In a chemical reaction, one unit of compound $$ y $$ and one unit of compound $$ Z $$ are comvertes into a singlie umit of compound $$ x $$. Let $$ x $$ be the amount of compound $$ x $$ tonmed. The rate of and $$ Z $$. From this equation you obtain the following. $$ \int \frac{1}{\left(y_{0}-x\right)\left(z_{0}-x\right)} d x=\int t d t $$ (a) Solve for $$ * $$ as a function of $$ t $$. $$ x=\square $$ (b) Use the result of part (a) to find $$ x $$ as $$ t \rightarrow \infty $$ if $$ y_{0}z_{50} $$ and $$ y_{0}=z_{0} $$. (4) $$ \begin{array}{l} Y_{0}+t_{0} \ \lim _{x \rightarrow-\infty} x=\square \end{array} $$ (ii) $$ \begin{array}{l} y_{0}z_{0} \ x-m=\square \end{array} $$ (iv) $$ \begin{array}{l} y_{0}=2_{0} \ -\infty x=\square=\square \end{array} $$

Question

In a chemical reaction, one unit of compound $$ y $$ and one unit of compound $$ Z $$ are comvertes into a singlie umit of compound $$ x $$. Let $$ x $$ be the amount of compound $$ x $$ tonmed. The rate of and $$ Z $$. From this equation you obtain the following. $$ \int \frac{1}{\left(y_{0}-x\right)\left(z_{0}-x\right)} d x=\int t d t $$ (a) Solve for $$ * $$ as a function of $$ t $$. $$ x=\square $$ (b) Use the result of part (a) to find $$ x $$ as $$ t \rightarrow \infty $$ if $$ y_{0}z_{50} $$ and $$ y_{0}=z_{0} $$. (4) $$ \begin{array}{l} Y_{0}+t_{0} \ \lim _{x \rightarrow-\infty} x=\square \end{array} $$ (ii) $$ \begin{array}{l} y_{0}>z_{0} \ x-m=\square \end{array} $$ (iv) $$ \begin{array}{l} y_{0}=2_{0} \ -\infty x=\square=\square \end{array} $$

UpStudy AI · Accepted Answer

To solve the problem, we will follow the steps outlined in the prompt. ### Step 1: Extract Known Conditions We have the integral equation: $$ \int \frac{1}{(y_{0}-x)(z_{0}-x)} dx = \int t dt $$ We need to solve for $$ x $$ as a function of $$ t $$ and analyze the limits as $$ t \rightarrow \infty $$ under different conditions for $$ y_{0} $$ and $$ z_{0} $$. ### Step 2: Solve the Integral Equation We will first solve the left-hand side integral: $$ \int \frac{1}{(y_{0}-x)(z_{0}-x)} dx $$ This can be solved using partial fraction decomposition. We can express: $$ \frac{1}{(y_{0}-x)(z_{0}-x)} = \frac{A}{y_{0}-x} + \frac{B}{z_{0}-x} $$ Multiplying through by the denominator $$(y_{0}-x)(z_{0}-x)$$ gives: $$ 1 = A(z_{0}-x) + B(y_{0}-x) $$ Setting $$ x = y_{0} $$ gives $$ 1 = A(z_{0}-y_{0}) $$ and setting $$ x = z_{0} $$ gives $$ 1 = B(y_{0}-z_{0}) $$. Solving these equations for $$ A $$ and $$ B $$: $$ A = \frac{1}{z_{0}-y_{0}}, \quad B = \frac{1}{y_{0}-z_{0}} $$ Thus, we can rewrite the integral: $$ \int \left( \frac{1}{z_{0}-y_{0}} \cdot \frac{1}{y_{0}-x} - \frac{1}{z_{0}-y_{0}} \cdot \frac{1}{z_{0}-x} \right) dx $$ This leads to: $$ \frac{1}{z_{0}-y_{0}} \left( -\ln|y_{0}-x| + \ln|z_{0}-x| \right) = \frac{1}{z_{0}-y_{0}} \ln \left| \frac{z_{0}-x}{y_{0}-x} \right| $$ Now, we can integrate the right-hand side: $$ \int t dt = \frac{t^2}{2} + C $$ Setting both integrals equal gives: $$ \frac{1}{z_{0}-y_{0}} \ln \left| \frac{z_{0}-x}{y_{0}-x} \right| = \frac{t^2}{2} + C $$ ### Step 3: Solve for $$ x $$ as a Function of $$ t $$ To isolate $$ x $$, we can exponentiate both sides: $$ \left| \frac{z_{0}-x}{y_{0}-x} \right| = e^{(z_{0}-y_{0})\left(\frac{t^2}{2} + C\right)} $$ This leads to: $$ \frac{z_{0}-x}{y_{0}-x} = K e^{(z_{0}-y_{0})\frac{t^2}{2}} $$ where $$ K = e^{(z_{0}-y_{0})C} $$. Now, we can solve for $$ x $$: $$ z_{0} - x = K e^{(z_{0}-y_{0})\frac{t^2}{2}}(y_{0} - x) $$ Rearranging gives: $$ z_{0} - x + K e^{(z_{0}-y_{0})\frac{t^2}{2}} x = K e^{(z_{0}-y_{0})\frac{t^2}{2}} y_{0} $$ Factoring out $$ x $$: $$ x(1 + K e^{(z_{0}-y_{0})\frac{t^2}{2}}) = z_{0} - K e^{(z_{0}-y_{0})\frac{t^2}{2}} y_{0} $$ Thus, we have: $$ x = \frac{z_{0} - K e^{(z_{0}-y_{0})\frac{t^2}{2}} y_{0}}{1 + K e^{(z_{0}-y_{0})\frac{t^2}{2}}} $$ ### Step 4: Analyze Limits as $$ t \rightarrow \infty $$ Now we will analyze the limits for different cases of $$ y_{0} $$ and $$ z_{0} $$. 1. **If $$ y_{0} < z_{0} $$**: As $$ t \rightarrow \infty $$, $$ e^{(z_{0}-y_{0})\frac{t^2}{2}} \rightarrow \infty $$: $$ x \rightarrow \frac{z_{0}}{K e^{(z_{0}-y_{0})\frac{t^2}{2}}} \rightarrow 0 $$ 2. **If $$ y_{0} > z_{0} $$**: Similarly, as $$ t \rightarrow \infty $$: $$ x \rightarrow \frac{z_{0}}{K e^{(z_{0}-y_{0})\frac{t^2}{2}}} \rightarrow z_{0} $$ 3. **If $$ y_{0} = z_{0} $$**: The equation becomes indeterminate, but intuitively: $$ x \rightarrow y_{0} \text{ or } z_{0} $$ ### Final Answers (a) The function $$ x $$ as a function of $$ t $$ is: $$ x = \frac{z_{0} - K e^{(z_{0}-y_{0})\frac{t^2}{2}} y_{0}}{1 + K e^{(z_{0}-y_{0})\frac{t^2}{2}}} $$ (b) The limits as $$ t \rightarrow \infty $$ are: - If $$ y_{0} < z_{0} $$: $$ x \rightarrow 0 $$ - If $$ y_{0} > z_{0} $$: $$ x \rightarrow z_{0} $$ - If $$ y_{0} = z_{0} $$: $$ x \rightarrow y_{0} $$ or $$ z_{0} $$ (indeterminate). (a) $$ x = \frac{z_{0} - K e^{(z_{0}-y_{0})\frac{t^2}{2}} y_{0}}{1 + K e^{(z_{0}-y_{0})\frac{t^2}{2}}} $$ (b) - If $$ y_{0} < z_{0} $$, then $$ x \rightarrow 0 $$ as $$ t \rightarrow \infty $$. - If $$ y_{0} > z_{0} $$, then $$ x \rightarrow z_{0} $$ as $$ t \rightarrow \infty $$. - If $$ y_{0} = z_{0} $$, then $$ x \rightarrow y_{0} $$ or $$ z_{0} $$ as $$ t \rightarrow \infty $$.

Solución de inteligencia artificial de Upstudy

Responder

Solución

The Deep Dive

preguntas relacionadas

Latest Chemistry Questions