Theorem: Law of Mass Action

Let $\AA$ and $\BB$ be two chemical substances in a solution $C$ which are involved in a second-order reaction.
Let $x$ grams of $\CC$ contain $a x$ grams of $\AA$ and $b x$ grams of $\BB$, where $a + b = 1$.
Let there be $a A$ grams of $\AA$ and $b B$ grams of $\BB$ at time $t = t_0$, at which time $x = 0$.
Then:
:<nowiki>$x = \begin{cases}
\dfrac {k A^2 a b t} {k A a b t + 1} & : A = B \\
& \\
\dfrac {A B e^{-k \paren {A - B} a b t}} {A - B e^{-k \paren {A - B} a b t}} & : A \ne B
\end{cases}$</nowiki>
for some positive real constant $k$.
This is known as the '''law of mass action'''.

Proof:
By the definition of a second-order reaction:
:The rate of formation of $\CC$ is jointly proportional to the quantities of $\AA$ and $\BB$ which have not yet transformed.
By definition of joint proportion:
:$\dfrac {\d x} {\d t} \propto \paren {A - x} a \paren {B - x} b$
or:
:$\dfrac {\d x} {\d t} = k a b \paren {A - x} \paren {B - x}$
for some positive real constant $k$.
Thus:
{{begin-eqn}}
{{eqn | l = \int \d t
      | r = \int \frac {\d x} {k a b \paren {A - x} \paren {B - x} }
      | c = Solution to Separable Differential Equation
}}
{{eqn | ll= \leadsto
      | l = k a b t
      | r = \int \frac {\d x} {\paren {A - B} \paren {B - x} } + \int \frac {\d x} {\paren {B - A} \paren {A - x} }
      | c = Partial Fractions
}}
{{eqn | ll= \leadsto
      | l = \paren {A - B} k a b t
      | r = \int \frac {\d x} {A - x} - \int \frac {\d x} {B - x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {A - B} k a b t
      | r = \map \ln {\frac {A - x} {B - x} } + C_1
      | c = $C_1$ is an arbitrary constant
}}
{{eqn | ll= \leadsto
      | l = -k \paren {A - B} a b t
      | r = \map \ln {\frac {B - x} {A - x} } + C_2
      | c = $C_2$ is another arbitrary constant: $C_2 = -C_1$
}}
{{eqn | ll= \leadsto
      | l = C e^{-k \paren {A - B} a b t}
      | r = \frac {B - x} {A - x}
      | c = $C$ is another arbitrary constant
}}
{{eqn | ll= \leadsto
      | l = C \paren {A - x} e^{-k \paren {A - B} a b t}
      | r = B - x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x \paren {1 - C e^{-k \paren {A - B} a b t} }
      | r = B - A C e^{-k \paren {A - B}) a b t}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \frac {B - A C e^{-k \paren {A - B} a b t} } {1 - C e^{-k \paren {A - B} a b t} }
      | c = 
}}
{{end-eqn}}
Note that in the above, we have to assume that $A \ne B$ or the integrals on the {{RHS}} will not be defined.
We will look later at how we handle the situation when $A = B$.
We are given the initial condition $x = 0$ at $t = 0$.
Thus:
{{begin-eqn}}
{{eqn | l = 0
      | r = \frac {B - A C} {1 - C}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = B
      | r = A C
      | c = (assuming $C \ne 1$)
}}
{{eqn | ll= \leadsto
      | l = C
      | r = \frac B A
      | c = 
}}
{{end-eqn}}
Our assumption that $C \ne 1$ is justified, because that only happens when $A = B$, and we have established that this is not the case.
So, we now have:
{{begin-eqn}}
{{eqn | l = x
      | r = \frac {B - A \paren {B / A} e^{-k \paren {A - B} a b t} } {1 - \paren {B / A} e^{-k \paren {A - B} a b t} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \frac {A B e^{-k \paren {A - B} a b t} } {A - B e^{-k \paren {A - B} a b t} }
      | c = 
}}
{{end-eqn}}
Now we can investigate what happens when $A = B$.
We need to solve:
:$\dfrac {\d x} {\d t} = k a b \paren {A - x} \paren {A - x} = k a b \paren {A - x}^2$
So:
{{begin-eqn}}
{{eqn | l = \frac {\d x} {\d t}
      | r = k a b \paren {A - x}^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int k a b \rd t
      | r = \int \frac {\d x} {\paren {A - x}^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = k a b t
      | r = \frac 1 {A - x} + C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {A - x} k a b t
      | r = 1 + C \paren {A - x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = A k a b t - x k a b t
      | r = 1 + C A - C x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x \paren {C - k a b t}
      | r = 1 + C A - A k a b t
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \frac {1 + C A - A k a b t} {C - k a b t}
      | c = 
}}
{{end-eqn}}
We are given the initial condition $x = 0$ at $t = 0$.
Thus:
{{begin-eqn}}
{{eqn | l = 0
      | r = \frac {1 + C A - 0} {C - 0}
      | c = assuming $C \ne 0$
}}
{{eqn | ll= \leadsto
      | l = 1 + C A
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = C
      | r = -1 / A
      | c = confirming that $C \ne 0$ as we had assumed
}}
{{end-eqn}}
This gives us:
{{begin-eqn}}
{{eqn | l = x
      | r = \frac {1 + \paren {-1 / A} A - A k a b t} {\paren {-1 / A} - k a b t}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \frac {A k a b t} {\frac 1 A + k a b t}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = \frac {k A^2 a b t} {k A a b t + 1}
      | c = 
}}
{{end-eqn}}
So:
:<nowiki>$x = \begin{cases}
\dfrac {k A^2 a b t} {k A a b t + 1} & : A = B \\
& \\
\dfrac {A B e^{-k \paren {A - B} a b t} } {A - B e^{-k \paren {A - B} a b t} } & : A \ne B
\end{cases}</nowiki>$
{{Qed}}