Law of Mass Action

Theorem
Let $\mathcal A$ and $\mathcal B$ be two chemical substances in solution which react together to form a compound $\mathcal C$.

Let the reaction occur by means of the molecules of $\mathcal A$ and $\mathcal B$ colliding and interacting as a result.

Then the rate of formation of $\mathcal C$ is proportional to the number of collisions in unit time.

This in turn is jointly proportional to the quantities of $\mathcal A$ and $\mathcal B$ which have not yet transformed.

Such a chemical reaction is called a second-order reaction, and the law just stated which governs its rate is called the law of mass action.

Let $x$ grams of $\mathcal C$ contain $a x$ grams of $\mathcal A$ and $b x$ grams of $\mathcal B$, where $a + b = 1$.

Let there be $a A$ grams of $\mathcal A$ and $b B$ grams of $\mathcal B$ at time $t = t_0$, at which time $x = 0$.

Then:


 * $x = \begin{cases}

\dfrac {kA^2abt} {kAabt + 1} & : A = B \\ & \\ \dfrac {AB e^{-k \left({A - B}\right) abt}} {A - B e^{-k \left({A - B}\right) abt}} & : A \ne B \end{cases}$

Proof
We have:
 * $\displaystyle \frac{\mathrm d x}{\mathrm d t} \propto \left({A - x}\right) a \left({B - x}\right) b$

or:
 * $\displaystyle \frac{\mathrm d x}{\mathrm d t} = k a b \left({A - x}\right) \left({B - x}\right)$

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:


 * $\frac{\mathrm d x}{\mathrm d t} = k a b \left({A - x}\right) \left({A - x}\right) = k a b \left({A - x}\right)^2$

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 \left({A - B}\right) a b t} } {A - B e^{-k \left({A - B}\right) a b t} } & : A \ne B \end{cases} $