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

MATH101 April 2009

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Question 08 (a)

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. (a) Solve the initial-value problem to find x(t) explicitly.

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Hint
Start off by cross multiplying. Then integrate (you may want to use partial fractions to integrate). I hope you ate your Wheaties because this is a long one...

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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. Cross multiplying our equation, we have that ${\frac {1}{(0.1-x)(0.2-x)}}\,dx=k\,dt$ Now we integrate. The right hand side integrates to $\int k\,dt=kt$ As for the left hand side, we integrate by partial fractions. ${\begin{aligned}{\frac {1}{(0.1-x)(0.2-x)}}&={\frac {A}{(0.1-x)}}+{\frac {B}{(0.2-x)}}\\&={\frac {A(0.2-x)+B(0.1-x)}{(0.1-x)(0.2-x)}}\\&={\frac {(-A-B)x+A(0.2)+B(0.1)}{(0.1-x)(0.2-x)}}\end{aligned}}$ Now we can either compare coefficients or plug in values for x. I will choose to plug in values for x. The above tells us that $\displaystyle 1=A(0.2-x)+B(0.1-x)$ when $x=0.2$, we have that $\displaystyle 1=A(0.2-0.2)+B(0.1-0.2)=B(-0.1)$ and so $B=-10$. When $x=0.1$, we have $\displaystyle 1=A(0.2-0.1)+B(0.1-0.1)=A(0.1)$ and so $A=10$. Hence, we have ${\begin{aligned}\int {\frac {1}{(0.1-x)(0.2-x)}}\,dx&=\int {\frac {10}{(0.1-x)}}\,dx-\int {\frac {10}{(0.2-x)}}\,dx\\&=-10\ln |0.1-x|+10\ln |0.2-x|.\end{aligned}}$ (technically the above should add a constant but we can include this in a minute.) Hence $\int {\frac {1}{(0.1-x)(0.2-x)}}\,dx=\int k\,dt$ becomes $\displaystyle 10\ln |0.2-x|-10\ln |0.1-x|=kt+C$ Using the initial condition of $x(0)=0$, we see that our constant is $\displaystyle 10\ln |0.2|-10\ln |0.1|=C$ and simplifying, this becomes $\displaystyle C=10(\ln {\tfrac {0.2}{0.1}})=10\ln(2)$ Our equation simplifies to $\displaystyle 10\ln \left|{\frac {0.2-x}{0.1-x}}\right|=10\ln |0.2-x|-10\ln |0.1-x|=kt+10\ln(2)$ Exponentiating the far left and the far right hand sides of the above gives $\left|{\frac {0.2-x}{0.1-x}}\right|^{10}=2^{10}e^{kt}$ Taking the tenth root of both sides yields $\left|{\frac {0.2-x}{0.1-x}}\right|=2e^{\tfrac {kt}{10}}$ Removing the absolute values, we have ${\frac {0.2-x}{0.1-x}}=\pm 2e^{\tfrac {kt}{10}}$ Plugging in the initial condition of $x(0)=0$, we see that only the positive value above works. Hence ${\frac {0.2-x}{0.1-x}}=2e^{\tfrac {kt}{10}}$ Now, we cross multiply to see that $\displaystyle 0.2-x=2e^{\tfrac {kt}{10}}(0.1-x)=0.2e^{\tfrac {kt}{10}}-x(2e^{\tfrac {kt}{10}})$ Bringing the $x$ values to one side gives $\displaystyle x(2e^{\tfrac {kt}{10}})-x=0.2e^{\tfrac {kt}{10}}-0.2$ Lastly, isolating for $x$ gives $\displaystyle x={\frac {0.2e^{\tfrac {kt}{10}}-0.2}{2e^{\tfrac {kt}{10}}-1}}$ completing the question.

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