Vapor Pressure and Volatile Solutes: Ideal Solutions
The key point to remember about solutions with two (or more) volatile components?
All the components are represented in the vapor that is in contact with the
solution.
Got that?
ALL the components.
By the way, at this introductory level, we will only discuss solutions with two
volatile components. More complex solutions (with three or more volatile
components) are discussed at a level beyond the scope of the ChemTeam's mission.
Also, note the presence of the word 'ideal' in the title. We will discuss the real
behavior of solutions in a different tutorial.
The key equation to use is Raoult's Law, but in a slightly expanded form from how
it is first presented:
Psolution = P°A χA + P°B χB
The subscripts A and B stand for the two different volatile substances in the
solution.
Example #1: A solution is composed of 1.40 mol cyclohexane (P°cy = 97.6 torr) and
2.50 mol acetone (P°ac = 229.5 torr). What is Ptotal, the total vapor pressure, above
this solution?
Solution:
1) Calculate the mole fraction of each substance:
cyclohexane ⇒ 1.40 mol / (1.40 mol + 2.50 mol) = 0.358974359
acetone ⇒ 2.50 mol / (1.40 mol + 2.50 mol) = 0.641025641
2) Use Raoult's Law:
Ptotal = P°cy χcy + P°ac χac
x = (97.6 torr) (0.358974359) + (229.5) (0.641025641)
x = 35.03589744 + 147.1153846 = 182.15 torr = 182 torr (to 3 sf)
There is something very important about the above calculation for the vapor
pressure of the solution. It is that we also know the vapor pressures of the two
components of the vapor.
This is important because it allows us to calculate the composition (using mole
fractions) of the vapor. How? Divide each component's vapor pressure by the total
vapor pressure. Here's the calculation for the example:
χ for cyclohexane ⇒ 35.0359 torr / 182.15 torr = 0.19
χ for acetone ⇒ 147.1154 torr / 182.15 torr = 0.81
(By the way, we know that pressure is directly proportional to the number of moles
from consideration of the Ideal Gas Law.)
A key point about the above result is this: the vapor is richer in the component
with the higher vapor pressure than the solution. In our example, the acetone in
the solution had χ = 0.64, but acetone's mole fraction in the vapor equals 0.81.
The component with the higher vapor pressure? There's more of it in the vapor
(when using mole fractions) than in the solution.
If we were to constantly sweep away the vapor from above the solution, then more
of the solution would vaporize. However, what happens then is that we would
remove (from the solution) more and more of the component with the higher vapor
pressure. This fact allows us the means to separate two volatile components in a
solution and obtain each substance in a (relatively) pure form.
The general name of the separation process that exploits volatility differences is
distillation. It is an extremely important research and industrial process.
However, please be aware that the reality is much more complex than the
discussion just above. That discussion is for another time, another class.
You might be interested to know that there are some solutions that resist
distillation. These solutions are called azeotropes. Here is the begining of the
Wikipedia entry:
"An azeotrope is a mixture of two or more liquids (chemicals) in such a ratio that
its composition cannot be changed by simple distillation. This occurs because, when
an azeotrope is boiled, the resulting vapor has the same ratio of constituents as
the original mixture."
Notice how the composition of the solution and the vapor are the same. This means
that it is extremely difficult (although not impossible in some cases) to distill an
azeotrope.
Example #2: A solution is prepared by mixing 0.0400 mol CH2Cl2 and 0.0800 mol
of CH2Br2 at 25 °C. Assuming the solution is ideal, calculate the composition of the
vapor (in terms of mole fractions) at 25 °C. At 25 °C, the vapor pressures of pure
CH2Cl2 and pure CH2Br2 are 133 and 11.4 torr, respectively.
Solution:
1) Calculate the mole fraction of each substance:
CH2Cl2 ⇒ 0.0400 mol / (0.0400 mol + 0.0800 mol) = 0.333
CH2Br2 ⇒ 0.0800 mol / (0.0400 mol + 0.0800 mol) = 0.667
2) Use Raoult's Law:
Psolution = P°Cl χCl + P°Br χBr
Psolution = (133 torr) (0.333) + (11.4 torr) (0.667) = 44.289 torr + 7.604 torr = 51.893
torr
I won't round off here because I want to go to the mole fractions in the vapor.
3) Calculate mole fractions in vapor:
CH2Cl2 ⇒ 44.289 / 51.893 = 0.8535
CH2Br2 ⇒ 7.604 / 51.893 = 0.1465
I actually did the second mole fraction by subtraction (1 minus 0.8535).
Example #3: C10H16 and C10H18O are two of the many compounds used in perfumes
and cosmetics to provide a fresh pine scent. At 69.0 °C the pure substances have
vapor pressures of 100.3 torr and 9.8 torr, respectively. What is the composition
of the vapor in terms of mole fractions above a solution containing equal masses of
these compounds at 69.0 °C? Assume ideal behavior.
Solution:
1) Calculate the mole fraction of each substance:
Assume 50.00 g of each substance. (You can pick any mass of each substance, just
as long as they are equal. Just like Henry Ford and the Model A: you can have any
color of car, just as long as it was black.)
C10H16 ⇒ 50.00 g / 136.2364 g/mol = 0.36701 mol
C10H18O ⇒ 50.00 g / 154.2512 g/mol = 0.32415 mol
χC10H16 = 0.36701 mol / (0.36701 mol + 0.32415 mol) = 0.531
χC10H18O = 0.32415 mol / (0.36701 mol + 0.32415 mol) = 0.469
2) Use Raoult's Law:
Psolution = P°C10H16 χC10H16 + P°C10H18O χC10H18O
Psolution = (100.3 torr) (0.531) + (9.8 torr) (0.469) = 53.2593 torr + 4.5962 torr =
57.8555 torr
3) Calculate mole fractions in vapor:
C10H16 ⇒ 53.2593 / 57.8555 = 0.9206
C10H18O ⇒ 4.5962 / 57.8555 = 0.0794