Analytical solution for kinetics of bimolecular reaction

Question

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?

Answer 1

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

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

Answer 2

You need to enforce the boundary conditions of your problem to evaluate the integral. I think that might be the material point here.

But let's start at the beginning: You have the rate equation

\begin{align} - \frac{\mathrm{d}c_{\mathrm{A}}}{\mathrm{d}t} = k c_{\mathrm{A}} c_{\mathrm{B}} \ . \end{align}

Now, introduce the conversion $x = c_{\mathrm{A}0} - c_{\mathrm{A}} = c_{\mathrm{B}0} - c_{\mathrm{B}}$. This means that an infinitesimal change in the conversion can be written as $\mathrm{d} x = -\mathrm{d} c_{\mathrm{A}} = -\mathrm{d} c_{\mathrm{A}}$. Substituting this into the rate equation yields

\begin{align} \frac{\mathrm{d}x}{\mathrm{d}t} = k ( c_{\mathrm{A}0} - x ) ( c_{\mathrm{B}0} - x ) \ . \end{align}

To solve this differential equation it is sufficient to separate the variables and integrate both sides of the equation. And this is the point where the boundary conditions enter the game. To evaluate the integrals fully, i.e. without leaving a constant of integration $C$, you need to use some limits for your integrals for which you know the values of you variables beforehand. So the question is: What is known beforehand? You know that at the start of the reaction, and we will define this point in time as $t=0$, none of you reactants have reacted with one another. So, $c_{\mathrm{A}}(t\!=\!0) = c_{\mathrm{A}0}$ or in terms of the conversion $x(t\!=\!0) = 0$. This will set the lower limit of the integrals. For the upper limit you can use the value that you actually want to calculate, i.e. the concentration $c_{\mathrm{A}}(t)$ or conversion $x(t)$ at some time $t$. Using all this leads to

\begin{align} \int \limits_{x(t=0) = 0}^{x(t)} \frac{1}{( c_{\mathrm{A}0} - x ) ( c_{\mathrm{B}0} - x )}\mathrm{d}x = \int \limits_{t=0}^{t} k \, \mathrm{d}t \ . \end{align}

For the integral on the left-hand side you can use the hint already given to (although calculating it yourself isn't that difficult if you use partial-fraction decomposition) but you can leave out the constant $C$ as this will be fixed by setting explicit integration limits. The integral on the right-hand side is no problem.

\begin{align} \int \limits_{x(t=0)}^{x(t)} \frac{1}{( c_{\mathrm{A}0} - x ) ( c_{\mathrm{B}0} - x )}\mathrm{d}x &= \int \limits_{0}^{t} k \, \mathrm{d}t \\ \left[ \frac{1}{ c_{\mathrm{B}0} - c_{\mathrm{A}0} } \ln\left( \frac{c_{\mathrm{B}0} - x }{c_{\mathrm{A}0} - x} \right) \right]^{x(t)}_{0} &= \Bigl[ k t \Bigr]^{t}_{0} \\ \frac{1}{ c_{\mathrm{B}0} - c_{\mathrm{A}0} } \ln\left( \frac{c_{\mathrm{A}0} (c_{\mathrm{B}0} - x) }{c_{\mathrm{B}0} (c_{\mathrm{A}0} - x)} \right) = k t \end{align}

Exponentiating this gives nearly the result that you are after:

\begin{align} \frac{c_{\mathrm{A}0} (c_{\mathrm{B}0} - x) }{c_{\mathrm{B}0} (c_{\mathrm{A}0} - x)} = \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr) \end{align}

Now, the only work that is left is to make the back-substitution $x(t) = c_{\mathrm{A}0} - c_{\mathrm{A}}(t)$ and solve the resulting equation for $c_{\mathrm{A}}(t)$:

\begin{align} \frac{c_{\mathrm{A}0} (c_{\mathrm{B}0} - c_{\mathrm{A}0} + c_{\mathrm{A}}(t)) }{c_{\mathrm{B}0} c_{\mathrm{A}}(t)} &= \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr) \\ c_{\mathrm{A}0} (c_{\mathrm{B}0} - c_{\mathrm{A}0} + c_{\mathrm{A}}(t)) &= c_{\mathrm{B}0} c_{\mathrm{A}}(t) \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr) \\ c_{\mathrm{A}0} c_{\mathrm{B}0} - c_{\mathrm{A}0}^{2} &= c_{\mathrm{A}}(t) \left( c_{\mathrm{B}0} \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr) - c_{\mathrm{A}0} \right) \\ c_{\mathrm{B}0} - c_{\mathrm{A}0} &= c_{\mathrm{A}}(t) \left( \frac{c_{\mathrm{B}0}}{c_{\mathrm{A}0}} \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr) - 1 \right) \\ \Rightarrow \qquad c_{\mathrm{A}}(t) &= \frac{c_{\mathrm{B}0} - c_{\mathrm{A}0}}{\frac{c_{\mathrm{B}0}}{c_{\mathrm{A}0}} \exp\bigl( (c_{\mathrm{B}0} - c_{\mathrm{A}0} ) k t \bigr) - 1} \end{align}