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Consider two chemicals, $\ce{A}$ and $\ce{B}$ that react with each other to make $\ce{C}$ with a reaction rate $k$. The reaction can be expressed as $$\ce{A + B->C}$$ The equation expressing the rate of the reactions can be expressed as $$\frac{d[\ce{A}]}{dt}=\frac{d[\ce{B}]}{dt}=-\frac{d[\ce{C}]}{dt}=-k[\ce{A}][\ce{B}]$$

I can separate this equation to make a system of differential equations.$$\frac{d[\ce{A}]}{dt}=-k[\ce{A}][\ce{B}]$$ $$\frac{d[\ce{B}]}{dt}=-k[\ce{A}][\ce{B}]$$

With these two equations, I note that they are similar and will only work with one of these equations for the time being. Therefore, we can write one of these equations as $$\frac{d\ln([\ce{A}])}{dt}=-k[\ce{B}]$$ and by taking another derivative $$\frac{d^2 \ln([\ce{A}])}{dt}=-k\frac{[d\ce{B}]}{dt}=-k\frac{[d\ce{A}]}{dt}$$

I solved this equation using Wolfram Alpha (QED) $$[\ce{A}](t)=\frac{c_1 \exp[c_1(t+c_2)]}{k \exp[c_1(t+c_2)]-1}$$ Therefore the rate of reaction can is $$[\ce{A}]'(t)=\frac{c_1^2 \exp[c_1(t+c_2)]}{k \exp[c_1(t+c_2)]-1}-\frac{k c_1^2 \exp^2[c_1(t+c_2)]}{(k \exp[c_1(t+c_2)]-1)^2}$$

I observed that the rate of change can be written as $$[\ce{A}]'(t)= c_1 [\ce{A}](t)-k [\ce{A}](t)^2$$ so that $c_1$ may be solved, given the initial conditions of $[\ce{A}](0)$ and $[\ce{A}]'(0)$ such that $$c_1=\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2}{[\ce{A}](0)}$$

Substituting the definition of $c_1$ into the equation of $[A](t)$ and $[A]'(t)$ an equation for $c_2$ can be found.

$$c_2=\frac{1}{c_1} \ln(1-\frac{c_1 k}{[\ce{A}](0)}) $$$$ c_2= \frac{[\ce{A}](0)}{[\ce{A}]'(0)+k[\ce{A}](0)^2} \ln(1-\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2 }{[\ce{A}](0)^2}k) $$

Using the equations for $c_1$ and $c_2$ an explicit equation for $[\ce{A}](t)$ can be found.

$$ [\ce{A}](t)=\frac{\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2}{[\ce{A}](0)} \exp[\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2}{[\ce{A}](0)}(t+\frac{[\ce{A}](0)}{[\ce{A}]'(0)+k[\ce{A}](0)^2} \ln(1-\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2 }{[\ce{A}](0)^2}k))]}{k \exp[\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2}{[\ce{A}](0)}(t+\frac{[\ce{A}](0)}{[\ce{A}]'(0)+k[\ce{A}](0)^2} \ln(1-\frac{[\ce{A}]'(0)+k[\ce{A}](0)^2 }{[\ce{A}](0)^2}k))]-1} $$

Side note: since $[\ce{A}]'(0)=[\ce{B}]'(0)= -k[\ce{A}](0)[\ce{B}](0)$ then $c_1$ can be rewritten as $$c_1=\frac{-k[\ce{A}](0)[\ce{B}](0)+k[\ce{A}](0)^2}{[\ce{A}](0)}=k([\ce{A}](0)-[\ce{B}](0))$$

this simplifies $c_2$ to

$$c_2=\frac{k^{-1}}{[\ce{A}](0)-[\ce{B}](0)} \ln(\frac{[k^2 \ce{B}](0) }{[\ce{A}](0)})$$

which simplifes the equation for $[\ce{A}](t)$ to

$$[\ce{A}](t)=\frac{k([\ce{A}](0)-[\ce{B}](0)) \exp[k([\ce{A}](0)-[\ce{B}](0))(t+\frac{k^{-1}}{[\ce{A}](0)-[\ce{B}](0)} \ln(\frac{[k^2 \ce{B}](0) }{[\ce{A}](0)}))]}{k \exp[k([\ce{A}](0)-[\ce{B}](0))(t+\frac{k^{-1}}{[\ce{A}](0)-[\ce{B}](0)} \ln(\frac{[k^2 \ce{B}](0) }{[\ce{A}](0)}))]-1}$$

with a similar equation for $[\ce{B}](t)$

My quesion is: Is this a valid mathematical model for a bimolecular reaction? If not, what is commonly used?

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    $\begingroup$ I think it would be easier to take $[\ce A]_t = [\ce A]_0 - x$ and $[\ce B]_t = [\ce B]_0 - x$ (this must be true because of the stoichiometry of the reaction), then find an expression for $\mathrm dx/\mathrm dt$, then find $x(t)$. $\endgroup$ – orthocresol Oct 3 '17 at 23:52
  • $\begingroup$ At first glance, I see nothing wrong with this. I haven't checked your computations, but I believe they should be alright. @orthocresol's suggestion is also a valid, and probably easier approach to the problem. $\endgroup$ – getafix Oct 4 '17 at 0:53
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    $\begingroup$ If you assume that $[A] \gg [B]$, do you get a pseudo-first-order (i.e. single-exponential) expression? This could be a good check on your work. $\endgroup$ – Curt F. Oct 4 '17 at 4:12
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    $\begingroup$ Except that you've made it look absurdly complex for no reason at all, it seems OK. Expand $c_1(t+c_2)$ as $c_1t+c_1c_2$, and everything will simplify a great deal. $\endgroup$ – Ivan Neretin Oct 4 '17 at 5:11
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Assuming that the bimolecular chemical reaction $\ce{A + B ->[\kappa] C}$ has mass action kinetics, we have the following pair of coupled ODEs

$$\begin{array}{rl} \dot a &= - \kappa \, a \, b\\ \dot b &= - \kappa \, a \, b\end{array}$$

where $\kappa > 0$ is the rate constant, $a := [\ce{A}]$ and $b := [\ce{B}]$. Since $\dot a = \dot b$, we have $\frac{\mathrm d}{\mathrm d t} \left( a - b \right) = 0$ and, thus, integrating, we obtain

$$a (t) - b (t) = a_0 - b_0$$

where $a_0 > 0$ and $b_0 > 0$ are the initial concentrations. Since $b (t) = a (t) - (a_0 - b_0)$, the 1st ODE can be decoupled from the 2nd, as follows

$$\dot a = - \kappa \, a \, \left( a - (a_0 - b_0) \right)$$

which can be rewritten in the form

$$\frac{\mathrm d a}{a \, \left( a - (a_0 - b_0) \right)} = - \kappa \, \mathrm d t$$

Assuming that $a_0 \neq b_0$, we have the following partial fraction expansion

$$\left( \frac{1}{a - (a_0 - b_0)} - \frac{1}{a} \right) \mathrm d a = - \kappa \, (a_0 - b_0) \, \mathrm d t$$

Integrating, we obtain

$$\ln \left( \frac{a (t) - (a_0 - b_0)}{a_0 - (a_0 - b_0)} \right) - \ln \left( \frac{a (t)}{a_0} \right) = - \kappa \, (a_0 - b_0) \, t$$

which can be rewritten as follows

$$\ln \left( \frac{a (t) - (a_0 - b_0)}{a (t)} \right) = \ln \left( \frac{b_0}{a_0} \right) - \kappa \, (a_0 - b_0) \, t$$

Exponentiating both sides, we obtain

$$\frac{a (t) - (a_0 - b_0)}{a (t)} = \frac{b (t)}{a (t)} = \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0 - b_0) \, t)$$

and, eventually, we obtain

$$\boxed{\begin{array}{rl} &\\ a (t) &= \dfrac{a_0 - b_0}{1 - \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0 - b_0) \, t)}\\\\ b (t) &= \dfrac{(a_0 - b_0) \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0 - b_0) \, t)}{1 - \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0 - b_0) \, t)}\\ & \end{array}}$$

Taking the limit,

$$\lim_{t \to \infty} a (t) = \begin{cases} a_0 - b_0 & \text{if } a_0 > b_0\\\\ 0 & \text{if } a_0 < b_0\end{cases}$$

$$\\$$

$$\lim_{t \to \infty} b (t) = \begin{cases} 0 & \text{if } a_0 > b_0\\\\ b_0 - a_0 & \text{if } a_0 < b_0\end{cases}$$


What if $a_0 = b_0$?

Previously, we assumed that $a_0 \neq b_0$. If $a_0 = b_0$, then

$$\frac{\mathrm d a}{a \, \left( a - (a_0 - b_0) \right)} = - \kappa \, \mathrm d t$$

becomes

$$-\frac{\mathrm d a}{a^2} = \kappa \, \mathrm d t$$

Integrating, we obtain

$$\frac{1}{a (t)} - \frac{1}{a_0} = \kappa \, t$$

and, eventually, we obtain

$$\boxed{ a (t) = \frac{a_0}{1 + a_0 \, \kappa \, t} = b (t)} $$

In this case, both reactants are eventually exhausted

$$\lim_{t \to \infty} a (t) = \lim_{t \to \infty} b (t) = 0$$

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