Final answer:
The student's question involves finding the mole fractions in the vapor and liquid phases for a hydrocarbon mixture at equilibrium using Raoult's Law and known vapor pressures for n-butane, n-pentane, and n-hexane.
Step-by-step explanation:
The question is centered around the concept of vapor-liquid equilibrium and the calculation of corresponding mole fractions in both phases for a system of hydrocarbons (n-butane, n-pentane, and n-hexane) at a given temperature and pressure. According to Raoult's Law, the partial pressure of a component in a mixture is equal to the mole fraction of the component in the liquid phase times its vapor pressure when pure.
For the given mixture at 120 deg F and 20 psia, with a liquid composition of 0.10 mole fraction n-butane, we utilize Raoult's Law and the known vapor pressures of each component to find the partial pressures. The total vapor pressure is the sum of these partial pressures. The composition of the vapor phase can then be found by dividing the partial pressure of each component by the total pressure, giving us their respective mole fractions in the vapor phase.
Using an example from a reference, suppose we have a mixture of heptane and octane forming an ideal solution at 20.0℃, with respective vapor pressures P1 = 0.0562 atm for pure heptane and P2 = 0.0145 atm for pure octane. If the solution is 0.300 mol heptane and 0.700 mol octane, using Raoult's Law, we can calculate the mole fraction of heptane vapor over the solution.
This approach applies similarly to the query on the composition of liquid and vapor for the n-butane, n-pentane, and n-hexane mixture. Note that due to the complexities of hydrocarbon behavior and the potential for non-ideal mixing, experimental data or activity coefficients may be required for actual industrial calculations.