4.10 – Practice Problem 1

= Question

A vessel contains a mixture of benzene, $C_{6}H_{6}$ , and toluene, $C_{7}H_{8}$ . At 373K the pressure in the vessel is 1.53 bar(abs). Using the data provided, and any others necessary, find the mass fractions of benzene in the liquid and vapour. Assume the liquids and vapours behave ideally. Recall the formula for Antoine's equation: $\log _{10}p^{*}=A-{\frac {B}{T+C}}$ . Where T is in Kelvin for this question.

Component			Antoine Constants
	A	B	C
Benzene	4.72583	1660.652	-1.461
Toluene	4.07827	1343.943	-53.773

Answer

From Antoine's equation, we can find the vapour partial pressure of toluene and benzene.

$p_{b}^{*}=10^{{\text{ }}A-{\frac {B}{T+C}}}=10^{{\text{ }}4.72583-{\frac {1660.652}{373-1.461}}}=1.8037{\text{ }}{\text{bar}}$

$p_{t}^{*}=10^{{\text{ }}A-{\frac {B}{T+C}}}=10^{{\text{ }}4.07827-{\frac {1343.943}{373-53.773}}}=0.7384{\text{ }}{\text{bar}}$

Now we can solve for the liquid mole fractions of the two substances with two equations and two unknowns using Raoult's law and the fact that the mole fractions must add to one:

$x_{b}+x_{t}=1$
$P=x_{b}\cdot p_{b}^{*}+x_{t}\cdot p_{t}^{*}$

$x_{t}=1-x_{b}$

$P=x_{b}\cdot p_{b}^{*}+(1-x_{b})\cdot p_{t}^{*}$

$1.53{\text{ }}{\text{bar}}=x_{b}\cdot (1.8037{\text{ }}{\text{bar}})+(1-x_{b})\cdot (0.7384{\text{ }}{\text{bar}})$

$x_{b}=0.743$

$\therefore {\text{ }}x_{t}=0.257$

Next we can solve for the vapour mole fractions using Raoult's law:

$y_{b}={\frac {x_{b}\cdot p_{b}^{*}}{P}}={\frac {1.340}{1.53}}=0.876$

$y_{t}=1-y_{b}=0.124$

Finally we can solve for the mass fractions using the molecular weight of the benzne and toluene and a basis of 1 mole:

$MW_{b}=78.11{\text{ }}{\frac {g}{mol}}$

$MW_{t}=92.14{\text{ }}{\frac {g}{mol}}$

$m_{b}=x_{b}\times MW_{b}=0.743{\text{ }}mol\times 78.11{\text{ }}{\frac {g}{mol}}=58.03{\text{ }}g$

$m_{t}=x_{t}\times MW_{t}=0.257{\text{ }}mol\times 92.14{\text{ }}{\frac {g}{mol}}=23.68{\text{ }}g$

$\therefore {\text{ }}w_{b}={\frac {m_{b}}{m_{b}+m_{t}}}={\frac {58.03}{58.03+23.68}}=0.710$

and we will do the same for the vapour mass fractions:

$m_{b}=y_{b}\times MW_{b}=0.876{\text{ }}mol\times 78.11{\text{ }}{\frac {g}{mol}}=68.42{\text{ }}g$

$m_{t}=y_{t}\times MW_{t}=0.124{\text{ }}mol\times 92.14{\text{ }}{\frac {g}{mol}}=11.43{\text{ }}g$

$\therefore {\text{ }}w_{b}={\frac {m_{b}}{m_{b}+m_{t}}}={\frac {68.42}{68.42+11.43}}=0.857$