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Physical Chemistry Formula Sheet - Formula Sheet OF
Physical Chemistry
NEET (ALLEN Career Institute)
Studocu is not sponsored or endorsed by any college or university
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1
Some Basic Concept of Chemistry
W(g)  N A
GMM
1.
Number of molecules in W(g) of substance =
2.
Molality (m) = No. of moles of solute
Mass of solvent in kg
3.
Number of molecules in V litre of gas at S.T.P. =
4.
5.
6.
7.
VNA
22.4
W(g)
(GAM → gram atomic mass)
GAM
W(g)
Number of gram molecules =
Gram molecular mass
Dilution formula: M1V1 = M2V2
For mixing two solutions of the same substance
M1V1 + M2V2 = M3(V1 + V2)
Molarity can be directly calculated from % by mass (w/w) if density is known
% 10 d
Molarity =
GMM
Mass of 1 atom of element = GAM
NA
Number of gram atoms =
8.
Mass of 1 molecule of substance = MM (MM → Molar mass)
NA
9.
T(K) = T(C) + 273.15
Mass of an atom of the element
1
 Mass of an atom of carbon (C-12)
12
11. Number of molecules in n moles of substance = n × NA
Mass of that element in 1 mole of the compound
12. Mass % of an element in a compound =
Molar mass of the compound
10. Relative atomic mass =
Mass of solute
100
Mass of solution
XB
Molality × M A
14.
=
where MA - mass of solvent
1− X B
1000
13. Mass percent =
No. of moles of solute
mole / L
Volume of solution in litres
16. Avogadro’s No. NA = 6.022 × 1023
9
17. T(F) = T(C) + 32
5
18. Molecular mass = 2 × vapour density
No. of moles of A
19. Mole fraction of A =
No. of moles of solution
15. Molarity (M) =
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Structure of Atom
1.
Wavelength of matter wave
=
=
h
mv
h
p
h
2Em
Where, E = Kinetic energy
=
2.
Total number of nodes = n – 1
Radial nodes = n – l – 1
Angular nodes = l
3.
4.
Number of neutrons = A – Z
Number of subshells in nth shell = n
Number of orbitals in nth shell = n2
Number of electrons in nth shell = 2n2
Number of orbitals in subshell = 2l + 1
Number of electrons in subshell = 2 (2l + 1)
5.
Energy of quantum of radiation according to Planck’s quantum theory
E = hv
6.
7.
1
hv = hv 0+ m ve 2
2
Einstein’s photoelectric equation.
Line spectrum of hydrogen
1
1
1 
v =109677 2 − 2 cm −1 where v is wave number and v =

 n1 n 2 
Where:
n1 = 1, 2, …..
n2 = n1 + 1, n1 + 2, …….
8.
Bohr’s model of hydrogen atom
(a) Frequency of radiation absorbed or emitted during transition; v =
E
h
E 2 − E1
h
E1 = Energy of lower energy state
v=
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E2 = Energy of higher energy state.
(b) Orbit angular momentum of an electron, m e vr = n.
h
2
Where, n = 1, 2, 3,…..
 Z2 
(c) Energy of stationary states E n = −2.1810 −18  2  J
n 
n 
r = 52.9  2
(d) Radii of the stationary states/orbits n
  pm
 Z
9.
 1
1 
Energy gap between the two orbits E = R H  2 − 2 
 ni n f 
Where RH = 2.18 × 10–18
Where, ni = initial orbit
Nf = final orbit
10. Atomic number (Z)
= Number of protons in the nucleus of an atom
= Number of electrons in a neutral atom
11. Heisenberg’s uncertainty principle
x  p 
h
;
4
x  mv 
h
4
12. Speed of light is equal to the product of frequency and wavelength of light c = V
13. Mass Number (A) = Number of protons + Number of neutrons
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State of Matter
1.
2.
3.
4.
5.
6.
7.
8.
9.

an2 
Van der Waals Equation: P + 2  (V − nb) = nRT
V 

P1V1 P2V2
=
; (n, R constant)
T1
T2
Dalton’s Law of partial pressures: PTotal = P1 + P2 + P3 + …….
Ideal gas equation: PV = nRT
dRT
M=
P
V V
Charles’s Law: 1 = 2 (at constant P and n)
T1 T2
Avogadro’s Law: V = kn (at constant P and T)
Partial Pressure in Terms of Mole Fraction: Pi = xi PTotal
Boyle’s Law: P1V1 = P2V2 (at constant T and n)
PV
Compressibility factor Z =
(for 1 mole of gas)
RT
Combined Gas Equation:
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Thermodynamics
➢
⬥
⬥
⬥
THERMODYNAMICS:
Study of heat and work interaction between system and surrounding.
A macroscopic science.
Thermodynamic laws are experimentally verified.
➢
⬥
⬥
⬥
⬥
Important terms and concepts in thermodynamics.
System - Portion of universe under investigation.
Surrounding - Anything apart from system.
Boundary - Real or hypothetical line or surface between system and surrounding.
Wall - A real boundary.
Rigid wall - Immovable wall (w = 0)
Non-rigid wall - Movable wall (w  0)
Adiabatic wall - Insulated wall (q = 0)
Diathermic wall - Non-insulated wall (q  0)
State variable - Variable which defines state of system.
State of system - A condition defined by fixed value of state variables.
State of thermodynamic equilibrium - A condition in which state variables do not vary with time.
⬥
⬥
⬥
➢
Extensive state variable : State variable whose value depends upon size of system.
Examples - mass, volume, charge, mole etc.
➢
Intensive state variable : State variable whose value does not depends upon size of system.
Examples - concentration, density, temperature etc.
➢
⬥
⬥
Path variable :
Heat : Mode of energy transfer between system and surrounding due to temperature difference.
Work : Mode of energy transfer between system and surrounding due to difference in generalized
force.(Net force).
THE FIRST LAW
(i) Energy of universe is conserved
(ii) Internal energy (U) of a system is state function.
(iii) U = q + w
U = Increase in internal energy of system.
q = Heat absorbed by the system
w = work done on the system
(iv) In a cyclic process  U = 0
Cyclic
If a cyclic process involves n steps with heat absorbed and work done on the system, qi and wi
respectively, then –

Cyclic
i=n
i=n
i=n
i=1
i=1
i=1
U =  (q i + w i ) =  q i +  w i = 0
 Qnet = –Wnet (in a cyclic process)
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(v) If two states 1 and 2 are connected by n paths involving qi and wi, heat and work respectively,
then
U = q1 + w1 = q2 + w2 = .............. qn + wn
(vi) q and w are path dependent quantities (indefinite quantities) but there sum is a definite
quantity (U).
⬥
⬥
Enthalpy : A state function defined by first law
H = U + PV
(i) Enthalpy is (pressure volume energy + internal energy of system)
(ii) Enthalpy is also called heat content of system.
Heat absorbed at constant volume and constant pressure.
q v = U Heat absorbed by a system in isochoric process is equal to change in internal energy of
system.
q p = H Heat absorbed at constant pressure by a system is equal to change in enthalpy.
⬥
Enthalpy change :
For General process –
H = U + P2V2 – P1V1
For Isobaric change –
H = U + PV
For Isochoric change –
H = U + V(P)
For a differential change
dH = dU + PdV + VdP
...........(i)
..........(ii)
.........(iii)
.........(iv)
➢
⬥
Ideal gas processes : (See table page no. 11)
Enthalpy of phase transition
Hvap = heat absorbed at constant temperature and pressure to convert one mole liquid into it’s
vapours.
= molar enthalpy of vapourisation.
Hfusion = heat absorbed at constant temperature and pressure to convert one mole solid into liquid.
= molar enthalpy of fusion.
Hsublimation = heat absorbed at constant temperature and pressure to convert one mole solid into it’s
vapours.
= molar enthalpy of sublimation.
H = U + P(Vf – Vi) since phase transitions are isobaric and isothermal processes.
⬥
Relationship between H and U for phase transitions.
For vapourisation Hvap = Uvap + RT
For sublimation Hsublimation = Usublimation + RT
For fusion Hfusion  Ufusion
⬥
Heating curve at constant pressure:
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Enthalpy of reaction (rH) : The enthalpy of reaction is heat exchanged at constant pressure and
temperature to convert the stoichiometric amount of reactant into product with specified physical
state according to balanced chemical reaction at constant temperature and pressure.
for aA + bB ⎯→ cC + dD
rH = qP = enthalpy of reaction
rH = (cHC + dHD – aHA – bHB) where HA, HB, HC, HD are molar enthalpies of A, B, C and D.
⬥
Relationship between rH and rU
(for ideal gas)
rH = rU + ngRT
(for non ideal conditions)
rH = rU + P(Vf – Vi)
⬥
The stoichiometric coefficient of solids and liquids in not considered in calculation of ng
(because VS ~ VL << Vg)
⬥
➢
•
•
➢
•
•
Standard state for
(i) Ideal gas : 1 bar pressure ; any temperature.
(ii) Solid / Liquid : 1 bar pressure ; any temperature.
(iii)Solute : Molar concentration of 1 mole/L at P = 1 bar.
Standard enthalpy, internal energy change for reaction.
rH and rU are change in thermodynamics function of a system under standard conditions.
SECOND LAW
Spontaneous process :
A process which takes place on it’s own without any external help.
Spontaneous process  Irreversible process  Natural process.
Second law : During a spontaneous process.
Suniverse > 0
 Ssystem + Ssurr. > 0
S is a state function. S is measure of disorder of a system.
(A) Change in entropy of system is given by :
dq
dSsystem = rev.
T
(i) Entropy change for ideal gas process:
T
V

S = nC v ln 2 + nR ln 2
T1
V1
(ii) Entropy change for system in phase transition:
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Hvap.

Svap. =

Sfusion. =
Tb
Hfusion
Tf
Hsublimation
TSub.
(iii) Entropy change of system for a chemical reaction:
For a reaction –
 aA + bB ⎯→ cC + dD
 r S = cSc + dSD − aSA − bSB

Ssub lim ation =
SA, SB, SC and SD are molar absolute entropies which is obtained by third law.
(B) Entropy change in surrounding:
−q
(i) Ideal gas process : S surr. = actual
T
−H
(ii) Phase transition : S surr. =
T
H
(iii) Chemical reaction : S surr. = − r
T
For reversible processes :  system + Ssurr. = 0
system = −Ssurr.
For irreversible processes : system + Ssurr.  0
 total  0
⬥
Prediction of sign of rS from inspection :
(i) If ng > 0 ; rS > 0
Solid ⎯⎯
→liquid
(ii) If
 rS  0
liquid ⎯⎯
→gas 
(iii) If cyclisation taken place rS < 0.
➢
Gibb’s function : G = H – TS
G = H – TS → For isobaric change
G = –T(STotal) → Process spontaneous
 (G)T,P  0
(A) Change in G for phase transition :
(i) For reversible phase transitions : G = 0.
(ii) For irreversible phase transition : GP,T = HP,T – TSP,T
(B) Change in G for chemical reaction :
aA + bB ⎯→ cC + Dd
.....(i)
rG = cGC + dGD – aGA – bGB
.....(ii)
rG = rH – TrS
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rG = rG + RT ln Q
Where, Q  Reaction quotient
.....(iii)
G / G and state of chemical equilibrium :
At equilibrium :
G = 0  Gproduct = Greactant
G = –RT ln Keq.
At equilibrium the system gibb’s function is at minimum value.
➢
•
•
•
➢
Difference between rG and rG :
rG = change in Gibb’s function when all the reactants and products have arbitrary activities.
rG = change in Gibb's function when all the reactants and products are at unit activities.
 All gases at 1 bar pressure.
 All solute at molar concentration 1 M.
➢
Factors on which rG depends (i) Stoichiometric coefficients of a balanced chemical reaction.
(ii) the temperature.
(iii)the rG is independent of actual pressure or concentration of reactants or products.
➢
Gibb’s function and non-PV work :
–(G)T, P = Wmax
decrease in Gibb's function at constant temperature and pressure is equal to maximum non-PV
work obtainable from system reversibly.
–rG = –rH + TrS
Decrease in Gibb’s function = heat given out to surrounding + TrS.
Process
Reversible
isothermal
process
Expression for
w
w = −nRTln
= −nRTln
V2
V1
P1
P2
Irreversible
isothermal
process
IDEAL GAS PROCESSES:
Expression for
U
H
q
 V2 
V 
 1
 P1 
q = nRTln
P 
 2
q = nRTln
w = –Pext (V2 – V1)
 nRT nRT 
= Pext 
−
P
P  q = Pext (V2 – V1)
 2
1 
0
0
0
0
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Work on PV-graph
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Isobaric
process
w = –Pext (V2 – V1)
q = H = nCPT
H =
U =
nCVT nCPT
Isochoric
process
w=0
q = U = nCVT
U =
H =
nCVT nCPT
w = nC V (T2 − T1 )
q=0
= constant
TV–1 = constant
TP1–/ = constant
U =
H =
nCVT nCPT
Reversible
adiabatic
process
=
Irreversible
adiabatic
process
w = NcV(T2 – T1)
P2 V2 − P1V1
 −1
Polytropic
process
P V −PV
w= 2 2 1 1
n −1
R(T2 − T1 )
w=
(n −1)
P2 V2 − P1V1
 −1
PV
T2
q =  C V dT
T1
T2
R
+ 1− n dT
H =
U =
nCVT nCPT
T1
V2 = Final volume
P2 = Final pressure
V1 = Initial volume
P1 = Final pressure
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Chemical Kinetics
➢
Rate of reaction (ROR) =
➢
For a reaction:
aA + bB ⎯→ cC + dD
➢
▪

Rate of disappearance of reactant (appear ance of products )
Stoichiometric coefficient of reactant (p roducts )
1  d A 
1  d B  1  d C  1  d D
Instantaneous rate : – 
= – 
= 
= 

a  dt 
b  dt  c  dt  d  dt 
Relationship between rate of reaction and rate of disappearance of reactant (rate of appearance of
product).
Average rate : – 1  Δ A  = – 1  Δ B = 1  Δ C  = 1  Δ D 
b  Δt  c  Δt  d  dΔt 
a  Δt 






Graphical method for determining rate :
(
 R 2 – R 1  P 2 – P 1
Avg. Rate = – 
=
t
–t
t 2 –t1
2
1


)
OA'
OA 
Instantaneous rate = – 
= ± slope of tangent
= +
OB
OB'


➢
Important kinetic expression for reaction of type A ⎯→ B :
Order
Differential
rate law
Integrated
rate law
Zero
1st
2nd
Rate = k
Rate= k[A]
[A0 ]–[A]= kt
kt = In
A
Half life
(t1/2)
t1/2 =
(t3/4)
t3/4=1.5 t 1/2
0
2k
t =
1/2
A
A
In2
k
t 3/4= 2 t1/2
0
nth
Rate = k[A]2
kt =
t =
1/2
1
1
–
A A
Rate = k[A]n
kt =
0
1
A k
0
t 3/4 = 3 t1/2
t1/2 =

1  1
1
n–1 –
( n–1)   A  A0 n–1 


1  2n–1 –1 
k ( n–1)   A
 0
n–1
t 3/4= (2 + 1) t1/2
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


n–1
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➢
Graphs of various order
Order
Rate vs [A]
[A] vs t
log [A] vs t
1
t A
vs
Zero order
First order
Second order
Where
[A]0  initial concentration
[A]  concentration at time t
t1/2  time taken for initial concentration of reactant to finish by 50%
t3/4  time taken for initial concentration of reactant to finish by 75%
➢
e.g.
Monitoring Kinetics Experimently :
The kinetics of reaction can be followed (i.e. order, rate constant etc. can be established) by
measuring a property which changes with time.
(i) Total pressure in a gaseous reaction.
(ii) Volume of a reagent (Acidic, Basic, oxidising or reducing agent)
(iii) Volume of a gaseous mixture (V)
(iv)Optical rotation (R)
For a Reaction An ⎯ → nB
t=0
c0  conc. at t = 0
0
c
t = t
c–x
nx
ct  conc. at t = t
t=
0
nc
c  conc. at t = 
For any measurable property X proportional to the concentration of reaction mixture at various
times, following relations can be expressed.
In terms of –
(i) X0 and x
(ii) X0 and Xt
(iii) X and Xt
(iv) X0, Xt and X
1
X0
1 (n −1)X 0
1 (n −1)X  t
1  X − X0 
k = ln
k = ln
k = ln
k = ln  
t
X −X 
t X0 − x
t nX 0 − X t
n(X  − X t )
 
t 
where
x  amount of reactant reacted in time 't'.
X0  measured property at t = 0
Xt  measured property at t = t
X  measured property at t = 
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➢
(i)
Examples : (For Monitoring Kinetics Experimently)
Inversion of cane sugar :
+
C12 H 22 O11 (aq) + H 2O ⎯H⎯→C6H 12O 6(aq)+ C 6H 12O 6 (aq)
Sucrose
dextro-rotatory
(+66.5)
Glucose
dextro-rotatory
(+52.5)
Fructose
laevo-rotatory
(–92)
(Laevo−rotatory)
k=
2.303  r − r0 
log 

t
 r − rt 
r0 = rotation at time, t = 0
rt = rotation at time, t = t
r = rotation at time, t = 
(ii)
Acidic hydrolysis of ethyl acetate :
H+
CH 3 COOC 2 H 5 + H 2 O ⎯⎯→ CH 3 COOH + C 2 H 5 OH
2.303  V − V0 
log 

t
 V − Vt 
V0 = Volume of NaOH solution used at time, t = 0
Vt = Volume of NaOH solution used at time, t = t
V = Volume of NaOH solution used at time, t = 
Note: Here NaOH acts as a reagent. Acetic acid is one of the product the amount of which can be
found by titration against standard NaOH solution. But being an acid-catalysed reaction, the acid
present originally as catalyst, also reacts with NaOH solution.
K=
➢


Important characteristics of first order reaction :
t1/2 is independent of initial concentration.
In equal time interval, reactions finishes by equal fraction.
t = 2t
t = 3t ....
t=0
t=t
a0 x2
a0 x3........
a0 x
Reactant conc. a0
x = fraction by which reaction complete in time 't'.
k
2.303

Graph of ln[A] vs t is straight line with slope =

Graph of [A] vs t is exponentially decreasing.
➢
Zero order :
t1/2 of zero order is directly proportional to initial concentration.
In equal time interval, reaction finishes by equal amount.
t=0
t=t
t = 2t
t = 3t .....
C0
C0 – x C0 – 2x
C0 – 3x ....
Graph of [A] vs t is straight line.
[A]0
A zero order reaction finishes in t =
k
Temperature dependence :
Arrhenious equation : k = A.e–Ea/RT
Ea = minimum energy over and above the avg. energy of reactant which must be possessed by
reacting molecule for collision to be successful.
A = frequency factor - proportional to number of collisions per unit volume per second.



➢
–
–
–
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–
–
e–Ea/RT = Fraction of collision in which energy is greater than Ea.
A and Ea are constant i.e. do not vary with temperature.
E
ln k = ln A − a
RT
Graph : Graphical determination of Ea.
Temperature coefficient =
kT+10
kT
By default T = 298 K
Variation of rate constant with temperature  ln
➢
k2 Ea  1 1 
=
−

k1 R T1 T2 
Endothermic and exothermic reactions :
H = E af − E ab
➢
Parallel reaction :
Rate = (k1 + k2) [A] – (differential rate law)
k1 [B]
=
(ii)
k2 [C]
(i)
(iii) t 1/ 2 =
0.693
k1 + k 2
k1
k2
100 ; % of C =
100
k1 + k 2
k1 + k 2
(v) [A] = [A] 0 e −(k1 +k 2 )t
(iv) % of B =
➢
Pseudo-order reaction :
Rate law → rate = k [A]m [B]n
Pseudo rate law :
rate = k1 [A]m
[B] assumed constant in two cases :
(i) B in large excess
(ii)
B → CATALYST
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Solid State
➢
Density of Cubic Crystal System
(or d) =
Z M
g.cm–3
a3  NA
Where, Z = number of atoms per unit cell
NA = Avogadro's Number
M = Gram atomic weight of element (g. mol–1)
a = edge – length
➢
Radius-ratio in ionic solids
Voids
Radius-ratio
Triangular
+
0.155  r  0.225
r−
Tetrahedra
+
0.225  r  0.414
r−
Octahedral
+
0.414  r  0.732
r−
Cubical
+
0.732  r  0.999
r−
➢
Crystal
Radius of atom (r)
Simple cubic
a
2
a
22
F.C.C.
B.C.C.
➢
No. of atoms per unit
cell
1
52.4%
2
74%
2
68%
3
a
4
Packing Fraction (p. f.) =
Volume occupied by particles (per unit cell)
Volume of unit cell
Contribution of each atom present on the corner =
1
8
Contribution of each atom present on the centre of face =
1
2
Contribution of each atom present on the body centre = 1
Contribution of each atom present on the edge centre =
1
4
➢
Seven crystal system with dimensions:
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p.f.
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(a) Cubic:  =  =  = 90; a = b = c
(b) Tetragonal:  =  =  = 90; a = b  c
(c) Orthorhombic:  =  =  = 90; a  b  c
(d) Monoclinic:  =  = 90,  = 90; a = b = c
(e) Hexagonal:  =  = 90,  = 90; a = b = c
(f) Rhombohedral or trigonal:  =  =  = 90; a = b = c
(g) Triclinic:       90; a  b  c
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Solutions
➢
Depression in freezing point
Tf = K f  m
Kf =
➢
RT02
M A RT02
=
1000Ifusion 1000Hfusion
Dissociation of solute
(A) n → nA
i=
1+ (n −1)
( = degree of dissocition)
1
PAo X A + PBo X B 
 Raoult’s law
(PBo − PAo )X B + PA0 
➢
Psol =
➢
Osmotic pressure
 = CRT
For isotonic solution, 1 = 2
➢
1 YA YB
=
+
Psol PAo PNo
➢
Relative lowering of vapour pressure
PAo − P
=
PAo
➢
YA =
B
( B = mole fraction of solute)
PA
P
; YB = B
PA
PT
Where YA and YB are mole fraction in vapour phase and P A = PAo X A : PB = PBo X B
➢
Elevation of boiling point
Tb = K b  m
Kb =
RT02
M ART02
=
100I vapour 1000H vapour
Where, m = molality
MA = molecular mass of solvent
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➢
Molality (m) = Number of moles of solute mole/kg
Weight of solvent (kg)
Molarity (M) = Number of moles of solute mole/L
Volume of solution (L)
Normality (N) = Number of gram equivalent of solute g equiv/L
Voume of solution (L)
Formality (F) = Number of gram of gormula mass
Volume of solution (L)
➢
Van’t Hoff factor
i=
Experimental colligative property (observed)
Calculate (normal) colligative property
PAB − P
(i  n B )
=
B
PA
(i  n B ) + n A
Tb = i  K b  m
Tf = i  K f  m
 = iCRT
➢
Association of solute
nA → (A)n
(1 − ) +
i=

n
1
➢
( = degree of association)
Henry’s Law
p = partial pressure of gas in vapour phase

 = K H .X  K H = Henry's law constant

 X = Mole fraction of gas
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IONIC EQUILIBRIUM
➢
1.
ACCORDING TO STRENGTH IONIC CONDUCTORS ARE OF 2 TYPES :
Strong electrolyte : Those ionic conductors which are completely ionized in aqueous solution are
called as strong electrolyte.
Ex. Na+Cl–, K+Cl–, etc.
(a) Strong acid → H2SO4, HCl, HNO3 HClO4, H2SO5, HBr, HI
(b) Strong base → KOH, NaOH, Ba(OH)2 CsOH, RbOH
(c) All Salts → NaCl, KCl, CuSO4..........
2.
Weak electrolytes : Those electrolytes which are partially ionized in aqueous solution are called as
weak electrolytes. For weak electrolytes the value of  is less than one.
Ex.
(a) Weak acid → HCN, CH3COOH, HCOOH, H2CO3, H3PO3, H3PO2, B(OH)3
(b) Weak base → NH4OH, Cu(OH)2, Zn(OH)2, Fe(OH)3, Al(OH)3
➢
ACIDS BASES AND SALTS :
Arrhenius concept :
Arrhenius Acid : Substance which gives H+ ion on dissolving in water (H+ donor)
Ex. HNO3, HClO4, HCl, HI, HBr, H2SO4, H3PO4 etc.
⬥ H3BO3 is not Arrhenius acid.
➢
Arrhenius base : Any substance which releases OH– (hydroxyl) ion in water (OH– ion donor).
⬥ First group elements (except Li.) form strong bases
➢
Bronsted - Lowery concept : (Conjugate acid - base concept) (Protonic concept)
Acid : substances which donate H+ are Bronsted Lowery acids (H+ donor)
Base : substances which accept H+ are Bronsted Lowery bases (H+ acceptor)
➢
Conjugate acid - base pairs :
In a typical acid base reaction
Ex:
Acid
HCl
H2SO4
HSO −
4
H2 O
Conjugate base
Cl–
HSO −
Base
NH3
Conjugate acid
NH+
H2 O
H O+
SO
OH–
RNH2
RNH+
4
2−
4
4
3
3
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➢
LEWIS CONCEPT (electronic concept) :
An acid is a molecule/ion which can accept an electron pair with the formation of a coordinate bond.
Acid → e– pair acceptor
Ex. Electron deficient molecules
: BF3, AlCl3
Cations
: H+, Fe2+, Na+
Molecules with vacant orbitals : SF4, PF3
A base is any molecule/ion which has a lone pair of electrons which can be donated.
Base → (One electron pair donor)
Ex. Molecules with lone pairs : NH3, PH3, H2O, CH3OH
Anions
: –OH, H–, –NH2
➢
IONIC PRODUCT OF WATER :
According to arrhenius concept
H2 O
H+ + OH– so, ionic product of water, Kw = [H+][OH–] = 10–14 at 25° (exp.)
Dissociation of water is endothermic, so on increasing temperature Kw increases.
Kw increases with increase in temperature.
Now pH = –log[H+] = 7 and pOH = –log[OH–] = 7 for water at 25°C (experimental)
pH = 7 = pOH
 neutral
pH  7 or pOH  7  acidic  at 25 C
pH  7 or pOH  7  Basic 
⬥
➢
Ionic product of water is always a constant whatever has been dissolved in water since its an
equilibrium constant so will be dependent only on temperature.
Degree of dissociation of water :
no. of moles dissociated
HO
H + + OH −   =
2
Total no. of moles initially taken
−7
10
=
= 18 10−10 or 1.8 10−7%
55.55
[at 25 C]
➢
Absolute dissociation constant of water :
+
−
−7
−7
HO
H + + OH−
K = k = [H ][OH ] = 10 10 = 1.8 10 −16
2
a
b
[H 2 O]
55.55
–16
So, pKa = pKb = – log(1.8 × 10 ) = 16 – log1.8 = 15.74
❑
ACIDITY AND pH SCALE :
Acidic strength means the tendency of an acid to give H3O+ or H+ ions in water.
So greater then tendency to give H+, more will be the acidic strength of the substance.
Basic strength means the tendency of a base to give OH– ions in water.
So greater the tendency to give OH– ions, more will be basic strength of the substance.
The concentration of H+ ions is written in a simplified form introduced by Sorenson known as pH
scale.
pH is defined as negative logarithm of activity of H+ ions.
 pH = −log a + (where a + is the activity of H+ ions)
H
H
Activity of H+ ions is the concentration of free H+ ions or H3O+ ions in a dilute solution.
The pH scale was marked from 0 to 14 with central point at 7 at 25 °C taking water as solvent.
If the temperature and the solvent are changed, the pH range of the scale will also change. For
example
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0 – 14 at 25 C (Kw = 10–14)
0 – 13 at 80 C (Kw = 10–13)
pH can also be negative or > 14
Neutral point, pH = 7
Neutral point, pH = 6.5
➢ pH Calculation of different Types of solutions :
(a) Strong acid solution :
(i) If concentration is greater than 10–6 M.
In this case H+ ions coming from water can be neglected,
so [H+] = normality of strong acid solution
(ii) If concentration is less than 10–6 M
In this case H+ ions coming from water cannot be neglected.
So [H+] = normality of strong acid + H+ ions coming from water in presence of this strong acid
(b) pH of a weak acid (monoprotic) Solution :
⬥ Weak acid does not dissociated 100 % therefore we have to calculate the percentage
dissociation using Ka dissociation constant of the acid.
⬥ We have to use Ostwald's Dilution law (as have been derived earlier)
HA
H + + A−
t=0 C
0
0
[H + ][A− ] C2
=
Ka =
teq C(1− ) C C
[HA]
1− 
If a <<1  (1 – )  1  Ka  C2   =
Ka
(is valid if  < 0.1 or 10%)
C
Ka
= K a C So pH = 1 (pKa − log C )
C
2
on increasing the dilution  C =  and [H+]  pH
[H + ] = C = C
(c) pH of a mixture of weak acid (monoprotic) and a strong acid solution :
⬥ Weak acid and Strong acid both will contribute H+ ion.
⬥ For the first approximation we can neglect the H+ ions coming from the weak acid solution and
calculate the pH of the solution from the concentration of the strong acid only.
⬥ To calculate exact pH, we have to take the effect of presence of strong acid on the dissociation
equilibrium of the weak acid.
⬥ If the total [H+] from the acid is more than 10–6 M, then contribution from the water can be
neglected, if not then we have to take [H+] from the water also.
➢
Relative strength of weak acids and bases :
For two acids of equimolar concentrations.
Ka
Strength of acid (I)
1
=
Strength of acid (II)
Ka 2
(d) pH of a mixture of two weak acid (both monoprotic) solution :
⬥ Both acids will dissociate partially.
⬥ Let the acid are HA1 & HA2 and their final concentrations are C1 & C2 respectively, then
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H+ +
HA1
0
t=0
C1
At eq. C1 (1− 1 ) C11 + C 2 2
C  (C  + C2 2 )
K a1 = 1 1 1 1
C1 (1 − 1 )
A1−
0
C11
H+ +
A−2
0
0
C 2 2 + C11 C 2 2
(C  + C11 )C 2 2
K a2 = 2 2
C 2 (1 −  2 )
HA 2
C2
C 2 (1−  2 )
(Since 1, 2 both are small in comparision to unity)
Ka

K a1 = (C11 + C2 2 )1; K a2 = (C11 + C 2 2 ) 2  1 = 1
K a2  2
C 1Ka
[H + ] = C1 1 + C 2 2 =
⬥
1
C1K a1 + C 2 K a2
+
C2Ka
2
C1K a1 + C 2 K a2
 [H + ] = C1K a1 + C 2K a 2
If the dissociation constant of one of the acid is very much greater than that of the second acid
then contribution from the second acid can be neglected.
(e) pH of a solution of a polyprotic weak acid :
⬥ Diprotic acid is the one, which is capable of giving 2 protons per molecule in water. Let us take
a weak diprotic acid (H2A) in water whose concentration is c M.
In an aqueous solution, following equilbria exist.
If
K a = first ionisation constant of H2A
1 = degree of ionization of H2A in presence of HA–
1
2 = degree of ionisation of
HA–
in presence of H2A
K a = second ionisation constant of H2A
2
I step
HA −
+ H 3O +
H2 A + H2O
c1 (1 −  2 ) (c1 + c1 2 )
at eq. c(1 − 1 )
 H 3 O +   HA − 
= Ka
(K eq )1 H 2 O=
1
H 2 A 
 K a1 =
=
(c1 + c1 2 )[c1 (1 −  2 )]
c(1 − 1 )
[c1 (1 +  2 )][1 (1−  2 )]
1− 1
.....(i)
II step
A2 −
HA + H 2O
+ H 3O +
at eq. c1 (1 −  2 )
c1 2 (c1 + c1 2 )
H
O
 3 +   A 2− 
(K eq )2 H 2 O=  − = K a 2
HA 
(c1 + c1 2 )[c1 2 )]
K a2 =
c1 (1 −  2 )
[c1 (1 +  2 )] 2
=
.....(ii)
1−  2
−
Knowing the values of K a , K a and c, the values of 1 and 2 can be calculated using equations (i)
1
2
and
(ii) After getting the values of 1 and 2, [H3O+] can be calculated as
[H3O+]T = c1 + c12
Finally, for calculation of pH
⬥ If the total [H3O+] < 10–6 M, the contribution of H3O+ from water should be added.
⬥ If the total [H3O+] > 10–6 M, then [H3O+] contribution from water can be ignored.
Using this [H3O+], pH of the solution can be calculated.
Approximation :
For diprotic acids, K a  K a and 2 would be even smaller than 1
2
1
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
1 – 2  1 and 1 + 2  1
C1  1
1− 1
This is expression similar to the expression for a weak monoprotic acid.
⬥ Hence, for a diprotic acid (or a polyprotic acid) the [H3O+] can be calculated from its first
equilibrium constant expression alone provided K a 2  K a 1
SALTS :
Salts are the ionic compounds formed when its positive part (Cation) come from a base and its
negative part (Anion) come from an acid.
Classification of salts :
(1) Simple salts
(2) Normal salt :
(i) Acid salts
(ii) Basic salts
(3) Double salts
(4) Complex salts
(5) Mixed salts
Thus, equation (i) can be reduced to K a1 =
➢
➢
TYPES OF SALT HYDROLYSIS :
(1) Hydrolysis of strong acid – weak base [SA - WB] type salt –
Ex: CaSO 4 , NH4 Cl, (NH 4 ) 2 SO 4 , Ca(NO 3 ) 2 , ZnCl2 , CuCl2 , CaCl2
NH 4 Cl + H 2 O
NH 4 OH + HCl
NH + Cl + H O
+
NH OH + H + + Cl −
−
4
2
4
NH+ + H O
4
➢
NH OH + H +
2
4
Similarly :
K
(1) K h = w
Kb
(2) h =
Kh
Kw
=
C
Kb  C
Kw  C
Kb
(3)  H +  = Ch =
(4) pH = –log [H+]
1
1
pH = 7 − pK b − logC
2
2
(2) Hydrolysis of [WA – SB] type salt –
Ex: KCN, NaCN, K2CO3, BaCO3, K3PO4
NaCN + H 2 O
NaOH + HCN
Na + CN + H O
+
−
2
CN − + H O
2
➢
Na + + OH − + HCN
HCN + OH −
Summary :
K
(1) K h = w
Ka
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(2) h =
Kh
Kw
=
C
Ka  C
(3) [OH − ] = Ch =
Kw  C
Ka
K w  Ka
C
(5) pH = –log [H+]
1
1
pH = 7 + pK a + logC
2
2
(3) Hydrolysis of (WA – WB) type salt:
Ex: NH4CN, CaCO3, (NH4)2 CO3, ZnHPO3
(4) [H + ] =
➢
Summary :
(1) K h =
Kw
Ka  Kb
(2) h = K h =
Kw
Ka  Kb
Kw  Ka
= K a .h
Kb
(4) pH = –log [H+]
1
1
pH = 7 + pK a − pK b
2
2
(3) [H + ] =
(4) Hydrolysis of [SA – SB] type salt –
Ex. NaCl, BaCl2, Na2SO4, KClO4 etc.
(i) Hydrolysis of salt of [SA – SB] is not possible
(ii) Solution is neutral in nature (pH = pOH = 7)
(iii) pH of the solution is 7
➢
BUFFER SOLUTIONS :
A solution that resists change in pH value upon addition of small amount of strong acid or base (less
than 1 %) or when solution is diluted is called buffer solution.
The capacity of a solution to resist alteration in its pH value is known as buffer capacity and the
mechanism of buffer solution is called buffer action.
Types of buffer solutions
(A) Simple buffer solution
(B) Mixed buffer solution
➢
SIMPLE BUFFER SOLUTION :
A salt of weak acid and weak base in water e.g. CH3COONH4, HCOONH4, AgCN, NH4CN.
Buffer action of simple buffer solution
1
1
pH = 7 + pk a − pk b
2
2
➢
MIXED BUFFER SOLUTIONS :
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(a) Acidic buffer solution :
[A − ]
pH = pK a + log
[HA]
pH = pK a
+log [Salt]
[Acid]
(b) Basic buffer solution :
A basic buffer solution consists of a mixture of a weak base and its salt with strong acid. The
best known example is a mixture of NH4OH and NH4Cl.
⬥
Condition for maximum buffer action :
[NH 4 OH] : [NH 4 Cl]
1
pOH = pK b + log
1
1
1
pOH = pK b and pH = 14 − pK b
➢
SOLUBILITY (s) AND SOLUBILITY PRODUCT (Ksp) :
This is generally used for sparingly soluble salts. We will be dealing with the solubilities in the
following type of solution.
Solubility product (Ksp) is a type of equilibrium constant, so will be dependent only on
temperature for a particular salt.
⬥ Simple solubility
Let the salt is Ax By, in solution in water, let the solubility in H2O = 's' M, then
Ax B y
xAy+ + yB−x
 K sp = (xs) x (ys) y = x x , y y  (s) x + y
−
xs
ys
⬥ Condition of precipitation
⬥ For precipitation ionic product [IP] should be greater than solubility product ksp.
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Electro Chemistry
➢
Electrolytic cell : Converts electrical energy into chemical energy
Cathode: Na +(aq.) + e ⎯⎯
→ Na(s)
1
Anode: Cl−( aq.) ⎯⎯
→ Cl 2 (g) + e
2
➢
Deposition of material at any electrode follow faraday's law of electrolysis.
Faraday's Ist Law:
w = Z it
w=
M
it
n − factor  96500
where w = mass deposite (gm)
M = molar mass
i = current (Amp.)
t = time (sec.)
Faraday's second law:
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➢
At any electrode for material deposited.
w1 w 2 w 3
=
=
E1 E 2 E 3
Note: Order of discharge potential.
Cathode: Au+3 > Ag+ > Cu+2 > Zn+2 > H2O > Al+3 > Mg+2 > Na+ > Li+
Anode: SO 2−  NO −  H O  Cl−  Br −  I−
4
3
2
PRODUCTS OF ELECTROLYSIS OF SOME ELECTROLYTES
S. NO.
Electrolyte
Electrode
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
Aqueous NaCl
Fused NaCl
Aqueous NaOH
Fused NaOH
Aqueous CuSO4
Dilute HCl
Dilute H2SO4
Aqueous AgNO3
Pt or Graphite
Pt or Graphite
Pt or Graphite
Pt or Graphite
Pt or Graphite
Pt or Graphite
Pt or Graphite
Pt of Graphite
Product obtained at
anode
Cl2
Cl2
O2
O2
O2
Cl2
O2
O2
➢
Electrochemical Cell: converts chemical energy into electrical energy.
➢
E cell = SRPcathode − SRPAnode
= SRPcathode + SOPat Anode
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Product obtained at
cathode
H2
Na
H2
Na
Cu
H2
H2
Ag
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Half-cell reaction:
Anode: Zn(s) ⎯⎯
→ Zn +2
( aq.) + 2e
+2
Cathode: Cu (aq)
+ 2e ⎯⎯
→ CuS
➢
Zn(s) + Cu +2 ⎯⎯
→ Zn +2 + Cu(s)
Cell reaction:
(aq.)
(aq.)
 Zn +2 
Q=
;n = 2
Cu +2 
➢
Nearest equation:
E cell = E cell = −
➢
0.059
logQ
n
at 298 K
Max electrical work done = nFE = –G
electrical work done = nFE = –G
DIFFERENT TYPE OF ELECTRODES/HALF CELL
➢
Type
Example
Half-cell reaction
Metal - Metal ion
M/Mn+
Mn+ + ne– → M(s)
Gas - ion
Pt / H2 (P atm) / H+
(XM)
H+ (aq) + e– → 1 H2 (P
Oxidation reduction
Pt / Fe2+, Fe3+
Fe3+ + e– → Fe2+
Metal - insoluble
salt Anion
Ag/AgCl, Cl–
AgCl (s) + e– → Ag
(s) + Cl–
Calomel electrode
Cl–(aq)/Hg/Hg2Cl2
2
atm)
Hg2Cl2(s) +
→
2Hg(l) + 2Cl–(aq.)
2e–
Electrode
potential(reduction)
0.0591
E=E+
log [Mn+]
n
PH
2
E = E – 0.0591 log
+
[H ]
[Fe2+ ]
E = E – 0.0591 log
[Fe3+ ]
E −
= E0 −
Cl /AgCl/ Ag
Cl /AgCl/ Ag
–0.0591 log [Cl–]
E= E –0.0591 log [Cl–]
Gibb's Helmhaltz equation:
G 
G = H + T + 
 T 
G 
 H = –nFE + nFT 
 T P
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lOMoARcPSD|16499330
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'THE ELECTROCHEMICAL SERIES'
Element
Li
K
Ba
Ca
Na
Mg
Al
Mn
H2 O
Zn
Cr
Fe
Cd
Ni
Sn
Pb
H2
Cu
I2
Hg
Ag
Hg
Br2
O2
Cl2
Au
F2
Electrode Reduction
Reaction
Li+ + e → Li
K+ + e– → K
Ba+2 + 2e– → Ba
Ca+2 + 2e– → Ca
Na+ + e– → Na
Mg+2 + 2e– → Mg
Al+3 + 3e– →Al
Mn+2 + 2e– →Mn
2H2O + 2e– →H2 + 2OH–
Zn+2 + 2e– → Zn
Cr+3 + 3e– →Cr
Fe+2 + 2e– → Fe
Cd+2 + 2e– → Cd
Ni+2 + 2e– → Ni
Sn+2 + 2e– → Sn
Pb+2 + 2e– → Pb
2H+ + 2e– → H2
Cu+2 + 2e– → Cu
I2 + 2e– → 2I–
Hg2+2 + 2e → 2Hg
Ag+ + e– → Ag
Hg+2 + 2e– → Hg
Br2 + 2e– → 2Br–
O2+ 4H+ + 4e– → 2H2O
Cl2 + 2e– → 2Cl–
Au+3 + 3e → Au
F2 + 2e– → 2F–
Standard electrode
Reduction potential E0, Volts
– 3.05
– 2.93
– 2.90
– 2.87
– 2.71
– 2.37
– 1.66
– 1.18
– 0.828
– 0.76
– 0.74
– 0.44
– 0.40
– 0.25
– 0.14
– 0.13
0
+ 0.34
+ 0.54
+ 0.79
+ 0.80
+ 0.85
+ 1.08
+ 1.229
+ 1.36
+ 1.50
+ 2.87
CONDUCTION IN ELECTROLYTES
Conductance
C
–1
conductance of volume
within electrode
Specific Conductivity
k
–1 cm2
conductance of unit
volume
Change with
concentration
decrease with decrease
in concentration
Decrease with decrease
in concentration
Formula
C=1
k = C × cell constant
Symbol Unit
Specific
𝑅
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Molar Conductivity
m
–1 cm2 mol–1
conductance of that
volume which contain
exactly one mole
Increase with in
decrease in
concentration
k = m K V
V = Volume of
solution contain 1
lOMoARcPSD|16499330
14
Factors
➢
(i) nature of electrolyte
(ii)concentration of
electrolyte
(iii) Type of cell.
(i) nature of electrolyte
(ii)concentration of
electrolyte
KOHLRAUSEH'S LAW:
  (A B ) = x  + y 
m
x
+
y
−
  (K SO ) = 2  +  
m
2
+
4
−
 (Na PO ) = 3 +  


m
3
+
4
−
  [Fe (SO ) ] = 2  + 3 
m
➢
2
4 3
+
−
FORMULA
(1) R =
A
(2)  m = k 
1000
M
(3)  eq = k 
1000
N
(4) for strong electrolyte  m =  m − b C

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mole of electrolyte
(i)nature of
electrolyte
(ii)concentration of
electrolyte
lOMoARcPSD|16499330
15
Surface Chemistry
➢
Freundlich Adsorption isotherm
1
 
x
m
  = Kp ; n  1
m
➢
➢
➢
Langmuir Adsorption isotherm
=
K'P A
1+ KPA
or
m 1 b
x
aP
=
+
=
or
x aP a
m 1 + bP
1
x
 
= KC n ; C = Concentration of solute in solution
m
Zeta potential, Z =
4
D
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