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 provioded, 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 Antoines equation:
\(\log_{10} p^* = A - \frac{B}{T + C}\). Where T is in kelvin for
this question.
| Component |
Are |
Cool |
Antoine Constants |
|---|
| |
A |
B |
C |
| Benzene |
4.72583 |
1660.652 |
-1.461 |
| Toluene |
4.07827 |
1343.943 |
-53.773 |
Answer
From antoines equation, we can find the vapour partial pressure of
toluene and benze.
\[p^*_b = 10^{\space A - \frac{B}{T + C}} = 10^{\space 4.72583 - \frac{1660.652}{373 - 1.461}} = 1.8037 \space \text{bar}\]
\[p^*_t = 10^{\space A - \frac{B}{T + C}} = 10^{\space 4.07827 - \frac{1343.943}{373 - 53.773}} = 0.7384 \space \text{bar}\]
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:
- \(x_b + x_t = 1\)
- \(P = x_b \cdot p^*_b + x_t \cdot p^*_t\)
\[x_t = 1 - x_b\]
\[P = x_b \cdot p^*_b + (1 - x_b)\cdot p^*_t\]
\[1.53 \space \text{bar} = x_b \cdot (1.8037 \space \text{bar}) + (1 - x_b)\cdot (0.7384 \space \text{bar})\]
\[x_b = 0.743\]
\[\therefore \space x_t = 0.257\]
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:
\[MW_b = 78.11 \space \frac{g}{mol}\]
\[MW_t = 92.14 \space \frac{g}{mol}\]
\[m_b = x_b \times MW_b = 0.743 \space mol \times 78.11 \space \frac{g}{mol} = 58.03 \space g\]
\[m_t = x_t \times MW_t = 0.257 \space mol \times 92.14 \space \frac{g}{mol} = 23.68 \space g\]
\[\therefore \space w_b = \frac{m_b}{m_b + m_t} = \frac{58.03}{58.03 + 23.68} = 0.710\]
and we will do the same for the vapour mass fractions:
\[m_b = y_b \times MW_b = 0.876 \space mol \times 78.11 \space \frac{g}{mol} = 68.42 \space g\]
\[m_t = y_t \times MW_t = 0.124 \space mol \times 92.14 \space \frac{g}{mol} = 11.43 \space g\]
\[\therefore \space w_b = \frac{m_b}{m_b + m_t} = \frac{68.42}{68.42 + 11.43} = 0.857\]