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SP CRP CH-1

ATOMIC STRUCTURE

1. Mass of one mole of electrons, protons and neutrons are respectively 0.55 mg, 1.007 g and 1.008 g

2. e/m ratio (specific charge) of electron is 1.76 108 C/g and of a proton is 9.58 104 C/g

3. If nucleus is assumed to be spherical, its density is

23 3

Mass of nucleus Mass number=

4Vol. of the nucleus 6.022×10 × r3

4. The limiting line for any series of line in hydrogen spectrum is when n2 in the Rydberg equation is

(infinity).

5. The quantum number not obtained from the solution of Schrödinger wave equation is spin quantum number (s). It is given to account for the spin of the electron.

6. Compton effect is the decrease in the wavelength of X- rays after scattering from the surface of an object.

7. No. of waves in an orbit = Circumference of that orbit

nWavelength

8. Velocity of electron in first orbit of hydrogen is 1/137th of the velocity of light.

9. Principal quantum number ‘n’ tells the main shell in which electron resides and gives the maximum number of electron a shell can accommodate i.e. 2n2 where n = principal quantum number.

Angular quantum number ‘ ’ represents the number of sub shells present in the main shell and tells the shape of the sub shell its value lies between 0 to n – 1.

Magnetic quantum number ‘m’ determine the number of preferred orientation of electron present in sub shell. Its value lies between –

to +

including 0.

Spin quantum number ‘s’ can have two values +1/2 and –1/2 i.e.

or as an electron can have a clock wise or anti clock wise spin.

10. The total no. of nodes in a sub shell (or orbital) are (n-1), out of which the spherical nodes are (n-l-1) and planar or conical nodes equal to l.

11. The intensity of spectral lines in a particular series decrease with the increase in the value of n.

12. As we go away from the nucleus the energy levels come closer i.e., with the increase in the value of 'n' the difference of energy between successive orbits decrease. Thus, E2 - E1 > E3 - E2 > E4 - E3 > E5 - E4 etc

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13. According to Planck’s quantum hypothesis, substances radiate or absorb energy discontinuously in the form of smallest packet of energy called quantum (plural quanta). The energy of a quantum given by, E = h .

14. Humphrey series is produced when the electron in the hydrogen atom jumps from outer shells to 6th shell i.e., for these lines

2 22

1 1v R

6 nwhere n2 = 7,8,9 …

These lines lie in the far infrared region

15. The number of spectral lines in hydrogen spectrum when the electron jumps from an upper level (n2) to the lower level (n1) are = n, where n = n2-n1.

16. The orbital angular momentum ‘L’ of an electron as derived quantum mechanically is as given below:

hL ( 1)

= azimuthal quantum number = 0, 1, 2, 3.... corresponding to s, p, d, f…types of orbitals respectively. The classical analogue of the same as given by Bohr is:

hmvr n

2

where n = 1, 2, 3…. Corresponding to 1st, 2nd, 3rd shells respectively

17.

= h h

2m,KE 2m.eV

18. Energy of an electron in any shell for hydrogen like species;

En = 2 4 2 11

2 2 2

2 me .Z 2.179 10 .Zn h n

ergs / atom

= 19 2

2 2

21.79 10 .Z 13.6 ZJ / atom eV / atom

n n

2

1312 ZkJ / mole

n

Where Z = atomic number.

19. Radius of orbit, 2 2 2

n 2 2

n .h nr 0.529A

ZZ.4 .me

rn(H) = r1 n2

20. Velocity of electron in any orbit, 6n

Zv 2.188 10 m / s

n

21. No. of revolution per second made by electron in an orbit = 2

16 13

Z1.326 10 s

2 r n

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22. Rydberg formula 22 21 2

1 1 1R.Z

n n

Where n1 = lower level and n2 = higher level and R is Rydberg constant of hydrogen and it is equal to 109677 cm–1. The expression for R as derived by Bohr is as follows:

2 4

3

2 meR

ch

and putting the values of universal constants, we get R = 109737 cm–1

23. Frequency 15 22 21 2

1 13.289 10 .Z

n n

24. In 1924, de Broglie proposed tht electron has dual nature i.e. particle as well as wave character and de Broglie wavelength of moving charged particle may, therefore, be

calculated as h hp 2eVm

.

25. If the wave is completely in phase, the circumference of the orbit must be equal to an integral multiple of wave length ( )

Therefore 2 r = n

where ‘n’ is an integer and ‘r’ is the radius of the orbit

But

= h/mv

2 r = nh /mv

(or) mvr = n h/2

which is Bohr’s postulate of angular momentum, where ‘n’ is the principal quantum number.

“Thus, the number of waves an electron makes in a particular Bohr orbit in one complete revolution is equal to the principal quantum number of the orbit”.

26. According to Heisenberg’s uncertainty principle, it is impossible to define both position and momentum of a microscopic particle simultaneously with accuracy i.e.

h h hx p , x m v or x v

4 4 4 m.

27. Aufbau Principle suggests that orbitals of lower energy are filled first, filling of electrons proceeds until the electrons equal to the atomic number are not accommodated i.e. a) The orbital for which the value of (n + l) is the lowest is filled first b) When two orbitals have same (n + l) value, then the orbital having lower value of n is

filled first.

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28. According to Pauli’s Exclusion Principle, no two electrons in an atom can have same values of all four quantum numbers. If two electrons have same values of three quantum numbers then the fourth particularly spin quantum number will be different.

29. According to Hund’s rule, electron pairing in p, d, and f orbitals cannot occur until each orbital of a given sub shell contains one electron each or is singly occupied.

30. Einstein’s Photoelectric equation is K.E. = h

– h 0, where 0 is the threshold

frequency.

31. The negative potential at which the photoelectric current becomes zero is called cut off potential or stopping potential.

STOICHIOMETRY

In this chapter we will discuss the calculations based on chemical equations. It has been classified into two parts:

1. Volumetric Analysis

2. Gravimetric Analysis

1. Volumetric Analysis

Equivalent Weight: It is the weight of a substance which accepts or donates one mole of electrons.

Equivalent weight = factorn

htIonic weig)or(ghtAtomic wei)or( weightecularMol

Calculation of ‘n’ Factor for Different Class of Compounds

1. Acids: n = basicity H3PO4 n = 3 H3PO3 n = 2 H3PO2 n = 1

2. Bases: n = acidity

eg. Ammonia and all amines are monoacidic bases

3. Salt: (Which does not undergo redox reactions) n factor = Total cationic or anionic charge

4. Oxidizing Agents or Reducing Agents : ‘n’ factor = change in oxidation number Or number of electron lost or gained from one mole of the compound.

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Law of Chemical Equivalents: In a chemical reaction the equivalents of all the species (reactants or products) are equal to each other provided none of these compound is in excess.

N1V1 = N2V2 ( when normalities and volumes are given)

EW

= 1000NV

(When weight of one substance in given)

Relation between percentage weight by weight (a), sp. gravity (d) and strength in gm / litre (s) S = 10 ad

Back Titration

This is a method in which a substance is taken in excess and some part of it has to react with another substance and the remaining part has to be titrated against standard reagent.

Double Titration

This is a titration of specific compound using different indicators. Let us consider a solid mixture of NaOH, Na2CO3 and inert impurities.

When the solution containing NaOH and Na2CO3 is titrated using phenolphalein indicator following reaction takes place at the phenolphthalein end point –

NaOH + HCl

NaCl + H2O

Na2CO3 + HCl

NaHCO3 + H2O

Here, eq. of NaOH + 21

eq. of )2n(

32CONa = eq. of HCl

When methyl orange is used, Na2CO3 is converted into NaCl + CO2 + H2O.

Hence, eq. of NaOH + eq. of )2n(

32CONa = eq. of HCl

Above titration can be carried out using phenolphthalein and methyl orange in continuation as well as separately. Accordingly we apply law of equivalents to calculate percentage composition of the mixture.

Note: Phenolpthlein indicates end point when Na2CO3 is converted into NaHCO3

Volume Strength of H2O2

x volume of H2O2 means x litre of O2 is liberated by 1 volume of H2O2 on decomposition

2H2O2

2H2O + O2

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68 gm 22. 4 lit at STP

x litre of O2 is released from 4.22x68

gm of H2O2 = 6.5x17

Strength, S = 5.617x

Normality = ES

= 176.5x17

=

5.6x

Molarity = 21

Normality = 11.2

x

Gravimetric Analysis

In gravimetric analysis we deal with different types of relations like weight - weight, weight - volume, or volume - volume relationship between reactants or products of the reaction.

CHEMICAL KINETICS

Rate of Reaction: Consider the general reaction

P + 2Q

3R.

In the above reaction, two moles of Q disappear for every mol of P reacting and three moles of R are formed for each mol of P disappears.

rate of reaction = .dt

]R[d31

dt]Q[d

21

dt]P[d

Negative sign indicates the consumption of the reactants and positive sign indicates the production of the product.

Characteristics of First Order Reaction

i) Unit of Rate constant : (time)-1

General unit : (mole/litre)1-n time-1

n = order of a reaction.

ii) Time required for the completion of same fraction of initial concentration is independent of initial concentration.

iii) Half life period : Time period in which half of the reaction is completed is known as half - life period.

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SP CRP CH-7

t½ = .k

693.0

The radioactive disintegration follows first order kinetics, thus for any radioactive disintegration

=

t2.303

log xa

a

= disintegration constant

Arrhenius Equation

In 1889, Arrhenius suggested a simple relationship between the rate constant k for a reaction and the temperature T.

k = Ae-Ea/RT

i) log k = log ART303.2

Ea

ii) log 21

12a

1

2

TTTT

R303.2E

kk

Kinetics of Some Complex Reactions

1. Side or Concurrent Elementary Reactions

Ck2

k1B

A

The differential rate expressions for the above two reactions are

]A[kdt

]B[dr 11

]A[kdt

]C[dr 22

The overall-rate of disappearance of A is given as

]A[k]A[kkdt

]A[drr 2121

----- (i)

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The first-order rate constant k is the sum of the two rate constants k1 and k2 of the two side reactions. Equation (i) can be integrated to get the integrate rate expression. Thus, we have

t]A[

0]A[

t

0

dtkdt

]A[d

or tkkkt]A[]A[

ln 21

t

0

Since the rate of formation of products B and C are given by

t1 ]A[kdt

]B[d

t2 ]A[kdt

]C[d

It follows that the ratio of concentrations of B and C at any instant will be given by

2

1

kk

]C[]B[

Hence, if the ratio of B and C is determined at any instant (or at the end of the experiment) and k1 + k2 is determined from the kinetic study, it is possible to determine the individual rate constant k1 and k2.

2. Consecutive or Sequential Reactions

Consider the following consecutive reaction:

BA 1k

C2k

Let the initial concentration of A be [A]0 and let after time t, the concentrations of A, B and C be [A], [B] and [C], respectively. It is obvious that

[A]0 = [A] + [B] + [C] … (1)

The differential rate expressions are

]A[kdt

]A[d1 . … (2)

]B[k]A[kdt

]B[d21

… (3)

]B[kdtCd

2

…(4)

On integrating Equation, (2) we get

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[A] = [A]0 exp ( k1t) …(5)

Substituting [A] from equation (5) into equation (3), we get

]B[k)tkexp(]A[kdt

]B[d2101

On integrating the above linear differential equation of first-order, we get

tkexptkexpkk

k]A[]B[ 21

12

10

… (6)

Substituting equations (5) and (6) in equation (1), we get

[A]0 = [A]0 exp( k1t) + [A]0 12

1

kkk

[exp( k1t)

exp( k2t)] + [C]

or [C] = [A]0 tkexpktkexpkkk

11 2112

12

Maximum Concentration of B:

tkexptkexpkk

k]A[]B[ 21

12

10

…(6)

At the maximum concentration of B, we will have 0dt

]B[d

Hence, differentiating equation (6) with respect to t, we get

tkexpktkexpkkk

k]A[

dt]B[d

2211

12

0

Now, equating it to zero, we get, 0tkexpktkexpk max22max11

or max21

2

1 tkkexpk

k

or max21

2

1 tkkkk

ln

… (7)

or tmax = 2

1

21 k

kln

kk1

…(6)

Substituting equation (7) in equation (6), we get

2k1k/2k

1

20max k

k[A][B]

IONIC EQUILIBRIUM

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SP CRP CH-10

1. Calculation of pH

pH of a Weak Monobasic Acid Solution

HA

H+ + A–

C 0 0 … initially (mole L–1) C(1– ) C

C

… At equilibrium (mole L–1

Ka = C)1(]HA[

]A][H[ 2

= 2C (

1)

(For this approximation to apply

0.1)

[H+] = C

= CK

C a = CkA

pH = –log (Ka C)1/2 = - 21

[log Ka + logc]

= 21

[pKa – logC]

pH + pOH = pKw

pOH of a weak mono acidic base solution BOH

B+ + OH–

C(1 – ) C

C

… At equilibrium

Kb = )1(

C2

2C

[OH–] = C

= CKb

pOH = 21

[pKb – logC]

pH 14 pOH

pH of a mixture of two weak mono basic acids HA

H+ + A– HB

H+ + B–

C1 (1 – 1) (C1 1 + C2 2) C1 1 C2(1 – 2) (C1 1 + C2 2) C2 2

Using same approximations, 1 and 2 can be calculated and hence H+ ion concentration of the solution also.

pH of a solution of strong acid/strong base Strong acid remains completely ionised in solution (

= 1) and hence. [H+] = molarity of solution

basicity of acid = Normality of solution.

Similarly, [OH–] = Molarity of the solution

acidity of strong base = Normality of solution

However, if the solution of acid or base be very dilute (Normality

10–7), it is imperative to add H+ ion concentration of water to that of the acid itself after considering common ion effect of acid or base on the ionization of water.

2. Buffer Mixture

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Two types: i) Acid Buffer, e.g., CH3COOH + CH3COONa ii) Base Buffer, e.g., NH4OH + NH4Cl

Henderson’s equation for calculating pH of buffer mixture

pH = pKa + ]Acid[]salt[

log … For acid buffer

pOH = pKb + ]Base[

]salt[log … For base buffer

Buffer Capacity Amount of acid or base added to buffer solution to charge its pH by unit. It is the capacity of a buffer solution to resist the change in pH on addition of acid or alkali.

Buffer capacity ( ) = )pH(d

)b(d

Where b is no.of mole(s) of acid or base added to one litre solution and d(pH) is change in pH. Buffer solution registers maximum buffer capacity when [acid] or [base] = [salt]

pKa and pKb for a conjugate acid – base pair HA

H+ + A–

Acid (weak) base conjugate (strong) A– + H2O

HA + OH–

Ka

Kb = ]A[

]OH][HA[]HA[

]H][A[

= [H+] [OH–] = Kw

So, pKa + pKb = pKw = 14 (at 25°C) For example, pKa of CH3COOH is 4.75 so pKb of CH3COO– ion will be 14 – 4.75 = 9.25.

3. Salt Hydrolysis

i) Salt of weak acid with a strong base like CH3COONa, Na2CO3 etc. It is the anion of the salt that interacts partially with solvent water to furnish tree OH–

ion in solution thereby rendering the solution faintly alkaline. A– + H2O

)acidweak(HA + OH–

Initially: C mole L–1, constant 0 0 At equilibrium: C (1 – h), constant Ch Ch

KH = ChC)h1(

h)h1(C

ChCh

]A[

]OH][HA[ 22

CK

h H

Other ionic equilibria existing side by side in this solution:

HA

H+ + A–, Ka = ]HA[

]A][H[

H2O

H+ + OH–, Kw = [H+] [OH–]

KH = a

w

KK

, h = CK

K

a

w , [OH–] = Ch = CKK

a

w

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pOH = – log[OH–] = – log

2/1

a

w CKK

or pOH =

21

[pKw – pKa – logC]

Using this equation pOH may be calculated.

ii) Salt of strong acid with a weak base like NH4Cl, FeCl3 etc. Hydrolysis if this salt means partial interaction of cation of the salt to furnish free H+ ion in solution thereby rendering the solution faintly acidic. B+ + H2O

)acidweak(BOH + H+

C(l-h) constant Ch Ch … At equilibrium

cK

h H , KH = b

w

KK

[H+] = Ch = CKK

b

w

pH = 21

[pKw – pKb – logC]

With the help of this equation, pH can be calculated

iii) Salt of Weak acid with a weak base like CH3COONH4

Both cation and anion of the salt interacts with solvent water to furnish free H+ and OH– ions in solution which combine to form freshly ionised water molecule. The resulting solution may be either faintly acidic, alkaline or neutral depending on [H+]

[OH–], [H+

[OH–] or [H+] = [OH–] i.e., Ka

Kb, Ka

Kb or Ka = Kb

CH3COO– + 4NH + H2O

CH3COOH + NH4OH At equilibrium: C(l-h) C(l-h) Ch Ch

KH = h1

h)hl(C)hl(C

ChCh 2

h2

Other ionic equilibria existing here are:

CH3COOH CH3COO– + H+, Ka = ]COOHCH[

]H][COOCH[

3

3

NH4OH 4NH + OH– , Kb = ]OHNH[

]OH][NH[

4

4

H2O H+ + OH–, Kw = [H+] [OH–]

KH = ba

w

KKK

,

ba

w

KKK

h

From acetic acid dissociation equilibrium

[H+] = hK)hl(C

ChK

]COOCH[

]COOHCH[Kaa

3

3q

= b

aw

KKK

pH = –log [H+] = 21

[pKw + pKa – pKb]

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SP CRP CH-13

pH of a solution of the salt of a weak acid with a weak base is independent of the concentration of salt solution. Knowing Kw, Ka and Kb, pH can be calculated.

When Ka = Kb, pH = wpK21

= 7

When Ka

Kb, pH

7

When Ka

Kb, pH

7

4. Solubility Product

In a saturated solution of AxBy (sparingly soluble salt), following equilibrium exists.

AxBy(s)

xA+ + yB–

Ksp = [A+]x [B–]y, Ksp = Solubility product of AxBy Thus, for satd solution of AgCl Ksp = [Ag+] [Cl–] In satd solution of Ag2SO4

Ksp = [Ag+]2 [ 24SO ]

Relation between solubility(s) and solubility product Ksp

Let S be the solubility of AxBy in mole L–1. If AxBy be sparingly soluble, whole of AxBy passing into solution will remain completely ionised so.

AxBy(s)

xsxA +

ysyB

In satd solution [A+]x [B–]y = Ksp

(xs)x (ys)y = Ksp

or xx yy sx+y = Ksp

or S = yxyx

sp

yx

K

Solubility Product Principle

For Unsaturated solution: Ionic product

Ksp

Saturated solution: Ionic product = Ksp

Super saturated solution: Ionic product

Ksp

Ionic product means [A+]x [B–]y

Since super saturation is the condition of precipitation so for precipitation of any salt, the ionic product must exceed the solubility product.

Solubility of a salt when Acid – Base Reaction occur simultaneously or the anion or cation of the salt gets hydrlysed.

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SP CRP CH-14

Let us consider CH3COOAg, a salt of weak acid with a strong base. The solubility of CH3COOAg be S mole L–1

CH3COOAg(S)

CH3COO– + Ag+

s s The acetate ion will get hydrolysed as

CH3COO– + H2O

CH3COOH + OH–-

Initially: S 0 0 At equilibrium: S (l – h) Sh Sh

S – x x x (Where Sh = x)

So, [CH3COO–] [Ag+] = Ksp

or (S – x) S = Ksp

KH = xs

x2

, or xs

xKK 2

a

w

Knowing Ksp, Kw and Ka, S and x can be calculated.

Solubility of a salt in a Buffer Solubility of a solute in acidic or basic buffer can be determined if value of Ksp is given. For example solubility of Mg(OH)2 basic buffer of pH = 12 i.e. [OH–] = 10–2g – ion L–1, is S mole L–1

(say) Ksp = [Mg2+] [OH–]2 = S

(10–2)2

S = 104 Ksp

Theory of acid-base indicator An acid-base indicator is a weak organic acid or base which possesses different colour in its ionised and unionised form due to change in structure from benzenoid to quinonoid upon ionisation in solution.

AColourHIn H+ +

BColourI

]HIn[]In][H[

= Kin

or [H+] = ]In[

]HIn[Kin

or pH = pKin – log]In[

]HIn[

Let for colour A to predominate: [HIn]

10 [In–] then pH

pKin + 1. Similarly, for collur B to prevail pH

pKin + 1Thus, an indicator will change its colour when pH of solution changes abruptly from pKin – 1 to pKin + 1, called indicator range or pH – range of the indicator (Kin = indicator constant, a constant at constant temperature)

CHEMICAL EQUILIBRIUM

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1. a) Equilibrium constant =

b

f

K

K

Kf = rate const for forward reaction Kb = rate const for backward reaction

b) Heat of reaction H = activation energy of forward reaction - activation energy of backward reaction

2. Relation between Kp and Kc is Kp(RT) n. When n = 0, Kp = Kc. When n = Positive, Kp > Kc and when n = Negative, Kp < Kc. The equilibrium constant has a constant value at constant temperature. Concentration, pressure, temperature and catalyst are the important factors which influence the equilibrium state.

3. If a system at equilibrium is subjected to a change of pressure, temperature or concentration, the equilibrium gests shifted in such a way as to counteract the effect of that change. This is called Le- Chatelier's principle.

4. Increase in the concentration of any reactant always favours the forward reaction and so the equilibrium gets shifted to the right. Similarly, decrease in concentration of the products favours backward reaction. Hence equilibrium can be shifted accordingly by increasing the concentrations of the reactants or by decreasing the concentration of the products so that it may proceed more nearly to completion.

5. Pressure has no effect on the reaction in which there is no change in the number of moles of the reactants and products. Increase in pressure at equilibrium favours the reaction which takes place with decrease in volume or number of moles. Increase in temperature at equilibrium favours endothermic reaction.

6. A catalyst increase forward reaction and the backward reaction to the same extent.

7. The value of equilibrium constant at two different temperature are related by the

following equation 211

2

T1

T1

R303.2H

KK

log T2 > T1

a) If a reaction is exothermic (it is attended by the evolution of heat), increase in temperature will shift the equilibrium in the direction of the reactants and in now characterised by a lower value of equilibrium constant K.

b) If a reaction is endothermic (it is attended by the absorption of heat), increase in temperature will shift the equilibrium in the direction of products and it is now attended by an increase in the value of equilibrium constant K.

c) Conversely, lowering of temperature cause the equilibrium constant K to increase in the case of exothermic reaction and lowering of temperature causes the equilibrium K to decrease in the case of endothermic reaction.

8. Addition of an inert gas such as He at constant volume has no effect upon the position of equilibrium. For example, total pressure in N2O4

2NO2 system could be changed without changing the volume by adding helium which would increase the total number of moles of the gas and hence the total pressure.

Since volume remains constant, we have Kc =VOnN

nNO

42

22 .

The ratio (nNO2)2/(nN2O4) remains constant. Thus position of equilibrium is not affected

by the addition of helium.

9. Standard free energy change of a reaction and its equilibrium constant are related to each other at temperature T by the following relation:

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pKlogRT303.2G

STHG

When G

= -ve, the value of equilibrium constant will be large positive quantity and

when G

is positive, the value of K is less than 1, i.e. low concentration of products at

equilibrium state.

10. Degree of dissociation from density measurements: Degree of dissociation in the case of reversible reactions in which there is increase in the number of molecules can be determined by measuring density of the reaction mixture at equilibrium. Let d be the observed density at a particular temperature when degree of dissociation is x and D be the vapour density when there is no dissociation.

d1ndD

x ; Where n is the number of molecules of products.

ELECTROCHEMISTRY

1. Faraday’s first law of electrolysis

Weight of substance liberated

quantity of electricity passed

W Q W ZQ W Zit

(Z= electrochemical equivalent)

equivalent weight)

EZ (E

96500

2. Faraday’s second law of electrolysis: For the same quantity of electricity passed through solutions of different electrolytic cells connected in series, weight of substance produced is directly proportional to their equivalent weight.

1 1

2 2

w EW E

w E

3. One faraday (96500 coulombs) of electricity deposits one gram equivalent weight of the substance.

weight depositedE 96500

quantity of electricity coulomb

WE 96500

Q

W Q itE 96500 96500

4. 1

Conduc tance (C)R

[units of conductance ohm–1 or mohos or siemens (s)]

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5. If ‘ ’ is the distance between the electrodes and ‘a’ is the area of cross section of

electrode.

R ( )a

(Where

is specific resistance or resistivity)

1 1C

R a a

Where

= specific conductance or conductivity a

is referred to as cell constant.

6. a) Equivalent conductance eq

1000Normality

b) Molar conductance m

1000Molarity

7. For sparingly soluble salt like AgCl, PbSO4, BeSO4 etc.

Solubility in mol/litre (s) = 0m

1000

8. According to Kohlrauch law for an electrolyte. AmBn

0 0 0c

0 0 0m m n

Where 0 0and

are the ionic conductivities at infinite dilution for cation and anion

respectively.

9. Degree of dissociation of an electrolyte at conc. (C),cm0m

10. 0 0 0cell RP(cathode) R.P.(anode)E E E

11. For the cell reaction aA bB mC nD

m0 m

cell cell a b

[C] [D]2.303RTE E log

nF [A] [B]

m n0

cell cell a b

0.0591 [C] [D]E E log at 298K

n [A] [B]

(Nernst Equation)

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12. 0 0

cell cell

2.303RT 0.0591E logk E logK (at 298K)

nF n

(K = eqm. Constant for cell reaction)

13. 0 0G nFE and G nFE

14. The decreasing order of the discharge potential or the increasing order of deposition for

some cations and anions are in the following order. Cations: 2 2 3 2 2 3K , Na , Ca , Mg , Al , Zn , H , Cu , Ag , Au

Anions 24 3SO , NO , OH , Cl , Br , I

15. If two half cell reactions having potentials 01E

and 02E

are combined to give third half cell

reaction having a potential 03E

then 0 0 03 1 2G G G

0 00 1 1 2 23

3

n E n EE

n

If 0 0 01 2 3 3 1 2n n n , E E E

16. Temp. co efficiency of the emf of the cell

cell cell

P

dE EHdT nFT T

17. Enthalpy change:

cellcell

P

dEH nF T E

dt

18. Entropy change cel

p

dES nF

dT

THERMOCHEMISTRY

Heat of reaction at constant pressure = H

Heat of reaction at constant volume = E

gH E n RT

gn = difference in number of moles of gaseous products and reactants.

Kirchoff’s Law: 2

1

T

2 1 pTH H C dT

= p 2 1C (T T ) , assuming pC

remains almost

constant within the small temperature range T1 to T2.

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p products reac tantsC C C

2

1

T

2 1 v v 2 1TE E C dT C (T T )

products reactantsv v vC = C - C

First Law of Thermodynamics

pQ E ( w)

ext final initial ext 2 1 extWork P V V P (V V ), P

= external pressure = final

pressure of gas.

Any reversible work = 2

1

v

vpdV

Isothermal and reversible work for n moles of an ideal gas

2 1

1 2

V PnRTln nRTln

V P

Adiabatic work = v i fnC (T T )

Adiabatic Reversible Relations

1-yr r-1 yPV = constant TV = constant TP = constant

For an ideal gas T

dE0

dV

For a real gas 2

T

dE adV V

Joule Thomson co efficient J T

Tp p

1 H 1 2ab

C P C RT

Coefficient of thermal expansion P

1 dV( )

V dT

Coefficient of compressibility T

1 dVV dP

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p vC C (for any substance solid, liquid or gas) =

2VT

for an ideal gas p vC C R

for one mole.

v 2 1H nC (T T )

… for n moles of an ideal gas undergoing adiabatic

and reversible change

Second Law of Thermodynamics

revqdqS ds

T T

Entropy change

i) During heating or cooling of solid or liquid of mass m and specific heat

2

1

TS mc ln

T

ii) During isothermal expansion of gases

2 1

1 2

V PS nRln nRln

V P

iii) Non isothermal expansion

2 2 2 2v p

1 1 1 1

T V T PS n C ln Rln n C ln Rln

T V T P

iv) Transition of state Lt

ST

[Lt = latent heat of transition]

v) Free mixing i iS XRln X

per mole

vi) S 0

for adiabatic process

G H T S

p

( G)G H T

T

Putting isothermal expansion of a gas

1 2T

2 1

V PG RTln RTln

V P

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During isothermal pressure change of solid (liquid)

T 2 1G V(P P )

Clausius

Clapeyron equation

2 v 2 1

1 1 2

P L T Tln

P R T T

Clapeyron equation t

2 1

LdPdT T(V V )

t

2 1

l

1 1T

Lt = latent heat of transition/ mole but lt is same in/grams, 2 and 1, are densities in state 2 and 1 respectively.

G H T S

At equilibrium under the condition of constant P and T i.e. the laboratory condition, G 0

and hence

H T S

HS

T

A reaction to be feasible: G 0 i.e. G

must be negative.

LIQUID SOLUTION

1. Solution: A solution is homogeneous mixture of two or more components whose concentration can be varied with in a limit or mixture of solute (dispersed phase + dispersion medium) and solvent is combindly called as solution.

2. Solubility: The amount of solute which is dissolved in 1000gm of a solvent is called as solubility to make a saturated solution.

solution lattice energy hydration energyH H H

If Hydration Energy

Lattice energy, solute goes in solution and solutionH

becomes

negative.

3. Vapour pressure of liquid: It is the pressure exerted by vapour when they are in equilibrium with the liquid at a given temperature. It depend on

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i) Nature of liquid ii) Temperature iii) Purity of the liquid

More is the force of attraction between liquid molecules, lower is the vapour pressure and vice versa.

4. Vapour pressure of a solution: When a miscible solute is added to a pure solvent it results in the formation of a solution. Some solute will replace the molecules of solvent from the surface, therefore escaping tendency of solvent molecule decreases. This causes a lowering in vapour pressure.

s 0p p

5. Ideal solution: These are the solutions in which solute

solute and solute solvent interactions are almost similar to solute solvent interactions. They obeys Raoult’s Law for all range of concentrations and temperatures.

mixH 0

mixV 0

e.g. Hexane + heptane

6. Non ideal: In such solutions solute solvent interactions are different than solute solute and solvent solvent interactions. They does not follow Raoult’s Law at all concentrations.

mixH 0

mixV 0

They shows two types of deviations: i) If mix mixH 0 and V 0

then non ideal solution show positive deviation.

ii) If mix mixH 0 and V 0 then non ideal solution show negative deviation.

7. Raoult’s Law

a) If components are volatile 0

A A AP X P

0B B BP X P

TotalP = A BP P

= 0 0A A B BX P X P

= 0 0A A A BX P (1 X )P

TotalP = 0 0 0A A B BX P P P

b) If only (A) component solvent is volatile vapour pressure of solution s 0A AP X P

c) Mole fraction in vapour phase

AA

T

Py

P

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8. Relative lowering in vapour pressure

0s

B0

P PX

P

(mole fraction of solute)

9. Elevation in Boiling Point

b bT K m

b solutesolute

b solvent

K W 1000M

T W

10. Calculation of molal elevation constant

2 20 0 0

bv v

RT M RTK

1000L 1000 H

T0 = Boiling point of pure liquid (solvent)

vL

= Latent heat of vaporisation per gm of solvent

vH = Latent heat of vaporization per mole of the solvent

0M

= Molecular mass of solvent

R = Gas constant = 1 1

v v8.314JK mol if L or H are in Joules

= 1 1v v2 cal deg mol if L orH are in calories

11. Calculation of molal depression constant

2 20 0 0

ff f

RT M RTK

1000 1000 H

0T

= Freezing point of pure liquid (solvent)

0M

= Molecular mass of solvent

fH = Latent heat of fusion per mole of the solvent

fT = Latent heat of fusion per gm of the solvent

12. Mole fraction of a component in the vapour phase

AA

T

PPartial pressure of that componenty =

Total vapour pressure P

13. Various relations of colligative property

0s

02

P P dRTMP

2M

= Molecular mass of solute

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b

b

dRTT

1000 K

f

f

dRTT

1000 K

0

b sb 0

0

1000K P PT

M P

0sf

f 01

P P1000KT

M P

Abbreviation

= Osmotic pressure

2M

= Molecular mass of solute

bT = Elevation in boiling point

fT = Depression in freezing point

0M

= Molecular mass of solvent 0P

Vapour pressure of solvent sP

= Vapour pressure of solution

14. Relation between mole fractions of one component of ideal solution in the vapour phase and liquid phase.

0 0 0A B A0 0

A AB B

P P P1 1X yP P

Graph of A A

1 1vs

X yshould be linear with

Slope = 0A0B

Pand

P intercept =

0 0B A

0B

P PP

Where AX

= mole fraction in liquid phase of component A.

BX

= mole fraction in liquid phase of component B 0BP

= vapour pressure of pure liquid B 0AP

= vapour pressure of pure liquid A

Ay

= mole fraction of component A in vapour phase

By

= mole fraction of compound B in vapour phase

15. Relation between total pressures mole fraction of one component in vapour phase. 0 0B A

A0 0 0Total B A B

P P1 1y

P P P P

Graph between ATotal

1vs y

P should be linear.

16. Van’t Hoff Factor (i) = It is accountant for anomalies

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Observed colligative property

i=Calculated colligative property

observed

1Colligative property

M

Calculated Molar Mass

iObserved Molar Mass

0

sB0

P PiX

P

BX

= mole fraction of solute

b bT i K m

m = molality

f fT i K m

V = volume of solution

soluteni RT

V

n = no. of moles

i = 1 non electrolytes (glucose, sucrose, urea, protein) i 1

solute which dissociates 3 2(KCl, KNO , CaCl )

i 1

Solute which associates 6 5(RCOOH,C H OH)

17. i = a(m – 1) + 1 For solutes which dissociates i 1

am 1

m = fraction of solute after dissociation

For e.g. KCl (m = 2)

18. For solutes which associates in medium i 1

a1/m 1

m = fragment which get associates

19. Silica garden: When a small crystal of CuSO4 or NiSO4 is added to 5% sodium silicate solution. This is to be known as chemical garden or silica garden. It is produced due to diffusion of metal ion from crystal to form precipitate of metal silicate. This precipitate act as a semi permeable membrane around the crystal. Osmosis takes place water flows from dilute sodium silicate to stronger salt solution. After sometime membrane burst and more metal ions are free, again process repeats and this growth of crystal looks like a plant.

20. Henry’s Law: The mass of gas dissolved by a unit volume of a solvent at constant temperature is directly proportional to the pressure of the gas with which it is in equilibrium

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Equ

ilibr

ium

pr

essu

re

Solubility

Graph between pressure and solubility

2x P

2 H 2x K x

2x mole fraction of gas dissolved

HK = proportionality constant

SOLID STATE

1. Amorphous substance are isotropic, while crystalline substances are said to be anisotropic.

2. The reverse of crystallisation is the fusion or melting of the solid.

3. A crystal may have one or more planes and one or more axes of symmetry, but it has only one centre of symmetry. A cubic crystal, such as NaCl possesses a total of 23 elements of symmetry.

4. The unit cell is the smallest unit of a crystal, which when repeated, in three dimensions, will generate the crystal.

5. In a cubic system of crystals, there are three types of Bravais lattices, depending upon the shapes of the unit cell viz., simple cubic, face centred and body centred lattices.

6. Close packing means an arrangement in which a given number of equal spheres occupy the minimum amount of space.

Limiting radius ratio (r+/r–) Co ordination number

0.155 2 0.155 – 0.225 3 0.225

0.414 4 0.414 – 0.732 4 0.414 – 0.732 6 0.732 – 0.999 8

7. Structural relationship for cubic lattices

Simple BCC FCC Lattice points per unit cell 1 2 4 No. of nearest neighbours 6 8 12 Distance between nearest neighbours a 3a

2

a

2

Atomic radius a2

3a4

a

2 2

Occupied volume (P.F.) 52.3% 68% 74%

8. Density of cubic crystal is given by

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3

A

M ZN a

M – molecular mass of the substance Z – no. of formula units per unit cell of the crystal NA – Avogadro’s number a

edge length of the cubic cell

9. Schottky defect is observed in the crystals, of KCl, NaCl, CsCl, KBr etc. Frenkel defect is observed in crystals of CaF2, ZnS, AgI and AgBr etc. and is less common than Schottky defect. Metal excess defect because of anion vacancies, occurs e.g. in yellow coloured NaCl and due to cation vacancies, occurs in NiO, FeO and FeS etc. Metal excess defect due to interstitial cation occurs in Zinc oxide (ZnO).

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