For the photochemical reaction between gases and to yield products and , two competing mechanisms ar — Chemical Kinetics Chemistry Question
Question
For the photochemical reaction between gases $A$ and $B$ to yield products $C$ and $D$, two competing mechanisms are proposed. Mechanism I: $A + B \rightarrow AB^* \rightarrow C + D$ (with overall rate constant $k_I = 1 \times 10^{-5} \text{ M}^{-1}\text{s}^{-1}$). Mechanism II: $A \rightarrow A^* \rightarrow E$ (slow, $k_1 = 1 \times 10^{-4} \text{ s}^{-1}$), followed by $E + B \rightarrow C + D$ (fast, $k_2 = 1 \times 10^{10} \text{ M}^{-1}\text{s}^{-1}$). Assuming these follow their respective rate laws, at what specific concentration of reactant $B$ will the rates of both independent mechanisms be exactly equal?
💡 Solution & Explanation
The overall rate for Mechanism I is dictated by the elementary bimolecular step: $\text{Rate}_I = k_I[A][B] = 10^{-5}[A][B]$. For Mechanism II, the slow, rate-determining step only involves reactant $A$, making it pseudo-first order: $\text{Rate}_{II} = k_1[A] = 10^{-4}[A]$. Equating the two rates to find the intersection point: $10^{-5}[A][B] = 10^{-4}[A]$. Canceling $[A]$ yields $[B] = 10^{-4} / 10^{-5} = 10 \text{ M}$.