Consider two chemicals, A and B that react with each other to make C with a reaction rate k. The reaction can be expressed as A+B⟶C The equation expressing the rate of the reactions can be expressed as d[A]dt=d[B]dt=−d[C]dt=−k[A][B]
I can separate this equation to make a system of differential equations.d[A]dt=−k[A][B] d[B]dt=−k[A][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 dln([A])dt=−k[B] and by taking another derivative d2ln([A])dt=−k[dB]dt=−k[dA]dt
I solved this equation using Wolfram Alpha (QED) [A](t)=c1exp[c1(t+c2)]kexp[c1(t+c2)]−1 Therefore the rate of reaction can is [A]′(t)=c21exp[c1(t+c2)]kexp[c1(t+c2)]−1−kc21exp2[c1(t+c2)](kexp[c1(t+c2)]−1)2
I observed that the rate of change can be written as [A]′(t)=c1[A](t)−k[A](t)2 so that c1 may be solved, given the initial conditions of [A](0) and [A]′(0) such that c1=[A]′(0)+k[A](0)2[A](0)
Substituting the definition of c1 into the equation of [A](t) and [A]′(t) an equation for c2 can be found.
c2=1c1ln(1−c1k[A](0))c2=[A](0)[A]′(0)+k[A](0)2ln(1−[A]′(0)+k[A](0)2[A](0)2k)
Using the equations for c1 and c2 an explicit equation for [A](t) can be found.
[A](t)=[A]′(0)+k[A](0)2[A](0)exp[[A]′(0)+k[A](0)2[A](0)(t+[A](0)[A]′(0)+k[A](0)2ln(1−[A]′(0)+k[A](0)2[A](0)2k))]kexp[[A]′(0)+k[A](0)2[A](0)(t+[A](0)[A]′(0)+k[A](0)2ln(1−[A]′(0)+k[A](0)2[A](0)2k))]−1
Side note: since [A]′(0)=[B]′(0)=−k[A](0)[B](0) then c1 can be rewritten as c1=−k[A](0)[B](0)+k[A](0)2[A](0)=k([A](0)−[B](0))
this simplifies c2 to
c2=k−1[A](0)−[B](0)ln([k2B](0)[A](0))
which simplifes the equation for [A](t) to
[A](t)=k([A](0)−[B](0))exp[k([A](0)−[B](0))(t+k−1[A](0)−[B](0)ln([k2B](0)[A](0)))]kexp[k([A](0)−[B](0))(t+k−1[A](0)−[B](0)ln([k2B](0)[A](0)))]−1
with a similar equation for [B](t)
My quesion is: Is this a valid mathematical model for a bimolecular reaction? If not, what is commonly used?
Answer
Assuming that the bimolecular chemical reaction A+Bκ→C has mass action kinetics, we have the following pair of coupled ODEs
˙a=−κab˙b=−κab
where κ>0 is the rate constant, a:=[A] and b:=[B]. Since ˙a=˙b, we have ddt(a−b)=0 and, thus, integrating, we obtain
a(t)−b(t)=a0−b0
where a0>0 and b0>0 are the initial concentrations. Since b(t)=a(t)−(a0−b0), the 1st ODE can be decoupled from the 2nd, as follows
˙a=−κa(a−(a0−b0))
which can be rewritten in the form
daa(a−(a0−b0))=−κdt
Assuming that a0≠b0, we have the following partial fraction expansion
(1a−(a0−b0)−1a)da=−κ(a0−b0)dt
Integrating, we obtain
ln(a(t)−(a0−b0)a0−(a0−b0))−ln(a(t)a0)=−κ(a0−b0)t
which can be rewritten as follows
ln(a(t)−(a0−b0)a(t))=ln(b0a0)−κ(a0−b0)t
Exponentiating both sides, we obtain
a(t)−(a0−b0)a(t)=b(t)a(t)=(b0a0)exp(−κ(a0−b0)t)
and, eventually, we obtain
a(t)=a0−b01−(b0a0)exp(−κ(a0−b0)t)b(t)=(a0−b0)(b0a0)exp(−κ(a0−b0)t)1−(b0a0)exp(−κ(a0−b0)t)
Taking the limit,
lim
\\
\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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