# 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:

 $\ds \int \d t$ $=$ $\ds \int \frac {\d x} {k a b \paren {A - x} \paren {B - x} }$ Solution to Separable Differential Equation $\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 condition $x = 0$ 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 condition $x = 0$ 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:

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

$\blacksquare$