Chemical reaction modeled by a differential equation

Question

I am badly stuck on the following question. Thus, I am asking for some help :)

Consider a chemical reaction in which compounds $A$ and $B$ combine to form a third compound $X$. The reaction can be written as

$$A + B \xrightarrow{k} X$$

If $2 \, \rm{g}$ of $A$ and $1 \, \rm{g}$ of $B$ are required to produce $3g$ of compound $X$, then the amount of compound $x$ at time $t$ satisfies the differential equation

$$ \frac{dx}{dt} = k \left(a - \frac{2}{3}x\right)\left(b - \frac{1}{3}x\right) $$

where $a$ and $b$ are the amounts of $A$ and $B$ at time $0$ (respectively), and initially none of compound $X$ is present (so $x(0) = 0$). Time is in units of minutes, and $k$ is the reaction rate, per minute per gram.

Use separation of variables (and integration by partial fractions) to show that solution can be expressed in the form

$$ \ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)} \cdot \frac{b}{a}\right) = ct $$

where the constant c depends on k,a,b. Now suppose that a = 15g, b = 20g and after 10min, 15g of compound X has been formed, find the amount of X after 20 mins.

So, first of all, showing this solution can be expressed as second equation, by using separation of variables ( and integration by partial fractions), i got

$$ \ln \left(\frac{a - \frac{2}{3}x}{b - \frac{1}{3}x}\right) * \frac{3}{a-2b} = kt + c $$

how can this can be expressed in the form of the second equation ? am I doing something wrong ?, I dont get how I should get $$ \frac{b}{a} $$ inside the ln... and on the RHS, how kt + c becomes just ct ?

and by the second euqation, """

where the constant $c$ depends on $k$, $a$ and $b$. Now suppose that a = 15g, b = 20g and after 10min, 15g of compound X has been formed, find the amount of X after 20 mins. """

how do we find the amount of $X$ after $20$ minutes?

Thank you !

Answer 1

The given ODE can be rewritten in the form

$$\dot x = \frac{2 \kappa}{9} \left( x - \frac{3 a}{2} \right) \left( x - 3 b \right)$$

or, alternatively, in the form

$$\left( \frac{1}{x - \frac{3 a}{2}} - \frac{1}{x - 3 b} \right) \, \mathrm{d} x = \left(\frac{a - 2 b}{3}\right) \kappa \,\mathrm{d} t$$

Integrating,

$$\ln \left[ \left( \frac{x - \frac{3 a}{2}}{x - 3 b} \right) \left( \frac{x_0 - 3 b}{x_0 - \frac{3 a}{2}} \right) \right] = \left(\frac{a - 2 b}{3}\right) \kappa t$$

If $x_0 = 0$, then

$$\ln \left[ \left( \frac{2 x - 3 a}{x - 3 b} \right) \frac{b}{a} \right] = \left(\frac{a - 2 b}{3}\right) \kappa t$$

Answer 2

Without checking if the two forms are both correct : $$\begin{cases} \ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)} \cdot \frac{b}{a}\right) = ct \qquad [1]\\ \ln \left(\frac{a - \frac{2}{3}x}{b - \frac{1}{3}x}\right) * \frac{3}{a-2b} = kt + c \qquad [2] \end{cases}$$ The relationship between $a,b,c,k$ can be derived :

$\begin{cases} \ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)} \cdot \frac{b}{a}\right) =\ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)}\right) +\ln\left(\frac{b}{a} \right) =ct \\ \ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)} \cdot \frac{b}{a}\right) = \frac{a-2b}{3}(kt+c) \end{cases} \quad\to\quad ct-\ln\left(\frac{b}{a} \right)=\frac{a-2b}{3}(kt+c)$ $$c=\frac{a-2b}{3}k=-\ln\left(\frac{b}{a} \right)$$

If this relationship is true, Eqs. $[1]$ and $[2]$ are equivalent.

$\begin{cases} a=\frac{3c}{k}\frac{e^c}{e^c-1}\\ b=\frac{3c}{k}\frac{1}{e^c-1}\\ \end{cases}$

Answer 3

The special case $a=2b$ is missing in the answers. In this case $$ x(t)=3b-\frac{9b}{2b\kappa t+3}. $$