Question $1[35$ Marks$]$

Hydrodesulfurization is a process used to remove sulfur from hydrocarbons, such as crude oil,

to reduce pollution and improve the quality of the end$-$product. Packed bed reactors can be

used for hydrodesulfurization using solid catalysts such as cobalt$-$molybdenum or nickel.

molybdenum. A typical hydrodesulfurization gas phase reaction is represented in Reaction

$(1).$

$R-SH+H_{2}RH+H_{2}S$

Where, $R-SH$ represents a sulfur$-$containing organic molecule, $RH$ represents the improved

end$-$product, and $H_{2}S$ is hydrogen sulfide. Reaction $(1)$ is to be carried out isothermally at $350$

$C$ and $50$atm in a packed bed reactor. The reaction is irreversible, and a $2^{nd}$ order reaction

with respect to R$-$SH and zero order with respect to hydrogen. The mole percentage of R$-$SH

in the feed to the reactor is $20\%$ and the entering feed flowrate is $5\frac{L}{s}.$ If $k=1\mathrm{.}600$

$d\frac{m^6}{\text{m}\text{o}\text{l}\text{.}g.\text{c}\text{a}\text{t}\text{.}s}$ at $800K$ and the reaction has an activation energy of $130K\frac{J}{m}ol$ :

$1\mathrm{.}1.$ Determine the weight of nickel$-$molybdenum catalyst required to convert $80\%$ of R$-$

SH$.$ Neglect any pressure drop and write a full logical solution starting with the mole

balance. $[25]$

$1\mathrm{.}2.$ If you were to consider pressure drop in the above calculation will the weight of

catalyst required be lesser than, equal to or greater than the value obtained in Part $1\mathrm{.}1?$

Give a short reason for your answer and use appropriate equation$($s$)$ to support your

explanation. $[10]$