Chemical reactions and law of masses 

*

*Two chemicals $\mathrm{A}$ and $\mathrm{B}$ are combined to form a chemical $\mathrm{C}$. The rate, or velocity, of the reaction is proportional to the product of the instantaneous amount of $\mathrm{A}$ and $\mathrm{B}$ not converted to chemical $\mathrm{C}$. Initially , there are $40\,\rm g$ of $\mathrm{A}$ and $50\,\rm g$ of $\mathrm{B}$, and for each gram of $\mathrm{B}$, $2\,\rm g$ of $\mathrm{A}$ is used. It is observed that $10\,\rm g$ of $\mathrm{C}$ is formed in $5\,\rm min$. How much is formed in $20\,\rm min$? what is the limiting amount of $\mathrm{C}$ after a long time? How much of chemicals $\mathrm{A}$ and $\mathrm{B}$ remain after a long time?


*Solve part one if $100\,\rm g$ of chemical $\mathrm{A}$ is present initially. At what time is chemical $\mathrm{C}$ half-formed?
There are $M$ parts of $\mathrm{A}$ and $N$ parts of $\mathrm{B}$ formed in the compound $\mathrm{C}$.
The general differential equation must be
$$\frac{\mathrm{d}X}{\mathrm{d}t} = k(\alpha - x)(\beta - x)$$
being $\alpha = a(M + N)/M$ and $\beta = b(M + N)/N$

The application problem says that the velocity of the reaction is proportional to the product of the instantaneous amount of $\mathrm{A}$ and $\mathrm{B}$. So, we can give a value to $AB = (\text{some}~k)AB = 2$, for instance. As for each gram of $\mathrm{B}$, $2$ grams of $\mathrm{A}$ is used well, $\mathrm{A = 2B}$. This system leads me to $\mathrm{B = 1}$ and $\mathrm{A = 2}$.
The differential equation must be (according to me)
$$\frac{\mathrm{d}X}{\mathrm{d}t} = k(40 - 2x)(50 - x),$$
but in the solution manual the differential equation is
$$\frac{\mathrm{d}X}{\mathrm{d}t} = k(120 - 2x)(50 - x).$$
 A: Here is to get you started. The quantities $A(t)$, $B(t)$ and $C(t)$ of constituents $A$, $B$ and $C$ at time $t$ are such that 
$$\frac13C'(t)=-\frac12A'(t)=-B'(t)=kA(t)B(t),
$$ 
for some unknown constant $k$. One is given $A(0)=A_0$ and $B(0)=B_0$ and one probably assumes that $C(0)=0$. 
Then $A(t)=A_0-\frac23C(t)$ and $B(t)=B_0-\frac13C(t)$ for every $t$ hence
$$
C'(t)=3kA(t)B(t)=\gamma(\alpha-C(t))(\beta-C(t)).
$$
with $\alpha=\frac32A_0$, $\beta=3B_0$ and $\gamma=\frac23k$. Thus,
$$
\gamma t=\int_0^{C(t)}\frac{\mathrm dx}{(\alpha-x)(\beta-x)}.
$$
Using the decomposition
$$
\frac{\beta-\alpha}{(\alpha-x)(\beta-x)}=\frac1{\alpha-x}-\frac1{\beta-x},
$$
one gets
$$
(\beta-\alpha)\gamma t=\log(1-C(t)/\beta)-\log(1-C(t)/\alpha),
$$
that is,
$$\frac{1-C(t)/\beta}{1-C(t)/\alpha}=\mathrm e^{\delta t},\qquad \delta=(\beta-\alpha)\gamma,
$$
or,
$$
C(t)=\alpha\beta\,\frac{\mathrm e^{\delta t}-1}{\beta\mathrm e^{\delta t}-\alpha}.
$$
Note that $A_0=40$ and $B_0=50$ yield $\alpha=60$ and $\beta=150$. Since $\beta\gt\alpha$, when $t\to\infty$, $\mathrm e^{\delta t}\to\infty$ and $C(t)\to\alpha$. Another consequence of the analysis above which one might use to solve one of the questions is that, for every $t$,
$$
\frac{1-C(4t)/\beta}{1-C(4t)/\alpha}=\left(\frac{1-C(t)/\beta}{1-C(t)/\alpha}\right)^4.
$$
