Question

Determine the mole fractions of each component in the vapor phase of the vapor in equilibrium with a 1:1 molar ratio of hexane (C6H14) and cyclohexane (C6H12). The equilibrium vapor pressures of hexane and cyclohexane are equal to 151.4 torr and 97.6 torr respectively

Answer 1

Answer: mole fraction of hexane = 0.61 and mole fraction of cyclohexane = 0.39

Step-by-step explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

`p_1=x_1p_1^0` and
`p_2=x_2P_2^0`

where, x = mole fraction

`p^0` = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

`p_(total)=p_1+p_2`
`p_(total)=x_Ap_A^0+x_BP_B^0`

`x_(hexane)=\frac{\text {moles of hexane}}{\text {moles of hexane+moles of cyclohexane}}=(1)/(1+1)=0.5`,

`x_(cyclohexane)=\frac{\text {moles of cyclohexane}}{\text {moles of hexane+moles of cyclohexane}}=(1)/(1+1)=0.5`,

`p_(hexane)^0=151.4torr`

`p_(cyclohexane)^0=97.6torr`

`p_(total)=0.5* 151.4torr+0.5* 97.6torr=124.5torr`

`y_(hexane)` = mole fraction of hexane in vapor phase
`y_(hexane)=(p_(hexane))/(p_(total))=(0.5* 151.4)/(124.5)=0.61`

`y_(cyclohexane)=1-y_(hexane)=1-0.61=0.39`

Thus the mole fraction of hexane in the vapor above the solution is 0.61 and mole fraction of cyclohexane in the vapor above the solution is 0.39