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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \int \mathrm d t\)	\(=\)	\(\displaystyle \int \frac {\mathrm d x} {k a b \left({A - x}\right) \left({B - x}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Separation of Variables
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle kabt\)	\(=\)	\(\displaystyle \int \frac {\mathrm d x} {\left({A - B}\right) \left({B - x}\right)} + \int \frac {\mathrm d x} {\left({B - A}\right) \left({A - x}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by Partial Fractions
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({A - B}\right) k a b t\)	\(=\)	\(\displaystyle \int \frac {\mathrm d x} {A - x} - \int \frac {\mathrm d x} {B - x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({A - B}\right) k a b t\)	\(=\)	\(\displaystyle \ln \left({\frac {A - x} {B - x} }\right) + C_1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	$C_1$ is an arbitrary constant
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle -k \left({A - B}\right) a b t\)	\(=\)	\(\displaystyle \ln \left({\frac {B - x} {A - x} }\right) + C_2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	$C_2$ is another arbitrary constant: $C_2 = -C_1$
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle C e^{-k \left({A - B}\right) a b t}\)	\(=\)	\(\displaystyle \frac {B - x} {A - x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	$C$ is another arbitrary constant
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle C \left({A - x}\right) e^{-k \left({A - B}\right) a b t}\)	\(=\)	\(\displaystyle B - x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x \left({1 - C e^{-k \left({A - B}\right) a b t} }\right)\)	\(=\)	\(\displaystyle B - A C e^{-k \left({A - B}\right) a b t}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle \frac {B - A C e^{-k \left({A - B}\right) a b t} } {1 - C e^{-k \left({A - B}\right) a b t} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 0\)	\(=\)	\(\displaystyle \frac {B - A C} {1 - C}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle B\)	\(=\)	\(\displaystyle A C\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	(assuming $C \ne 1$)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle C\)	\(=\)	\(\displaystyle \frac B A\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle \frac {B - A \left({B / A}\right) e^{-k \left({A - B}\right) a b t} } {1 - \left({B / A}\right) e^{-k \left({A - B}\right) a b t} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle \frac {A B e^{-k \left({A - B}\right) a b t} } {A - B e^{-k \left({A - B}\right) a b t} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Now we can investigate what happens when $A = B$. We need to solve:

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

So:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {\mathrm d x} {\mathrm d t}\)	\(=\)	\(\displaystyle k a b \left({A - x}\right)^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \int k a b \mathrm d t\)	\(=\)	\(\displaystyle \int \frac {\mathrm d x} {\left({A - x}\right)^2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle k a b t\)	\(=\)	\(\displaystyle \frac 1 {A - x} + C\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({A - x}\right) k a b t\)	\(=\)	\(\displaystyle 1 + C \left({A - x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle A k a b t - x k a b t\)	\(=\)	\(\displaystyle 1 + C A - Cx\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x \left({C - k a b t}\right)\)	\(=\)	\(\displaystyle 1 + C A - A k a bt\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle \frac {1 + C A - A k a b t} {C - k a b t}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 0\)	\(=\)	\(\displaystyle \frac {1 + CA - 0}{C - 0}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	assuming $C \ne 0$
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 1 + CA\)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle C\)	\(=\)	\(\displaystyle -1 / A\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	... good, $C \ne 0$ as we'd assumed

This gives us:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle \frac {1 + \left({-1/A}\right) A - A k a b t} {\left({-1/A}\right) - k a b t}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle \frac {A k a b t} {\frac 1 A + k a b t}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle \frac {k A^2 a b t} {k A a b t + 1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So:

$x = \begin{cases} \dfrac {k A^2 a b t} {k A a b t + 1} & : A = B \\ & \\ \dfrac {A B e^{-k \left({A - B}\right) a b t} } {A - B e^{-k \left({A - B}\right) a b t} } & : A \ne B \end{cases}$

$\blacksquare$

Sources

• George F. Simmons: Differential Equations (1972): $\S 4$: Problem $1$