Question

A certain reaction proceeds in a sequence of three elementary steps with rate constants k1, k2 and k3. If the observed rate constant, Kobs, of the reaction is expressed as Kobs = K3 [K1/K2]^1/2, then what is the observed energy of activation of the reaction?

• (A) $E_3+\frac{1}{2}\left(\frac{E_1}{E_2}\right)$
• (B) $E_3+\left(\frac{E_1-E_2}{2}\right)$
• (C) $E_3{\left(\frac{E_1}{E_2}\right)}^{1/2}$
• (D) ${\mathrm{E}}_3+\frac{E_1+E_2}{2}$

Answer 1

The Correct Option is B

Explanation:
${\mathrm{K}}_{\mathrm{obs}}={\mathrm{K}}_3{\left[\frac{{\mathrm{K}}_1}{{\mathrm{K}}_2}\right]}^{1/2}={\mathrm{A}}_3\cdot {\mathrm{e}}^{-E_3/\mathrm{RT}}\times {\left[\frac{{\mathrm{A}}_1{\mathrm{e}}^{-{\mathrm{E}}_1/\mathrm{R}T}}{{\mathrm{A}}_2{\mathrm{e}}^{-{\mathrm{E}}_3/\mathrm{RT}}}\right]}^{1/2}$${\mathrm{K}}_{\mathrm{obs}}={\mathrm{A}}_3\sqrt{\frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}}\times {\mathrm{e}}^{-[\frac{{\mathrm{E}}_2}{\mathrm{RT}}+\frac{1}{2}\left(\frac{{\mathrm{E}}_1-{\mathrm{E}}_2}{\mathrm{FT}}\right)]}\Rightarrow {\mathrm{K}}_{\mathrm{obs}}=\mathrm{A}\cdot {\mathrm{e}}^{-{\mathrm{E}}_{\mathrm{obs}}/\mathrm{Fr}}$$\therefore {\mathrm{E}}_{\mathrm{obs}}={\mathrm{E}}_3+\left(\frac{{\mathrm{E}}_1-{\mathrm{E}}_2}{2}\right)$

Answer 2

To determine the observed energy of activation (\(E_a\)) of the reaction based on the given rate constants and observed rate constant expression, we need to understand the relationship between the rate constants and their corresponding activation energies.
Given:
\[ K_{\text{obs}} = k_3 \left(\frac{k_1}{k_2}\right)^{1/2} \]
The Arrhenius equation for each elementary step can be written as:
\[ k_1 = A_1 e^{-\frac{E_{a1}}{RT}} \]
\[ k_2 = A_2 e^{-\frac{E_{a2}}{RT}} \]
\[ k_3 = A_3 e^{-\frac{E_{a3}}{RT}} \]
Here, \(A_1, A_2,\) and \(A_3\) are the pre-exponential factors, and \(E_{a1}, E_{a2},\) and \(E_{a3}\) are the activation energies for the elementary steps.
Let's rewrite the expression for \(K_{\text{obs}}\) using the Arrhenius expressions for \(k_1, k_2,\) and \(k_3\):
\[ K_{\text{obs}} = A_3 e^{-\frac{E_{a3}}{RT}} \left( \frac{A_1 e^{-\frac{E_{a1}}{RT}}}{A_2 e^{-\frac{E_{a2}}{RT}}} \right)^{1/2} \]
Simplify the expression:
\[ K_{\text{obs}} = A_3 \left( \frac{A_1}{A_2} \right)^{1/2} e^{-\frac{E_{a3}}{RT}} \left( e^{-\frac{E_{a1} - E_{a2}}{2RT}} \right) \]
Combining the exponents, we get:
\[ K_{\text{obs}} = A_3 \left( \frac{A_1}{A_2} \right)^{1/2} e^{-\left( \frac{E_{a3}}{RT} + \frac{E_{a1} - E_{a2}}{2RT} \right)} \]
The exponent can be further simplified:
\[ K_{\text{obs}} = A_3 \left( \frac{A_1}{A_2} \right)^{1/2} e^{-\left( \frac{2E_{a3} + E_{a1} - E_{a2}}{2RT} \right)} \]
Thus, the observed energy of activation \(E_a\) for the overall reaction can be identified as the term in the exponent:
\[ E_a = \frac{2E_{a3} + E_{a1} - E_{a2}}{2} \]
Therefore, the observed energy of activation for the reaction is:
\[ E_a = E_{a3} + \frac{E_{a1} - E_{a2}}{2} \]
So the correct Answer is Option (B): \[E_{3} + \frac{E_{1} - E_{2}}{2} \]