Question 6 of 6: n-Heptane / n-Octane Vapour–Liquid Equilibrium from Raoult’s Law

Question 6: n-Heptane / n-Octane Vapour–Liquid Equilibrium from Raoult’s Law (25 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Binary n-heptane ($A$, the more volatile component, normal boiling point 98.4 °C) and n-octane ($B$, normal boiling point 125.6 °C) at total pressure $P=101.325$ kPa. The table gives the pure-component vapour pressures $P_A^{\text{sat}}$, $P_B^{\text{sat}}$ at six temperatures spanning the two boiling points. Ideal liquid (Raoult) and ideal vapour (Dalton).

Find. The equilibrium $y$–$x$ relation (tabulated points and a closed-form relation) and the $x$–$y$ equilibrium diagram.

Approach. At each tabulated temperature the liquid is at its bubble point under $P$. Raoult plus Dalton gives $x_A$ from the pressure balance and $y_A$ from the partial-pressure ratio. The relative volatility $\alpha=P_A^{\text{sat}}/P_B^{\text{sat}}$ is nearly constant for this near-ideal paraffin pair, so an averaged $\alpha$ yields a single analytic equilibrium curve for plotting and stage calculations.

1. Raoult–Dalton relations. Ideal liquid: $p_A=x_AP_A^{\text{sat}}$, $p_B=(1-x_A)P_B^{\text{sat}}$. Ideal vapour: $y_A=p_A/P$. Because the partial pressures sum to $P$,$$x_A=\frac{P-P_B^{\text{sat}}}{P_A^{\text{sat}}-P_B^{\text{sat}}},\qquad y_A=\frac{x_AP_A^{\text{sat}}}{P}.$$
2. Worked point, 110 °C.$$x_A=\frac{101.325-64.528}{139.988-64.528}=\frac{36.797}{75.460}=0.488,\qquad y_A=\frac{0.488(139.988)}{101.325}=0.674,\qquad \alpha=\frac{139.988}{64.528}=2.17.$$
3. All tabulated points. Repeating at each temperature:

   $T$ (°C)	$x_A$	$y_A$	$\alpha=P_A^{\text{sat}}/P_B^{\text{sat}}$
   98.4	1.000	1.000	2.282
   105	0.656	0.811	2.254
   110	0.488	0.674	2.169
   115	0.311	0.492	2.139
   120	0.157	0.279	2.077
   125.6	0.000	0.000	2.026

   The end rows check the data: at 98.4 °C $P_A^{\text{sat}}=P$ gives pure heptane ($x=y=1$), and at 125.6 °C $P_B^{\text{sat}}=P$ gives pure octane.
4. Equilibrium relation. Since $y_A/x_A=\alpha\,(1-y_A)/(1-x_A)$, the ideal-system relation is$$y=\frac{\alpha x}{1+(\alpha-1)x}.$$$\alpha$ falls only from 2.28 to 2.03 across the column, so its geometric mean over the six temperatures, $\alpha_{\text{avg}}=2.16$, represents the system:$$\boxed{\,y=\frac{2.16\,x}{1+1.16\,x}\,}$$Check at $x=0.488$: $y=2.16(0.488)/[1+1.16(0.488)]=0.672$, against 0.674 from Raoult. The largest deviation from the tabulated points is 0.008 in $y$, at 120 °C.
5. Equilibrium diagram. Plot $y$ against $x$ on 0–1 axes with the $45^\circ$ line $y=x$ (Figure 6). The six Raoult points lie on the smooth constant-$\alpha$ curve, which bows well above the diagonal. That gap is what makes heptane/octane separable by distillation; a larger $\alpha$ would open it further and need fewer stages.