化学反应速率方程 是利用反应物 浓度 或分压 计算化学反应 的反应速率 的方程。[ 1] m A + n B → C {\displaystyle mA+nB\rightarrow C} [ 2]
r = − 1 m d [ A ] d t = k [ A ] x [ B ] y {\displaystyle r=-{\frac {1}{m}}{\frac {d[A]}{dt}}=k\ [A]^{x}[B]^{y}} 在这个方程中， [ X ] {\displaystyle [X]} X {\displaystyle X} 活度 ，单位通常为摩尔 每升（mol/L），但在实际计算中有时也用浓度 代替（若该反应物为气体，表示分压，单位为帕斯卡 (Pa）。 k {\displaystyle k} 速率常数 ，与温度 、离子活度 、光照 、固体反应物的接触面积、反应活化能 等因素有关，通常可通过阿累尼乌斯方程 计算出来，也可通过实验测定。
指数 x {\displaystyle x} y {\displaystyle y} 反应级数 ，取决于反应历程 。在基元反应 中，反应级数等于化学计量数 。但在非基元反应中，反应级数与化学计量数不一定相等。
复杂反应速率方程可能以更复杂的形式出现，包括含多项式的分母。
上述速率方程的一般形式是速率方程的微分 形式，它可以从反应机理导出，而且能明显表示出浓度对反应速率的影响，便于进行理论分析。将它积分 便得到速率方程的积分形式，即反应物／产物浓度 [ X ] {\displaystyle [X]} t {\displaystyle t}
表中， M {\displaystyle M} 莫耳浓度 （ m o l / L {\displaystyle mol/L} t {\displaystyle t} k {\displaystyle k} 速率常数 。所说的“二级反应”和“ n {\displaystyle n} n {\displaystyle \ n}
考慮到微分定律式：
− d [ A ] d t = k [ A ] {\displaystyle -{\frac {d[A]}{dt}}=k[A]} 經過一翻整理可得：
d [ A ] [ A ] = − k d t {\displaystyle {\frac {d[A]}{[A]}}=-{k\,dt}} ∫ d [ A ] [ A ] = − ∫ k d t {\displaystyle \int {\frac {d[A]}{[A]}}=-\int {k\,dt}} l n [ A ] = − k t + C {\displaystyle ln{[A]}=-kt+C} 其中 C {\displaystyle C} 不定積分常數 。接著，帶入初始條件 [ A ] ( 0 ) = [ A ] 0 {\displaystyle [A](0)=[A]_{0}} [ A ] 0 {\displaystyle [A]_{0}}
[ A ] = [ A ] 0 e − k t {\displaystyle [A]=[A]_{0}e^{-kt}} n {\displaystyle n} n ≠ 1 {\displaystyle n\neq 1} [ 编辑 ] 同樣地，考慮到微分定律式：
− d [ A ] d t = k [ A ] n {\displaystyle -{\frac {d[A]}{dt}}=k[A]^{n}} 而其中的 n ≠ 1 {\displaystyle n\neq 1}
d [ A ] [ A ] n = − k d t {\displaystyle {\frac {d[A]}{[A]^{n}}}=-{k\,dt}} ∫ d [ A ] [ A ] n = − ∫ k d t {\displaystyle \int {\frac {d[A]}{[A]^{n}}}=-\int {k\,dt}} [ A ] − n + 1 − n + 1 = − k t + C {\displaystyle {\frac {[A]^{-n+1}}{-n+1}}=-kt+C} 1 [ A ] n − 1 = ( n − 1 ) k t + C {\displaystyle {\frac {1}{[A]^{n-1}}}=(n-1)kt+C} 接著，帶入初始條件 [ A ] ( 0 ) = [ A ] 0 {\displaystyle [A](0)=[A]_{0}} [ A ] 0 {\displaystyle [A]_{0}}
1 [ A ] 0 n − 1 = ( n − 1 ) k ⋅ 0 + C {\displaystyle {\frac {1}{[A]_{0}^{n-1}}}=(n-1)k\cdot 0+C} C = 1 [ A ] 0 n − 1 {\displaystyle C={\frac {1}{[A]_{0}^{n-1}}}} 1 [ A ] n − 1 = 1 [ A ] 0 n − 1 + ( n − 1 ) k t {\displaystyle {\frac {1}{[A]^{n-1}}}={\frac {1}{{[A]_{0}}^{n-1}}}+(n-1)kt} 可逆反应 （又称平衡反应、对行反应、对峙反应）指的是反应物与产物形成化学平衡 的反应，其中正向和逆向反应同时进行，而且反应速率相等。它可以用下面的方程式来表示： s A + t B ⇌ u X + v Y {\displaystyle sA+tB\rightleftharpoons uX+vY} k 1 {\displaystyle k_{1}} k − 1 {\displaystyle k_{-1}}
r = k 1 [ A ] s [ B ] t − k − 1 [ X ] u [ Y ] v {\displaystyle r={k_{1}[A]^{s}[B]^{t}}-{k_{-1}[X]^{u}[Y]^{v}}} k 1 {\displaystyle k_{1}} k − 1 {\displaystyle k_{-1}} 平衡常数 K {\displaystyle K} k 1 [ A ] s [ B ] t = k − 1 [ X ] u [ Y ] v {\displaystyle {k_{1}[A]^{s}[B]^{t}=k_{-1}[X]^{u}[Y]^{v}}\,} K = [ X ] u [ Y ] v [ A ] s [ B ] t = k 1 k − 1 {\displaystyle K={\frac {[X]^{u}[Y]^{v}}{[A]^{s}[B]^{t}}}={\frac {k_{1}}{k_{-1}}}}
A ( A 0 = 0.25 M {\displaystyle \ A(A_{0}=0.25M} B {\displaystyle \ B} c − t {\displaystyle \ c-t} 2 m i n − 1 {\displaystyle \ 2min^{-1}} 1 m i n − 1 {\displaystyle \ 1min^{-1}} 下面讨论一个简单的单分子可逆一级反应： A ⇌ B {\displaystyle A\rightleftharpoons B} K {\displaystyle K} K = d e f k 1 k − 1 = [ B ] e [ A ] e {\displaystyle K\ {\stackrel {\mathrm {def} }{=}}\ {\frac {k_{1}}{k_{-1}}}={\frac {\left[B\right]_{e}}{\left[A\right]_{e}}}} [ A ] e {\displaystyle [A]_{e}} [ B ] e {\displaystyle [B]_{e}} A {\displaystyle A} B {\displaystyle B} B {\displaystyle B} [ B ] 0 = 0 {\displaystyle [B]_{0}=0} A {\displaystyle A} B {\displaystyle B} X {\displaystyle X} [ A ] e = [ A ] 0 − X {\displaystyle \ [A]_{e}=[A]_{0}-X} [ B ] e = X {\displaystyle \ [B]_{e}=X} k − 1 k 1 = [ A ] e [ B ] e = [ A ] 0 − X X {\displaystyle {\frac {k_{-1}}{k_{1}}}={\frac {[A]_{e}}{[B]_{e}}}={\frac {[A]_{0}-X}{X}}} [ B ] e = X = k 1 k 1 + k − 1 [ A ] 0 {\displaystyle \ [B]_{e}=X={\frac {k_{1}}{k_{1}+k_{-1}}}[A]_{0}} [ A ] e = [ A ] 0 − X = k − 1 k 1 + k − 1 [ A ] 0 {\displaystyle \ [A]_{e}=[A]_{0}-X={\frac {k_{-1}}{k_{1}+k_{-1}}}[A]_{0}} A {\displaystyle A} − d [ A ] d t = k 1 [ A ] − k − 1 [ B ] = k 1 [ A ] − k − 1 ( [ A ] 0 − [ A ] ) {\displaystyle -{\frac {d[A]}{dt}}=k_{1}[A]-k_{-1}[B]=k_{1}[A]-k_{-1}([A]_{0}-[A])} t = ∞ {\displaystyle t=\infty } A {\displaystyle A} − d [ A ] e d t = k 1 [ A ] e − k − 1 ( [ A ] 0 − [ A ] e ) = 0 {\displaystyle -{\frac {d[A]_{e}}{dt}}=k_{1}[A]_{e}-k_{-1}([A]_{0}-[A]_{e})=0} − d [ A ] d t = k 1 ( [ A ] − [ A ] e ) + k − 1 ( [ A ] − [ A ] e ) = ( k 1 + k − 1 ) ( [ A ] − [ A ] e ) {\displaystyle -{\frac {d[A]}{dt}}=k_{1}([A]-[A]_{e})+k_{-1}([A]-[A]_{e})=(k_{1}+k_{-1})([A]-[A]_{e})} [ A ] 0 {\displaystyle [A]_{0}} [ A ] e {\displaystyle [A]_{e}} d [ A ] d t = d ( [ A ] − [ A ] e ) d t {\displaystyle {\frac {d[A]}{dt}}={\frac {d([A]-[A]_{e})}{dt}}} − d ( [ A ] − [ A ] e ) d t = ( k 1 + k − 1 ) ( [ A ] − [ A ] e ) {\displaystyle -{\frac {d([A]-[A]_{e})}{dt}}=(k_{1}+k_{-1})([A]-[A]_{e})} [ A ] − [ A ] e = Δ [ A ] {\displaystyle [A]-[A]_{e}=\Delta [A]} A {\displaystyle A}
进行分离变数积分，可以得到： ∫ [ A ] 0 [ A ] d ( [ A ] − [ A ] e ) [ A ] − [ A ] e = ∫ 0 t ( k 1 + k − 1 ) d t {\displaystyle \int _{[A]_{0}}^{[A]}{\frac {d([A]-[A]_{e})}{[A]-[A]_{e}}}=\int _{0}^{t}(k_{1}+k_{-1})dt} l n ( [ A ] 0 − [ A ] e [ A ] t − [ A ] e ) = ( k 1 + k − 1 ) t … ( 1 ) {\displaystyle \ ln({\frac {[A]_{0}-[A]_{e}}{[A]_{t}-[A]_{e}}})=(k_{1}+k_{-1})t\qquad \qquad \ldots (1)} ln ( [ A ] − [ A ] e ) − t {\displaystyle \ln {([A]-[A]_{e})}-t} k 1 + k − 1 {\displaystyle k_{1}+k_{-1}} A e {\displaystyle A_{e}} B e {\displaystyle B_{e}} K {\displaystyle K} k 1 k − 1 {\displaystyle {\frac {k_{1}}{k_{-1}}}} k 1 + k − 1 {\displaystyle k_{1}+k_{-1}} k 1 {\displaystyle k_{1}} k − 1 {\displaystyle k_{-1}} [ 3]
或者，将 x {\displaystyle x} B {\displaystyle B} A {\displaystyle A} − d [ A ] d t = d x d t = k 1 [ A ] t − k − 1 [ B ] t = k 1 ( [ A ] 0 − x ) − k − 1 x {\displaystyle -{\frac {d[A]}{dt}}={\frac {dx}{dt}}=k_{1}[A]_{t}-k_{-1}[B]_{t}=k_{1}([A]_{0}-x)-k_{-1}x} A {\displaystyle A} [ A ] t {\displaystyle [A]_{t}} d t d x = 1 k 1 [ A ] 0 − x ( k 1 + k − 1 ) {\displaystyle {\frac {dt}{dx}}={\frac {1}{k_{1}[A]_{0}-x(k_{1}+k_{-1})}}} t = − 1 k 1 + k − 1 [ ln ( k 1 [ A ] 0 − x ( k 1 + k − 1 ) ) − ln ( k 1 [ A ] 0 ) ] {\displaystyle t=-{\frac {1}{k_{1}+k_{-1}}}[\ln {(k_{1}[A]_{0}-x(k_{1}+k_{-1}))-\ln {(k_{1}[A]_{0})}}]} [ A ] t = k 1 [ A ] 0 e − ( k 1 + k − 1 ) t + k − 1 [ A ] 0 k 1 + k − 1 … ( 2 ) {\displaystyle [A]_{t}={\frac {k_{1}[A]_{0}e^{-(k_{1}+k_{-1})t}+k_{-1}[A]_{0}}{k_{1}+k_{-1}}}\qquad \qquad \ldots (2)} t = ∞ {\displaystyle t=\infty } [ A ] e = k − 1 [ A ] 0 k 1 + k − 1 {\displaystyle [A]_{e}={\frac {k_{-1}[A]_{0}}{k_{1}+k_{-1}}}} ln ( [ A ] 0 − [ A ] e [ A ] t − [ A ] e ) {\displaystyle \ln({\frac {[A]_{0}-[A]_{e}}{[A]_{t}-[A]_{e}}})} ( k 1 + k − 1 ) t {\displaystyle (k_{1}+k_{-1})t}
为了求得反应的半衰期，令 [ A ] t = 1 2 [ A ] 0 {\displaystyle [A]_{t}={\frac {1}{2}}[A]_{0}} t 1 2 = 1 k 1 + k − 1 ln 2 k 1 k 1 − k − 1 {\displaystyle t_{\tfrac {1}{2}}={\frac {1}{k_{1}+k_{-1}}}\ln {\frac {2k_{1}}{k_{1}-k_{-1}}}} K {\displaystyle K} k 1 >> k − 1 {\displaystyle k_{1}>>k_{-1}} t 1 2 = 1 k 1 + k − 1 ln 2 {\displaystyle t_{\tfrac {1}{2}}={\frac {1}{k_{1}+k_{-1}}}\ln {2}}
可以看出，平衡反应中的半衰期与反应物的初始浓度无关。
对行反应的例子有：
乙酸 和乙醇 的酯化反应 ： CH 3 COOH + C 2 H 5 OH ↽ − − ⇀ CH 3 COOC 2 H 5 + H 2 O {\displaystyle {\ce {CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O}}} 邻苯二甲酸酐 与异辛醇 反应生成邻苯二甲酸二异辛酯 的反应。连续反应 （又称串联反应、连串反应）指的是如下类型的化学反应： A → k 1 B → k 2 C {\displaystyle A{\xrightarrow {k_{1}}}B{\xrightarrow {k_{2}}}C} B {\displaystyle B} C {\displaystyle C}
对反应物 A {\displaystyle A} d [ A ] d t = − k 1 [ A ] … ( 3 ) {\displaystyle {\frac {d[A]}{dt}}=-k_{1}[A]\qquad \qquad \ldots (3)} 对中间体 B {\displaystyle B} d [ B ] d t = k 1 [ A ] − k 2 [ B ] … ( 4 ) {\displaystyle {\frac {d[B]}{dt}}=k_{1}[A]-k_{2}[B]\qquad \qquad \ldots (4)} 对产物 C {\displaystyle C} d [ C ] d t = k 2 [ B ] … ( 5 ) {\displaystyle {\frac {d[C]}{dt}}=k_{2}[B]\qquad \qquad \ldots (5)} 对 ( 3 ) {\displaystyle (3)} [ A ] t = [ A ] 0 e − k 1 t … ( 6 ) {\displaystyle [A]_{t}=[A]_{0}e^{-k_{1}t}\qquad \qquad \ldots (6)}
将 ( 6 ) {\displaystyle (6)} ( 4 ) {\displaystyle (4)} d [ B ] d t + k 2 [ B ] = k 1 [ A ] 0 e − k 1 t {\displaystyle {\frac {d[B]}{dt}}+k_{2}[B]=k_{1}[A]_{0}e^{-k_{1}t}}
对其进行积分：
[ B ] e ∫ 0 t k 2 d t = ∫ 0 t k 1 [ A ] 0 e − k 1 t e ∫ 0 t k 2 d t d t {\textstyle [B]e^{\int _{0}^{t}k_{2}dt}=\int _{0}^{t}k_{1}[A]_{0}e^{-k_{1}t}e^{\int _{0}^{t}k_{2}dt}dt} [ B ] e k 2 t = ∫ 0 t k 1 [ A ] 0 e − k 1 t e k 2 t d t = k 1 [ A ] 0 k 2 − k 1 [ e ( k 2 − k 1 ) t − 1 ] {\displaystyle [B]e^{k_{2}t}=\int _{0}^{t}k_{1}[A]_{0}e^{-k_{1}t}e^{k_{2}t}dt={\frac {k_{1}[A]_{0}}{k_{2}-k_{1}}}[e^{(k_{2}-k_{1})t}-1]} [ B ] = [ A ] 0 k 1 k 2 − k 1 ( e − k 1 t − e − k 2 t ) … ( 7 ) {\displaystyle [B]=[A]_{0}\ {\frac {k_{1}}{k_{2}-k_{1}}}(e^{-k_{1}t}-e^{-k_{2}t})\qquad \qquad \ldots (7)} 由于
[ A ] + [ B ] + [ C ] = [ A ] 0 {\displaystyle [A]+[B]+[C]=[A]_{0}} [ C ] = [ A ] 0 − [ A ] − [ B ] {\displaystyle [C]=[A]_{0}-[A]-[B]} 将 ( 6 ) {\displaystyle (6)} ( 7 ) {\displaystyle (7)} [ C ] = [ A ] 0 k 2 − k 1 [ k 2 ( 1 − e − k 1 t ) − k 1 ( 1 − e − k 2 t ) ] = [ A ] 0 ( 1 + k 1 e − k 2 t − k 2 e − k 1 t k 2 − k 1 ) {\displaystyle [C]={\frac {[A]_{0}}{k_{2}-k_{1}}}[k_{2}(1-e^{-k_{1}t})-k_{1}(1-e^{-k_{2}t})]=[A]_{0}(1+{\frac {k_{1}e^{-k_{2}t}-k_{2}e^{-k_{1}t}}{k_{2}-k_{1}}})}
这样， [ A ] {\displaystyle [A]} [ B ] {\displaystyle [B]} [ C ] {\displaystyle [C]} B {\displaystyle B} [ B ] {\displaystyle [B]} ( 7 ) {\displaystyle (7)} t {\displaystyle t} B {\displaystyle B} t max {\displaystyle t_{\mbox{max}}} B {\displaystyle B} [ B ] max {\displaystyle [B]_{\mbox{max}}}
t max = ln k 1 k 2 k 1 − k 2 {\displaystyle t_{\mbox{max}}={\frac {\ln {\dfrac {k_{1}}{k_{2}}}}{k_{1}-k_{2}}}} [ B ] max = [ A ] 0 ( k 1 k 2 ) k 2 k 2 − k 1 {\displaystyle [B]_{\mbox{max}}=[A]_{0}\left({\frac {k_{1}}{k_{2}}}\right)^{\dfrac {k_{2}}{k_{2}-k_{1}}}} 用稳态近似 法分析也可以取得类似的结果。
连续反应的例子有：
丙烯 被氧气 氧化为丙酮 、乙酸 的反应： C 3 H 6 → O 2 CH 3 COCH 3 → O 2 CH 3 COOH {\displaystyle {\ce {C3H6 ->[O_2] CH3COCH3 ->[O_2] CH3COOH}}} 4-氨基偶氮苯 用发烟硫酸 磺化 时发生的反应： 4－氨基偶氮苯 → H 2 S O 4 {\displaystyle \ \ {\xrightarrow {H_{2}SO_{4}}}\ } → H 2 S O 4 {\displaystyle \ \ {\xrightarrow {H_{2}SO_{4}}}\ } 平行反应 （又称竞争反应）指的是同一反应物可以同时进行几种不同的反应，生成不同的产物。
A { → k 1 B → k 2 C {\displaystyle A{\begin{cases}\ {\xrightarrow {k_{1}}}\ B\\\ {\xrightarrow {k_{2}}}\ C\end{cases}}} 如果两个平行反应都是一级反应，则三个速率方程分别为：
− d [ A ] d t = ( k 1 + k 2 ) [ A ] {\displaystyle -{\frac {d[A]}{dt}}=(k_{1}+k_{2})[A]} d [ B ] d t = k 1 [ A ] {\displaystyle \ {\frac {d[B]}{dt}}=k_{1}[A]} d [ C ] d t = k 2 [ A ] {\displaystyle \ {\frac {d[C]}{dt}}=k_{2}[A]} 对其积分可以得到 [ A ] {\displaystyle [A]} [ B ] {\displaystyle [B]} [ C ] {\displaystyle [C]}
[ A ] = [ A ] 0 e − ( k 1 + k 2 ) t {\displaystyle \ [A]=[A]_{0}e^{-(k_{1}+k_{2})t}} [ B ] = k 1 k 1 + k 2 [ A ] 0 ( 1 − e − ( k 1 + k 2 ) t ) {\displaystyle [B]={\frac {k_{1}}{k_{1}+k_{2}}}[A]_{0}(1-e^{-(k_{1}+k_{2})t})} [ C ] = k 2 k 1 + k 2 [ A ] 0 ( 1 − e − ( k 1 + k 2 ) t ) {\displaystyle [C]={\frac {k_{2}}{k_{1}+k_{2}}}[A]_{0}(1-e^{-(k_{1}+k_{2})t})} 一个比较重要的关系式是： [ B ] [ C ] = k 1 k 2 {\displaystyle {\frac {[B]}{[C]}}={\frac {k_{1}}{k_{2}}}}
两个平行反应分别为一级和二级反应：[ 4]
想象下面的情形：反应物 A {\displaystyle A} A + R → C {\displaystyle A+R\rightarrow C} A {\displaystyle A} 水解 （可以看作准一级反应）： A + H 2 O → B {\displaystyle A+H_{2}O\rightarrow B}
d [ B ] d t = k 1 [ A ] [ H 2 O ] = k 1 ′ [ A ] {\displaystyle {\frac {d[B]}{dt}}=k_{1}[A][H_{2}O]=k_{1}'[A]} k 1 ′ {\displaystyle k_{1}'} d [ C ] d t = k 2 [ A ] [ R ] {\displaystyle {\frac {d[C]}{dt}}=k_{2}[A][R]} 通过假设 [ A ] 0 − [ C ] ≈ [ A ] 0 {\displaystyle [A]_{0}-[C]\approx [A]_{0}} C {\displaystyle C} [ C ] {\displaystyle [C]} B {\displaystyle B} [ B ] {\displaystyle [B]}
[ C ] = [ R ] 0 [ 1 − e − k 2 k 1 ′ [ A ] 0 ( 1 − e − k 1 ′ t ) ] {\displaystyle [C]=[R]_{0}[1-e^{-{\frac {k_{2}}{k_{1}'}}[A]_{0}(1-e^{-k_{1}'t})}]} l n [ R ] 0 [ R ] 0 − [ C ] = k 2 [ A ] 0 k 1 ′ ( 1 − e − k 1 ′ t ) {\displaystyle ln{\frac {[R]_{0}}{[R]_{0}-[C]}}={\frac {k_{2}[A]_{0}}{k_{1}'}}(1-e^{-k_{1}'t})} [ B ] = − k 1 ′ k 2 l n ( 1 − [ C ] [ R ] 0 ) {\displaystyle [B]=-{\frac {k_{1}'}{k_{2}}}ln(1-{\frac {[C]}{[R]_{0}}})} 以上只是几种基本的复合反应类型，除此以外，还有很多情况是上述几种基本复合反应的混合。请有兴趣的读者参见酶动力学 、米氏方程 和酶抑制剂 等文章。
速率方程的确定 主要有以下三种方式：
微分法：对 − d [ A ] d t = k [ A ] n {\displaystyle \ -{\frac {d[A]}{dt}}=k[A]^{n}} 对数 ，得到： ln ( d [ A ] d t ) = ln k + n ln [ A ] {\displaystyle \ln {\left({\frac {d[A]}{dt}}\right)}=\ln {k}+n\ln {[A]}} [ A ] 0 {\displaystyle [A]_{0}} [ A ] − t {\displaystyle [A]-t} d [ A ] 0 d t {\displaystyle {\frac {d[A]_{0}}{dt}}} ln ( − d [ A ] 0 d t ) − ln [ A ] 0 {\displaystyle \ln {\left(-{\frac {d[A]_{0}}{dt}}\right)}-\ln {[A]_{0}}} A {\displaystyle A} 尝试法／试差法：分别将某一化学反应的 [ A ] {\displaystyle [A]} t {\displaystyle t} 半衰期法：求得两个不同初始浓度下的反应半衰期，然后根据总结一节中的半衰期通式，求出反应级数。 ^ IUPAC 金皮书 对“反应速率”（rate law）的定义：[1] （页面存档备份 ，存于互联网档案馆 ）。^ Connors, Kenneth A. Chemical Kinetics, the study of reaction rates in solution. VCH Publishers. 1991. ^ Rushton, Gregory T.; William G. Burns, Judi M. Lavin, Yong S. Chong, Perry Pellechia, Ken D. Shimizu. Determination of the Rotational Barrier for Kinetically Stable Conformational Isomers via NMR and 2D TLC - An Introductory Organic Chemistry Experiment. J. Chem. Educ. 2007, 84 (1499). Abstract （页面存档备份 ，存于互联网档案馆 ）.^ Manso, José A.; et al. . A Kinetic Approach to the Alkylating Potential of Carcinogenic Lactones. Chem. Res. Toxicol. 2005, 18 (7): 1161-1166.
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![{\displaystyle [A]_{0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32488af10ddcce5bf90dbc659e6472d1279bd2f0)
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![{\displaystyle {\frac {d[A]}{[A]^{n}}}=-{k\,dt}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b49d82f6f2897ff343c673bef0c1aa2d54f8365)
![{\displaystyle \int {\frac {d[A]}{[A]^{n}}}=-\int {k\,dt}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab5178b060a6e9126f7c418633be4f28eafb25ad)
![{\displaystyle {\frac {[A]^{-n+1}}{-n+1}}=-kt+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/375daa94c42ae45a1047282171af09e2dfa574c6)
![{\displaystyle {\frac {1}{[A]^{n-1}}}=(n-1)kt+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2679513b8198f24ec3dc5107ff47f9660a72225c)
![{\displaystyle {\frac {1}{[A]_{0}^{n-1}}}=(n-1)k\cdot 0+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e562b02dca33ce8bb04d4c61f0a56cf204bf75b)
![{\displaystyle C={\frac {1}{[A]_{0}^{n-1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0bb29f85bb09933befaa63501b85389c73832ebd)
![{\displaystyle r={k_{1}[A]^{s}[B]^{t}}-{k_{-1}[X]^{u}[Y]^{v}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/afa5536f41148f92769c593b265660902f419648)
![{\displaystyle {k_{1}[A]^{s}[B]^{t}=k_{-1}[X]^{u}[Y]^{v}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/026603d76f35a997d6c746e91de6afefa1c5ec21)
![{\displaystyle K={\frac {[X]^{u}[Y]^{v}}{[A]^{s}[B]^{t}}}={\frac {k_{1}}{k_{-1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6dc3fb0b4bc9ac466541cc12cd0170417bc2af6f)
![{\displaystyle K\ {\stackrel {\mathrm {def} }{=}}\ {\frac {k_{1}}{k_{-1}}}={\frac {\left[B\right]_{e}}{\left[A\right]_{e}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/88e14c3292154b8fa16716b98e02f82c06bcb9fb)
![{\displaystyle [A]_{e}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f1203ac1ff71ffb6b2300a26f8590e12c151d32)
![{\displaystyle [B]_{e}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a7aa706dd2652248c8529b53515ea4f359c90c6)
![{\displaystyle [B]_{0}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4416f74ff55a09d3d0ead6dddabf9d41e9aa4f5f)
![{\displaystyle \ [A]_{e}=[A]_{0}-X}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7bd277d2d7eb956c76649d0476edb35689508ef2)
![{\displaystyle \ [B]_{e}=X}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02b480ab1be205ce034704ce318005431d2ed27d)
![{\displaystyle {\frac {k_{-1}}{k_{1}}}={\frac {[A]_{e}}{[B]_{e}}}={\frac {[A]_{0}-X}{X}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c653bddc078442aa53f5a53b7f243e20a3bb5b6)
![{\displaystyle \ [B]_{e}=X={\frac {k_{1}}{k_{1}+k_{-1}}}[A]_{0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a86251ae17b2e04ac0bea5d703d9597fa4580e3)
![{\displaystyle \ [A]_{e}=[A]_{0}-X={\frac {k_{-1}}{k_{1}+k_{-1}}}[A]_{0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b8c8aa2e3a2583b7e78cfea491470a0fca8cf22)
![{\displaystyle -{\frac {d[A]}{dt}}=k_{1}[A]-k_{-1}[B]=k_{1}[A]-k_{-1}([A]_{0}-[A])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de3b6b87b8fd5d61b898f489c69be7d59889809c)
![{\displaystyle -{\frac {d[A]_{e}}{dt}}=k_{1}[A]_{e}-k_{-1}([A]_{0}-[A]_{e})=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2cc782d006a097bdd38f2b540a41946594abfc3)
![{\displaystyle -{\frac {d[A]}{dt}}=k_{1}([A]-[A]_{e})+k_{-1}([A]-[A]_{e})=(k_{1}+k_{-1})([A]-[A]_{e})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/caa0acb125a099452b7c057a403841539b905b7d)
![{\displaystyle {\frac {d[A]}{dt}}={\frac {d([A]-[A]_{e})}{dt}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f53019b202c12cf209980aecf0613cd9cb323f99)
![{\displaystyle -{\frac {d([A]-[A]_{e})}{dt}}=(k_{1}+k_{-1})([A]-[A]_{e})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c1fa1bfd9ad25e72b719d6ada40e5831519f756)
![{\displaystyle [A]-[A]_{e}=\Delta [A]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d4a01071aeebc4dd77624cb42a84601b6acf444)
![{\displaystyle \int _{[A]_{0}}^{[A]}{\frac {d([A]-[A]_{e})}{[A]-[A]_{e}}}=\int _{0}^{t}(k_{1}+k_{-1})dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d263d2ade5d33e115610880bcec97823948d4f7e)
![{\displaystyle \ ln({\frac {[A]_{0}-[A]_{e}}{[A]_{t}-[A]_{e}}})=(k_{1}+k_{-1})t\qquad \qquad \ldots (1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/44e18e84f64e240b028d4e79605eccc07e8c8be6)
![{\displaystyle \ln {([A]-[A]_{e})}-t}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eeea255ceb22f9748121203ce0c4ddf0c6013f34)
![{\displaystyle -{\frac {d[A]}{dt}}={\frac {dx}{dt}}=k_{1}[A]_{t}-k_{-1}[B]_{t}=k_{1}([A]_{0}-x)-k_{-1}x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a2f07bbad1773cda47891d60d83a94ce200b53d1)
![{\displaystyle [A]_{t}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03d6eaf7fc064571a2d2e88332f923d25a436f1c)
![{\displaystyle {\frac {dt}{dx}}={\frac {1}{k_{1}[A]_{0}-x(k_{1}+k_{-1})}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed2cd252186924a33d05c9a4f07cc9087238477d)
![{\displaystyle t=-{\frac {1}{k_{1}+k_{-1}}}[\ln {(k_{1}[A]_{0}-x(k_{1}+k_{-1}))-\ln {(k_{1}[A]_{0})}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8fbfcf9ff0a498f810e96bd5073e163382249eb8)
![{\displaystyle [A]_{t}={\frac {k_{1}[A]_{0}e^{-(k_{1}+k_{-1})t}+k_{-1}[A]_{0}}{k_{1}+k_{-1}}}\qquad \qquad \ldots (2)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c608dea3eddd899feef34de5b68231d8d09ef2e4)
![{\displaystyle [A]_{e}={\frac {k_{-1}[A]_{0}}{k_{1}+k_{-1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1019f3a6e37323f112d7b3917ba09e1547e4c2a5)
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![{\displaystyle [A]_{t}={\frac {1}{2}}[A]_{0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/218a4660bbf2336f57a86ef14e9986f38ae2ca6c)
![{\displaystyle {\frac {d[A]}{dt}}=-k_{1}[A]\qquad \qquad \ldots (3)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bcf21f01ed756a9d22d854b77d2e785ee5ee036)
![{\displaystyle {\frac {d[B]}{dt}}=k_{1}[A]-k_{2}[B]\qquad \qquad \ldots (4)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c160a188e75486adf93b19a3756e816753f59ee4)
![{\displaystyle {\frac {d[C]}{dt}}=k_{2}[B]\qquad \qquad \ldots (5)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdc0b0e30ffed9df89e9b243cfcd22a1409efade)
![{\displaystyle [A]_{t}=[A]_{0}e^{-k_{1}t}\qquad \qquad \ldots (6)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6673e25fa8269cbe13265f39a495e18afac45294)
![{\displaystyle {\frac {d[B]}{dt}}+k_{2}[B]=k_{1}[A]_{0}e^{-k_{1}t}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d47eee63dd5fe79ec20a17806555c3daad761faa)
![{\textstyle [B]e^{\int _{0}^{t}k_{2}dt}=\int _{0}^{t}k_{1}[A]_{0}e^{-k_{1}t}e^{\int _{0}^{t}k_{2}dt}dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d81cabe7e9d69a17bfddbc7315a5f1a0f9e81793)
![{\displaystyle [B]e^{k_{2}t}=\int _{0}^{t}k_{1}[A]_{0}e^{-k_{1}t}e^{k_{2}t}dt={\frac {k_{1}[A]_{0}}{k_{2}-k_{1}}}[e^{(k_{2}-k_{1})t}-1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a2d9bd8e2d9ff9f6bc7707c5c35f96eb0a4d084e)
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![{\displaystyle [C]=[A]_{0}-[A]-[B]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1a4ec42646562cdc64626e533e94f91595ce395)
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![{\displaystyle [A]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79eaa334597b1861f1b08ca0c8fecb3858ebcb12)
![{\displaystyle [B]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e6c6caf9d188e4a608dfdac98bbcd140a9ea106)
![{\displaystyle [C]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/798adf333080234dd9da202633a4d63c4ea091aa)
![{\displaystyle [B]_{\mbox{max}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff7fc26b1c063a698666cb2ea233a18635eab4c1)
![{\displaystyle [B]_{\mbox{max}}=[A]_{0}\left({\frac {k_{1}}{k_{2}}}\right)^{\dfrac {k_{2}}{k_{2}-k_{1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7a493b44f1fad1a95d1df187603f772e85c1049)
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![{\displaystyle \ {\frac {d[B]}{dt}}=k_{1}[A]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20f00898f069554bc460d8ca2db7ab05bb74c3fd)
![{\displaystyle \ {\frac {d[C]}{dt}}=k_{2}[A]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/125c594e179d045aafaf10ba61c195f7ded1fc1b)
![{\displaystyle \ [A]=[A]_{0}e^{-(k_{1}+k_{2})t}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe7441c5ffc69f4c54c616b5478344a89192a56a)
![{\displaystyle [B]={\frac {k_{1}}{k_{1}+k_{2}}}[A]_{0}(1-e^{-(k_{1}+k_{2})t})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56b1791743c2e9b14423db8e256d1efd2d97586f)
![{\displaystyle [C]={\frac {k_{2}}{k_{1}+k_{2}}}[A]_{0}(1-e^{-(k_{1}+k_{2})t})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96e66cc320b1f522d1ff60dab89a79e9786c1083)
![{\displaystyle {\frac {[B]}{[C]}}={\frac {k_{1}}{k_{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7c639bb569eb7687c88fc26df54db6591242fbc)
![{\displaystyle {\frac {d[B]}{dt}}=k_{1}[A][H_{2}O]=k_{1}'[A]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4eb8afde68ff771829096c46101404d3ba5b653b)
![{\displaystyle {\frac {d[C]}{dt}}=k_{2}[A][R]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b1bde0fddb8b07b9af4142a4a425eb0a7ce6218)
![{\displaystyle [A]_{0}-[C]\approx [A]_{0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56419eeaa515513c74984c0d2ef1cdf1b2a6beaa)
![{\displaystyle [C]=[R]_{0}[1-e^{-{\frac {k_{2}}{k_{1}'}}[A]_{0}(1-e^{-k_{1}'t})}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0591e55b9408afef01ef67c41a56da907dd5e37)
![{\displaystyle ln{\frac {[R]_{0}}{[R]_{0}-[C]}}={\frac {k_{2}[A]_{0}}{k_{1}'}}(1-e^{-k_{1}'t})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae8fd16d45328e54c0a575983cbfe0e817270601)
![{\displaystyle [B]=-{\frac {k_{1}'}{k_{2}}}ln(1-{\frac {[C]}{[R]_{0}}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d88248959fbb77a4a538e4d05a50a16d6af9b91)
![{\displaystyle \ -{\frac {d[A]}{dt}}=k[A]^{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d96418aa64494e55ff8835b75302311138512b36)
![{\displaystyle \ln {\left({\frac {d[A]}{dt}}\right)}=\ln {k}+n\ln {[A]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0148ad94ad33587dafe2775d1e1282f12d0c91fd)
![{\displaystyle [A]-t}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e8beb6a079392c252daf229d38d460caf343c38)
![{\displaystyle {\frac {d[A]_{0}}{dt}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cdac4c5b9e69d5f3600d3d50cc105e70ffe8c2ac)
![{\displaystyle \ln {\left(-{\frac {d[A]_{0}}{dt}}\right)}-\ln {[A]_{0}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6023d39174c1ebf7fe31598070691d41d00b3e6)