Chemical Reaction In a chemical reaction, one unit of compound Y and one unit of

Chapter 8, Problem 50

(choose chapter or problem)

In a chemical reaction, one unit of compound Y and one unit of compound Z are converted into a single unit of compound X. Let  \(x\)  be the amount of compound X formed. 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  \(t \rightarrow \infty\)  for (1)  \(y_{0}<z_{0}\),  (2)  \(y_{0}>z_{0}\), and (3)  \(y_{0}=z_{0}\).

Text Transcription:

x

\(dx/ dt = k(y_{0} - x)(z_{0} - x)

y_0  

z_0

int 1 / (y_0 - x)(z_0 - x) dx = int k d t

t

t rightarrow infty

y_0 < z_0

y_0 > z_0

y_0 = z_0