Butadiene is prepared by the gas phase catalytic dehydration of 1-butene, at 900

Question

Butadiene is prepared by the gas phase catalytic dehydration of 1-butene, at 900 k and 1 bar. The equilibrium constant K at 900 K is 0.242. C_4H_8 (g) doubleheadarrow C_4H_6 (g) + H_2 (g) In order to suppress side reactions, the butane is diluted with steam before it passes into the reactor. Estimate the conversion of 1-butene for a feed consisting of 10 moles of steam per mole of 1-butene. Find the conversion if the inerts were absent and side reactions are ignored. Find the total pressure that would be required to obtain the same conversion as in (a) if no inerts were present.

Explanation / Answer

Let x be the conversion

At equilibrium

C4H8= 1-x Steam =10    C4H6= x and H2=x

K= [ H2] [C4H6] / C4H8] [Steam]

0.242= x2/10*(1-x),   x2/ (1-x)= 2.42 , x2= 2.42- 2.42x x2+2.42x- 2.42= 0 =0.76

When interts were absent   x2/(1-x)= 0.242, x2= 0.242-0.242x , x2+0.242x-0.242 =0 x= 0.386

Since the partial pressure is 1, pressure isnot taken into account

Let the pressure be = P

Partial pressures

C4H8=(1-x)*P C4H6= x*P and H2= x*P

C4H8= 0.24P C4H6= 0.76P and H2= 0.76P

0.242= [ 0.76P] 0.76P]/ 0.24P

0.242= 0.76*0.76*P/ 0.24

P= 0.242*0.24/(0.76*0.76)= 0.1 atm

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