Modeling a chemical reaction with differential equations

Question

The problem says:

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

$\ A_0$ = 40 g, $\ B_0$ = 50 g.

Well, first of all, $\alpha $ = $\ A_0 \frac{M+N}{M} $ and $\beta $ = $\ B_0 \frac{M+N}{N} $

and then our differential equation must be:

$\frac{dX}{dt} = k(\alpha-X)(\beta-X) $ which can be easily solved. In order to create $\ x$ part of the chemical C we will need 2 parts of $\ A$ and one part of $\ B$. This lead me to believe that $\ M$ = 2 and $\ N$ = 1 By calculating $\alpha$ = $\ 40 \frac{2+1}{2}$ = 60 and $\beta $=$\ 50 \frac{2+1}{1}$ = 150. The differential equation must become $\frac{dX}{dt} = k(\ 60-X)(\ 150-X) $ right?

I separate the variables and solved the equation $$ \int \frac{dx}{(60-x)(150-x)}\, = \int kdt $$

$$ \ln \frac{150-x}{60-x} = 90kt+C_1 $$ By using X(0)=0, $$ \frac{150-x}{60-x} = Ce^{90k0}, C=\frac{5}{2} $$ and using X(5)=10 and solving for k, $$ \frac{150-10}{60-10} = \frac{5}{2}e^{450k}, k= 2.5184X10^{-4} $$ and this is different to the solution of k in the solution manual which is $\ k$ = $\ 1.259X^{-4}$ and also, the differential equation is different, they obtain $\frac{dX}{dt} = k(\ 120-2X)(\ 150-X)$

And I'm wondering why! I assume my mistake is in the values for $\ M$ and $\ N$. Can you give me a hand with this?

Answer 1

The only difference is that your $k$ is their $2k$. And look, you get exactly double their answer. So you actually have the same answer.

Answer 2

Solving our equation for x: $$ \frac{150-x}{60-x}=C_1e^{90kt} $$ We obtain: $$ X(t)=\frac{60 C_1e^{90kt}-150}{C_1e^{90kt}-1} $$ And if we substitute $\ k=2.5184X10^{-4}$ and $\ C_1 = \frac{5}{2}$ this would be: $$ X(t)=\frac{150 e^{0.0226656t}-150}{\frac{5}{2}e^{0.0226656t}-1} $$ Now we're able to answer Part A: $$ X(20)=\frac{150 e^{0.0226656*20}-150}{\frac{5}{2}e^{0.0226656*20}-1}=29.32 $$ Part B: $$ \lim_{x \to \infty} f(x) = \frac{150 e^{0.0226656t}-150}{\frac{5}{2}e^{0.0226656t}-1} = 60g $$ <- This would be the maximum amount of C possible. And finally Part C: Chemical A remaining after a long time: $$ A = A_0-\frac{M}{M+N}(X) = 40-\frac{2}{3}(60) = 0 g $$ And Chemical B after a long time: $$ B = B_0-\frac{N}{M+N}(X) = 50-\frac{1}{3}(60) = 30 g $$