a=Edge of the unit cell
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1.SOLID STATE Number of atoms in a unit cell: Type of unit cell Simple (Primitive)cubic lattice Body centered unit cell Face cetred cubic cell No of atoms per unit cell(Z) 1 2 4 Relation between atomic radius(r) and the edge length (a): Simple cube : r= Face centred: r= Body centred: r= π 2 π 2√2 √3π 4 =0.3535a =0.433a Relationship betweenDensity and edge of cubic crystals: ρ= π π ππ΄ π3 Where ρ= density of the crystal in g mol-1 Z=Number of particles per unit cell M=Molar mass of the element. a=Edge of the unit cell 2. SOLUTIONS 1. Molarity = ππ’ππππ ππ πππππ ππ π‘βπ π πππ’π‘π ππππ’ππ ππ π‘βπ π πππ’π‘πππ ππ πΏ 2. Mole fraction = 3. Molality π1 π1 +π2 ππ’ππππ ππ πππππ ππ π πππ’π‘π = πππ π ππ π πππ£πππ‘ ππ πΎπ Henry’s law: The mass of a gas dissolved in a given volume of of the liquid at constant temperature is directly proportional to the pressure of the gas present in equilibrium with the liquid. m = kH p The widely used form of Henry’s Law is “The partial pressure of a gas in vapour phase is directly proportional to the mole fraction of the gas in the solution.” π«π= π¦β π³π Raoult’s law: “The vapour pressure of a solution containing non-volatile solute is directly proportional to the mole fraction of the solvent.” π« A=π« β¦Aπ³ A Raoult’s law as a special case of Henry”s law: Comparing the two laws, the following conclusion can be drawn: Pressure of the volatile component of a solution is directly proportional to its mole fraction. The two equations become identical when KH becomes equal to π« β¦A . Colligative properties: Those properties which depend only on the number of particles of the solute not on the nature of the solute particles. Relative lowering of vapour pressure: ππ = ππ π¨ ππ π¨−ππ¨ ππ π¨ = π§B = πΎ π© ⁄π΄ π© πΎπ¨ ⁄π΄π¨+ πΎπ© ⁄π΄π© Δp= Depression in vapour pressure π0 π΄=Vapour pressure of pure solvent π«π = πππππ’π ππππ π π’ππ ππ π πππ’π‘πππ π³β¬ = ππππ πππππ‘πππ ππ πππππππππ‘ π΅(ππππ’π‘π) π²A=Mass of component A(solvent) Elevation in boiling point: ΔTb=π¦ b m= π¦ b ππ΅ ×1000 ππ΅ ππ΄ ΔTb= Elevation in boiling point π¦ b = molal elvation constant in K Kg mol-1 m = molality( mol/g) WB= Mass of solute(g) MB= Molar mass of solute WA =Mass of solvent in g Kb = ππ΄ π ππ2 π₯π£ππ π»×1000 Kb = Molal elevation constant (K Kg mol-1) MA=molar mas of solvent R= Universal gas constant (8.314 J K-1mol-1; 8.314*10-2 L bar K-1mol-1; 8.21*10-2 L atm K-1mol-1) π₯π£ππ π»=Enthalpy of vaporization of solvent ππ = π΅ππππππ πππππ‘ ππ ππ’ππ π πππ£πππ‘ Depression in freezing point: ΔT f= K f m ΔTf= πΎπ ×ππ΅ ×1000 ππ΅ ππ΄ Kf= Molal freezing constant Determination of molar mass from osmotic pressure: π= ππ΅ ×π π πππ΅ (ππ ) π = πΆπ π Π = Osmotic pressure C= Molarity V= Volume of solution in L Units of pressure: 1 bar = 10 5 pascal 1atm = 1.01*10 5 Pa =760 torr = 1.01 bar = 760 mm Hg 1 mmHg= 1 torr 1 torr= 1.33*10 -3 bar VAN’T HOFF’S FACTOR: πππ πππ£ππ π£πππ’π ππ ππππππππ‘ππ£π πππππππ‘πππ i = ππππππ π£πππ’π ππ ππππππππ‘ππ£π πππππππ‘πππ (ππ π π’ππππ ππ ππ π ππ πππ‘πππ ππ πππ π ππ πππ‘πππ) π−1 πΌ= π−1 πΌ =the degree of dissociation i= Vant’ Hoff factor i=1 if the solute behaves normally i>1 if solute undergoes dissociation πΌ= π−1 π−1 i<1 if the solution undergoes association πΌ= 1−π 1 π 1− n = no of ions produced on dissociation/no of ions undergoing association. 3.ELECTROCHEMISTRY For the reaction aA+ bB cC +dD Nernst equation to calculate the electrode potential of the cell is E cell= Eβ¦cell - 0.0591 π log [πΆ]π [π·]π [π΄]π [π΅]π STANDARD ELECTRODE POTENTIAL: E0cell= E0cathode - E0anode AT EQUILLIBRIUM, Ecell= 0 ,then E0cell= π.πππ π log Kc GIBB’S FREE ENERGY/MAXIMUM WORK DONE ΔGo= -nF E0cell G= 1 π π ρ=R 1 π G*=R π π ×1000 πΆ Λm= π Λm=πΆ∗1000 C= mol L-1; πΌ= π¬ππ π¬°π G= Conductance, a= area, π π = where R= Resistance l=length π =conductivity; ρ=resistivity G*= cell constant Λm=molar conductivity in S cm2 mol-1 k=conductivity in S cm-1 C= molarity in mol/dm3 ( 1dm3=1000cm3) Λm =molar conductivity in S m2 mol-1 K = conductivity in S m-1 C =molarity in mol/m3 ( 1m3=1000L; 1L=1/1000m3;L-1 = 1000 m-3) π¬ππ =molar conducitivty of solutions at any concentration. π¬°π =limiting molar conductivity. 4. CHEMICAL KINETICS Rate= πβππππ ππ π‘βπ πππππππ‘πππ‘πππ ππ πππππ‘πππ‘ ππ πππππ’ππ‘ π‘πππ πΉππ π‘βπ πππππ‘πππ π΄ π΅ Rate= -d[R] = d[P] dt dt Rate= K [A]α[π΅]π½ Order=πΌ + π½ Unit for k: Zero order reaction =mol L-1 s-1 First order reaction =sec-1 Second order reaction =L mol-1 sec-1 Differential rate equation: πΉππ π‘βπ πππππ‘πππ π΄ − π[π ] ππ₯ π΅ = π[π ]π (For n order) Integrated rate equation: 1. Zero order 1 π = π‘ [[π ]0− [π ]] (Rate of reaction= Rate constant ) 2. First order reaction: k= 2.303 [π ]0 πππ π‘ [π ] 3. First ordeReaction for gaseous substances k= 2.303 ππ πππ π‘ (2ππ − ππ‘ ) where pi =initial pressure, pt = total pressure Half life period: Zero order: π‘1/2 = [π ]0 2π First order: π‘1 2= 0.693 π Arrhenius Equation: K= Aπ −πΈπ⁄π π π −πΈπ⁄π π = Fraction of molecules that have energy equal to or greater than Ea ππ Log k= logA - log π2 π1 = πΈπ 2.303 π π.πππππ [ π2− π1 π2 π1 ]
