Law of Mass Action

Theorem
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:


 * $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}$

This is known as the law of mass action

Proof
By the defintion 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:
 * $\dfrac {\d x} {\d t} = k a b \paren {A - x} \paren {B - x}$

Thus:

Note that in the above, we have to assume that $A \ne B$ or the integrals on the right hand side will not be defined.

We will look later at how we handle the situation when $A = B$.

We are given the initial conditions $x = 0$ at $t = 0$. Thus:

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:

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:

We are given the initial conditions $x = 0$ at $t = 0$.

Thus:

This gives us:

So:
 * $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}$