Question A2 (25 marks). The liquid phase reaction 2A → B is catalysed and follows the rate law rm = KAMCA². Here, Am is the specific area of the catalyst, and

I'm is in mol-s-¹-(kg-cat)-¹. The process takes place in a packed bed reactor. a) Show that the design equation of the reactor is X= 2kAm Co 1+2kAm CAD/V Ao v where X is specified conversion, mc is the mass of catalyst in the reactor, v is feed flow rate. [9 marks] b) The catalyst slowly loses activity, due to sintering, abrasion and wash-off of the material. The sintering leads to linear drop of the specific area with time: Am = Amo(1-t/ts), where Amo is the initial specific area and ts is the characteristic time of the sintering process. The abrasion leads to a decrease of the mass of the catalyst, and is also linear with time: mc = moc(1-t/ta). By specification, the catalyst in the reactor has to be changed once its activity drops to 75% of its initial value. Find how often that is (i.e. design the schedule of catalyst refilling - the period to.75 between two changes). [8 marks] c) Since the reaction is 2nd order (i.e. the rate depends strongly on the concentration), an easy way to compensate for the loss of catalytic activity is by increasing the feed concentration of A with time. In order to stabilize the operation of the separation stages that follow the reactor, it is required that the conversion X is kept constant despite the loss of catalytic activity, by compensating the drop in mc and Am via a scheduled increase in the concentration CAO. Find what the schedule Cao(t) should be in order for X to remain constant. If CAO(t=0) = 0.5 M, what will CAO(t = to.75) be right before the change of catalyst? [8 marks]