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Sunday, April 28, 2019

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


Saturday, April 27, 2019

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).

                Most recommended books of Physical Chemistry Books
(1) Physical Chemistry
 Book by O.P Tandon

 
(2) Physical chemistry
 Book by KS Verma (Cengage Publications)

(3) Physical Chemistry
Book by P.W. Atkins (Oxford publication)
 
(4) Physical Chemistry
Book by P. Bahadur
 
(5) Physical Chemistry
Book by Ranjeet Shahi (Arihant publication)
 
(6) Balaji Advanced problems in Physical chemistry
Book by Narendra Awasthi
 
 
(7) Modern Approach to Chemical Calculations
 Book by R.C. Mukherjee


FACE CENTRED UNIT CELL (FCC/CCP):

CCP or FCC has two lattice point corner as well as face centred:

Suppose ‘r’ be the radius of sphere and ‘a’ be the edge length of the cube As there are 4 sphere in FCC unit cell

(1) Relation between radius (r) and side (a)
In FCC, the corner spheres are in touch with the face centred sphere. Therefore, face diagonal AD is equal to four times the radius of sphere AC = 4r
But from the right angled triangle ABC:
(2)  Effective no. of atoms per unit cell (Z):
(3) Volume of four spheres (atoms):
(4) Volume of unit cube:
(5) Packing efficiency (PE): Percentage of space occupied by sphere
(6) Percentage Voids: 100- PE= 26 %
(7) Density of FCC(CCP):
(8) Coordination Numbers:
(9) LOCATION OF VOIDS: FCC/CCP UNIT CELL:

(A) Tetrahedral voids: The FCC/CCP unit cell has eight tetrahedral voids per unit cell. Just below every corner of the unit cell, there is one tetrahedral void. As there are eight corners, there are eight tetrahedral voids.

(B) Octahedral voids: In an FCC/CCP unit cell, there are four octahedral voids. They are present at all the edge centres and at the body centre. The contribution of the edge centre is 1/4
Hence, total number of octahedral voids:
In CCP/FCC:
 Rank (Z) = 4,
 Number of tetrahedral voids = 8 and
 Number of tetrahedral voids = 2 × Z
Number of tetrahedral voids in close packing = 2 × eff. no. of spheres.
Hence, there are two Tetrahedral Voids per sphere in closed packing arrangements.
In CCP/FCC:
 Z = 4
Number of octahedral voids = 4 
Number of octahedral voids = Z
There is exactly one OV per sphere in close packing.