For a complex reaction A xrightarrow{K} text{ | vedclass

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(D) The Arrhenius equation is given by $K = A e^{-E_a/RT}$.Substituting the expressions for $K_1, K_2, K_3$ into the given equation $K = (\frac{K_1 K_

Vedclass · Accepted Answer

$140$

Vedclass · Answer

(D) The Arrhenius equation is given by $K = A e^{-E_a/RT}$.Substituting the expressions for $K_1, K_2, K_3$ into the given equation $K = (\frac{K_1 K_2}{K_3})^{2/3}$:$K = (\frac{A_1 e^{-Ea_1/RT} \cdot A_2 e^{-Ea_2/RT}}{A_3 e^{-Ea_3/RT}})^{2/3}$.Taking the natural logarithm on both sides or comparing the exponents,the overall activation energy $E_a$ is given by:$E_a = \frac{2}{3} (Ea_1 + Ea_2 - Ea_3)$.Substituting the given values:$E_a = \frac{2}{3} (180 + 80 - 50) = \frac{2}{3} (210) = 140 \ kJ/mol$.

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