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We're covering fractional distillation in my physical chemistry class. I'm confused regarding this diagram: enter image description here

From your starting point, you heat the binary mixture (in this case benzene and toluene) to reach point a. Is the change in benzene mole fraction, point a to point b, then spontaneous? If so, why? If not, what triggers this rightward shift in mole fraction?

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    $\begingroup$ It should not be surprising that different substances have different volatility, leading to different compositions of liquid and gaseous phase. $\endgroup$
    – Poutnik
    Commented Nov 13, 2023 at 3:51

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Here is an interpretation of the diagram, implemented with two distillation setups (blue and orange):

enter image description here

In the first setup (blue), you have a large amount of your initial mixture, which you have to heat to about 370 K to boil (a). The vapor that boils off is enriched in benzene. You cool down the vapor to under about 355 K (c) to get liquid again.

In the second setup (orange), you have to heat this second mixture a bit again to get it to boil (c). The initial mixture boiling off is almost entirely benzene (d), and will liquify at about 350 K (e).

What is not shown is that as the distillation proceeds, the composition of the liquid changes, the boiling temperature will increase, and you will have more and more toluene in your distillate.

Is the change in benzene mole fraction, point a to point b, then spontaneous? If so, why?

The overall mole fraction can't change. You enrich the liquid in toluene while getting vapor that is enriched in benzene. If you turn the entire sample into vapor, there is no enrichment, of course.

If not, what triggers this rightward shift in mole fraction?

The same as the difference in boiling points of the two pure substances: It requires less energy to get a benzene molecule to boil or evaporate than a toluene molecule (in this particular system).

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    $\begingroup$ And we do this enrichment many times (using a fractionating column) to get a pure product (for example a Vigreux column). $\endgroup$
    – Mäßige
    Commented Nov 13, 2023 at 19:19
  • $\begingroup$ Yes, but that's not shown in the picture. Here is a short video showing how the composition and temperature change in a simple distillation (one theoretical stage): $\endgroup$
    – Karsten
    Commented Nov 13, 2023 at 22:26
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The use of an arrow in the diagram is probably misleading you. This diagram, for the most part, does not show how a system evolves over time.

This graph with the two red curves describes the equilibrium condition, at 1 atm, in a system containing only toluene and benzene. The top curve shows the equilibrium composition of the gas. The bottom curve shows the equilibrium composition of the liquid.

There is no movement from a to b. These two points represent the states achieved by the two phases in a single system at a single point in time. Point a represents the state of the liquid. Point b represents the state of the gas. The system is assumed to be at thermal equilibrium, so both points have the same temperature, but they have different compositions.

Similarly, points c and d both represent a single system at a single point in time, just at a different temperature from the a/b system.

In a fractional distillation, the idea is to create several "mini-systems" in your apparatus. The bottom flask is at 370 K, and conditions in that region of the apparatus are described by points a and b. Somewhere in the middle of the column, the temperature is 355 K, and the conditions there are described by points c and d. The whole a->d path exists in the system at a single time.

To make this more confusing, the diagram includes a point marked "Start." This point is on neither curve, and does not describe a benzene/toluene equilibrium at 1 atmosphere. This could indicate a system that is not at equilibrium, but in practice, for a fractional distillation, it means there is a third substance -- air -- in the system. As you heat the bottom flask to 370 to start the distillation, the air is driven out of the system by the vapor, and the bottom flask reaches the a/b state.

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