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: :$\dfrac {\d x} {\d t} = k a b \paren {A - x} \paren {B - x}$ Thus: {{begin-eqn}} {{eqn | l = \int \d t       | r = \int \frac {\d x} {k a b \paren {A - x} \paren {B - x} }       | c = Separation of Variables }} {{eqn | ll= \implies       | l = k a b t       | r = \int \frac {\d x} {\paren {A - B} \paren {B - x} } + \int \frac {\d x} {\paren {B - A} \paren {A - x} }       | c = Partial Fractions }} {{eqn | ll= \leadsto       | l = \paren {A - B} k a b t       | r = \int \frac {\d x} {A - x} - \int \frac {\d x} {B - x}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {A - B} k a b t       | r = \map \ln {\frac {A - x} {B - x} } + C_1       | c = $C_1$ is an arbitrary constant }} {{eqn | ll= \leadsto       | l = -k \paren {A - B} a b t       | r = \map \ln {\frac {B - x} {A - x} } + C_2       | c = $C_2$ is another arbitrary constant: $C_2 = -C_1$ }} {{eqn | ll= \leadsto       | l = C e^{-k \paren {A - B} a b t}       | r = \frac {B - x} {A - x}       | c = $C$ is another arbitrary constant }} {{eqn | ll= \leadsto       | l = C \paren {A - x} e^{-k \paren {A - B} a b t}       | r = B - x       | c =  }} {{eqn | ll= \leadsto       | l = x \paren {1 - C e^{-k \paren {A - B} a b t} }       | r = B - A C e^{-k \paren {A - B}) a b t}       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \frac {B - A C e^{-k \paren {A - B} a b t} } {1 - C e^{-k \paren {A - B} a b t} }       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = 0       | r = \frac {B - A C} {1 - C}       | c =  }} {{eqn | ll= \leadsto       | l = B       | r = A C       | c = (assuming $C \ne 1$) }} {{eqn | ll= \leadsto       | l = C       | r = \frac B A       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = x       | r = \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} }       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \frac {A B e^{-k \paren {A - B} a b t} } {A - B e^{-k \paren {A - B} a b t} }       | c =  }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d t}       | r = k a b \paren {A - x}^2       | c =  }} {{eqn | ll= \leadsto       | l = \int k a b \rd t       | r = \int \frac {\d x} {\paren {A - x}^2}       | c =  }} {{eqn | ll= \leadsto       | l = k a b t       | r = \frac 1 {A - x} + C       | c =  }} {{eqn | ll= \leadsto       | l = \paren {A - x} k a b t       | r = 1 + C \paren {A - x}       | c =  }} {{eqn | ll= \leadsto       | l = A k a b t - x k a b t       | r = 1 + C A - C x       | c =  }} {{eqn | ll= \leadsto       | l = x \paren {C - k a b t}       | r = 1 + C A - A k a b t       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \frac {1 + C A - A k a b t} {C - k a b t}       | c =  }} {{end-eqn}} We are given the initial conditions $x = 0$ at $t = 0$. Thus: {{begin-eqn}} {{eqn | l = 0       | r = \frac {1 + C A - 0} {C - 0}       | c = assuming $C \ne 0$ }} {{eqn | ll= \leadsto       | l = 1 + C A       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = C       | r = -1 / A       | c = confirming that $C \ne 0$ as we had assumed }} {{end-eqn}} This gives us: {{begin-eqn}} {{eqn | l = x       | r = \frac {1 + \paren {-1 / A} A - A k a b t} {\paren {-1 / A} - k a b t}       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \frac {A k a b t} {\frac 1 A + k a b t}       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \frac {k A^2 a b t} {k A a b t + 1}       | c =  }} {{end-eqn}} So: :$x = \begin{cases} \dfrac {k A^2 a b t} {k A a b t + 1} & : A = B \\ & \\ \dfrac {A B e^{-k \paren {A - B} a b t} } {A - B e^{-k \paren {A - B} a b t} } & : A \ne B \end{cases}$ {{Qed}}