The elementary irreversible gas-phase catalytic reaction A+B→kC+D
is to be carried out in a moving-bed reactor at constant temperature. The reactor contains 5 kg of catalyst. The feed is stoichiometric in A and B. The entering concentration of A is 0.2 mol/dm³. The catalyst decay law is zero order with k_D = 0.2 s^–1 and k = 1.0 dm6/(mol · kg-cat ·s) and the volumetric flow rate is ν0 = 1 dm³/s.
a. What conversion will be achieved for a catalyst feed rate of 0.5 kg/s?
b. Sketch the catalyst activity as a function of catalyst weight (i.e., distance) down the reactor length for a catalyst feed rate of 0.5 kg/s.
c. What is the maximum conversion that could be achieved (i.e., at an infinite catalyst loading rate)?
d. What catalyst loading rate is necessary to achieve 40% conversion? (Ans: US = 1.5 kg/s)
e. At what catalyst loading rate (kg/s) will the catalyst activity be exactly zero at the exit of the reactor?
f. What does an activity of zero mean? Can catalyst activity be less than zero?
g. How would your answer in part (a) change if the catalyst and reactant were fed at opposite ends? Compare with part (a).
h. Now consider the reaction to be zero order with
i = 0.2 mol/kg-cat · min. The economics:

The economics:

• The product sells for $160 per gram mole.
• The cost of operating the bed is $10 per kilogram of catalyst exiting the bed.

What is the feed rate of solids (kg/min) that will give the maximum profit?