Question

Data possibly useful to all problems: 1m2 -1000 dm2, R-0.082 (L*atm)/(mole*K)= 8.314 3/(mol K)- 1.987 cal/(mol K) Problem 1 (8 points out of 25) The first order gas phase reaction: A> 2B with k'-0.3 mole/(kg-catalyst*min* atm) takes place in isothermal packed bed reactor. The feed, which is 75% in A and 25 % inert, enters the reactor at 400 K and total pressure of 10 atm with the total flow rate of 40 mole/min. If there is no pressure drop along the length of the packed bed reactor, calculate the weight of catalyst needed to produce 36 mole/min of product B. an

Answer 1

For a packed bed reactor,

the design equation is

$\frac{W}{F_{A_{0}}} = \int_{0}^{X_{a}} \frac{dX_{a}}{-r_{a}}$

No of moles of A entering = 0.75 * 40 = 30 moles/min

1 mole of A gives 2 moles of B

Hence 18 moles of A gives 36 moles of B

Which means 12 moles of A is exiting the reactor.

Conversion = 1- 12/30 =0.6

Hence the integral becomes,

$\int_{0}^{0.6}\frac{dX_{A}}{k *P_{A}}$

$=\int_{0}^{0.6}\frac{dX_{A}}{k *P_{A_{0}}(1-X_{A})}$

= $=\frac{1}{7.5*0.3} [-ln(1-X_{A})]_{0 } ^{0.6}$

= 0.22703

W = 0.22703 X 30 = 6.811 g of Catalyst