### Related Formula From the Arrhenius equation, rate constants vary exponentially with temperature: k = A cdot e^-E_a / RT$k = A \cdot e^{-E_a / RT}$ When rate constants combine multiplicatively or via roots, their corresponding activation energies combine linearly. ### Core Logic Given the overall rate constant expression: k = left(frack_1 cdot k_3k_2 ight)^1/2$k = \left(\frac{k_1 \cdot k_3}{k_2} ight)^{1/2}$ Substitute the Arrhenius expression (k_i = A_i cdot e^-Eai/RT$k_i = A_i \cdot e^{-E{ai}/RT}$) for each rate constant: A cdot e^-E_a/RT = left[frac(A_1 cdot e^-Ea1/RT) cdot (A_3 cdot e^-Ea3/RT)A_2 cdot e^-Ea2/RT ight]^1/2$A \cdot e^{-E_a/RT} = \left[\frac{(A_1 \cdot e^{-E{a1}/RT}) \cdot (A_3 \cdot e^{-E{a3}/RT})}{A_2 \cdot e^{-E{a2}/RT}} ight]^{1/2}$ Equating the exponential terms yields the linear relationship for the overall activation energy (E_a$E_a$): fracE_aRT = frac12 left(fracE_a1RT + fracE_a3RT - fracE_a2RT ight)$\frac{E_a}{RT} = \frac{1}{2} \left(\frac{E_{a1}}{RT} + \frac{E_{a3}}{RT} - \frac{E_{a2}}{RT} ight)$ E_a = fracE_a1 + E_a3 - E_a22$E_a = \frac{E_{a1} + E_{a3} - E_{a2}}{2}$ Substitute the given activation energy values (E_a1 = 60, E_a2 = 30, E_a3 = 10text kJ/mol$E_{a1} = 60, E_{a2} = 30, E_{a3} = 10\text{ kJ/mol}$): E_a = frac60 + 10 - 302 = frac402 = 20text kJ mol^-1$E_a = \frac{60 + 10 - 30}{2} = \frac{40}{2} = 20\text{ kJ mol}^{-1}$ The overall activation energy is 20text kJ/mol$20\text{ kJ/mol}$. ### Pattern Recognition Shortcut: Convert the rate constant algebraic expression directly into an activation energy formula by swapping k$k$ for E_a$E_a$, turning multiplications into additions, divisions into subtractions, and powers into multipliers. Here, k = (k_1 k_3 / k_2)^1/2$k = (k_1 k_3 / k_2)^{1/2}$ translates directly to E_a = frac12(E_a1 + E_a3 - E_a2)$E_a = \frac{1}{2}(E_{a1} + E_{a3} - E_{a2})$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics