Q$1.$ Ventilation gases from underground coal mines contain methane. In the development

stages of a mine, methane concentration in ventilation gases can be high and this is a major

risk with explosions possible. Unfortunately, there are many mine disasters where this has

occurred. During the operational stage of a mine, the ventilation should be sufficient to keep

the $\frac{v}{v}\%$ of methane below explosive limits $(5-15\frac{v}{v}\%)$ and the mine workers safe from

the risk of methane gas explosions. Even though the $\frac{v}{v}\%$ of methane in ventilation streams is

low $($typically $1\frac{v}{v}\%$ or less$),$ it is a major contributor to methane emissions and greenhouse

emissions more broadly. One way to reduce greenhouse gas emissions is catalytic oxidation of

methane $($remember greenhouse potential of methane is much greater than carbon dioxide$).$

$C{H}_4(g)+2O_2(g)C{O}_2(g)+2H_{2}O(g)$

Why catalytic oxidation and not simply burning the methane? Because methane is below the

flammability limit of $5\frac{v}{v}\%$ so will not burn $($same reason it is safe from explosions$).$ This

reaction is irreversible and, when the concentration of methane is very low, behaves like a

first order reaction with respect to methane on the catalyst surface.

A catalyst is being tested for catalytic oxidation of methane. The catalyst particle is porous

with a diameter of $6mm,$ a solid density of $2\mathrm{.}2\frac{g}{c}{m}^3,$ a surface area of $200\frac{m^2}{g},$ and a

porosity of $0\mathrm{.}4.$ This catalyst is packed into a bed and has a bed voidage of $0\mathrm{.}35.$ The bed is fed

with $0\mathrm{.}5\frac{v}{v}\%$ of methane with the rest being air.

a$)$ If the conversion of methane is $98\mathrm{.}9\%$ with a residence time of $2$ seconds, what is the

effective rate constant for this reaction on a bed volume basis, in ${?}^{-1})?$

b$)$ What is the effective rate constant on a mass of catalyst basis, $k_e^{\text{'}}($in L$/$kg$.$s$),$ and on a

catalyst surface area basis, $k_e^{\text{'}\text{'}}($in $\frac{L}{m^2}.s)?$

c$)$ When the flowrate is increased by a factor of $10,$ conversion drops to $40\%.$ If external

mass transfer resistance is negligible at this flowrate, and other assumptions from part

$($a$)$ still hold, what is the surface rate constant, $k_s^{\text{'}\text{'}}($in $\frac{L}{m^2}.s)?$

d$)$ What is the mass transfer coefficient at the conditions in part$($a$),k_m($in $\frac{m}{s})?$ You can

use the result from the lecture notes:

$k_e=\frac{k_s^{\text{'}\text{'}}{k}_m{S}_g{}_p(1-{}_b)}{k_m+\frac{1}{3}{k}_s^{\text{'}\text{'}}{S}_g{}_p{r}_p}$