Compound A, with concentration CA, diffuses through a 4-cm long tube and reacts as it diffuses. The equation governing diffusion with reaction is given by D d 2CA dz2 − kCA = 0 (7) At one end of the tube, there is a large source of A at a concentration of CA=0.1M. At the other end of the tube there is an adsorbent material that quickly absorbs any A, making the concentration 0 M. If D =1.5×10−6 cm2/s and k =5×10−6 s −1 , what is the concentration of A as a function of distance in the tube

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Answer:

[tex]C_{A}=\frac{e^{\sqrt{\frac{10}{3} }z}}{10}[/tex]

Step-by-step explanation:

First you need to rearrange the original equation in order to have an equation with a general form:

[tex]D \frac{d^{2}C_{A}}{d^{2}z} -kC_{A}=0[/tex]

We divide everything by D, so we have:

[tex]\frac{d^{2}C_{A}}{d^{2}z} -\frac{k}{D} C_{A}=0[/tex]

We know that k and D are constants, so we can leave them as just one constant that gives as result:

[tex]\frac{10}{3} =\frac{k}{D}[/tex]

Then we proceed to change terms in the equation son it looks friendlier:

[tex]y''-\frac{10}{3}y=0[/tex]

Knowing that [tex]y=C_{A}[/tex]

Then we give the general solution for the differential equation of second order:

[tex]y=C_{1}e^{\sqrt{\frac{10}{3} } z} + C_{2}e^{-\sqrt{\frac{10}{3} } z}[/tex]

The following is knowing the frontier values so we can calculate the constants of the equation:

[tex]z=0; y=0.1\\z=z; y=0[/tex]

With these values, we can replace in the general solution to find the final equation:

[tex]0.1=C_{1}e^{\sqrt{\frac{10}{3} }0} + C_{2}e^{-\sqrt{\frac{10}{3} }0}\\0.1=C_{1} + C_{2}[/tex]

And with the other values:

[tex]0=C_{1}e^{\sqrt{\frac{10}{3} }z} + C_{2}e^{-\sqrt{\frac{10}{3} }z}[/tex]

Then we solve the equation system, as we have two unknowns, the constants, and two equations:

[tex]C_{1}=0.1-C_{2}\\0=(0.1-C_{2})e^{\sqrt{\frac{10}{3} }z} +C_{2}e^{-\sqrt{\frac{10}{3} }z}[/tex]

After solving both equations, we have;[tex]C_{A}=\frac{e^{\sqrt{\frac{10}{3} } z}}{10}((1-\frac{e^{\sqrt{\frac{10}{3} }z}}{{-e^{\sqrt{\frac{10}{3} }z}}+{e^{-\sqrt{\frac{10}{3} }z}}} ) + (\frac{e^{\sqrt{\frac{10}{3} }z}}{{-e^{\sqrt{\frac{10}{3} }z}}+{e^{-\sqrt{\frac{10}{3} }z}}}))[/tex]

The we see that we can cancel terms, so we have the final solution that is:

[tex]C_{A}=\frac{e^{\sqrt{\frac{10}{3} }z}}{10}[/tex]

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