ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Chavi on June 30, 2010, 03:17:50 pm

Title: Extended response DE
Post by: Chavi on June 30, 2010, 03:17:50 pm: Hey, can I please get some help with setting up this DE?

2 Chemicals, A and B, are put together in a solution where they react to form a compound, X. The rate of increase of the mass, x kg, of X is proportional to the product of the masses of unreacted A and B present at time t minutes. It takes 1kg of A and 3kg of B to form 4kg of X. Initial 2kg of A and 3kg of B are put together in solution. 1kg of X forms in 1 min.

What do you do??
Title: Re: Extended response DE
Post by: Mao on July 01, 2010, 01:19:27 am: Mass ratio: A + 3B --> X

Thus,
$X = A_{reacted} = \frac{1}{3} B_{reacted}$

Thus, $A_{remaining} = A_i - A_{reacted} = A_i - X$

$B_{remaining} = B_i - B_{reacted} = B_i - 3X$

The mass of X changes with time, that is to say, X is a function of time. A_i and B_i are both constants.

From the description, we also know that $\frac{dX}{dt} = k A_{remaining} B_{remaining} = k (A_i - X)(B_i - 3X)$

Thus, for this application, $\frac{dX}{dt} = k (2-X)(3-3X)$ , where k is some positive number (rate of increase of X has to be positive)

$\implies \frac{dt}{dX} = \frac{1}{3k} \frac{1}{X^2 - 3X + 2} = \frac{1}{3k} \left( \frac{1}{X-2} - \frac{1}{X-1} \right)$

$\implies t = \frac{1}{3k} \log_e \left| \frac{X-2}{X-1} \right| + C$

Given that when t=0, X=0

$\implies C = -\frac{1}{3k} \log_e (2)$

$\implies t = \frac{1}{3k} \log_e \left| \frac{X-2}{2(X-1)} \right| = \frac{1}{3k} \log_e \left| \frac{1}{2}\left( 1-\frac{1}{X-1} \right) \right|$

$\implies X = \frac{1}{1-2e^{3kt}} + 1$

k is unknown, as the second boundary condition is pointless (1kg of X cannot form, as that will require ALL of B to react, when in reality, you'll observe asymptotic behaviour).
Title: Re: Extended response DE
Post by: Chavi on July 01, 2010, 07:11:16 pm: Thanks for the help Mao.
For some reason, in the answers they had:

$\frac{dx}{dt} = \frac{3k}{16} (8 - x)(4 -x)$
Title: Re: Extended response DE
Post by: Mao on July 02, 2010, 02:38:44 am: Ahh yes, of course. I didn't read the question properly.

The relationship is A + 3B --> 4C

The method is the same as before.