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DALTON'S LAW VERSES RAOULT'S LAW:

Determination of composition in vapour phase:
The composition of the vapour in equilibrium with the solution can be calculated applying Daltons’ law of partial pressures. Let the mole fractions of vapours A and B be YA and YB respectively. Let PA and PB be the partial pressure of vapours A and B respectively and total pressure PT.
In Vapours phase:
YA= mole fraction of A in vapour phase
YB = mole fraction of B in vapour phase
                   (YA+YB =1)
In liquid solution phase:
XA = mole fraction of A in liquid phase
XB = mole fraction of B in liquid phase
                  (XA + XB = 1)

According to Raoult’s Law: The partial pressure of any volatile component of a solution at any temperature is equal to the vapour pressure of the pure component multiplied by the mole fraction of that component in the solution.
      Where XA­ and XB is the mole fraction of the component A and B in liquid phase respectively

According to Dalton’s Law:
The vapour behaves like an ideal gas, then according to Dalton’s law of partial pressures, the total pressure PT is given by:
 Partial pressure of the gas = Total pressure x Mole fraction
                                        PA = PT YA and PB =PT YB
Where YA­ and YB is the mole fraction of the component A and B in gas phase respectively
Combination of Raoult’s and Dalton’s Law:
(3) Thus, in case of ideal solution the vapour phase is phase is richer with more volatile component i.e., the one having relatively greater vapour pressure

Graph Between 1/YA Vs 1/XA:
According to Dalton’s law of partial pressures, the total pressure PT is given by:
 Partial pressure of the gas = Total pressure x Mole fraction
Where YA­ and YB is the mole fraction of the component A and B in gas phase respectively
According to Raoult’s law:
On rearrangement of this equation we get a straight line equation:

VAPOUR PRESSURE AND RAOULT’S LAW:

VAPOUR PRESSURE:
(1) If a sample of water in its liquid phase is placed in an empty container, some of it will vaporize to form gaseous of water. This change is called evaporation.
(2) The pressure exerted by the vapour (molecules in the vapour phase) over the surface of the liquid at the equilibrium at given temperature is called the vapour pressure of the liquid.
OR
(3) It is the pressure exerted by the vapour when vapours are equilibrium with the liquid.
(4) The pressure exerted by vapours is called unsaturated vapour pressure or partial vapour at non equilibrium condition. 
Factors affecting vapour pressure:
(A) Temperature:.
(1) The temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure is called its boiling point.
(2) Vapour pressure is directly proportional to the Temperature so that on increasing temperature the rate of evaporation increases and rate of condensation decreases and hence vapour pressure increases.
(3) The dependence of vapour pressure and temperature is given by CLASIUS CLAPERON equation.

(4) Vapour pressure of a particular liquid system is only the function of temperature only. It is independent from all other factors like surface area, amount of liquid, available space etc.

(A) Nature of liquid:
Vapour pressure of liquid =1/the strength of intermolecular forces acting between molecules 
For example: CCl4 has higher vapour pressure because of the weak intermolecular forces acting between its molecules than water which has stronger intermolecular forces acting between water molecules of volatile liquid has lower boiling point than a non-volatile liquid.
 Note:
(1) Relative lowering of vapour pressure of a solvent is a colligative property equal to the vapour pressure of the pure solvent minus the vapour pressure of the solution.
(2) For example: water at 20°C has a vapour pressure of 17.54 mmHg. Ethylene glycol is a liquid whose vapour pressure at 20°C is relatively low, an aqueous solution containing 0.010 mole fraction of ethylene glycol has a vapour pressure of 17.36 mmHg. Thus the vapour pressure lowering, DP = 17.54 mmHg ¾ 17.36 mmHg = 0.18 mmHg.

RAOULT’S LAW:
(1) Vapour pressure of a number of binary solutions of volatile liquids such as benzene and toluene at constant temperature gave the following generalization which is known as the Raoult’s law.
(2) Raoult’s law states thatThe partial pressure of any volatile component of a solution at any temperature is equal to the vapour pressure of the pure component multiplied by the mole fraction of that component in the solution
(A) Vapour pressure of liquid-liquid Solution:
(3) Suppose a binary solution contains nA moles of a volatile liquid A and nB moles of a volatile liquid B, if PA and PB are partial pressure of the two liquid components, the according to Raoult’s law
(4) If the vapour behaves like an ideal gas, then according to Dalton’s law of partial pressures, the total pressure P is given by 
Graphical representation of Raoult’s law:
(5) The relationship between vapour pressure and mole fraction of an ideal solution at constant temperature is shown. The dashed lines 1 and 2 represent the partial pressure of the components. The total vapour pressure is given by 3rd line in the above figure.

(B) Vapour pressure of Solid-liquid Solution:
(1) Vapor pressure, when a small amount of a non-volatile solute (solid) is added to the liquid (solvent). It is found that the vapour pressure of the solution is less than that of the pure solvent.
(2) The lowering of vapour pressure is due to the fact that the solute particles occupy a certain surface area and evaporation takes place from the surface only. and
(3) The particles of the solvent will have a less tendency to change into vapour i.e. the vapour pressure of the solution will be less than that of the pure solvent and it is termed as lowering of vapour pressure.
For a solution of non-volatile solute with volatile solvent.
ILLUSTRATIVE EXAMPLE (1): The vapour pressure of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40g of methanol. Calculate total vapour pressure of the solution.
SOLUTION:     
ILLUSTRATIVE EXAMPLE (2): What is the composition of the vapour which is in equilibrium at 30°C with a benzene-toluene solution with a mole fraction of benzene of 0.400?                    
SOLUTION: 
ILLUSTRATIVE EXAMPLE (3): The composition of vapour over a binary ideal solution is determined by the composition of the liquid. If XA    and YA are the mole-fraction of A in the liquid and vapour, respectively find the value of XA for which YA-X has a minimum. What is the value of the pressure at this composition?
SOLUTION
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ILLUSTRATIVE EXAMPLE (4): One mole of a non-volatile solute is dissolved in two moles of water. The vapour pressure of the solution relative to that of water is
SOLUTION:
Mole fraction of solute in solution Or
                                                                 
Raoult's Law  v/s Dalton's Law: Determination of composition in vapour phase: Coming soon..

IDEAL AND NON IDEAL SOLUTIONS:


An ideal solution of substance A and B is one in which both substances follow Raoult’s law for all values of mole fractions. Such solutions occurs when the substances are chemically similar so that the intermolecular forces between A and B molecules are similar to those between two A molecules or between two B molecules. The total vapour pressure over an ideal solution equals the sum of the partial vapour pressures, each of which is given by Raoult’s law.

(1) Condition for a solution to be Ideal
Two liquids on mixing form an ideal solution only when
(1) Both have similar structures and polarity so that they have similar molecular environment.
(2) Both have similar molecular sizes.
(3) Both have identical intermolecular forces.
(4) The liquid should not dissociate or associate each other.
(5) For the solid solute, solution must be extremely diluted.

(2) Characteristic of Ideal Solution:

(3) Examples of Ideal Solution: Solution of benzene C6H6 and toluene C6H5CH3 are ideal. Note the similarity in their structural formula.
Suppose a solution is 0.70 mole fraction in benzene and 0.30 mole fraction in toluene. The vapour pressure of pure benzene and pure toluene are 75 mmHg and 22 mmHg respectively. Hence the total vapour pressure is 
Other Example of Ideal Solutions
(1) Benzene + Toluene,
 (2) n hexane + n Heptane;
 (3) Chlorobenzene + Bromobenzene
(4) Ethyl bromide + Ethyl iodide;
(4) n-Butyl chloride + n-Butyl bromide
 (5) Ethyl alcohol + Methyl alcohol
(6) Tetrachloromethane + Tetrachlorosilane

REAL OR NON - IDEAL SOLUTIONS
Those solution which do not obey Raoult’s law over entire range of composition and deviate from ideal behaviour, are real or non – ideal solution.
Distinction between Ideal and Non Ideal Solutions:

Types of non-ideal solutions:
(1) Non ideal solutions showing positive deviation
(2) Non ideal solutions showing negative deviation.

(1) Non ideal solutions showing positive deviation:
(A)Characters for positive deviation:
When two liquids A and B on mixing form this type of solution and show following characters
(1) Non-ideal solution showing positive deviation from Raoult’s law.
(2) A—B attractive force should be weaker than A—A and B—B attractive forces.
(3) ‘A’ and ‘B’ have different shape, size and character.
(4) ‘A’ and ‘B’ escape easily showing higher vapour pressure than the expected value.
(5) The solution showing positive deviations from ideal behaviour for those type of solutions,
(B) Condition of positive deviation:
(C) Graph of Positive deviation:
(D) Examples and cause of Positive Deviation:

(A) Difference in extent of association in two liquids
(1) H2O and CH3OH (Methanol)
(2) H2O and ROR’ (Ester)
(3) H2O and CHCl3 (Chloroform)
Explanation: Mixture of above pair produces a high distorted curve with maximum  vapour pressure.  
(B) Association in one of the liquids through H-bonding
(4)  C2H5OH and C6H6 (Benzene)    
(5) ROH and ROR’ (Ester)
(6) ROH and CHCl3 (Chloroform)
(7) ROH CH3COCH3 (Acetone)
(8) ROH and C6H12 (Cyclohexane)
Explanation:           
(C) Greater difference in length of hydrocarbon part of members of same homologous series
(9) n-butane and n-Heptane
(D) Difference in polarity of liquids: General Examples are when one is polar and other is non polar 
(10) CCl4 and CHCl3
Explanation:           
(5) Greater Difference in molar mass of non-polar molecules.
(11) CCl4 and C6H6
(13) CCl4 and Toluene
(14) Acetone and Benzene
(15) CS2 and Acetone
(16) CH3OH and Benzene
Explanation:    
   

(2) Non ideal solutions showing negative deviation

(A) Characters for Negative deviation:
When two liquids A and B on mixing form this type of solution and shows following character
(1) Non-ideal solution showing negative deviation from Raoult’s law.
(2) A—B attractive force should be greater than A—A and B—B attractive forces.
(3) ‘A’ and ‘B’ have different shape, size and character
(4) Escaping tendency of both components ‘A’ and ‘B ’is lowered showing lower vapour pressure than expected ideally.
(B) Condition of Negative deviation:
The solution showing large negative deviations from ideal behaviour and the vapour pressure of each component is considerably less than that predicted by Raoult’s law, for these type of solutions.
(C) Graph of Negative deviation:


(D) Examples and cause Positive Deviation:


(1) An acidic & a basic liquid: Due to strong intermolecular hydrogen Bonding between the proton of the acid & lone   pair of the donor atom of the basic liquid (C6H5OH & C6H5NH2)
(2) Haloalkanes (like chloroform) with more electronegative atoms: like oxygen or nitrogen or   fluorine containing liquid (like ketones, esters, ethers, amines etc) due to formation of Hydrogen – bonding between these.
Exception: Excluding ALCOHOLS which are highly associated and would show positive deviations.
(3) Aqueous solutions of strong volatile acids and water:  For example sulfuric acid, nitric   acid etc., which give non-volatile ions with water

Newly form hydronium ions and Sulphate ions strongly associated hence these solution show negative deviation


HEXAGONAL CLOSE PACKING (HCP):

Step-(1) In order to develop three dimensional close packing take a 2D hexagonal close packing sheet as  first layer (A- layer).

Step-(2) Another 2D hexagonal close sheet (B-layer) is taken and it is just over the depression (Pit) of the first layer (A) .When the second layer is placed in such a way that its spheres find place in the ‘b’ voids of the first layer, the ‘c’ voids will be left unoccupied. Since under this arrangement no sphere can be placed in them, (c voids), i.e. only half (50%) the triangular voids in the first layer are occupied by spheres of the second layer (i.e. either b or c)

Step-(3) There are two alternative ways in which spheres in the third layer can be arranged over the second layer
(1)  When a third layer is placed over the second layer in such a way that the spheres cover the tetrahedral or ‘a’ voids; a three dimensional closest packing is obtained where the spheres in every third or alternate layers are vertically aligned (i.e. the third layer is directly above the first, the fourth above the second layer and so on) calling the first layer A and second layer as layer B, the arrangement is called ABAB …………. pattern or hexagonal close packing (HCP) as it has hexagonal symmetry.   
ANALYSIS OF HCP UNIT CELL:
(1) Number of effective atoms in HCP unit cell (Z):


Lattice point:  corner- total 12 carbon contribute 1/6 to the unit cell
Lattice point:   face- total face 2 contribute ½
Lattice point: body centre- total atom 3 (100% contribution)
(2) Radius of atom in HCP unit cell:

Let the edge of hexagonal base =( a) And the height of hexagon =( h) And radius of sphere =( r)
(3) Area of hexagon:
 Area of hexagonal can be divided into six equilateral triangles with side 2r
(4) Height of HCP unit cell:
(5)  Volume of hexagon = area of base x height
(6)  Volume of spheres:
(7) Packing efficiency: Percentage of space occupied by sphere.   
(8) Voids %: 100 - PE= 26 %
(9) Coordination Numbers:  12 (each spheres touches 6 spheres in its layer,3 above and 3 below).

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