Science:Math Exam Resources/Courses/MATH101/April 2009/Question 08 (b)

MATH101 April 2009

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Question 08 (b)

Consider the chemical reaction $\displaystyle A+B\rightarrow C+D.$ Suppose at time t = 0 sec the concentration of chemical A is 0.1 mol/L, the concentration of chemical B is 0.2 mol/L, and the concentrations of chemicals C and D are both 0. For t ≥ 0, let x(t) be the concentration of chemical D in mol/L. It can be shown that x(t) is the solution to the initial-value problem ${\frac {dx}{dt}}=k(0.1-x)(0.2-x),\quad x(0)=0,$ where k is a positive constant whole value can be determined by experiment. (b) What value does the concentration of chemical D approach as t approaches $\infty$?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Use the previous part to get the equation. Then take the limit. You may want to factor out a $e^{\tfrac {kt}{10}}$ from the numerator and denominator.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. From the previous part, we know that $\displaystyle x(t)={\frac {0.2e^{\tfrac {kt}{10}}-0.2}{2e^{\tfrac {kt}{10}}-1}}$ taking the limit as t tends to infinity yields $\lim _{t\to \infty }{\frac {0.2e^{\tfrac {kt}{10}}-0.2}{2e^{\tfrac {kt}{10}}-1}}=\lim _{t\to \infty }{\frac {0.2-0.2e^{\tfrac {-kt}{10}}}{2-e^{\tfrac {-kt}{10}}}}={\frac {0.2}{2}}={\frac {1}{10}}$ Thus, the concentration tends to ${\tfrac {1}{10}}=0.1$ as t tends to infinity.

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