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

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

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