Discuss about vapour pressure for ideal and nonideal solutions?
Answers
Explanation:
Since the interactions between solute and solvent are weaker than solute-solute and solvent-solvent interactions, the vapour pressure of the solution is higher than that predicted by Raoult's law. The molecules of A or B will find it easier to escape than in their pure state
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Answer:
The liquid-liquid solutions can be divided into ideal and non-ideal (real) solutions based on whether they obey Raoult's law exactly or not. It can tell us about the type of forces present in pure solvent before mixing and after the addition of solute.
Ideal solutions
A solution is said to be ideal if each of its components obeys Raoult's law for the entire range of concentration (composition). In the preparation of ideal solution, no thermal change is observed. i.e.
△Hmix=0
The force of interaction between
A−A
,
B−B
and
A−B
are of same order. The volume of mixing
(△Vmix)
is also zero. i.e. The volume of the solution will be equal to sum of the volumes of the two components.
For example, the solution of benzene in CCl4 shows same attractive forces as are present in benzene and CCl4 separately. Another examples are benzene and toluene, bromoethane and chloroethane.
Non-Ideal Solutions
Solutions are non-ideal if they do not obey Raoult's law. i.e.
△Hmix≠0
and
△Vmix≠0
. Therefore these solution deviate from ideality and depending on type of deviation from ideal behaviour, non-ideal solutions may be classified as showing negative deviation or positive deviation.
i) Negative deviation:
In such solutions of components
A
and
B
, the
A−B
interactions are stronger than
A−A
and
B−B
. i.e, The interactions between solute and solvent are (stronger) greater than solute-solute and solvent-solvent interactions. As a result, the vapour pressure of solution will be lower than that predicted by Raoult's law. For example, water and acetone when form solution, attract each other more strongly than acetone-acetone and water-water. This is due to the formation of hydrogen bonds by water molecules with acetone molecules in the solution, whereas acetone molecule cannot form hydrogen bond with another acetone molecule.
Another example is that of phenol and aniline. Here hydrogen bonding is more stronger between acidic Hydrogen of phenol with Nitrogen of aniline having lone pair.
ii) Positive deivation:
In such solutions of component
A
and
B
, the
A−B
interactions are weaker than
A−A
and
B−B
interactions. Since the interactions between solute and solvent are weaker than solute-solute and solvent-solvent interactions, the vapour pressure of the solution is higher than that predicted by Raoult's law. The molecules of A or B will find it easier to escape than in their pure state.
For example, mixture of ethanol and hexane where ethanol molecules have intermolecular hydrogen bonding among themselves and on the other hand when hexane is added to ethanol, hexane breaks some of the hydrogen bonds to fit in between the ethanol molecules. Therefore the escaping tendency of ethanol molecules is more in the mixture (solution) as compared to when in pure state
(△H=+ve)
. Another example is acetone and carbon disulphide.
Azeotropic Mixtures
Some liquid solutions formed by two (or more) components in a particular composition, boils at a constant temperature and distill as single component. i.e. It is not possible to separate the two components of the solution by fractional distillation also. Such solution are azeotropic or constant boiling solution.
The solutions which show a large positive deviation from Raoult's law from minimum boiling azeotrope at a specific composition. The solutions which show large negative deviation from maximum boiling azeotrope