Propylbenzene (an alkylbenzene) is one of many constituents in gasoline. Shown b
ID: 891554 • Letter: P
Question
Propylbenzene (an alkylbenzene) is one of many constituents in gasoline. Shown below are some solubility data for propylbenzene at three different temperatures.
Propylbenzene: Tm = -99.5 0C
Tb = 159.2 0C
Mw = 120.2
Temperature (0C)
Solubility (ppm)
10
53
20
60
30
62
At each temperature, calculate the solubility of propylbenzene in mol/L and mole fraction units.
At each temperature, calculate the partial molar excess free energy of solution and aqueous activity coefficient of propylbenzene in water. Use the equation:
RTlngwxw = RTlngpurexpure
where partial molar excess free energy = RTlngw. Assume ideal behavior in pure liquid.
Explanation / Answer
moles = g/molar mass
ppm = mg/L = 0.001 g/L
molarity = mol/L = (0.001)(ppm value)/L x molar mass
Say for 53 ppm = 0.053 g/L
moles = 0.053/120.2 = 0.00044 moles
molarity = 0.053 / 120.2 x 1L = 0.00044 mol/L
mole fraction at 10 oC = moles at 10 oC/total moles at all temperatures
Data table:
temperature (oC) Solubility (ppm) moles molarity (mol/L) mole fraction
10 53 0.00044 0.00044 0.301
20 60 0.00050 0.00050 0.342
30 62 0.00052 0.00052 0.356
Total moles = 0.00146 mols
Calculation for partial molar excess energy
For ideal solutions = acitivity coefficient = 1
So, partial molar excess free energy = RTlng/mole fraction
g = H-TS = -38.4 x 1000 - (159.2+273) x 288 = 86073.6 J/mol
Say we have 10 oC = 10 + 273 = 283 K
R = 8.314 J/K.mol
So,
For T = 10 oC, partial molar free energy = 8.314 x 283 ln 86073.6/0.301 = 29.56 kJ/mol
For T = 20 oC, partial molar free energy = 30293.88 kJ/mol
For T = 30 oC, partial molar free energy = 31226.74 kJ/mol
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