The
derivation of the Thiele Modulus (from Hill) begins with a
material balance on the catalyst
pore. For a first-order irreversible reaction in a straight cylindrical pore at steady state: {\pi} r^2\left ( -D_c \frac{dC}{dx} \right )_x = {\pi}r^2\left ( -D_c \frac{dC}{dx} \right )_{x+{\Delta}x}+\left ( 2{\pi}r{\Delta}x \right )\left ( k_1C \right ) where D_c is a
diffusivity constant, and k_1 is the
rate constant. Then, turning the equation into a
differential by dividing by {\Delta}x and taking the
limit as {\Delta}x approaches 0, D_c\left (\frac{d^2C}{dx^2} \right ) = \frac{2k_1C}{r} This differential equation with the following
boundary conditions: C=C_o \text{ at } x=0 and \frac{dC}{dx} = 0 \text{ at } x= L where the first boundary condition indicates a constant external
concentration on one end of the pore and the second boundary condition indicates that there is no
flow out of the other end of the pore. Plugging in these boundary conditions, we have \frac{d^2C}{d(x/L)^2} = \left (\frac{2k_1L^2}{rD_c} \right) C The
term on the right side multiplied by C represents the square of the Thiele Modulus, which we now see rises naturally out of the material balance. Then the Thiele modulus for a
first order reaction is: h^2_T=\frac{2k_1L^2}{rD_c} From this relation it is evident that with large values of h_T, the rate term dominates and the
reaction is fast, while slow diffusion limits the overall rate. Smaller values of the Thiele modulus represent slow reactions with fast diffusion. == Other forms ==