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A student is separating $\ce{CHCl3}$ (bp = $\pu{61 ^{\circ}C}$) from $\ce{CHCl2CHCl2}$ (bp = $\pu{146 ^{\circ}C}$) by distillation. She has just begun to collect the first distillate in the receiving flask. At what position in the apparatus will the temperature be $\pu{61 ^{\circ}C}$?

Distillation apparatus, with heated flask labeled A, space above fractionating column labeled B, space after condenser labeled C and receiving flask labeled D

The answer to this questions is B. I narrowed down the options to either A or B, because I know that the temperature at A would be slightly higher than $\pu{61 ^{\circ}C}$, but lower than $\pu{146 ^{\circ}C}$, and the temperature at B would be slightly cooler than the temperature at A. C and D would be much lower in comparison.

What makes the answer to this questions B rather than A? I think that the two temperatures should be somewhat close to each other, but is there some intricacy to this that I am missing?

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    $\begingroup$ You want to find the place where chloroform is present both as pure liquid and as pure gas at normal pressure. If you boil pure chloroform, this would be at the liquid gas interface of the bubbles rising in the liquid. $\endgroup$
    – Karsten ♦
    Mar 18, 2019 at 5:47

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The temperature in area A can well exceed B all right, especially if the distillation adapter is long or if there is a fractioning column or another condenser before the three-way adapter. However, no target component is going to make it through the condenser unless its vapors reach the side arm (the one with the joint to condenser).

Apparently, it's only possible if the temperature of that area coincides with the boiling point of the target, that's why the thermocouple or thermometer probe is installed exactly at that area (B).

As you rightfully suggested the temperatures in areas C and D are arbitrary: it only should be above the m.p. to prevent clogging and less than b.p. so that the target is not evaporating away.

P. S. there is a vacuum adapter shown on the picture (area C), so there also might be distillation at lower pressure, which would also lower the b.p. of the mix constituents. Since the question doesn't mention its usage, atmospheric pressure is assumed.

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  • $\begingroup$ A vacuum adapter shown on the picture (in area C) is for pressure equilibrium in this case due to the closed circuit otherwise. $\endgroup$
    – Mathew Mahindaratne
    Mar 18, 2019 at 20:35
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What makes the answer to this questions B rather than A? I think that the two temperatures should be somewhat close to each other, but is there some intricacy to this that I am missing?

The factor that you're missing is heat loss. Because the apparatus is in the air it is losing heat to the atmosphere. As the vapor rises through the three way adapter it cools and drips back into the pot. So there is a temperature gradient from the pot temperature to the condenser. So Point B will be colder than the liquid or the vapor just above it.

The other thing here is that you are not going to get perfect separation since 1,1,2,2 tetracholoroethane has an appreciable vapor pressure at $\pu{61 ^{\circ}C}$. The separation that can be achieved is measured by the number of plates of the distillation apparatus. So the greater number of plates, the better the separation.

Since the condensate will be a mixture, you want the thermometer where the gas phase will start traveling down the condenser. This will give you the best information about the composition of condensate.

What you are trying to do is to adjust the applied heat to keep Point B at or just above $\pu{61 ^{\circ}C}$ so that the condensate is as rich as possible in trichloromethane.

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