Respuesta :
Answer: 45. 78 g is formed in 16 minutes.
Explanation:
Let [tex]A_{o} =40 g[/tex] and [tex]B_{o} =50 g[/tex]
we know that [tex]\alpha =A_{o}\frac{M+N}{M}\, and \, \beta =B_{o}\frac{M+N}{N}[/tex]
According to the question to create [tex]x[/tex] part of the chemical C we will need 2 part of A and one part of B.
Therefore, M = 2 and N = 1.
Now, we can easily solve for the value of α and β.
[tex]\alpha =40\frac{2+1}{2}=60\\\\\, and \, \beta =50\frac{2+1}{1}=150[/tex]
Now, the differential equation must be: [tex]\frac{dX}{dt}=k\left ( \alpha -X \right )\left ( \beta -X \right )[/tex]
I separate the variable and solve the equation
[tex]\int \frac{dx}{\left ( 60-x \right )\left ( 150-x \right )}=\int k\, dt[/tex]
[tex]ln\frac{150-x}{60-x}=90kt+C_{1}[/tex]
By using [tex]X(0) =0[/tex],
[tex]\frac{150-x}{60-x}=Ce^{90k_{o}},C=\frac{5}{2}[/tex]
and using [tex]X (8)=25[/tex] and solving for the value of 'k'
[tex]\Rightarrow \frac{150-25}{60-25}=\frac{5}{2}e^{450k}\\\\\Rightarrow 3.6\times \frac{2}{5}=e^{450k}\\\\\Rightarrow 1.4=e^{450k}[/tex]
Taking 'ln' both side, we get
[tex]\Rightarrow k=7.4716\times 10^{-4}[/tex]
We obtain:[tex]X(t)=\frac{60Ce^{90kt}-150}{Ce^{90kt}-1}[/tex]
Now, for the 16 min
[tex]\Rightarrow X(16)=\frac{150e^{1.07559}-150}{2.5e^{1.0759}-1}[/tex]
[tex]\Rightarrow X(16)=\frac{439.7583-150}{7.3293-1}\\\\\Rightarrow X(16)=\frac{289.7583}{6.3293}[/tex]
[tex]\Rightarrow X(16)=45.78 g[/tex]
