化學反應速率方程 是利用反應物 濃度 或分壓 計算化學反應 的反應速率 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}}
在這個方程中，
[
X
]
{\displaystyle [X]}
X
{\displaystyle X}
活度 ，單位通常為摩爾 每升（mol/L），但在實際計算中有時也用濃度 代替（若該反應物為氣體，表示分壓，單位為帕斯卡 (Pa）。
k
{\displaystyle k}
速率常數 ，與溫度 、離子活度 、光照 、固體反應物的接觸面積、反應活化能 等因素有關，通常可通過阿累尼烏斯方程 計算出來，也可通過實驗測定。
指數
x
{\displaystyle x}
y
{\displaystyle y}
反應級數 ，取決於反應歷程 。在基元反應 中，反應級數等於化學計量數 。但在非基元反應中，反應級數與化學計量數不一定相等。
複雜反應速率方程可能以更複雜的形式出現，包括含多項式的分母。
上述速率方程的一般形式是速率方程的微分 形式，它可以從反應機理導出，而且能明顯表示出濃度對反應速率的影響，便於進行理論分析。將它積分 便得到速率方程的積分形式，即反應物／產物濃度
[
X
]
{\displaystyle [X]}
t
{\displaystyle t}
各級反應 [ 編輯 ] 表中，
M
{\displaystyle M}
莫耳濃度 （
m
o
l
/
L
{\displaystyle mol/L}
t
{\displaystyle t}
k
{\displaystyle k}
速率常數 。所說的「二級反應」和「
n
{\displaystyle n}
n
{\displaystyle \ n}
積分速率定律式證明 [ 編輯 ] 一級反應 [ 編輯 ] 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 。接着，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}
[ 編輯 ] 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 ([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}
接着，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}
可逆反應 [ 編輯 ] 可逆反應 （又稱平衡反應、對行反應、對峙反應）指的是反應物與產物形成化學平衡 的反應，其中正向和逆向反應同時進行，而且反應速率相等。它可以用下面的方程式來表示：
s
A
+
t
B
⇌
u
X
+
v
Y
{\displaystyle sA+tB\rightleftharpoons uX+vY}
假設正向反應速率之速率常數為
k
1
{\displaystyle k_{1}}
，逆向反應速率之速率常數為
k
−
1
{\displaystyle k_{-1}}
。因此，正向反應的淨速率是正、逆反應速率的代數和，即：
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}}
平衡常數
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}}
下面討論一個簡單的單分子可逆一級反應：
A
⇌
B
{\displaystyle A\rightleftharpoons B}
達到平衡時的平衡常數
K
{\displaystyle K}
可以表示為：
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}}
和
[
B
]
e
{\displaystyle [B]_{e}}
分別為
A
{\displaystyle A}
和
B
{\displaystyle B}
的平衡濃度。假設，
B
{\displaystyle B}
的初始濃度
[
B
]
0
=
0
{\displaystyle [B]_{0}=0}
，而且達到平衡時，反應物
A
{\displaystyle A}
已反應成
B
{\displaystyle B}
的濃度為
X
{\displaystyle X}
，於是有：
[
A
]
e
=
[
A
]
0
−
X
{\displaystyle \ [A]_{e}=[A]_{0}-X}
[
B
]
e
=
X
{\displaystyle \ [B]_{e}=X}
代入上面的定義式：
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}
的淨消耗速率為：
−
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 }
時，反應達到平衡，
A
{\displaystyle A}
的淨消耗速率為零，即：
−
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}
上上式減上式，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})}
當
[
A
]
0
{\displaystyle [A]_{0}}
一定時，
[
A
]
e
{\displaystyle [A]_{e}}
為常量，於是有：
d
[
A
]
d
t
=
d
(
[
A
]
−
[
A
]
e
)
d
t
{\displaystyle {\frac {d[A]}{dt}}={\frac {d([A]-[A]_{e})}{dt}}}
將上式代入上上式，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]}
稱為反應物
A
{\displaystyle A}
的距平衡濃度差。
進行分離變數積分，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)}
從上式可以看出，平衡反應的
ln
(
[
A
]
−
[
A
]
e
)
−
t
{\displaystyle \ln {([A]-[A]_{e})}-t}
圖為一直線，由直線斜率可求出
k
1
+
k
−
1
{\displaystyle k_{1}+k_{-1}}
，然後再根據實驗測得的
A
e
{\displaystyle A_{e}}
／
B
e
{\displaystyle B_{e}}
，就可以求得
K
{\displaystyle K}
和
k
1
k
−
1
{\displaystyle {\frac {k_{1}}{k_{-1}}}}
，與上面的
k
1
+
k
−
1
{\displaystyle k_{1}+k_{-1}}
聯立便可以解出
k
1
{\displaystyle k_{1}}
和
k
−
1
{\displaystyle k_{-1}}
。
[3] 或者，將
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，可得某一時刻反應物
A
{\displaystyle A}
的濃度
[
A
]
t
{\displaystyle [A]_{t}}
的表達式：
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 }
時，得到
[
A
]
e
=
k
−
1
[
A
]
0
k
1
+
k
−
1
{\displaystyle [A]_{e}={\frac {k_{-1}[A]_{0}}{k_{1}+k_{-1}}}}
，與上面的結果相吻合。將這些結果代入
ln
(
[
A
]
0
−
[
A
]
e
[
A
]
t
−
[
A
]
e
)
{\displaystyle \ln({\frac {[A]_{0}-[A]_{e}}{[A]_{t}-[A]_{e}}})}
，化簡之後得到
(
k
1
+
k
−
1
)
t
{\displaystyle (k_{1}+k_{-1})t}
，與上面(1)式結果相同。
為了求得反應的半衰期，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}}}}
如果反應的
K
{\displaystyle K}
值很大，那麼
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，平衡反應中的半衰期與反應物的初始濃度無關。
對行反應的例子有：
乙酸 和乙醇 的酯化反應 ：
CH
3
COOH
+
C
2
H
5
OH
↽
−
−
⇀
CH
3
COOC
2
H
5
+
H
2
O
{\displaystyle {\ce {CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O}}}
鄰苯二甲酸酐 與異辛醇 反應生成鄰苯二甲酸二異辛酯 的反應。連續反應 [ 編輯 ] 連續反應 （又稱串聯反應、連串反應）指的是如下類型的化學反應：
A
→
k
1
B
→
k
2
C
{\displaystyle A{\xrightarrow {k_{1}}}B{\xrightarrow {k_{2}}}C}
第一個反應產生的
B
{\displaystyle B}
又可以再起反應生成
C
{\displaystyle C}
。反應的速率方程一般形式為：
對反應物
A
{\displaystyle A}
d
[
A
]
d
t
=
−
k
1
[
A
]
…
(
3
)
{\displaystyle {\frac {d[A]}{dt}}=-k_{1}[A]\qquad \qquad \ldots (3)}
對中間體
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)}
對產物
C
{\displaystyle C}
d
[
C
]
d
t
=
k
2
[
B
]
…
(
5
)
{\displaystyle {\frac {d[C]}{dt}}=k_{2}[B]\qquad \qquad \ldots (5)}
對
(
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)}
將
(
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}}
對其進行積分：
[
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)}
由於
[
A
]
+
[
B
]
+
[
C
]
=
[
A
]
0
{\displaystyle [A]+[B]+[C]=[A]_{0}}
[
C
]
=
[
A
]
0
−
[
A
]
−
[
B
]
{\displaystyle [C]=[A]_{0}-[A]-[B]}
將
(
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}}})}
這樣，
[
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}}}}
用穩態近似 法分析也可以取得類似的結果。
連續反應的例子有：
丙烯 被氧氣 氧化為丙酮 、乙酸 的反應：
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 用發煙硫酸 磺化 時發生的反應： 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}}}\ }
平行反應 [ 編輯 ] 平行反應 （又稱競爭反應）指的是同一反應物可以同時進行幾種不同的反應，生成不同的產物。
A
{
→
k
1
B
→
k
2
C
{\displaystyle A{\begin{cases}\ {\xrightarrow {k_{1}}}\ B\\\ {\xrightarrow {k_{2}}}\ C\end{cases}}}
如果兩個平行反應都是一級反應，則三個速率方程分別為：
−
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]}
對其積分可以得到
[
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})}
一個比較重要的關係式是：
[
B
]
[
C
]
=
k
1
k
2
{\displaystyle {\frac {[B]}{[C]}}={\frac {k_{1}}{k_{2}}}}
兩個平行反應分別為一級和二級反應：[4]
想像下面的情形：反應物
A
{\displaystyle A}
A
+
R
→
C
{\displaystyle A+R\rightarrow C}
A
{\displaystyle A}
水解 （可以看作準一級反應）：
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]}
通過假設
[
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}}})}
以上只是幾種基本的複合反應類型，除此以外，還有很多情況是上述幾種基本複合反應的混合。請有興趣的讀者參見酶動力學 、米氏方程 和酶抑制劑 等文章。
速率方程的確定 [ 編輯 ] 速率方程的確定 主要有以下三種方式：
微分法：對
−
d
[
A
]
d
t
=
k
[
A
]
n
{\displaystyle \ -{\frac {d[A]}{dt}}=k[A]^{n}}
對數 ，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}
嘗試法／試差法：分別將某一化學反應的
[
A
]
{\displaystyle [A]}
t
{\displaystyle t}
半衰期法：求得兩個不同初始濃度下的反應半衰期，然後根據總結一節中的半衰期通式，求出反應級數。 參考資料 [ 編輯 ]
^ IUPAC 金皮書 對「反應速率」（rate law）的定義：[1] （頁面存檔備份 ，存於互聯網檔案館 ）。^ 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 （頁面存檔備份 ，存於互聯網檔案館 ）.^ Manso, José A.; et al. . A Kinetic Approach to the Alkylating Potential of Carcinogenic Lactones. Chem. Res. Toxicol. 2005, 18 (7): 1161-1166.
外部連結 [ 編輯 ] 
![{\displaystyle r=-{\frac {1}{m}}{\frac {d[A]}{dt}}=k\ [A]^{x}[B]^{y}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d1eb2250234345d3360cc32d75a3e4ef0b73d1ae)
![{\displaystyle [X]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fadb3aef0836cb1d004479f470703a45972bf8fe)
![{\displaystyle -{\frac {d[A]}{dt}}=k}](https://wikimedia.org/api/rest_v1/media/math/render/svg/70a61912097064a45b863e32cf06027bc6de0f02)
![{\displaystyle -{\frac {d[A]}{dt}}=k[A]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0120f46f6b6ae9ca799958e775a1aa77f393a7a6)
![{\displaystyle -{\frac {d[A]}{dt}}=k[A]^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db5d7530cf1c29b4ec501ad7b1d2e02db26d2138)
![{\displaystyle -{\frac {d[A]}{dt}}=k[A]^{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0f0cb169df3e2f6a03c31542ff60657e958453b)
![{\displaystyle \ [A]=[A]_{0}-kt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/432cafdd1188e2e75adc8399ff2cdc62504ff564)
![{\displaystyle \ [A]=[A]_{0}e^{-kt}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ad77c4a6dcbefac1eebffd6f437ea55be739706)
![{\displaystyle {\frac {1}{[A]}}={\frac {1}{[A]_{0}}}+kt}](https://wikimedia.org/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/api/rest_v1/media/math/render/svg/b9b307524a4bd8aab4ee959bcf20a42d92624801)
![{\displaystyle [A]\ -\ t}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1d16a31201dd038cabcde73c456f80d27d220b6)
![{\displaystyle \ln([A])\ -\ t}](https://wikimedia.org/api/rest_v1/media/math/render/svg/660ab5e561863672310787028735c7aedf7e7eb2)
![{\displaystyle {\frac {1}{[A]}}\ -\ t}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1304417d32da6e66eca6c89402898bb519d11740)
![{\displaystyle {\frac {1}{[A]^{n-1}}}\ -\ t}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7ad27987a9f100647defd007e5b955b95096ca8)
![{\displaystyle t_{\frac {1}{2}}={\frac {[A]_{0}}{2k}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9627a98d73d4461d0d02de833436bbcba2b3921)
![{\displaystyle t_{\frac {1}{2}}={\frac {1}{[A]_{0}k}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e187aecb0a91741c254d7369bcffb79eb729d6d)
![{\displaystyle t_{\frac {1}{2}}={\frac {2^{n-1}-1}{(n-1)k{[A]_{0}}^{n-1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/441302e0ee48731be96f0c51f366019a1a47b938)
![{\displaystyle {\frac {d[A]}{[A]}}=-{k\,dt}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0dfa41ca59c0b69dd4435b5094b0ddd1101b015d)
![{\displaystyle \int {\frac {d[A]}{[A]}}=-\int {k\,dt}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b2cb4a0a5ab8d129d34dec85e4ac8271a7f9664)
![{\displaystyle ln{[A]}=-kt+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9b59e6243030b5074f35c34ece5bb8ef3d7550c)
=[A]_{0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b9666b18f0193086bfec59bb7a62db97eb4f7d9)
![{\displaystyle [A]_{0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32488af10ddcce5bf90dbc659e6472d1279bd2f0)
![{\displaystyle [A]=[A]_{0}e^{-kt}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f5dd76b176247a3f070324d682579ee855693a0)
![{\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 ([A]^{-n+1})(-n+1)=-kt+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d41b1d7b368163390df132793683d3ad452cb916)
![{\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)
![{\displaystyle \ln({\frac {[A]_{0}-[A]_{e}}{[A]_{t}-[A]_{e}}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/edfbda42aaeb044d16da878a15aa8fe55db06608)
![{\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)
![{\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/api/rest_v1/media/math/render/svg/8a658ee1aae58be4240800304888a3e02a4cffb0)
![{\displaystyle [A]+[B]+[C]=[A]_{0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a7bcb70deda2b099aafbd5650e06ce1e37e206b)
![{\displaystyle [C]=[A]_{0}-[A]-[B]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1a4ec42646562cdc64626e533e94f91595ce395)
![{\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/api/rest_v1/media/math/render/svg/c0b8dad3fad961e7b7f00c5544a4356b59e6fb44)
![{\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)
![{\displaystyle {\ce {C3H6 ->[O_2] CH3COCH3 ->[O_2] CH3COOH}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bfa702c656dae3ff2ddece4f6e2e2532116dd14e)
![{\displaystyle -{\frac {d[A]}{dt}}=(k_{1}+k_{2})[A]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f3f0a738f74eff107066bef0bea6a45c7abf3e82)
![{\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)