\[[\mathrm B] = \frac{k_1[\mathrm A]_0}{k_2 - k_1}\left(e^{-k_1t} - e^{-k_2t}\right),\qquad t_\text{max} = \frac{\ln(k_2/k_1)}{k_2 - k_1}\]
含义Meaning
A 经中间体 B 生成 C。B 先增后减，在 t_max 时最大。若要获得中间产物 B（如部分氧化、部分加氢），必须在 t_max 附近终止反应。
A goes to C via B; B rises then falls, peaking at t_max. To isolate B (partial oxidation or hydrogenation) the reaction must be stopped near t_max.
符号说明Symbols
| 符号Symbol | 含义 Meaning | 单位 Unit |
|---|---|---|
| \(k_1, k_2\) | 两步一级速率常数first-order rate constants of the two steps | s⁻¹ |
| \(t_\text{max}\) | B 浓度最大的时刻time of maximum [B] | s |
推导Derivation
\([\mathrm A] = [\mathrm A]_0e^{-k_1t}\)；\(d[\mathrm B]/dt = k_1[\mathrm A] - k_2[\mathrm B]\)。
[A] decays exponentially; d[B]/dt = k₁[A] − k₂[B].
一阶线性微分方程，用积分因子 \(e^{k_2t}\) 求解。
Solve the linear ODE with integrating factor e^{k₂t}.
\([\mathrm C] = [\mathrm A]_0 - [\mathrm A] - [\mathrm B]\)；令 \(d[\mathrm B]/dt = 0\) 得 t_max。
[C] by mass balance; d[B]/dt = 0 gives t_max.
适用条件与局限Assumptions & limitations
两步都是一级且不可逆；k₁ = k₂ 时需取极限 \([\mathrm B] = k[\mathrm A]_0te^{-kt}\)。
Both steps first order and irreversible; k₁ = k₂ needs the limit.