e^{-E_{a1} / RT} \] \[ k_2 = A_2 e^{-E_{a2} / RT} \] \[ k_3 = A_3 e^{-E_{a3} / RT} \] Step 2: Substitute the Arrhenius expressions into the given overall rate constant equation.
The given relation is: \[ K = \frac{k_1 k_3}{k_2} \] Substituting the Arrhenius forms: \[ A e^{-E_a / RT} = \frac{(A_1 e^{-E_{a1} / RT}) (A_3 e^{-E_{a3} / RT})}{(A_2 e^{-E_{a2} / RT})} \] Assuming \( A = \frac{A_1 A_3}{A_2} \), we can focus on the exponential parts: \[ e^{-E_a / RT} = \frac{e^{-E_{a1} / RT} \cdot e^{-E_{a3} / RT}}{e^{-E_{a2} / RT}} \] Using the rule \(e^a \cdot e^b = e^{a+b}\) and \(e^a / e^b = e^{a-b}\): \[ e^{-E_a / RT} = e^{(-E_{a1} / RT) + (-E_{a3} / RT) - (-E_{a2} / RT)} \] \[ e^{-E_a / RT} = e^{(-E_{a1} - E_{a3} + E_{a2}) / RT} \] Step 3: Equate the exponents to find the overall activation energy (E\(_a\)).
For the equality to hold, the exponents must be equal: \[ -\frac{E_a}{RT} = \frac{-E_{a1} - E_{a3} + E_{a2}}{RT} \] Multiplying by -RT, we get: \[ E_a = E_{a1} + E_{a3} - E_{a2} \] Step 4: Substitute the given values and calculate E\(_a\).
We are given:

• E\(_a_1\) = 40 kJ mol\(^{-1}\)
• E\(_a_2\) = 50 kJ mol\(^{-1}\)
• E\(_a_3\) = 60 kJ mol\(^{-1}\)

\[ E_a = (40) + (60) - (50) \] \[ E_a = 100 - 50 = 50 \text{ kJ mol}^{-1} \] Step 5: Final Answer.
The overall activation energy of the reaction is 50 kJ mol\(^{-1}\).