Butadiene sulfone (MW=118) can be produced by the following irreversible liquid-
ID: 952327 • Letter: B
Question
Butadiene sulfone (MW=118) can be produced by the following irreversible liquid-phase reaction at 190 degree F and 160 psia: Butadiene + SO_2 rightarrow Butadiene Sufone Pure SO_2 is fed to the reactor in one stream. Pure butadiene (MW=54) is fed separately into the reactor at a molar flow rate that is 25% more than that required to react with all of the SO_2, and 70% of the entering butadiene is converted into product. The density of the stream leaving the reactor is 42.2 lb_m/ft^3. For an SO_2 flow rate of 100 Ibmol/hr, determine the volume of the CSTR required to achieve the specified 70% conversion. The rate of reaction is described by r_reaction, butadiene = kr C butadiene c SO_2 where k_r = 4.44ft^3/l bmol hr. Would the addition of an inert (non-reacting) liquid to the butadiene feed increase the required reactor size, or decrease it? Why?Explanation / Answer
Butadiene + SO2 --> Butadiene sulfone
Let x lbmol/hr of SO2 enters the reactor.
Flow rate of Butadiene entering, Fo = 1.25 x lbmol/hr
Exit conditions at 70% conversion (XA = 0.7):
Butadiene sulfone flow rate = 0.7 * 1.25 x = 0.875 x lbmol/hr = 0.875 x * 118 lb/hr = 103.25 lb/hr
SO2 flow rate = (1 - 0.875) x = 0.125 x lbmol/hr = 0.125 x * 64 lb/hr = 8 lb/hr
Butadiene flow rate, F = (1.25 - 0.875) x = 0.375 x lbmol/hr = 0.375 x * 54 lb/hr = 20.25 lb/hr
rA = 4.44 * (0.125 x /VCSTR) * (0.375 x /VCSTR) ft3 lbmol/hr = 0.208 x2 / VCSTR2 lbmol/hr-ft3
VCSTR = Fo XA / rA (CSTR reaction rate)
= 1.25 x * 0.7 VCSTR2 / (0.208 x2) = 4.2 VCSTR2 / x
VCSTR = x / 4.2 ft3
a.
x = 100 lbmol/hr
VCSTR = 100 / 4.2 ft3 = 23.8 ft3
b.
VCSTR = Fo XA / rA
With the addition of an inert liquid, the concentration of butadiene and SO2 in the outlet stream will decrease and hence the rate of reaction rA.
This would require a reactor of larger size.
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