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}
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}
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}
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)}
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，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}
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}
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}}}}
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。 
![{\displaystyle r=-{\frac {1}{m}}{\frac {d[A]}{dt}}=k\ [A]^{x}[B]^{y}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/d1eb2250234345d3360cc32d75a3e4ef0b73d1ae)
![{\displaystyle [X]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/fadb3aef0836cb1d004479f470703a45972bf8fe)
, 
...![{\displaystyle -{\frac {d[A]}{dt}}=k[A]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/0120f46f6b6ae9ca799958e775a1aa77f393a7a6)
![{\displaystyle -{\frac {d[A]}{dt}}=k[A]^{2}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/db5d7530cf1c29b4ec501ad7b1d2e02db26d2138)
![{\displaystyle -{\frac {d[A]}{dt}}=k[A]^{n}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/e0f0cb169df3e2f6a03c31542ff60657e958453b)
![{\displaystyle \ [A]=[A]_{0}-kt}](http://wikimedia.org/api/rest_v1/media/math/render/svg/432cafdd1188e2e75adc8399ff2cdc62504ff564)
![{\displaystyle \ [A]=[A]_{0}e^{-kt}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/5ad77c4a6dcbefac1eebffd6f437ea55be739706)
![{\displaystyle {\frac {1}{[A]}}={\frac {1}{[A]_{0}}}+kt}](http://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}](http://wikimedia.org/api/rest_v1/media/math/render/svg/b9b307524a4bd8aab4ee959bcf20a42d92624801)
![{\displaystyle [A]\ -\ t}](http://wikimedia.org/api/rest_v1/media/math/render/svg/f1d16a31201dd038cabcde73c456f80d27d220b6)
![{\displaystyle \ln([A])\ -\ t}](http://wikimedia.org/api/rest_v1/media/math/render/svg/660ab5e561863672310787028735c7aedf7e7eb2)
![{\displaystyle {\frac {1}{[A]}}\ -\ t}](http://wikimedia.org/api/rest_v1/media/math/render/svg/1304417d32da6e66eca6c89402898bb519d11740)
![{\displaystyle {\frac {1}{[A]^{n-1}}}\ -\ t}](http://wikimedia.org/api/rest_v1/media/math/render/svg/d7ad27987a9f100647defd007e5b955b95096ca8)
![{\displaystyle t_{\frac {1}{2}}={\frac {[A]_{0}}{2k}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/f9627a98d73d4461d0d02de833436bbcba2b3921)
![{\displaystyle t_{\frac {1}{2}}={\frac {1}{[A]_{0}k}}}](http://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}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/441302e0ee48731be96f0c51f366019a1a47b938)
![{\displaystyle {\frac {d[A]}{[A]}}=-{k\,dt}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/0dfa41ca59c0b69dd4435b5094b0ddd1101b015d)
![{\displaystyle \int {\frac {d[A]}{[A]}}=-\int {k\,dt}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/1b2cb4a0a5ab8d129d34dec85e4ac8271a7f9664)
![{\displaystyle ln{[A]}=-kt+C}](http://wikimedia.org/api/rest_v1/media/math/render/svg/f9b59e6243030b5074f35c34ece5bb8ef3d7550c)
=[A]_{0}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/8b9666b18f0193086bfec59bb7a62db97eb4f7d9)
![{\displaystyle [A]_{0}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/32488af10ddcce5bf90dbc659e6472d1279bd2f0)
![{\displaystyle [A]=[A]_{0}e^{-kt}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/1f5dd76b176247a3f070324d682579ee855693a0)
![{\displaystyle {\frac {d[A]}{[A]^{n}}}=-{k\,dt}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/2b49d82f6f2897ff343c673bef0c1aa2d54f8365)
![{\displaystyle \int {\frac {d[A]}{[A]^{n}}}=-\int {k\,dt}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/ab5178b060a6e9126f7c418633be4f28eafb25ad)
![{\displaystyle {\frac {[A]^{-n+1}}{-n+1}}=-kt+C}](http://wikimedia.org/api/rest_v1/media/math/render/svg/375daa94c42ae45a1047282171af09e2dfa574c6)
![{\displaystyle {\frac {1}{[A]^{n-1}}}=(n-1)kt+C}](http://wikimedia.org/api/rest_v1/media/math/render/svg/2679513b8198f24ec3dc5107ff47f9660a72225c)
![{\displaystyle {\frac {1}{[A]_{0}^{n-1}}}=(n-1)k\cdot 0+C}](http://wikimedia.org/api/rest_v1/media/math/render/svg/3e562b02dca33ce8bb04d4c61f0a56cf204bf75b)
![{\displaystyle C={\frac {1}{[A]_{0}^{n-1}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/0bb29f85bb09933befaa63501b85389c73832ebd)
![{\displaystyle r={k_{1}[A]^{s}[B]^{t}}-{k_{-1}[X]^{u}[Y]^{v}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/afa5536f41148f92769c593b265660902f419648)
![{\displaystyle {k_{1}[A]^{s}[B]^{t}=k_{-1}[X]^{u}[Y]^{v}}\,}](http://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}}}}](http://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}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/88e14c3292154b8fa16716b98e02f82c06bcb9fb)
![{\displaystyle [A]_{e}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/6f1203ac1ff71ffb6b2300a26f8590e12c151d32)
![{\displaystyle [B]_{e}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/6a7aa706dd2652248c8529b53515ea4f359c90c6)
![{\displaystyle [B]_{0}=0}](http://wikimedia.org/api/rest_v1/media/math/render/svg/4416f74ff55a09d3d0ead6dddabf9d41e9aa4f5f)
![{\displaystyle \ [A]_{e}=[A]_{0}-X}](http://wikimedia.org/api/rest_v1/media/math/render/svg/7bd277d2d7eb956c76649d0476edb35689508ef2)
![{\displaystyle \ [B]_{e}=X}](http://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}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/3c653bddc078442aa53f5a53b7f243e20a3bb5b6)
![{\displaystyle \ [B]_{e}=X={\frac {k_{1}}{k_{1}+k_{-1}}}[A]_{0}}](http://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}}](http://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])}](http://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}](http://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})}](http://wikimedia.org/api/rest_v1/media/math/render/svg/caa0acb125a099452b7c057a403841539b905b7d)
![{\displaystyle {\frac {d[A]}{dt}}={\frac {d([A]-[A]_{e})}{dt}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/f53019b202c12cf209980aecf0613cd9cb323f99)
![{\displaystyle -{\frac {d([A]-[A]_{e})}{dt}}=(k_{1}+k_{-1})([A]-[A]_{e})}](http://wikimedia.org/api/rest_v1/media/math/render/svg/5c1fa1bfd9ad25e72b719d6ada40e5831519f756)
![{\displaystyle [A]-[A]_{e}=\Delta [A]}](http://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}](http://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)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/44e18e84f64e240b028d4e79605eccc07e8c8be6)
![{\displaystyle \ln {([A]-[A]_{e})}-t}](http://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}](http://wikimedia.org/api/rest_v1/media/math/render/svg/a2f07bbad1773cda47891d60d83a94ce200b53d1)
![{\displaystyle [A]_{t}}](http://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})}}}](http://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})}}]}](http://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)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/c608dea3eddd899feef34de5b68231d8d09ef2e4)
![{\displaystyle [A]_{e}={\frac {k_{-1}[A]_{0}}{k_{1}+k_{-1}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/1019f3a6e37323f112d7b3917ba09e1547e4c2a5)
![{\displaystyle \ln({\frac {[A]_{0}-[A]_{e}}{[A]_{t}-[A]_{e}}})}](http://wikimedia.org/api/rest_v1/media/math/render/svg/edfbda42aaeb044d16da878a15aa8fe55db06608)
![{\displaystyle [A]_{t}={\frac {1}{2}}[A]_{0}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/218a4660bbf2336f57a86ef14e9986f38ae2ca6c)
![{\displaystyle {\frac {d[A]}{dt}}=-k_{1}[A]\qquad \qquad \ldots (3)}](http://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)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/c160a188e75486adf93b19a3756e816753f59ee4)
![{\displaystyle {\frac {d[C]}{dt}}=k_{2}[B]\qquad \qquad \ldots (5)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/fdc0b0e30ffed9df89e9b243cfcd22a1409efade)
![{\displaystyle [A]_{t}=[A]_{0}e^{-k_{1}t}\qquad \qquad \ldots (6)}](http://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}}](http://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}](http://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]}](http://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)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/8a658ee1aae58be4240800304888a3e02a4cffb0)
![{\displaystyle [A]+[B]+[C]=[A]_{0}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/0a7bcb70deda2b099aafbd5650e06ce1e37e206b)
![{\displaystyle [C]=[A]_{0}-[A]-[B]}](http://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}}})}](http://wikimedia.org/api/rest_v1/media/math/render/svg/c0b8dad3fad961e7b7f00c5544a4356b59e6fb44)
![{\displaystyle [A]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/79eaa334597b1861f1b08ca0c8fecb3858ebcb12)
![{\displaystyle [B]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/6e6c6caf9d188e4a608dfdac98bbcd140a9ea106)
![{\displaystyle [C]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/798adf333080234dd9da202633a4d63c4ea091aa)
![{\displaystyle [B]_{\mbox{max}}}](http://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}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/a7a493b44f1fad1a95d1df187603f772e85c1049)
![{\displaystyle {\ce {C3H6 ->[O_2] CH3COCH3 ->[O_2] CH3COOH}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/bfa702c656dae3ff2ddece4f6e2e2532116dd14e)
![{\displaystyle -{\frac {d[A]}{dt}}=(k_{1}+k_{2})[A]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/f3f0a738f74eff107066bef0bea6a45c7abf3e82)
![{\displaystyle \ {\frac {d[B]}{dt}}=k_{1}[A]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/20f00898f069554bc460d8ca2db7ab05bb74c3fd)
![{\displaystyle \ {\frac {d[C]}{dt}}=k_{2}[A]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/125c594e179d045aafaf10ba61c195f7ded1fc1b)
![{\displaystyle \ [A]=[A]_{0}e^{-(k_{1}+k_{2})t}}](http://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})}](http://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})}](http://wikimedia.org/api/rest_v1/media/math/render/svg/96e66cc320b1f522d1ff60dab89a79e9786c1083)
![{\displaystyle {\frac {[B]}{[C]}}={\frac {k_{1}}{k_{2}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/c7c639bb569eb7687c88fc26df54db6591242fbc)
![{\displaystyle {\frac {d[B]}{dt}}=k_{1}[A][H_{2}O]=k_{1}'[A]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/4eb8afde68ff771829096c46101404d3ba5b653b)
![{\displaystyle {\frac {d[C]}{dt}}=k_{2}[A][R]}](http://wikimedia.org/api/rest_v1/media/math/render/svg/7b1bde0fddb8b07b9af4142a4a425eb0a7ce6218)
![{\displaystyle [A]_{0}-[C]\approx [A]_{0}}](http://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})}]}](http://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})}](http://wikimedia.org/api/rest_v1/media/math/render/svg/ae8fd16d45328e54c0a575983cbfe0e817270601)
![{\displaystyle [B]=-{\frac {k_{1}'}{k_{2}}}ln(1-{\frac {[C]}{[R]_{0}}})}](http://wikimedia.org/api/rest_v1/media/math/render/svg/0d88248959fbb77a4a538e4d05a50a16d6af9b91)
![{\displaystyle \ -{\frac {d[A]}{dt}}=k[A]^{n}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/d96418aa64494e55ff8835b75302311138512b36)
![{\displaystyle \ln {\left({\frac {d[A]}{dt}}\right)}=\ln {k}+n\ln {[A]}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/0148ad94ad33587dafe2775d1e1282f12d0c91fd)
![{\displaystyle [A]-t}](http://wikimedia.org/api/rest_v1/media/math/render/svg/2e8beb6a079392c252daf229d38d460caf343c38)
![{\displaystyle {\frac {d[A]_{0}}{dt}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/cdac4c5b9e69d5f3600d3d50cc105e70ffe8c2ac)
![{\displaystyle \ln {\left(-{\frac {d[A]_{0}}{dt}}\right)}-\ln {[A]_{0}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/a6023d39174c1ebf7fe31598070691d41d00b3e6)
