[confusedstud]

When distilling ethanol and water solution, the low boiling point ethanol will boil first and rise up while the fractionating column prevents any water from rising as it will condense along the way.

Why is that so? If the water has enough energy to boil then shouldn't it have enough energy to rise up the column as well? When I learnt about heat capacities and evaporation more detailedly in physics, I realised that the water molecules which evaporate have enough energy to escape the forces of attraction (van der waals) in the water. So it has as much energy as boiling water just that it's mass is really small. Something like Q=mcΔT+ml where m is very small? So shouldn't the evaporated water molecules also have enough energy to rise up the column without condensing?

Secondly, when we distill a salt water solution, why is the thermometer's reading 100 degrees? Since the boiling point increases so shouldn't the reading also be higher? If the boiling range is around 120 degrees then won't the temperature be 120 too? What causes the temperature of the water vapour to drop to 100?

[curiouscat]

Are you talking about a single stage distillation or a multi-stage fractionation with reflux?

In either case, there's always a tiny number of H2O molecules that make it to the top.

[discodermolide]

Water and ethanol form an azeotrope.
How can the boiling point be more than 100°C, salt does not come with the water.

[confusedstud]

Yes the boiling point of ethanol and water is smaller. But evaporation of water can take place in any temperature. So some water can still evaporate right?

Salt doesn't go with the vapour but when the salt water boils its around 120 degrees. So shouldn't the reading also be 120 degrees?

[curiouscat]

But evaporation of water can take place in any temperature. So some water can still evaporate right?

Yes. And it does.

[discodermolide]

Yes the boiling point of ethanol and water is smaller. But evaporation of water can take place in any temperature. So some water can still evaporate right?

Salt doesn't go with the vapour but when the salt water boils its around 120 degrees. So shouldn't the reading also be 120 degrees?

So can some of the ethanol evaporate, probably faster as it's vapour pressure is higher.
It you stuck a thermometer in the flask containing the salt solution it may well read 120°C. But at the still-head, where the water vapour starts to come over it will be pure water and it will read 100°C,

[fledarmus]

You are perfectly correct. A simple distillation is rarely sufficient to purify a compound from a mixture, unless one of the components of the mixture is non-volatile. For example, Discodermolide's salt-water example.

Look at a phase diagram of ethanol and water - for example, http://www.chemguide.co.uk/physical/phaseeqia/bpcompn1.gif. Let's assume that you are starting with a mixture that is 1:1, right in the middle of the diagram. You start to heat the pot, and the temperature rises. What happens?

Draw a vertical line from the 50% by mass point and note where it crosses the liquid composition line. This is where your solution begins to boil. The vapors that are being collected above that liquid, however, do not have the same composition as the liquid. If you draw a horizontal line now from where your vertical line crossed the liquid composition line until it crosses the vapor composition line, you will find the vapors above that liquid are enriched in ethanol. This is the composition of the vapors that you are condensing. As you noted, there is still water in this mixture, but the mixture is enriched in ethanol. You could take this mixture and distill it further to generate purer ethanol, and keep going until you reach the azeotrope.

In practice, this is carried out by fractional distillation. In theory, you are setting up a whole series of boiling pots, starting with one at the boiling point of water and ending with one at the boiling point of your azeotrope. You do this by setting up a long column over your pot, and maintaining a temperature differential along the column. The bottom of the column is at 100° and the top at 78°. When you heat up your pot, pure water will condense at the very bottom of the column, and drip back into the pot to be boiled again. Slightly higher than that, 10% ethanol in water will condense, and as it starts to drip down the column, it will get hot enough to boil again, sending the ethanol further up the column and the water further down.

Here is a page that describes the process with some better visual aids...

http://www.chemguide.co.uk/physical/phaseeqia/idealfract.html

Hope this helps

[confusedstud]

Oh I think I sort of understand this better. As you boil a solution the condensed product will have more of the volatile substance than the non volatile substance. But can my O levels explanation be "the evaporated water lose its energy and condenses while the ethanol is able to rise up" but actually what allows the water (less volatile) to lose its energy during this process? And what allows the more volatile liquid to not lose its energy and rise up?

Also, for the salt water case why does the temperature of the water suddenly drop from. 120 to 100? I'm thinking that the specific heat capacity of salt water is lower then pure water, so when some vapour forms there is no more salt so for the same amount of thermal energy supplied the water will be of a lower temperature?

Thanks for the help guys! And man these vapour and raoults law stuff is really hard to understand haha I'm only 16 this year.

[discodermolide]

A substance boils when its vapour pressure reaches that of the surroundings, in this case atmospheric pressure.
It takes less energy to vaporise ethanol than water, therefore it distills first.

In the salt case salt is NOT volatile and does not distil. While the temp. in the pot may be 120°C, at the still head where the vapour starts to condense it is pure water. 100°C.

[confusedstud]

Oh, so the temperature is 100 degrees because of the condensation and the highest temperature for condensation is 100 degrees? I always thought that it was the gas's temperature not the condensed substance. Does this mean that a lot of water already condenses there? Since if it doesn't condense it might measure the temperature of the water vapour gas which is above 100 degrees (maybe 120degrees)

For the fractionating column: oh it makes sense now, the less volatile substance require more energy to boil than the more volatile substance. So we boil and collect the gas above it, the compositions by moles will be different whereby the gas has a higher percenentage of the volatile substance. So the fractionating column allows that boil and re boil process to repeat until there is practically no more of the less volatile substance?

Thanks for the help guys!

[discodermolide]

You said "Since if it doesn't condense it might measure the temperature of the water vapour gas which is above 100 degrees (maybe 120degrees)".

At the top of the still head there is a thermometer which measures the temperature of the gas/liquid condensing there. In this case water, that temperature will be 100°C no higher. Get this 120°C out of your mind.
If you measure the temperature of the pot, i.e. where the salt water mixture is being heated, in there you may have 120°C but NOT after distillation.

[confusedstud]

But what causes that drop in temperature? I mean when we see a heating curve, the emeprture of the gs can continues to heat above 100 degrees right? Thanks!

[discodermolide]

The gas temperature cannot heat above the boiling point of water, 100°C, end of story.
All you are doing is distilling the water out of the solution. So the pot temperature can be higher due to the dissolved salt but the vapour/liquid distilling out is water, BPt. 100°C.
There is no mystery here.

[fledarmus]

Not to quibble, but yes, you can in fact have enough heat that the gas is superheated and the still head temperature is higher than 100°C. This is the wrong way to distill. If you are distilling properly, you want the vapor to be condensing right at the top of the still head, and the bulb of the thermometer to be completely immersed in that condensing vapor. If you are putting enough heat into the system that your head temperature is higher than the boiling point (which is also the condensation point) of your pure product, then you will blow impure vapor into your condensor and down into your collector. Again, this is the wrong way to distill.

Once you have collected all of your low-boiling component, the region of condensation will start to drop below the still head and below the bulb of the thermometer. The thermometer actually registers a drop in temperature at this point, because it is no longer immersed in condensing vapor. When that happens, you change collectors and increase the pot temperature until the highest region of condensation is once again at the still head, and start collecting the next fraction.

[curiouscat]

@fledarmus

Nice explaination!

My question to @confusedstud is: Where exactly is your thermometer? Still head or not? Otherwise we might be arguing at cross purposes.