- #1
Dustinsfl
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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?
dx/dt = k*(50-\frac{2}{5}*x)*(80-\frac{3}{5}*x)
After separation and solving for partial fractions, I obtain:
\int\frac{1}{10-2*x} - \frac{3}{2}\int\frac{1}{16-3*x} = k*t+c
Which then yields:
\frac{16-3*x}{10-2*x} = C*e^{2*k*t}
C=8/5
k=\frac{ln(71/76)}{20}
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.
dx/dt = k*(50-\frac{2}{5}*x)*(80-\frac{3}{5}*x)
After separation and solving for partial fractions, I obtain:
\int\frac{1}{10-2*x} - \frac{3}{2}\int\frac{1}{16-3*x} = k*t+c
Which then yields:
\frac{16-3*x}{10-2*x} = C*e^{2*k*t}
C=8/5
k=\frac{ln(71/76)}{20}
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.
The equlibibrium is all towards product since there is no mention of back reaction in the kinetic laws given. 50g of A reacting in a 2:3 ratio with B will after a long time give you 125g of C leaving 0 A and 5g of B it seems to me.