Law of Mass Action

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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:

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

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:

\(\ds \int \d t\) \(=\) \(\ds \int \frac {\d x} {k a b \paren {A - x} \paren {B - x} }\) Separation of Variables
\(\ds \leadsto \ \ \) \(\ds k a b t\) \(=\) \(\ds \int \frac {\d x} {\paren {A - B} \paren {B - x} } + \int \frac {\d x} {\paren {B - A} \paren {A - x} }\) Partial Fractions
\(\ds \leadsto \ \ \) \(\ds \paren {A - B} k a b t\) \(=\) \(\ds \int \frac {\d x} {A - x} - \int \frac {\d x} {B - x}\)
\(\ds \leadsto \ \ \) \(\ds \paren {A - B} k a b t\) \(=\) \(\ds \map \ln {\frac {A - x} {B - x} } + C_1\) $C_1$ is an arbitrary constant
\(\ds \leadsto \ \ \) \(\ds -k \paren {A - B} a b t\) \(=\) \(\ds \map \ln {\frac {B - x} {A - x} } + C_2\) $C_2$ is another arbitrary constant: $C_2 = -C_1$
\(\ds \leadsto \ \ \) \(\ds C e^{-k \paren {A - B} a b t}\) \(=\) \(\ds \frac {B - x} {A - x}\) $C$ is another arbitrary constant
\(\ds \leadsto \ \ \) \(\ds C \paren {A - x} e^{-k \paren {A - B} a b t}\) \(=\) \(\ds B - x\)
\(\ds \leadsto \ \ \) \(\ds x \paren {1 - C e^{-k \paren {A - B} a b t} }\) \(=\) \(\ds B - A C e^{-k \paren {A - B}) a b t}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \frac {B - A C e^{-k \paren {A - B} a b t} } {1 - C e^{-k \paren {A - B} a b t} }\)


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:

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


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:

\(\ds x\) \(=\) \(\ds \frac {B - A \paren {B / A} e^{-k \paren {A - B} a b t} } {1 - \paren {B / A} e^{-k \paren {A - B} a b t} }\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \frac {A B e^{-k \paren {A - B} a b t} } {A - B e^{-k \paren {A - B} a b t} }\)


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:

\(\ds \frac {\d x} {\d t}\) \(=\) \(\ds k a b \paren {A - x}^2\)
\(\ds \leadsto \ \ \) \(\ds \int k a b \rd t\) \(=\) \(\ds \int \frac {\d x} {\paren {A - x}^2}\)
\(\ds \leadsto \ \ \) \(\ds k a b t\) \(=\) \(\ds \frac 1 {A - x} + C\)
\(\ds \leadsto \ \ \) \(\ds \paren {A - x} k a b t\) \(=\) \(\ds 1 + C \paren {A - x}\)
\(\ds \leadsto \ \ \) \(\ds A k a b t - x k a b t\) \(=\) \(\ds 1 + C A - C x\)
\(\ds \leadsto \ \ \) \(\ds x \paren {C - k a b t}\) \(=\) \(\ds 1 + C A - A k a b t\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \frac {1 + C A - A k a b t} {C - k a b t}\)


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

Thus:

\(\ds 0\) \(=\) \(\ds \frac {1 + C A - 0} {C - 0}\) assuming $C \ne 0$
\(\ds \leadsto \ \ \) \(\ds 1 + C A\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds C\) \(=\) \(\ds -1 / A\) confirming that $C \ne 0$ as we had assumed


This gives us:

\(\ds x\) \(=\) \(\ds \frac {1 + \paren {-1 / A} A - A k a b t} {\paren {-1 / A} - k a b t}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \frac {A k a b t} {\frac 1 A + k a b t}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \frac {k A^2 a b t} {k A a b t + 1}\)


So:

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

$\blacksquare$


Sources

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