Question

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

Answer 1

Answer:

$C_{A}=\frac{e^{\sqrt{\frac{10}{3} }z}}{10}$

Step-by-step explanation:

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

$D \frac{d^{2}C_{A}}{d^{2}z} -kC_{A}=0$

We divide everything by D, so we have:

$\frac{d^{2}C_{A}}{d^{2}z} -\frac{k}{D} C_{A}=0$

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

$\frac{10}{3} =\frac{k}{D}$

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

$y''-\frac{10}{3}y=0$

Knowing that $y=C_{A}$

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

$y=C_{1}e^{\sqrt{\frac{10}{3} } z} + C_{2}e^{-\sqrt{\frac{10}{3} } z}$

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

$z=0; y=0.1\\z=z; y=0$

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

$0.1=C_{1}e^{\sqrt{\frac{10}{3} }0} + C_{2}e^{-\sqrt{\frac{10}{3} }0}\\0.1=C_{1} + C_{2}$

And with the other values:

$0=C_{1}e^{\sqrt{\frac{10}{3} }z} + C_{2}e^{-\sqrt{\frac{10}{3} }z}$

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

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

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

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

$C_{A}=\frac{e^{\sqrt{\frac{10}{3} }z}}{10}$