Problem 33 Suppose that a diffusing substance disappears in a chemical reaction and that the rate at which it disappears is proportional to the concentration $- k C$. Write down the Fick's second law in this case. Show what the equation becomes if one makes the substitution $C ( x, y, z, t )=C^{\prime} ( x, y, z, t ) e^{-k t}$ ?

Options:
A. $$
{\frac{\partial C} {\partial t}}=D {\bigg(} {\frac{\partial^{2} C} {\partial x^{2}}}+{\frac{\partial^{2} C} {\partial y^{2}}} {\bigg)}-k^2C
$$
$$
{\frac{\partial C^{\prime}} {\partial t}}=C^{\prime} \left( {\frac{\partial^{2} C^{\prime}} {\partial z^{2}}}+{k D} \right)
$$
B. $$
{\frac{\partial C} {\partial t}}=D {\bigg(} {\frac{\partial^{2} C} {\partial z^{2}}} {\bigg)}
$$
$$
{\frac{\partial C^{\prime}} {\partial t}}=C \! \left( {\frac{\partial^{2} C^{\prime}} {\partial x^{2}}}+{\frac{\partial^{2} C^{\prime}} {\partial z^{2}}} \right)
$$
C. $$
{\frac{\partial C} {\partial t}}=D {\bigg(} \frac{\partial^{2} C} {\partial y^{2}}+{\frac{\partial^{2} C} {\partial z^{2}}} {\bigg)}
$$
$$
{\frac{\partial C^{\prime}} {\partial t}}=D^2 \left( {\frac{\partial^{2} C^{\prime}} {\partial x^{2}}} \right)
$$
D. $$
{\frac{\partial C} {\partial t}}=D^2 {\bigg(} {\frac{\partial^{2} C} {\partial x^{2}}}+{\frac{\partial^{2} C} {\partial y^{2}}}+{\frac{\partial^{2} C} {\partial z^{2}}} \bigg)
$$
$$
{\frac{\partial C^{\prime}} {\partial t}}=D \! \left( {\frac{\partial^{2} C^{\prime}} {\partial y^{2}}}-k C^{\prime} \right)
$$
E. $$
{\frac{\partial C} {\partial t}}=C^2 {\bigg(} {\frac{\partial^{2} C} {\partial x^{2}}}+{\frac{\partial^{2} C} {\partial y^{2}}} {\bigg)}-k D
$$
$$
{\frac{\partial C^{\prime}} {\partial t}}=D \! \left( {\frac{\partial^{2} C^{\prime}} {\partial z^{2}}} \right)-k
$$
F. $$
{\frac{\partial C} {\partial t}}=D {\bigg(} {\frac{\partial^{2} C} {\partial x^{2}}}+{\frac{\partial^{2} C} {\partial y^{2}}}+{\frac{\partial^{2} C} {\partial z^{2}}} {\bigg)}
$$
$$
{\frac{\partial C^{\prime}} {\partial t}}=D \left( {\frac{\partial^{2} C^{\prime}} {\partial x^{2}}}-k C^{\prime} \right)
$$
G. $$
{\frac{\partial C} {\partial t}}=k {\bigg(} {\frac{\partial^{2} C} {\partial x^{2}}} \bigg)-D C
$$
$$
{\frac{\partial C^{\prime}} {\partial t}}=D \! \left( {\frac{\partial^{2} C^{\prime}} {\partial y^{2}}}+{\frac{\partial^{2} C^{\prime}} {\partial z^{2}}} \right)
$$
H. $$
{\frac{\partial C} {\partial t}}=D {\bigg(} {\frac{\partial^{2} C} {\partial x^{2}}}+{\frac{\partial^{2} C} {\partial y^{2}}}+{\frac{\partial^{2} C} {\partial z^{2}}} \bigg)+k^2
$$
$$
{\frac{\partial C^{\prime}} {\partial t}}=C^{\prime} \! \left( {\frac{\partial^{2} C} {\partial x^{2}}}+{\frac{\partial^{2} C^{\prime}} {\partial y^{2}}} \right)
$$
I. $$
{\frac{\partial C} {\partial t}}=D {\bigg(} {\frac{\partial^{2} C} {\partial y^{2}}}+{\frac{\partial^{2} C} {\partial z^{2}}} {\bigg)}
$$
$$
{\frac{\partial C^{\prime}} {\partial t}}=D \! \left( {\frac{\partial^{2} C^{\prime}} {\partial x^{2}}}+k \right)
$$
J. $$
{\frac{\partial C} {\partial t}}=D {\bigg(} {\frac{\partial^{2} C} {\partial x^{2}}}+{\frac{\partial^{2} C} {\partial y^{2}}}+{\frac{\partial^{2} C} {\partial z^{2}}} {\bigg)}-k C
$$
$$
{\frac{\partial C^{\prime}} {\partial t}}=D \! \left( {\frac{\partial^{2} C^{\prime}} {\partial x^{2}}}+{\frac{\partial^{2} C^{\prime}} {\partial y^{2}}}+{\frac{\partial^{2} C^{\prime}} {\partial z^{2}}} \right)
$$
Answer: J