Question

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 25 grams of C is formed in 8 minutes. How much (in grams) is formed in 16 minutes

Answer 1

Answer: 45. 78 g is formed in 16 minutes.

Step-by-step explanation:

Let
`A_(o) =40 g` and
`B_(o) =50 g`

we know that
`\alpha =A_(o)(M+N)/(M)\, and \, \beta =B_(o)(M+N)/(N)`

According to the question to create
`x` 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.

`\alpha =40(2+1)/(2)=60\\\\\, and \, \beta =50(2+1)/(1)=150`

Now, the differential equation must be:
`(dX)/(dt)=k\left ( \alpha -X \right )\left ( \beta -X \right )`

I separate the variable and solve the equation

`\int (dx)/(\left ( 60-x \right )\left ( 150-x \right ))=\int k\, dt`

`ln(150-x)/(60-x)=90kt+C_(1)`

By using
`X(0) =0`,

`(150-x)/(60-x)=Ce^{90k_(o)},C=(5)/(2)`

and using
`X (8)=25` and solving for the value of 'k'

`\Rightarrow (150-25)/(60-25)=(5)/(2)e^(450k)\\\\\Rightarrow 3.6* (2)/(5)=e^(450k)\\\\\Rightarrow 1.4=e^(450k)`

Taking 'ln' both side, we get

`\Rightarrow k=7.4716* 10^(-4)`

We obtain:
`X(t)=(60Ce^(90kt)-150)/(Ce^(90kt)-1)`

Now, for the 16 min

`\Rightarrow X(16)=(150e^(1.07559)-150)/(2.5e^(1.0759)-1)`

`\Rightarrow X(16)=(439.7583-150)/(7.3293-1)\\\\\Rightarrow X(16)=(289.7583)/(6.3293)`

`\Rightarrow X(16)=45.78 g`