Separable ODEs in chemical reaction kinetics

Ordinary differential equations: Separable differential equations

Separable ODEs in chemical reaction kinetics

The ODE that describes the reaction rate of the chemical reaction $2\,\text{A}\longrightarrow \textit{B}$ is $-\frac{1}{2}\frac{\dd [\text{A}]}{\dd t} = k\cdot [\text{A}]^2$ when elementary chemical kinetics laws are applicable. Using symbol $y$ instead of the concentration $[\text{A}]$ we can rewrrite the so-called rate equation in differential form: $\dd y=-2\,k\,y^2\,\dd t$ By separating variables we get: $-\frac{1}{y^2}\,\dd y=2\,k\,\dd t$ Integrating both sides $-\int\frac{1}{y^2}\,\dd y=2\int k\,\dd t$ gives $\frac{1}{y}= 2\,k\,t+c$ where the integration constant $c$ can be determined from the initial value $y(0)=y_0$ as $\frac{1}{y_0}= c$ So: $\frac{1}{y}=\frac{1}{y_0} +2\,k\,t$ This can be rewritten (check this!) as $y=\frac{y_0}{1+2\,k\,y_0t}$ Thus: $[\text{A}]=\frac{[\text{A}]_0}{1+2\,k\,[\text{A}]_{{}_0}\,t}$ where $[\text{A}]_{{}_0}$ is the initial concentration at time $t=0$ . When we look at the implicit solution, i.e., the relationship $\frac{1}{[\text{A}]}=\frac{1}{[\text{A}]_{{}_0}}+2\,k\,t$ then we can understand that a plot of $\dfrac{1}{[\text{A}]}$ against time $t$ is a straight line whose slope is $2k$ and whose intercept with the vertical axis is $\dfrac{1}{[\text{A}]_{{}_0}}$ .

Simulation The simulation below shows the concentration profile of A.

0,0

0.005

$k$

$[A]$

The rate equation of the chemical reaction $\text{A}+\text{B}\longrightarrow \textit{C}$ is: $-\frac{\dd [\text{A}]}{\dd t} = -\frac{\dd [\text{B}]}{\dd t} = k\cdot [\text{A}]\cdot [\text{B}]$ when elementary chemical kinetics laws are applicable. Let the initial concentrations of $\text{A}$ and $\text{B}$ be $a$ and $b$ , respectively, and let the symbol $y$ be defined as $[\text{A}]=a-y$ . Then: $[\text{B}]=b-y$ and the ODE for $y$ is (check this!) as follows: $\frac{\dd y}{\dd t}=k\,(a-y)\,(b-y)$ Two cases are distinguished.

In this case, we can write the rate equation in differential form as $\dd y=k\,(a-y)^2\,\dd t$ Separation of variables gives: $\frac{1}{(a-y)^2}\,\dd y=k\,\dd t$ Integrating both sides $\int\frac{1}{(a-y)^2}\,\dd y=\int k\,\dd t$ gives $\frac{1}{a-y}= k\,t+c$ where the integration constant $c$ can be determined from the initial value $y(0)=0$ as $\frac{1}{a}= c$ So: $\frac{1}{a-y}= \frac{1}{a}+ k\,t$ This can be rewritten (check this!) as $[\text{A}]=\frac{[\text{A}]_{{}_0}}{1+k[\text{A}]_{{}_0}t}$ where $[\text{A}]_{{}_0}$ is the initial concentration at time $t=0$ . When we look at the implicit solution, i.e., the relationship $\frac{1}{[\text{A}]}=\frac{1}{[\text{A}]_{{}_0}}+k\,t$ then we can understand that a plot of $\dfrac{1}{[\text{A}]}$ against time $t$ is a straight line whose slope is $k$ and whose intercept with the vertical axis is $\dfrac{1}{[\text{A}]_{{}_0}}$ .

In this case, one of the two reagents remains. In this case we must apply partial fraction decomposition to solve separable ODE $\frac{1}{(a-y)(b-y)}\,\dd y=k\,\dd t$ This leads (check this!) to $\frac{1}{(a-y)(b-y)}=\frac{1}{b-a}\left(\frac{1}{a-y}-\frac{1}{b-y}\right)$ So $\begin{aligned}\int \frac{1}{(a-y)(b-y)}\,\dd y &= \frac{1}{b-a}\int\left(\frac{1}{a-y}-\frac{1}{b-y}\right)\dd y \\ \\ &= \frac{1}{b-a}\bigl(-\ln(a-y)+\ln(b-y)\bigr)+c\\ \\ &= \frac{1}{b-a}\ln\left(\frac{b-y}{a-y}\right)+c\end{aligned}$ with some constant $c$ , and this should be equal to the integration result $k\,t$ of the right-hand side of the rate equation in differential form. At the beginning of the reaction, $y(0)=0$ and therefore $c= -\frac{1}{b-a}\ln\left(\frac{b}{a}\right)$ We can finally write (check this!) $\frac{1}{b-a}\ln\left(\frac{a\,(b-y)}{b\,(a-y)}\right)=k\,t$ In terms of concentrations, it can be rewritten as: $\frac{1}{[\text{B}]_0-[\text{A}]_0}\ln\left(\frac{[\text{A}]_{{}_0}\cdot [\text{B}]}{[\text{B}]_0\cdot [\text{A}]}\right)=k\,t$ A plot of $\displaystyle\ln\left(\frac{ [\text{B}]}{ [\text{A}]}\right)$ expand against time $t$ is a straight line whose slope is $k$ .

Simulation The simulation below shows the concentration profiles of A and B.

0,0

0.40

$k$

$[A]$

$[B]$

We consider the kinetics of the following autocatalytic chemical reaction $\text{A}+\text{B}\longrightarrow 2\,\text{B}$ in which the reaction rate is proportional to the product of the concentrations of substances A and B, say $\frac{\dd [\text{A}]}{\dd t}=-k\, [\text{A}]\, [\text{B}]$ for some reaction rate constant $k$ . The formation rate of substance B satisfies the differential equation $\frac{\dd [\text{B}]}{\dd t}=k\, [\text{A}]\, [\text{B}]$ Because of the stoichiometry of the reaction it is true that the sum of the concentrations of the substances A and B is constant, say $[\text{A}]+[\text{B}]=K$ Then we can rewrite the differential equation for the formation rate of substance B as $\frac{\dd\,[\text{B}]}{\dd t}=k\, [\text{B}]\bigl(K-[\text{B}]\bigr)$ The concentration of substance B is herewith a solution of a logistic differential equation and is explicitly described by a logistic function.

Simulation Below is a simulation of the kinetic model. Play with the parameter values. Check whether the concentration of B is bounded from above and that this boundary is reached, even when there is initially only a lit bit of substance B present in the reaction vessel.

0,0

0.20

$k$

$[A]$

$[B]$