1. ## chemical reaction....

ok im not sure how to set this problem up

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. ho much is formed in 20 minutes? what is the limiting amount of C after a long time?

i know im suppose to be using this formula...

$\frac{dX}{dt}=k(\alpha -X)(\beta -X)$

i also know that X(5)=10, the relationship between A and B is 2 to 1 or 2X to X, in jus not sure how to get the alpha or beta....thanks in advance.

2. ok my book has the setup...

$\frac{dX}{dt}=k(\120 -2X)(\150 -X)$

im not sure how it came up with the 120 and the 150?

3. Originally Posted by slapmaxwell1
ok my book has the setup...

$\frac{dX}{dt}=k(\120 -2X)(\150 -X)$

im not sure how it came up with the 120 and the 150?
You are correct that the above post is wrong, I had a serious brain cramp!

$\displaystyle \displaystyle \alpha=\frac{a(M+N)}{M}, \text{ and } \beta=\frac{b(M+N)}{N}$

Where $\displaystyle a$ is the intial amount of A
Where $\displaystyle b$ is the intial amount of B
M is the proportion of A and N is the proportion of B

$\displaystyle \displaystyle \alpha=\frac{a(M+N)}{M}=\frac{40(2+1)}{2}$

$\displaystyle \displaystyle \beta=\frac{b(M+N)}{N}=\frac{50(2+1)}{1}$

Now we get

$\displaystyle \displaystyle \left( \frac{40(2+1)}{2}-X\right) \left( \frac{50(2+1)-X}{1}\right)=\frac{1}{2}(120-2X)(150-X)$

Again sorry for the brain cramp!

4. Originally Posted by TheEmptySet
You are correct that the above post is wrong, I had a serious brain cramp!

$\displaystyle \displaystyle \alpha=\frac{a(M+N)}{M}, \text{ and } \beta=\frac{b(M+N)}{N}$

Where $\displaystyle a$ is the intial amount of A
Where $\displaystyle b$ is the intial amount of B
M is the proportion of A and N is the proportion of B

$\displaystyle \displaystyle \alpha=\frac{a(M+N)}{M}=\frac{40(2+1)}{2}$

$\displaystyle \displaystyle \beta=\frac{b(M+N)}{N}=\frac{50(2+1)}{1}$

Now we get

$\displaystyle \displaystyle \left( \frac{40(2+1)}{2}-X\right) \left( \frac{50(2+1)-X}{1}\right)=\frac{1}{2}(120-2X)(150-X)$

Again sorry for the brain cramp!
ok so what happens to the 1/2? thanks in advance!

5. Originally Posted by slapmaxwell1
ok so what happens to the 1/2? thanks in advance!
Don't forget the $\displaystyle k$

So you get

$\displaystyle \displaystyle \frac{dX}{dt}=k\left( \frac{1}{2}\right)(120-2X)(150-X)= \hat{k}(120-2X)(150-X)$

Since $\displaystyle k$ is determined by the $\displaystyle X(5)=10$

We can relabel and use a different arbitrary constant.