# Azeotropic mixtures and Raoult's Law

So I find this everywhere:
"Mixtures showing large deviations from Rault's law form azetropic mixtures at specific compositions"

My questions are:
(a)Why is it so?Why deviation from Raoult's law is needed at all.
Wikipedia simply skips over the answer stating its a mathematical consequence. (Quoting wikipedia: When the deviation is great enough to cause a maximum or minimum in the vapor pressure versus composition function, it is a mathematical consequence that at that point, the vapor will have the same composition as the liquid, resulting in an azeotrope.)

(b) why can't ideal mixtures form azeotropes?

P.S: (a)A more intuitive answer would be appreciated although maths may be included if necessary.
(b)I did my research but nowhere the reason was to be found. It's presented as a very obvious consequence everywhere.

• Can someone pls tell why was it downvoted the moment i posted it...? – JustCuriois Apr 27 at 13:13
• About the downvote, it can be said only by the user who did so, as it is fully anonymous. The reason can be you have not paid much effort to think about the topic. Generally, if there is a sloped line, it has to significantly deviate from a line to have maximum or minimum. – Poutnik Apr 27 at 13:59
• Also you can use Raoult Law as a sort of definition for ideal mixtures. And there is no space for azeotropes in it. – Alchimista Apr 27 at 14:05
• @Justcuriois Possible Duplicate Here. – Rishi Apr 27 at 14:53
• @Everyone... pls see the follow up question i put under #Dario Miric's reply. Thanks(Also i was wondering if i am asking such follow ups how will others be notified...?) – JustCuriois Apr 27 at 16:15

To understand why non - ideal mixtures form azeotropes, we need to understand what is ideal mixture. If you have a liquid mixture of two components A and B, ideal mixture is a mixture in which heterogeneous interactions (A - B) are the same as the average homogeneous (A - A and B - B). This means when you mix components A and B, they don't "feel" like their enviroment has changed at all and thus no component has higher nor lower tendency to escape the liquid phase and go into vapor phase. When interactions are largely different, in most cases heterogeneous interactions are less favourable than homogeneous (molecules of A in most cases like to be around molecules of A rather than B) than component will have higher tendency to escape from liquid phase to vapor since in vapor interactions are much less pronounced due to much bigger average distance of the molecules. If this scenario happens you got yourself a lower boiling azeotrope since due to higher tendency of molecules to escape liquid phase, vapor pressure is higher than in ideal case (Raoult's Law). In some cases heterogeneous interactions can be very favourable and this is usually due to some specific interactions which mostly originate from hydrogen bonding, if so molecules will tend to stay in liquid and you got yourself higher boiling azeotrope. Example of such azeotrope is HNO3 and water. If you have any questions, fire.

• Very well explained. But i have a question, azotropes have constant boiling temperature (so components can't be separated by fractional distillation).//(a)Why is it not possible for an ideal solution that it has constant boiling temperature..? //(b)Say the two components of ideal binary solution are equally volatile (i think they can be equally volatile...correct me if i am wrong) and so can't be separated by fractional distillation..i.e this ideal solution has constant boiling temperature...will it not qualify such an ideal solution as an azotrope...? – JustCuriois Apr 27 at 15:36
• [This was the picture i had in mind while asking why ideal solitions can't be azotropes....] //(c)Finally...all the examples of binary ideal solutions i have seen have taken one of the component(say A) to be more volatile than the other (B)....is it necessary..? – JustCuriois Apr 27 at 15:37
• Reason is if mixture is ideal than vapor pressure follows Raoult's Law which means that partial pressure of each component follows a STRAIGHT LINE since Raoult's Law posits linear relation between partial pressure and composition. If we regard vapor as ideal (which you can at low vapor pressure) than vapor pressure is a sum of partial pressure of each component which means that vapor pressure follows a LINE in dependence on composition. Line by definiton doesn't have local maximum or minimum. Take a look on Wikipedia article section Principles en.m.wikipedia.org/wiki/Raoult%27s_law – Dario Mirić Apr 27 at 16:48
• Yes, they can. In such case vapor pressures of pure components are the same (or approximately the same). This situation happens when you have isomers of some component which have very similiar boiling point and are needed to be separated. If pure components have exactly the same vapor pressure than mixture would boil at the same temperature at any compositon since vapor pressure would always be equal to vapor pressure of pure component. But, such mixtures aren't azeotropes since they would be ideal and because azeotropes don't boil at the same temperature regardless of composition. – Dario Mirić Apr 27 at 17:26
• Azeotropes have a composition at which they boil at minimum or maximum temperature at which volatility of each component is exactly the same and you can't separate components by simple distillation. This is why they are said to be constant boiling since you can't change composition of neither liquid nor vapor when you reach azeotropic point, but this doesn't mean their boiling point is independent on composition like it is in your example with two components with the same vapor pressure which acts ideally. – Dario Mirić Apr 27 at 17:31

In his answer Dario Miric has explained the basis of azeotropes clearly, this is a small addition to his answer. Raoult determined experimentally that the partial vapour pressure over some liquid mixtures is directly proportional to the mole fraction $$x$$ of each component, and thus the total pressure is

$$P=x_1p_1^\text{o}+x_2p_2^\text{o}=x_1p_1^\text{o}+(1-x_1)p_2^\text{o}$$

where $$p^\text{o}$$ is the vapour pressure of the pure liquid. An ideal solution can be defined as one that obeys Raoult's law at all temperatures. Solutions obeying Raoult's law are, for example, dichloroethane - benzene or ethylene bromide - propylene bromide. As the pressure is the sum of the partial pressures the two species cannot interact to any great extent that would cause their vapour pressures to differ from that expected from the pure solution. Of course this is only true in a few cases.

If the two types of molecule interact with one another then the potential energy of the mixture has an additional term describing this interaction. The result of this interaction is that the total pressure becomes

$$P=x_1p_1^\text{o}e^{\alpha}+(1-x_1)p_2^\text{o}e^{\beta}$$

where $$\alpha =(1-x_1)^2)\Delta U/RT$$ and $$\beta =x_1^2\Delta U/RT$$ and $$\Delta U$$ is a measure of the average difference in interaction energy. When $$\Delta U$$ is small vs $$RT$$ only a slight deviation from Raoult's law (pressure $$p_{1,2}/p_{1,2}^\text{o}$$,vs. $$x$$) is evident and instead of being a straight line is slightly bent, convex of concave, for example methanol in water. If the plot of total pressure vs mole fraction passes through a maximum or minimum then this is the Azeotrope and the vapour has the same composition as the mixture. Additionally as $$dP/dx=0$$ at the azeotrope and as the mole fraction is known an estimate of $$\Delta U$$ can be obtained. Example of azeotropes are dichloromethane - acetone, carbon disulphide - dimethoxymethane and CS$$_2$$-CHCl$$_3$$