Question

Two linear hydrocarbons, Hexane (C6H14) and Heptane (C7H16), form pretty much an ideal solution at any composition. A solution is made at 25°C that contains 463.96 g of Hexane in 667.71 g Heptane: Characterise the vapour above this solution, and answer, What is the mole fraction of Hexane in the vapour?

Answer 1

y = 0.92

Step-by-step explanation:

According to this problem, we have an ideal solution of hexane and heptane. This solution was made at 25 °C. If this is an ideal solution, this means that the solution has a vapour above it, so the first thing we need to know is the pressure of hexane and heptane at 25 °C. These pressures are reported and tabuled in differents literatures and handbook.

P₁ (Hexane) = 0.1989 atm

P₂ (Heptane) = 0.0139 atm

Now, to calcule the mole fraction of hexane in the vapour of this solution we need to apply the fugacity concept to the hexane, which is the following expression:

fv hex = fl hex (1)

This means that the fugacity of hexane in vapour would be the same in the liquid state. This concept can then be relationed to the mole fraction. in the liquid phase we have:

fl hex = x * P₁

and in the vapour, the mole fraction is relationed to the total pressure of the solution so:

fv hex = y * Pt

Where "y" would be the mole fraction of hexane in the vapour.

Using these two expression into (1) we have:

y * Pt = x₁ * P₁ (2)

And then, we can solve for y:

y = x * P₁/Pt (3)

So, all we have to do now is calculate the total pressure, and the mole fraction of hexane.

To get the mole fraction we need the moles of each reactant:

n₁ = 463.96 / 86 = 5.395 moles of hexane

n₂ = 667.71 / 100 = 6.677 moles of heptane

The mole fraction is:

x₁ = n₁ / n₁ + n₂

x₁ = 5.395 / (5.395+6.677) = 0.447

x₂ = 1 - 0.447 = 0.553

We have the mole fraction of hexane, let's calculate the total pressure:

Pt = x₁P₁ + x₂P₂

Pt = (0.447 * 0.1989) + (0.553 * 0.0.0139) = 0.0966 atm

Finally, let's use expression (3) to get the mole fraction in the vapour:

y = 0.447 * 0.1989/0.0966

y = 0.92

Answer 2

`y_{C_6H_(14)}=0.92`

Step-by-step explanation:

Hello,

At first, we compute liquid-phase molar fractions:

`n_{C_6H_(14)}=463.96 g*(1mol)/(86g) =5.3949molC_6H_(14)\\n_{C_7H_(16)}=667.71 g*(1mol)/(100g) =6.6771molC_7H_(16)\\x_{C_6H_(14)}=(5.3949)/(5.3949+6.6771) =0.447\\x_{C_7H_(16)}=1-x_{C_6H_(14)}=0.553`

Now, by means of the fugacity concept, for hexane, for instance, we have:

`f_{C_6H_(14)}^V=f_{C_6H_(14)}^L\\y_{C_6H_(14)}p_T=x_{C_6H_(14)}p_{C_6H_(14)}`

In this manner, at 25 °C the vapor pressure of hexane and heptane are 0.198946 atm and 0.013912 atm repectively, thus, the total pressure is:

`p_T=x_{C_6H_(14)}p_{C_6H_(14)}+x_{C_7H_(16)}p_{C_7H_(16)}\\p_T=0.447*0.198946 atm +0.553*0.013912 atm=0.096622atm`

Finally, from the hexane's fugacity equation, we find its mole fraction in the vapour as:

`y_{C_6H_(14)}=\frac{x_{C_6H_(14)}p_{C_6H_(14)}}{p_T}=(0.447*0.198946 atm)/(0.096622atm) \\y_{C_6H_(14)}=0.92`

Best regards.