ATAR Notes: Forum
Archived Discussion => 2009 => Mid-year exams => Exam Discussion => Victoria => Chemistry => Topic started by: Mao on June 10, 2009, 05:56:20 pm
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The actual question:
A liquid mixture of 50% ethanol and 50% water was distilled in the apparatus shown above (a laboratory distillation apparatus, see solutions book for actual picture). The boiling point of ethanol is 78 degrees, and that of water is 100 degrees.
As the mixture was heated the temperature shown by the thermometer initially rose but then remained constant at 78 degrees for some time.
Which of the following statements about percentage of ethanol in the vapours shown at point X (just above liquid level), Y (half way up fractionating column) and Z (top of fractionating column) when the temperature is at a constant 78 degrees, is true?
A. The percentage of ethanol in the vapours at X is equal to 50%
B. The percentages of ethanol in the vapours increase in order at positions X, Y and Z
C. The percentages of ethanol in the vapours at Y and Z are equal but greater than at X
D. The percentages of ethanol in the vapours at X, Y and Z are equal but greater than 50%
I would like to invite anyone who would like to have a crack at this. It is my understanding that there is a dispute between B and D.
Posts from other thread:
it was out of 73.
Hey, was answer to q19 MC really D? because i got that and ppl said it was wrong.
I got B but I don't doubt Mao too much aha.
Technically, B would be correct. Some of the water would vaporise and then condense as it rose up the apparatus. However, the amount of vaporised water would be very small. Practically, the difference in percentage concentration of ethanol would be so small as to be insignificant.
This should have been an explanation question.
Some ethanol forms an azeotrope with water (94~95% ethanol, 5~6% water). That is, this mixture does NOT separate when boiled. Hence in the vapour, there will definitely be some water.
For example, if a 50/50 mixture of ethanol and water is distilled once, the distillate will be 80% ethanol and 20% water (see ethanol data page), which is closer to the azeotropic mixture than the original. Distilling the 80/20% mixture produces a distillate that is 87% ethanol and 13% water. Further repeated distillations will produce mixtures that are progressively closer to the azeotropic ratio of 95.5/4.5%. No number of distillations, however, will ever result in a distillate that exceeds the azeotropic ratio. Likewise when distilling a mixture of ethanol and water that is richer in ethanol than the azeotrope, the distillate (contrary to intuition) will be poorer in ethanol than the original but slightly richer than the azeotrope.[2]
However, this is much beyond VCE level. And with VCE knowledge, D would seem the most obvious answer..
[note, the question itself is incorrect. The boiling point will be less than 78 degrees]
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Hmm I don't think they can expect us to have a detailed knowledge of azeotropes
I think the way to interpret this question at VCE level is to just assume no water is present as vapour. The boiling point of water is given, even though we should all know that. It's almost as if they're stressing the fact that no water will be vapourised.
Maybe they will award B and D
Maybe not. At my school, one kid got marked down for talking about multiplets (which are beyond the course) in a question regarding the NMR of propanol.
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This information about the azeotropes tells us that when ethanol is distilled, some water will vaporise and end up in the distillation tank with the ethanol. What this does not tell us is whether or not some of this water vapor will then condense back to liquid as it rises up the distillation column.
We know that water vapour and liquid water form a dynamic equilibrium dependent on the temperature. As the temperature rises, the dynamic equilibrium favours the vapour form more and more. In a distillation unit, the water temperature will be greatest at the bottom of the unit, where the flame is. Thus some water will be vaporised, even at 78C. However, as this vapour rises, the temperature decreases up the column (it is further away from the flame and heat is lost through the glass walls of the column). This therefore means that some of the water will condense back to liquid form.
So
-maximum water vapour concentration occurs at the bottom of the distillation colum
-water vapour conencentration decreases further up the colum and hence ethanol concentration increases
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Ethanol should also condense. It may in fact be that ethanol will condense at a greater rate than water, since it has a lower specific enthalpy of vaporisation. Thus, the concentration of ethanol may DECREASE up the column.
All these variables lead me to suspect that D is the correct answer, because all these considerations are far beyong Unit 3 knowledge.
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Upon re-thinking, I am leaning towards B. The actual 50/50 solution boils at 81 degrees according to http://en.wikipedia.org/wiki/Ethanol_(data_page)#Boiling_points_of_aqueous_solutions , at this temperature, vapour of 50%~80% ethanol would boil off, and as they rise up the fractionating column, only the most volatile, 80% ethanol mixture, would remain as vapour.
(http://homedistiller.org/graphics/Alcohol_curve.jpg)
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Ethanol should also condense. It may in fact be that ethanol will condense at a greater rate than water, since it has a lower specific enthalpy of vaporisation. Thus, the concentration of ethanol may DECREASE up the column.
All these variables lead me to suspect that D is the correct answer, because all these considerations are far beyong Unit 3 knowledge.
I agree that D is what VCAA has in mind, as it falls within the assumptions we 'should' have for this question.
However, if we can help in granting a few more people that extra mark, I think we should try.
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Mao, I'm confused by what that graph is actually showing. Why is alcohol % by volume 0 for both liquid and vapour at 100C?
I think HYSYS would really help with this one ;D
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yea go Mao. maybe you should take this straight to VCAA (might save me a mark .lol)
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Mao, I'm confused by what that graph is actually showing. Why is alcohol % by volume 0 for both liquid and vapour at 100C?
I think HYSYS would really help with this one ;D
Hahaha
Below the liquid line, the mixture can only be found as liquid, above the vapour line, the mixture can only be found as vapour.
In between those lines, you get an 'intermediate', where it can be both liquid and vapour [kinda like 'saturation'] depending on transition. Liquids starts boiling at the 'liquid' line, and vapours condense at the 'vapour' line.
Since pure compounds don't have this 'transition' state, at 0% alcohol (100% water), it goes from liquid straight to vapour.
[hope that made sense]
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i put B - wouldnt the concentration of the ethanol in the vapours increase as the mixture gets higher - although the average energy required to boil water is 100degress, some would certianly boil at 78 degrees - ambigiguity!
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I am almost completely convinced that the answer is B now. If you look at this page the explanation is very clear: as the vapour rises up the column it becomes richer in the component with the lower boiling point (ethanol).
http://www.chemguide.co.uk/physical/phaseeqia/idealfract.html#top
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We did no work on fractional distillation at my school :D
I was like...pick C!
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We did no work on fractional distillation at my school :D
I was like...pick C!
hi 5
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So D will not be accepted?
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Why can't C also be considered correct? When I did this (admittedly rushedly) I thought that the water the evaporates would rapidly condense on the packed glass beads in the column. Hence the change from midway through the column to the top would be very little. This mixture is just ethanol and water. Nobody is discussing C but I think A is truly the only one that can be discounted.
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I did B :] made sense to mee
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I talked to my teacher about it and he believes B is the correct answer
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Why can't C also be considered correct? When I did this (admittedly rushedly) I thought that the water the evaporates would rapidly condense on the packed glass beads in the column. Hence the change from midway through the column to the top would be very little. This mixture is just ethanol and water. Nobody is discussing C but I think A is truly the only one that can be discounted.
Think about it this way: as the water vapour moves up the column, it condenses, and then starts to trickle back down the beads. However, a fraction of this condensate is then vaporised again by the hot gases rising from below, and will continue moving up the colum again, where it will condense once more. Then as it trickles down a fraction will again be vaporised, and so on and so forth.
This practically results in a lower and lower concentration of water as you move towards the top, discounting option C.
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I picked B because that's what Heinemann says. lol.
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D makes more sense to me now :(
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Why can't C also be considered correct? When I did this (admittedly rushedly) I thought that the water the evaporates would rapidly condense on the packed glass beads in the column. Hence the change from midway through the column to the top would be very little. This mixture is just ethanol and water. Nobody is discussing C but I think A is truly the only one that can be discounted.
Think about it this way: as the water vapour moves up the column, it condenses, and then starts to trickle back down the beads. However, a fraction of this condensate is then vaporised again by the hot gases rising from below, and will continue moving up the colum again, where it will condense once more. Then as it trickles down a fraction will again be vaporised, and so on and so forth.
This practically results in a lower and lower concentration of water as you move towards the top, discounting option C.
Actually now that I look at it, are you sure there are glass beads in that column? I know there are meant to be but it looks more like a thermometer..... :S
In any case, while B may be considered the best answer. What I'm saying is that if B and D are both accepted, then should C be accepted too as it is essentially an intermediate between the two?
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I think D won't be accepted now.
Most likely, the examiners would assume that we have the Heinemann knowledge.
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And also just to ask a question that was nagging in the back of my mind on the other contentious MC: the hydrogen as a carrier gas one. Without meaning to beat on about something that's long over and can't be changed, wouldn't hydrogen react with the stationary phase (generally high boiling point liquid hydrocarbons like methyl esters) at such high temperatures. And would hydrogen blow up the flame ionization detector by any chance (that has oxygen)?
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Mate, who knows? The exam is over, just wait until the grades come out and inspect your paper, nobody knows for sure the right answers, nobody here is the chief VCAA examiner.
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my examiner teacher said its B.........
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Ok so is it definite that B is one of if not the only answer? And also I don't understand why people are getting worked up about this 1 mark MC question when there were so many short answer questions that had much more of an impact XD
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every mark counts mate.
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My teacher said he expects only around 10% of state to get this question right and if you go by what he said, this question won't have that much of an impact in the scheme of things - even though every mark counts...
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i said B since the beads are in the fractionating colomn specifically to recondense any vapour that isnt wanted there, so water would be recondensed more as the height incresed, so there would be more ethanol as a percentage
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I hope the examiners made a 'mistake' on the exam somewhere so VCAA will pressure them to make exam 2 easier.
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I don't know...if the next exam is too easy then this exam will have a huge effect in differentiating between us an determining our SS. So if you fully owned this exam and got full marks then it would be better to get an easy exam next time. It's like going in 40 points ahead at half time and all you need to do is hold onto the lead. So a good performance on an exam where everyone performs good and an awesome performance on an exam where everyone does not perform as well would be enough. But if you aren't satisfied with your performance in the sense that you believe that some better time-management or doing a couple of things right here and there could have significantly boosted your mark; then getting another hard exam next time may be just what you need to get that SS higher. In any case, I wouldn't bet on an easy exam for U4, if you look at subjects like Accounting that were in the 2nd year of the study design last year both their midyear and endyears were considered fairly difficult.
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I hope the examiners made a 'mistake' on the exam somewhere so VCAA will pressure them to make exam 2 easier.
They have already made the unit 4 exams even before the actual examination of unit 3. So you can not assume that they going to make exam 2 easier because they did not know the impact of the unit 3 exam on the students when they made it only last years reaction to their paper. Who knows exam 2 can be just as hard! The only hope for an easier exam is for 2010 when they have gotten the results for both 09 exams and can adjust the difficulty or length of the exam.
I dont mean to shatter your hope or anything but exam 2 might not be made easier, im in the same position as you and yea i got told the harsh truth by my teacher. However, it allowed me to try even harder for exam 2 and concentrate more, it was a learning experience. Hopefully you can see this in the same light, and study even harder for this exam to redeem the marks lost in unit 3.
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Who cares, as long as you get an exam just do it.
Hard exam, easy exam, they're all exams.
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I hope the examiners made a 'mistake' on the exam somewhere so VCAA will pressure them to make exam 2 easier.
Exam was already written long time ago (February)!
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Who cares, as long as you get an exam just do it.
Hard exam, easy exam, they're all exams.
Yeah but what I'm saying is it's a lot easier to catch up on a harder exam (if you study really hard and own it)
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That's true. I learned that in Accounting last year, your own preparation is key. I found the first exam much more challenging than the second even though their actual difficulty was roughly the same, which reflected my preparation.