chemical reaction....

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April 4th 2011, 02:07 PM
slapmaxwell1

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...

http://latex.codecogs.com/gif.latex?...X)(\beta%20-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.
April 4th 2011, 03:21 PM
slapmaxwell1

ok my book has the setup...

http://latex.codecogs.com/gif.latex?...2X)(\150%20-X)

im not sure how it came up with the 120 and the 150?
April 4th 2011, 03:38 PM
TheEmptySet

Quote:

Originally Posted by slapmaxwell1

ok my book has the setup...

http://latex.codecogs.com/gif.latex?...2X)(\150%20-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!

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

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

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

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

Now we get

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

Again sorry for the brain cramp!
April 4th 2011, 06:27 PM
slapmaxwell1

Quote:

Originally Posted by TheEmptySet

You are correct that the above post is wrong, I had a serious brain cramp!

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

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

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

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

Now we get

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

Again sorry for the brain cramp!

ok so what happens to the 1/2? thanks in advance!
April 4th 2011, 07:06 PM
TheEmptySet

Quote:

Originally Posted by slapmaxwell1

ok so what happens to the 1/2? thanks in advance!

Don't forget the $k$

So you get

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

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

We can relabel and use a different arbitrary constant.