In a chemical reaction, one unit of compound Y and one unit of compound Z are converted

Chapter 8, Problem 68

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Chemical Reactions In a chemical reaction, one unit of compound Y and one unit of compound Z are converted into a single unit of compound X, x is the amount of compound X formed, and the rate of formation of X is proportional to the product of the amounts of unconverted compounds Y and Z. So, \(d x / d t=k\left(y_{0}-x\right)\left(z_{0}-x\right)\) where \(y_{0}\) and \(z_{0}\) are the initial amounts of compounds Y and Z. From this equation you obtain

\(\int \frac{1}{\left(y_{0}-x\right)\left(z_{0}-x\right)} d x=\int k \ d t \)

(a) Perform the two integrations and solve for x in terms of t.

(b) Use the result of part (a) to find x as \(t \rightarrow \infty\)  if  (1)  \(y_{0}<z_{0}\) , (2) \(y_{0}>z_{0}\) , and (3) \(y_{0}=z_{0}\).

Text Transcription:

dx/dt = k(y_0 - x)(z_0 - x)

y_0

z_0

t rightarrow infty

y_0 < z_0

y_0>z_0

y_0=z_0

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