Chemistry – Analytical solution for kinetics of bimolecular reaction

Solution 1:

Assuming that the bimolecular chemical reaction $ce{A + B ->[kappa] C}$ has mass action kinetics, we have the following pair of coupled ODEs

$$begin{array}{rl} dot a &= – kappa , a , b\ dot b &= – kappa , a , bend{array}$$

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,

$$lim_{t to infty} a (t) = begin{cases} a_0 – b_0 & text{if } a_0 > b_0\\ 0 & text{if } a_0 < b_0end{cases}$$

$$\$$

$$lim_{t to infty} b (t) = begin{cases} 0 & text{if } a_0 > b_0\\ b_0 – a_0 & text{if } a_0 < b_0end{cases}$$


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

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

Solution 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} } lnleft( 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} } lnleft( 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)} = expbigl( (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)} &= expbigl( (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) expbigl( (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} expbigl( (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}} expbigl( (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}} expbigl( (c_{mathrm{B}0} – c_{mathrm{A}0} ) k t bigr) – 1}
end{align}

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