Question:3 Benzene, Toluene, Ethyl Benzene, and Xylene (BTEX) is the common cons
ID: 1017479 • Letter: Q
Question
Question:3 Benzene, Toluene, Ethyl Benzene, and Xylene (BTEX) is the common constituent of gasoline. Vapor pressures of the pure liquids are, 0.126, 0.0380, 0.0126, and 0.0117 atm. at 25oC. Assuming an equimolar mixture of these liquids obeys Raoult’s law, calculate the vapor pressure exerted by each chemical and the total vapor pressure exerted by mixture.
a. If the mixture of BTEX is in equilibrium with water, estimate the solubility of benzene in water if the Henry’s law constant for benzene at 25oC is 0.18 M/atm?
Explanation / Answer
Vapour pressure of pure benzene at 25oC=0.126 atm
Vapour pressure of pure Toluene at 25oC=0.0380 atm
Vapour pressure of pure Ethyl Benzene at 25oC=0.0126 atm
Vapour pressure of pure Xylene at 25oC= 0.0117 atm
According to Rault’s law, partial vapour pressure of a component in a mixture is equal to the vapour pressure of the pure component at that temperature multiplied by its mole fraction in the mixture.
In the present case, the partial vapour pressure of benzene (PB)=PoBXB
PoB is the partial pressure of pure benzene which is 0.126 atm.
XB is the mole fraction of benzene.
The partial vapour pressure of toluene (PT)=PoTXT
PoT is the partial pressure of pure toluene which is 0.0380 atm.
XT is the mole fraction of toluene.
The partial vapour pressure of Ethyl Benzene (PE)=PoEXE
PoE is the partial pressure of pure Ethyl Benzene which is 0.0126 atm.
XE is the mole fraction of Ethyl Benzene.
The partial vapour pressure of Xylene (PX)=PoXXX
PoX is the partial pressure of pure Xylene which is 0.0117 atm.
XX is the mole fraction of Xylene.
The total pressure of BTEX (PBTEX) is equal to the partial pressure exerted by each of the individual components.
That means PBTEX= PB+PT+PE+PX = PoBXB+PoTXT+PoEXE+PoXXX
Assuming X to be the number of moles of benzene and given that BTEX contain equimolar mixture of benzene, toluene, Ethyl Benzene, and Xylene. Therefore total number of moles is 4X.
XB=number of moles of benzene/total number of moles
=X/4X=1/4=0.25
Similarly XT=XE=XX=1/4=0.25
PBTEX=PoBXB+PoTXT+PoEXE+PoXXX
=0.126 x 0.25+0.0380x0.25+0.0126x0.25+0.0117x0.25
=0.0315+0.0095+0.00315+0.002925
Total pressure of BTEX (PBTEX)=0.04708 atm
The partial vapour pressure of benzene (PB)=PBXB=0.126 x 0.25=0.0315 atm
The partial vapour pressure of toluene (PT)= PoTXT=0.0380x0.25=0.0095 atm
The partial vapour pressure of Ethyl Benzene (PE)=PoEXE = 0.0126x0.25=0.00315 atm
The partial vapour pressure of Xylene (PX)=PoXXX=0.0117x0.25=0.002925 atm
Where C is the solubility of dissolved gas, k is hentry’s constant, Pgas is the pressure of gas.
In the present case, the partial pressure of benzene in the mixture of BTEX as calculated above is 0.0315 atm and it is given that hentry’s constant for benzene is 0.18 M/atm.
Knowing the above, the solubility of dissolved benzene can be calculated using hentry’s law equation.
The solubility of dissolved benzene (C)= kPB=0.18X0.0315=0.0567M
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