二级反应 Second-order reaction — 公式推导、符号与局限

\[\frac{1}{[\mathrm A]} = \frac{1}{[\mathrm A]_0} + kt,\qquad t_{1/2} = \frac{1}{k[\mathrm A]_0}\]

A + B → P 且初始浓度不等A + B → P with unequal initial concentrations

\[\frac{1}{[\mathrm B]_0 - [\mathrm A]_0}\ln\frac{[\mathrm B][\mathrm A]_0}{[\mathrm A][\mathrm B]_0} = kt\]

含义Meaning

速率与浓度平方（或两种反应物浓度之积）成正比。1/[A] 对 t 为直线；半衰期与初始浓度成反比，越稀反应越慢。

Rate proportional to the square of concentration (or the product of two). 1/[A] is linear in t, and half-life is inversely proportional to initial concentration.

符号说明Symbols

符号Symbol	含义 Meaning	单位 Unit
\(k\)	二级速率常数second-order rate constant	L·mol⁻¹·s⁻¹

推导Derivation

\(-d[\mathrm A]/dt = k[\mathrm A]^2\)，分离变量积分 \(\int d[\mathrm A]/[\mathrm A]^2 = -\int k\,dt\)。
Integrate \(d[\mathrm A]/[\mathrm A]^2 = -k\,dt\).
得 \(1/[\mathrm A] - 1/[\mathrm A]_0 = kt\)。
Gives \(1/[\mathrm A] - 1/[\mathrm A]_0 = kt\).
A + B 型用部分分式积分得 alt 式。
A + B with unequal concentrations: partial fractions.

适用条件与局限Assumptions & limitations

判断级数需比较多种作图的线性，误差大时可能难以区分。
Distinguishing orders by linearity can be ambiguous with noisy data.

例题Worked example

2NO₂ → 2NO + O₂ 在 300 °C 为二级。乙酸乙酯皂化（等浓度）是经典的二级反应实验。

2NO₂ → 2NO + O₂ at 300 °C is second order; saponification of ethyl acetate is the classic lab example.