8.(8) Consider a process for the partial oxidation of benzene to produce maleic

Question

8.(8) Consider a process for the partial oxidation of benzene to produce maleic anhydride (C4H203) using air There are two reactions to be concerned with: The first partially oxidizes the Benzene to produce maleic acid along with carbon dioxide and water. The second, undesired reaction, fully oxidizes the benzene. The process used is as follows: Benzene Purification: The benzene flows into a pressurized still with water at 23.2 psia. The liquid in the still is 80% benzene by mass. Energy is supplied to keep the mixture at its boiling point (180°F) resulting in a vapor mixture of pure water and pure benzene. The contaminates along with waste benzene flow out as a liquid 1) 2) Reaction: The vapor stream is mixed with enough air (T = 150°F, 1 atm, 50% relative humidity) to provide an 18/1 molar ratio of oxygen/benzene entering the reactor. At the exit to the reactor an analysis of the gasses (mole, dry basis) shows 0.76% maleic acid (C4H0.) and 80.0% nitrogen 3) Stripping: The exit gasses from the reactor flow into a water scrubber operating at 1 atm. Pure water enters the top of the column and the reactor outlet gas enters the bottom. The water "scrubs" out the maleic acid where it exits as a liquid at the bottom of the column with a concentration of (mole %) 12% C4H404, 88% water. Exiting the top of the column as a gas are all of the unreacted benzene, oxygen, carbon dioxide, and nitrogen. This gas stream exits at 142°F saturated with water 4) Drying: The liquid stream from the scrubber goes to a dryer where the water is driven off leaving pure maleic anhydride What is (use pound moles) a) The moles of benzene that are completely oxidized in the reactor per mole of benzene fed into the b) c) d) reactor? (i.e. the moles benzene used in the second reaction per mole fed) The pounds of water leaving the dryer per mole of benzene fed into the reactor? The composition of the gas leaving the scrubber (mole percent, wet basis)? The pounds of water added to the top of the scrubber per mole of benzene fed into the reactor?

Explanation / Answer

180 F = 92.5 C

At 92.5 C, Benzene saturation Vapor Pressure, Pb = 1.31 bar

At 92.5 C, Water saturation vapor Pressure, Pw = 0.74 bar

Still Pressure, P = 23.2 psia = 1.6 bar

Given, In still, Benzene liquid mass fraction =80%=0.8

Since Impurity is unknown, Considering only water and benzene in liquid phase to calculate benzene liquid mole fraction

Benzene liquid mole fraction = (Benzene liquid mass fraction/Benzene Molar mass)/(Benzene liquid mass fraction/Benzene Molar mass+(1-Benzene liquid mass fraction)/Water Molar mass)

Benzene Molar mass=78 g/mol

Water Molar mass=18g/mol

=>Benzene liquid mole fraction,xb=0.8/78/(.8/78+.2/18)=0.48

By Raoult's law, Benzene still vapor mole fraction, yb=xb*Pb/P=.48*1.31/1.6=0.39

=> Water still vapor mole fraction,yw = 1-Benzene still vapor mole fraction=1-.39=0.61

=> For 1 mole of benzene, mole of water in still vapor = 1/.39*.61=1.56

In reaction zone,

T=150 F= 65.5 C

Water saturation vapor pressure at 65.5 C,Ps = 0.24 bar

Given, Relative humidity, RH = 50%

Water content in air, mole/mol = RH*Ps/760=0.5*.24/760=.00015

1 mole of air contains 0.21 mole O2 and 0.21 mole N2

Given, for 1 mole of benzene in feed, 18 mole of O2 is entered

=> Moles of air in feed = 18/.21=85.71 moles

=> Moles of N2 in feed air = 85.71*.79=67.7 moles

=> Moles of water in feed air = 0.00015*85.71=0.012 moles

Desired reaction, C6H6+4.5O2=>C4H4O4+2CO2+H2O

Undesired reaction, C6H6+7.5O2=>6CO2+3H2O

Let Feed benzene converts x fraction in desired reaction and (1-x) fraction in undesired reaction

Dry basis content of reaction outlet consist of Nitrogen, Oxygen, CO2, C4H4O4

Hypothesis: Benzene reacting completely

=> Total dry content moles in reaction outlet = x+2*x+6*(1-x)+(18-x*4.5-(1-x)*7.5)+67.7=84.2

=> Total moles of Water in reaction outlet = x+3*x+.00015*85.71+1.56=4*x+1.57

Mole fraction of N2 in outlet = 67.7/84.4 = .8=80% = Given in question

Hence, Benzene reacting completely

Given, Mole fraction of C4H4O4 in reaction outlet = 0.76%

=>Mole fraction of C4H4O4 in reaction outlet = x/84.2 = 0.76/100

=>x=0.63

a) Moles of benzene oxidized completely per mole of benzene inlet = (1-x)=1-.63=0.37

c)

Stripping zone

In stripper, all gases other than C4H4O4 goes into Gas Phase i.e. N2, O2, CO2 goes into vapor space

Total Moles of N2+O2+CO2 in Stripper gas outlet = 2*x+6*(1-x)+(18-x*4.5-(1-x)*7.5)+67.7=83.57

142 F = 61.1 C

Vapor Pressure of Water at 61.1 C = 0.2 bar

Since, Outlet gas is saturated with water =>Moles of Water in Stripper Vapor outlet = (Vapor Pressure of Water at 61.1 C)/(Outlet gas total Pressure)*(otal Moles of N2+O2+CO2 in Stripper gas outlet)

Given, Outlet gas total Pressure=1 atm =~1bar

=>Moles of Water in Stripper Vapor outlet=0.2/1*83.57=16.71 mole

=>Total Moles in Stripper Vapor outlet=Moles of Water in Stripper Vapor outlet+Total Moles of N2+O2+CO2 in Stripper gas outlet =16.71+83.57=100.28 moles

=>In stripper outlet, On wet basis,

Mole fraction of N2 = Moles of N2/Total Moles = 67.7/100.28=67%

Mole fraction of O2 = Moles of O2/Total Moles = (18-x*4.5-(1-x)*7.5)/100.28=12%

Mole fraction of CO2 = Moles of CO2/Total Moles = 2*x+6*(1-x)/100.28=5%

Mole fraction of H2O = Moles of H2O/Total Moles = 16.71/100.28=16%

Stripper bottom consists of 12% C4H4O4 and 88% water,

Since whole C4H4O4 feeding in stripper going in bottom

=> Moles of Water in bottom = Moles of C4H4O4 in feed/.12*.88=x/.12*.88=.63/.12*.88=4.62 moles

d) Moles of water feeding into scrubber per mole of benzene= Moles of Water in bottom+Moles of Water in Stripper Vapor outlet-Moles of water in feed=4.62+16.71-(4*x+1.57)=4.62+16.71-(4*.63+1.57)=17.24 moles/mole of benzene

Pounds of water feeding into scrubber per mole of benzene=Moles of water feeding into scrubber per mole of benzene*18/453=17.24*18/453=0.68

Dryer

C4H4O4=>C4H2O3+H2O

=>Moles of C4H4O4 in dryer inlet = Moles of H2O produce in dryer

Moles of water leaving the dryer per mole of benzene = Moles of Water in dryer inlet + Moles of C4H4O4 in dryer inlet

=>Moles of water leaving the dryer per mole of benzene=4.62+0.63=5.25

Moles of water leaving the dryer per mole of benzene=Moles of water leaving the dryer per mole of benzene*Molar mass of water/453

=> Pounds of water leaving the dryer per mole of benzene=5.25*18/453=0.2

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