Question

What type of deviation is shown by a mixture of ethanol and acetone? What type of azeotrope is formed by mixing ethanol and acetone?

Answer 1

Hint: Deviations from Raoult’s law totally depends upon the attracting forces acting between different constituents of the solution. When attractive forces between A and B will be greater than A-A attractive forces and B-B attractive forces then their solution will show negative deviation from Raoult’s law. When attractive forces between A and B will be weaker than A-A attractive forces and B-B attractive forces then their solution will show positive deviation from Raoult’s law.

Complete answer:
An azeotrope or a constant boiling point mixture is a mixture of two or more liquids whose proportions cannot be altered or changed by simple distillation. This happens because when an azeotrope is boiled, the vapour has the same proportions of constituents as the unboiled mixture. If ${{N}_{A}}$ moles of liquid A is mixed into ${{N}_{b}}$ moles of liquid B and their mole fractions are respectively ${{X}_{A}}$ and ${{X}_{B}}$ then according to Raoult’s law
Partial pressure of A ${{P}_{A}}$=\[{{X}_{A}}P_{A}^{0}\]
Partial pressure of B ${{P}_{B}}={{X}_{B}}P_{B}^{0}$
And total pressure of the system will be
\[P={{X}_{A}}P_{A}^{0}+{{X}_{B}}P_{B}^{0}\]
Based on Raoult law there will be two types of Non-ideal solutions:
-Negative deviation from Raoult’s law: when attractive forces between A and B will be greater than A-A attractive forces and B-B attractive forces then there solution will show negative deviation from Raoult’s law; Vapour pressure of each component will be
\[{{P}_{A}}<{{X}_{A}}P_{A}^{0}\] and ${{P}_{B}}<{{X}_{B}}P_{B}^{0}$
Also total vapor pressure will be \[P<{{X}_{A}}P_{A}^{0}+{{X}_{B}}P_{B}^{0}\]
Following are a few examples of solutions those show negative deviation from Raoult’s law
Acetone + Aniline
Acetone + Chloroform
Water + HCl
Acetic Acid + Pyridine etc.
-Positive deviation from Raoult’s law: when attractive forces between A and B will be weaker than A-A attractive forces and B-B attractive forces then there solution will show positive deviation from Raoult’s law;
\[{{P}_{A}}>{{X}_{A}}P_{A}^{0}\] and ${{P}_{B}}>{{X}_{B}}P_{B}^{0}$
Also total pressure will also be \[P>{{X}_{A}}P_{A}^{0}+{{X}_{B}}P_{B}^{0}\]
Following are a few examples of solutions those show positive deviation deviation from Raoult’s law
Acetone + Ethanol
Acetone + Carbon disulphide
Water + Ethanol
Cyclohexane Ethanol etc.
We can clearly see that Acetone + Ethanol solution will show positive deviation because in ethanol hydrogen bonding exists which is a strong intermolecular attraction force while there is no such force between Ethanol and Acetone as in acetone there is no hydrogen that is attached to oxygen. So B-B attractive forces here are stronger than A-B attractive forces.
Solution of Acetone + Ethanol is going to make minimum boiling azeotrope because here total vapor pressure of the system will be more than the ideal solutions and we know that boiling point is the temperature where vapor pressure of the system becomes equal to the atmospheric pressure and here high pressure of the solution will make sure that pressure of the system will become equal to the atmospheric pressure at lesser temperature in comparison to the ideal solution.

Additional information: Solutions that show negative deviation from Raoult’s law will form maximum boiling azeotrope.

Note: For the solutions that show positive deviation from Raoult’s law, enthalpy change for the mixing process is positive, it means it is an endothermic process and also volume of the system increases after the dissolution than expected.