A chemical reaction mechanism strictly exhibits an overall Arrhenius pre-exponential factor identica — Chemical Kinetics Chemistry Question
Question
A chemical reaction mechanism strictly exhibits an overall Arrhenius pre-exponential factor identically structured as $A_{eff} = A_1 \cdot A_2 / A_3$. Simultaneously, its derived effective activation energy is confirmed as $E_{a(eff)} = 40 \text{ kJ mol}^{-1}$. If $E_{a1} = 20 \text{ kJ mol}^{-1}$ and $E_{a2} = 50 \text{ kJ mol}^{-1}$, calculate the exact hidden value of $E_{a3}$ in $\text{kJ mol}^{-1}$.
💡 Solution & Explanation
The overall rate constant mimics the arrangement of the pre-exponential terms, meaning $k_{eff} = \frac{k_1 k_2}{k_3}$. Applying the standard Arrhenius substitution translates these multipliers and divisors into additive and subtractive energy components respectively: $E_{a(eff)} = E_{a1} + E_{a2} - E_{a3}$. Given $E_{a(eff)} = 40$, $E_{a1} = 20$, and $E_{a2} = 50$, we substitute: $40 = 20 + 50 - E_{a3}$. This rearranges to $E_{a3} = 70 - 40 = 30 \text{ kJ mol}^{-1}$.