Two chemicals, A and B, are combined, forming chemical C. The rate of the reaction is jointly proportional to the amounts of A and B not yet converted to C. Initially, there are 50 grams of A and 80 grams of B, and, during the reaction, for each two grams of A used up in the conversion, there are three grams of B used up. An experiment shows that 100 grams of C are produced in the first 10 minutes. After a long period of time, how much of A and B remains, and how much of C has been reproduced?(adsbygoogle = window.adsbygoogle || []).push({});

dx/dt = k*(50-[tex]\frac{2}{5}[/tex]*x)*(80-[tex]\frac{3}{5}[/tex]*x)

After separation and solving for partial fractions, I obtain:

[tex]\int\frac{1}{10-2*x}[/tex] - [tex]\frac{3}{2}[/tex][tex]\int\frac{1}{16-3*x}[/tex] = k*t+c

Which then yields:

[tex]\frac{16-3*x}{10-2*x}[/tex] = C*e[tex]^{2*k*t}[/tex]

C=8/5

k=[tex]\frac{ln(71/76)}{20}[/tex]

However, something is wrong with my final equation solved for x(t) due to x(0) doesn't = 0 and x(10) doesn't = 100.

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# Homework Help: DE Chemical Combinations

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