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
]
m
[
B
]
n
{\displaystyle r\;=-{\frac {1}{m}}{\frac {d[A]}{dt}}=\;k\ [A]^{m}[B]^{n}}
在这个方程中，
[
X
]
{\displaystyle [X]\,}
X
{\displaystyle X\,}
活度 ，单位通常为摩尔 每升（
m
o
l
/
L
{\displaystyle mol/L\,}
浓度 代替（若该反应物为气体，表示分压，单位为帕斯卡 (
P
a
{\displaystyle 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
m
{\displaystyle m\,}
n
{\displaystyle n\,}
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}
摩尔浓度
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}
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}
假设正向反应速率为
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}
达到平衡时的平衡常数
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}}}}
其中，
[
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}
代入上面的定义式：
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}}}
整理得：
[
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}
上上式减上式，可得：
−
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})}
当
[
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}}}
将上式代入上上式，得：
−
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)}
从上式可以看出，平衡反应的
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}
整理，可得某一时刻反应物
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}}}}
如果反应的
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 ：
C
H
3
C
O
O
H
+
C
2
H
5
O
H
⇌
C
H
3
C
O
O
C
2
H
5
+
H
2
O
{\displaystyle \ CH_{3}COOH+C_{2}H_{5}OH\rightleftharpoons CH_{3}COOC_{2}H_{5}+H_{2}O}
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}
第一个反应产生的
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
{\displaystyle \ [B]e^{\int _{0}^{t}k_{2}dt}=\int _{0}^{t}k_{1}[A]_{0}e^{-k_{1}te^{\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： 丙烯
→
O
2
{\displaystyle \ \ {\xrightarrow {O_{2}}}\ }
→
O
2
{\displaystyle \ \ {\xrightarrow {O_{2}}}\ }
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}}
ln
(
d
[
A
]
d
t
)
=
ln
k
+
n
ln
[
A
]
{\displaystyle \ \ln {({\frac {d[A]}{dt}})}=\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 {(-{\frac {d[A]_{0}}{dt}})}-\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] 。^ 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 .^ Manso, José A.; et al. . A Kinetic Approach to the Alkylating Potential of Carcinogenic Lactones. Chem. Res. Toxicol. 2005, 18 (7): 1161-1166.
![{\displaystyle r\;=-{\frac {1}{m}}{\frac {d[A]}{dt}}=\;k\ [A]^{m}[B]^{n}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/a20adbaf050e0dbe84dc6789b6385ef86661532d)
![{\displaystyle [X]\,}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/b21a1ebb162d37ac91d23e1562dd75a2ff218d32)
![{\displaystyle -{\frac {d[A]}{dt}}=k}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/70a61912097064a45b863e32cf06027bc6de0f02)
![{\displaystyle -{\frac {d[A]}{dt}}=k[A]}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/0120f46f6b6ae9ca799958e775a1aa77f393a7a6)
![{\displaystyle -{\frac {d[A]}{dt}}=k[A]^{2}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/db5d7530cf1c29b4ec501ad7b1d2e02db26d2138)
![{\displaystyle -{\frac {d[A]}{dt}}=k[A]^{n}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/e0f0cb169df3e2f6a03c31542ff60657e958453b)
![{\displaystyle \ [A]=[A]_{0}-kt}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/432cafdd1188e2e75adc8399ff2cdc62504ff564)
![{\displaystyle \ [A]=[A]_{0}e^{-kt}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/5ad77c4a6dcbefac1eebffd6f437ea55be739706)
![{\displaystyle {\frac {1}{[A]}}={\frac {1}{[A]_{0}}}+kt}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/79c54169f180a95a38379eaa20ecfed27bb94877)
![{\displaystyle {\frac {1}{[A]^{n-1}}}={\frac {1}{{[A]_{0}}^{n-1}}}+(n-1)kt}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/b9b307524a4bd8aab4ee959bcf20a42d92624801)
![{\displaystyle [A]\ -\ t}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/f1d16a31201dd038cabcde73c456f80d27d220b6)
![{\displaystyle \ln([A])\ -\ t}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/660ab5e561863672310787028735c7aedf7e7eb2)
![{\displaystyle {\frac {1}{[A]}}\ -\ t}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/1304417d32da6e66eca6c89402898bb519d11740)
![{\displaystyle {\frac {1}{[A]^{n-1}}}\ -\ t}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/d7ad27987a9f100647defd007e5b955b95096ca8)
![{\displaystyle t_{1/2}={\frac {[A]_{0}}{2k}}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/406e22f160a25df415f5b82c4ffcba55d2cad9fa)
![{\displaystyle t_{1/2}={\frac {1}{[A]_{0}k}}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/46fdcfbff63344657fe4dd259c77c07c7a522af4)
![{\displaystyle t_{1/2}={\frac {2^{n-1}-1}{(n-1)k{[A]_{0}}^{n-1}}}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/0be16ea0bf01b1b7b6f2ac546d72c51442bbec69)
![{\displaystyle r={k_{1}[A]^{s}[B]^{t}}-{k_{-1}[X]^{u}[Y]^{v}}\,}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/60eb49567a2517352154210a5e6a0e634b00c146)
![{\displaystyle {k_{1}[A]^{s}[B]^{t}=k_{-1}[X]^{u}[Y]^{v}}\,}](https://wikimedia.org/zhwiki/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/zhwiki/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/zhwiki/api/rest_v1/media/math/render/svg/88e14c3292154b8fa16716b98e02f82c06bcb9fb)
![{\displaystyle \ [A]_{e}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/157c791eb39fb0f38181cc26816cc85467f141a3)
![{\displaystyle \ [B]_{e}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/2b0d002ac0aaa13f7d59f87840017c662ac07ad2)
![{\displaystyle \ [B]_{0}=0}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/821e7eef825cb80a068596b0f10fc685ffc86c06)
![{\displaystyle \ [A]_{e}=[A]_{0}-X}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/7bd277d2d7eb956c76649d0476edb35689508ef2)
![{\displaystyle \ [B]_{e}=X}](https://wikimedia.org/zhwiki/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/zhwiki/api/rest_v1/media/math/render/svg/4f46ffd7bb9890d5a5de69d00a144632112ffdda)
![{\displaystyle \ [B]_{e}=X={\frac {k_{1}}{k_{1}+k_{-1}}}[A]_{0}}](https://wikimedia.org/zhwiki/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/zhwiki/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/zhwiki/api/rest_v1/media/math/render/svg/ddb02d2629a76c791025f25645ef75e78a61d0a7)
![{\displaystyle \ -{\frac {d[A]_{e}}{dt}}=k_{1}[A]_{e}-k_{-1}([A]_{0}-[A]_{e})=0}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/cfe2f4ff5d6f7a7821b0171c90d9ee743b7a62ea)
![{\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/zhwiki/api/rest_v1/media/math/render/svg/679f1e8ec4bf7601f12c2a92957479be41adafbb)
![{\displaystyle \ [A]_{0}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/cdff0636a4029d32ad8513b828f9d8062586f899)
![{\displaystyle \ {\frac {d[A]}{dt}}={\frac {d([A]-[A]_{e})}{dt}}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/da54523938a664cdc3c793de71ac4d5ef66d2e0e)
![{\displaystyle \ -{\frac {d([A]-[A]_{e})}{dt}}=(k_{1}+k_{-1})([A]-[A]_{e})}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/12b8ceddfa6d92c759cfedcfd9e20b634bc3000c)
![{\displaystyle \ [A]-[A]_{e}=\Delta [A]}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/d1e7975c1fd41a8812b81294eeb6210aad61f0eb)
![{\displaystyle \ \int _{[A]_{0}}^{[A]}{\frac {d([A]-[A]_{e})}{[A]-[A]_{e}}}=\int _{0}^{t}(k_{1}+k_{-1})dt}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/c0b561e71adc34a29dfd667676118fcde8751a0a)
![{\displaystyle \ ln({\frac {[A]_{0}-[A]_{e}}{[A]_{t}-[A]_{e}}})=(k_{1}+k_{-1})t\qquad \qquad \ldots (1)}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/44e18e84f64e240b028d4e79605eccc07e8c8be6)
![{\displaystyle \ \ln {([A]-[A]_{e})}-t}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/5b0764b8d7f6297e687862658b8afafcaa3ba5c6)
![{\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/zhwiki/api/rest_v1/media/math/render/svg/328084f4a38a9ec4a788c7f2d927ab3a05b78c7e)
![{\displaystyle \ [A]_{t}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/05ccb022c501d0efdbb2722c145db1f03bc0260e)
![{\displaystyle \ {\frac {dt}{dx}}={\frac {1}{k_{1}[A]_{0}-x(k_{1}+k_{-1})}}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/3d1c1477250258c3c1b08d1c7d5192f6ea382f6b)
![{\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/zhwiki/api/rest_v1/media/math/render/svg/942ac7f56cf034ac8cde81e24e139dccdab84268)
![{\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/zhwiki/api/rest_v1/media/math/render/svg/759cf9fb1a2b6618b99cc1620edddf30e4fa3bd0)
![{\displaystyle \ [A]_{e}={\frac {k_{-1}[A]_{0}}{k_{1}+k_{-1}}}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/a61253845e0ac6040dd0c9c92bb7af3588148ff9)
![{\displaystyle \ \ln({\frac {[A]_{0}-[A]_{e}}{[A]_{t}-[A]_{e}}})}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/59da87988f9aab7561b16d2663a879b3d32263b2)
![{\displaystyle \ [A]_{t}={\frac {1}{2}}[A]_{0}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/5fdec9ecec7c8dce0ade7c1f97665a8874b717d9)
![{\displaystyle {\frac {d[A]}{dt}}=-k_{1}[A]\qquad \qquad \ldots (3)}](https://wikimedia.org/zhwiki/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/zhwiki/api/rest_v1/media/math/render/svg/c160a188e75486adf93b19a3756e816753f59ee4)
![{\displaystyle {\frac {d[C]}{dt}}=k_{2}[B]\qquad \qquad \ldots (5)}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/fdc0b0e30ffed9df89e9b243cfcd22a1409efade)
![{\displaystyle \ [A]_{t}=[A]_{0}e^{-k_{1}t}\qquad \qquad \ldots (6)}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/ab7edb598b12b496879655f5d25a5956f7399ade)
![{\displaystyle \ {\frac {d[B]}{dt}}+k_{2}[B]=k_{1}[A]_{0}e^{-k_{1}t}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/1f02cada53806ee433a170ed647d88f144c4f761)
![{\displaystyle \ [B]e^{\int _{0}^{t}k_{2}dt}=\int _{0}^{t}k_{1}[A]_{0}e^{-k_{1}te^{\int _{0}^{t}k_{2}dt}}dt}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/407bf53eb457fe9343b2d8e38c3968e4e6aec825)
![{\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/zhwiki/api/rest_v1/media/math/render/svg/be21835c2897c7110b801181257f9d47df90928b)
![{\displaystyle \ [B]=[A]_{0}\ {\frac {k_{1}}{k_{2}-k_{1}}}(e^{-k_{1}t}-e^{-k_{2}t})\qquad \qquad \ldots (7)}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/e8f077e37337d184ddd2f3fe0588a94ca475dce5)
![{\displaystyle \ [A]+[B]+[C]=[A]_{0}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/ec6370b064013c50aa78156f522b09205205bb64)
![{\displaystyle \ [C]=[A]_{0}-[A]-[B]}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/518355e07ae1498400e2ddb00ca8e1824fd4610c)
![{\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}}})}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/c0b8dad3fad961e7b7f00c5544a4356b59e6fb44)
![{\displaystyle \ [A]}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/9cf05e14412b943a2a9c3e2b1ca59e4a4e976bbe)
![{\displaystyle \ [B]}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/05079c247f3537b3fea48bd5276422d8e195a96c)
![{\displaystyle \ [C]}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/dbcaff4744a270eddc3494979d8bba45466cdbac)
![{\displaystyle \ [B]_{\mbox{max}}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/3847dfc8018b431da2de90db68f8119da0ea5c18)
![{\displaystyle \ [B]_{\mbox{max}}=[A]_{0}\left({\frac {k_{1}}{k_{2}}}\right)^{\dfrac {k_{2}}{k_{2}-k_{1}}}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/8f877c1f6f33ece8c6ccb187ae1cb111413598f9)
![{\displaystyle \ -{\frac {d[A]}{dt}}=(k_{1}+k_{2})[A]}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/22648ff14885987b027bcd95b0e3cfc55978dc3e)
![{\displaystyle \ {\frac {d[B]}{dt}}=k_{1}[A]}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/20f00898f069554bc460d8ca2db7ab05bb74c3fd)
![{\displaystyle \ {\frac {d[C]}{dt}}=k_{2}[A]}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/125c594e179d045aafaf10ba61c195f7ded1fc1b)
![{\displaystyle \ [A]=[A]_{0}e^{-(k_{1}+k_{2})t}}](https://wikimedia.org/zhwiki/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/zhwiki/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/zhwiki/api/rest_v1/media/math/render/svg/96e66cc320b1f522d1ff60dab89a79e9786c1083)
![{\displaystyle {\frac {[B]}{[C]}}={\frac {k_{1}}{k_{2}}}}](https://wikimedia.org/zhwiki/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/zhwiki/api/rest_v1/media/math/render/svg/e8f46e08ceaa1c63f3aaf9a9aa54226d4e43ced4)
![{\displaystyle \ {\frac {d[C]}{dt}}=k_{2}[A][R]}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/8377fc3b141013451144184709fc5b26b0177cf1)
![{\displaystyle [A]_{0}-[C]\approx \;[A]_{0}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/e959d9294f395f483ca051442fc3d54f355c62c9)
![{\displaystyle [C]=[R]_{0}[1-e^{-{\frac {k_{2}}{k_{1}'}}[A]_{0}(1-e^{-k_{1}'t})}]}](https://wikimedia.org/zhwiki/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/zhwiki/api/rest_v1/media/math/render/svg/ae8fd16d45328e54c0a575983cbfe0e817270601)
![{\displaystyle [B]=-{\frac {k_{1}'}{k_{2}}}ln(1-{\frac {[C]}{[R]_{0}}})}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/0d88248959fbb77a4a538e4d05a50a16d6af9b91)
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![{\displaystyle \ \ln {({\frac {d[A]}{dt}})}=\ln {k}+n\ln {[A]}}](https://wikimedia.org/zhwiki/api/rest_v1/media/math/render/svg/668ff07df6364402e330d713b11f001f6c809429)
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