raoult law question

8/3/2019 Raoult Law Question

1/20

B r a insmi ths E ducat ion P vt. L td.

1

===================================================================================

12th

MHBOARD

CHEMISTRY-PART-I

==========================================================

CLASS : XII

Chemistry Paper I

No. NameMarks without

optionMarks with

option

1 Solution & Colligative properties 7 10

2 Chemical Thermodynamics & Energetics 9 12

3 Ionic Euquilibria 7 10

4 Electrochemistry 6 10

5 Nuclear and Radiochemistry 4 6

6 Chemical Kinetics 3 6

7 d-block Elements 4 6

Total 40 60

CHEMISTRY PAPER-1

Chapter 1_Solution & Colligative Properties

Q 1. Define colligative property. Give examples.Ans 1. Colligative property: A property which depends upon the number of solute particles (atoms, molecules or ions) present in

solution and is independent of the size or chemical nature of the particles is called colligative property.

Examples:i) Relative lowering of vapour pressure.

ii) Elevation of boiling point

iii) Depression of freezing point.

iv) Osmotic pressure.

Q 2. State Raoults law of lowering of vapour pressure.Ans 2. Statement : The relative lowering of the vapour pressure of a dilute solution containing non -volatile solute is equal to the

mole fraction of the solute present in the solution.

Mathematically, it is expressed as

0

0

P P n

P n N

Where, P0 is vapour pressure of pure solvent

P = vapour pressure of solutionn = number of moles of solute.

N = Number of moles of solvent.

Q 3. Decribe Ostwald and Walkers method for determination of molecular weight of a non-volatile solute by measuring the

lowering of vapour pressure.

Ans 3. Principle: The principle behind this experiment is that when one gas is passed through another gas, the later diffuses intoformer. This means that when dry air is bubbled through solution, the vapour present in solution diffuses into air, the diffusion

takes place till the pressure of the vapour in air becomes equal to that of the vapours in the container therefore there will be

loss of weight in the container.

Apparatus : It consists of two sets of bulbs as follows:

i) Set A containing the solution.

ii) Set B containing the pure solvent.Bulb B is further connected to U-tube containing CaCl2 solution if water is used as solvent.

Procedure:

i) Set A and B bulbs are first weighted separately.


2/20


2

ii) A slow stream of dry air is passed through these bulbs, there is loss in weight of two sets of bulbs. These loss in weights

are recorded.iii) If water is used as solvent, calcium chloride absorbs water, gain in weight of calcium chloride is also recorded.

The temperature is kept constant during measurements.

Calculations:i) When the air passes through set A bulbs it is saturated upto the vapour pressure of solution P.

There is loss in weight in bulb A, let it is W1

Loss in weight Vapour pressure

of bulb A of solution P

w1 P (i)

ii) When the same moistened air passes through set B bulbs it get saturated up to the vapour pressure of solvent P 0. There is

again loss in weight in bulb B. Let, it is W2.

0

Difference in vapourThe loss in

pressure of solvent andweight

vapour pressure ofof bulb B

solution (P P)

w2 P0P (ii)

Adding equation (i) and (ii) we have

w1 + w2 P + P0 - P

w1 + w2 P0 (iii)

Dividing equation (ii) by (iii) we get

2

1 2

w

w w= 2

1 2

w

w w (iv)

Hence, knowing loss in weights of bulbs A and B, relative lowering of vapour pressure can be calculated.

Determination of molecular weight:According to Raoult law we know the relation between relative lowering of vapour pressure and molecular weight.

0

0

P P

P=

w M

m W (v)

Comparing equation (iv) and (v) we get,

2

1 2

w

w w=

w M

m W

The molecular weight of solute (m) in dilute solution is given as,

m = 1 2

2

w ww M

W w

Where, w = weight of solute

W = weight of solvent

M = mol. Wt. of solvent

w1 = loss in wt. in Bulb A

w2 = loss in wt. in Bulb BAs a check on the value of total decrease in weight of two sets of bulbs an adsorption unit containing CaCl2 is fixed at the end

of the apparatus all the water from air is absorbed by CaCl2 and hence there is gain in weight of CaCl2. It is given by

0

2 0

P Ploss in wt. of Bulb B(solvent)

Gaininwt.ofCaCl tubes P

2

1 2

w

w w= 0

0

P P

P

Q 4. What is the elevation of boiling point? On what factor does it depend?Ans. i) The boiling point of a liquid is the temperature oat which its vapour pressure becomes equal to the atmospheric pressure.

ii) As we know that the vapour pressure of the solution containing non-volatile solute is less than the solvent thereforesolution has to be heated to a higher temperature so that its vapour pressure becomes equal to the atmospheric pressure.

iii) Thus the boiling point of the solution is always higher than that of the pure solvent.

The difference in the boiling point of the solution (Ts) and boiling point of pure solvent (T0) is called elevation in boiling

point ( Tb).

Tb = TsT0

iv) Elevation in boiling point depends upon,

(a) Concentration of solute particles.

(b) The nature of solvent.

Q 5. What do you understand by depression in freezing point?

Ans. i) Freezing point of solution is the temperature at which the liquid and its solid exist in a state of equilibrium.

ii) Since, the addition of non-volatile solute always lowers the vapour pressure of solvent therefore it will be in equilibrium

with solid phase at a lower pressure and hence at a lower temperature.ii) Since, the addition of non-volatile solute always lowers the vapour pressure of solvent therefore it will be in equilibrium

with solid phase at a lower pressure and hence at a lower temperature.

iii) Thus freezing point of pure solvent is decreased by adding non volatile solute in it. This is called depression in freezingpoint.

iv) The difference between the freezing point of pure solvent (T0) and the freezing point of solution (Ts) is called depression in


3/20


4/20


4

Chapter 2_Chemical thermodynamics and EnergeticsQ 1. Distinguish between isothermal process and adiabatic process.Ans.

Isothermal Process Adiabatic process

1. Temperature of system remains constant. ( T = 0) 1. Temperature of system changes ( T 0)

2. Internal energy remains constant. ( E = 0) 2. Internal energy changes. ( E 0)

3. System exchanges heat with surroundings. 3. System does not exchanges heat with surroundings.

4. Enthalpy remains constant ( H = 0) 4. Enthalpy changes. ( H 0)

5. System is not thermally isolated. E.g. Boiling of water 5. System is not thermally isolated. e.g. Chemical reaction taking placein insulated vessel

Q 2. Distinguish between reversible process and irreversible process.

Ans 2.

Reversible process Irreversible process

1. Driving and opposing forces differ by a infinitesimally

small value.

1. Driving and opposing forces differ by large value.

2. It is an imaginary process. 2. It is an natural process.

3. It is slow process. 3. It is fast process.

4. There is equilibrium at every stage. 4. Equilibrium is attained only when process is completed.

5. Maximum work can be obtained. 5. Maximum work can not be obtained.

6. The direction of the process can be reversed by

increasing opposing force. e.g. Isothermal reversible

expansion of gas.

6. The direction of the process can not be reversed by increased by

small change of driving and opposing forces. e.g. Flow of water from

high to low level.

Q 3. Derive the expression for the work done when an ideal gas expands from isothermally and reversibly from a volume of V1dm3

to V2dm3.

OR

Derive the expression for work done in an isothermal and reversible expansion of an ideal gas.

Ans.

1. Let 'n' moles of an ideal gas be enclosed in a cylinder at constant temperature T.2. The cylinder is fitted with weightless and frictionless piston.

3. Let the pressure of gas be 'P'4. Let the gas expand isothermally and reversibly at constant temperature T. For this let the small change in volume be dV dueto small decrease in pressure by dP.

5. Thus, small work done (dW) obtained in this expansion process is,

dW = - (P - dP).dVdW = - P.dV + dRdV

6. As dP and dV are very small, the product dP.dV will be still smaller and hence can be neglected.

The above expression becomes, dW = - P.dV .......(1)7. If the gas expands from initial volume Vjdm3 to final volume V2 dm

3 strictly

under reversible and isothermal conditions, then total maximum work Wmax is obtained by taking integration of dW between

volume limits V1 to V2

Wmax = dW Wmax =2

1

v

v

P.dV

For 'n' moles of an ideal gas is PV = nRT P =nRT

V

Putting this value of P in above eqn, we have Wmax =2

1

v

v

nRTnRT dV

V

2

1

v

v

dVnRT

V= - nRT 2

1

v

e vlog V

= - nRT[logeV2 - logeV1] = - nRTloge 2

1

V

V

Wmax = -2.303nRTlog10 2

1

V

V (vlogex = 2.303 log10x)


5/20


5

According to Boyle's law at constant temperature, PV = PV 1 1

2 2

P V

P V

Hence,

Wmax= - 2.303nRTlog10 1

2

P

P

Note : For 1 mole of the gas, n = 1. Wmax = - 2.303 RTlog10 2

1

V

Vand Wmax = -2.303 RT log10 1

2

P

P

Q 4. Derive the expression, H = E + P VOR

Derive the expression for enthalpy change.Ans 4. Let H1 be the enthalpy of a sytem in initial state and H2 be the enthalpy of the system in the final state

Then change in enthalpy ( H) is given as, H = H2 - H1 ........(i)

But, by the definition of enthalpy H = E + PV

H1 = E1 + P1V1 .........for initial state

H2 = E2 + P2V2.........for final state of system.

Substituting these values of H1 and H2 in equation (i)

H = (E2 + P2V2) - (E1 + P1V1) = (E2 - E1) + (P2V2 - P1V1)

But, at constant pressure P1 = P2 = P

H - (E2 - E1) + (PV2 - PV1)

H = (E2 - E1) + P(V2 - V1)

But, (E2 - E1) = E and (V2 - V1) = V

Thus, change in enthalpy at constant pressure is equal to the sum of change in internal energy and P - V type work.

H is positive in endothermic reaction. H is negative in exothermic reaction.

Q 5. Explain the terms.

1. Heat of reaction

2. Heat of reaction at constant pressure

3. Heat of reaction at constant volume.Ans 5. 1. Heat of reaction: It is the difference between the sum of enthalpies of products and sum of the enthalpies of reactants at a

given temperature and at constant pressure or at constant volume when every substance is in its standard state.

H = (products ) ( reactants)H H

2(g ) (s) 2(g)CO C OH H H

= 395.4 - [0 + 0] =-395.4 kJ.

Since, enthalpy of all elements in their standrard states (stable physical state at room temperature and pressure) are taken zero.

2. Heat of reaction at constant pressure ( H): It is the difference between the sum of enthalpies of products and sum of the

enthalpies of reactants at given temperature and at constant pressure when every substance is in its standard state. It is denoted

by H.

3. Heat of reaction at constant volume ( E) : It is the difference between the sum of enthalpies of products and sum of

enthalpies of reactants at given temperature and at constant volume when every substance is in its standard state. It is desnoted

by E.

Q 6. Derive the equation showing effect of temperature on heat of reaction at constant pressure.

OR

Derive Kirchhoff's equation.

OR

Show that Cp =2 1

2 1

H H

T T

Ans 6. Kirchhoff's equation gives the effect of temperature on heat of reaction.

1. Consider the following general reaction

A B

Where, 'A' represents the initial state i.e. reactants & 'B' represents the final state i.e. products.

2. Let H1 and H2 be the enthalpies of initial and final states respectively.

Then change in enthalpy i.e. heat of reaction at constant pressure ( H) is given by

H = H2 - H1 .........(i)

3. On differentiating equation (ii)w.r.t. temperature at constant pressure, we get

2 1

P PP

dH dHd( H)

dT dT dT

But molar heat capacity at constant pressure (Cp) is given by the expression,

P

dH

dT

1

P

dH

dT=

1PC and 2

P

dH

dT=

2PC


6/20


6

P

d( H)

dT= Cp2 - Cp1

P

( H)d

dT= CP ..........(ii)

Where CP = difference in molar heat capacities of products and reactants.

The equation (ii) is known as Kirchhoff's equation.

Statement : The rate of change of heat of reaction with respect to temperature is equal to the difference an molar heat

capacities of products and reactants at constant pressure.4. For practical use, eqn(ii) is integrated between temperature limit T1 to T2

P

( H)d

dT= Cp

on rearrangement of above equation,

d( H) = Cp.dT2 2

1 1

H T

H T

d( H) Cp dT

( CP is assumed to remain constant in temperature range T1 and T2)2 2

1 1

H T

H TH Cp T

H2 - H1 = Cp(T2 - T1)

H2 = H1 + Cp (T2 - T1)

This is Kirchhoff's eq. at constant pressure.

It can also be written as,

Cp = 2 12 1

H HT T

Where,

H1 = heat of reaction at temp T1

H2 = heat of reaction at temp T2

CP = Difference in molar heat capacities of products and reactants (Cp of products -Cp of reactants)

Note : Similarly, Kirchhoff's equation at constant volume can be derived as,

E2 - E1 = Cv(T2 - T1)

E2 = E1 + Cv(T2 - T1)

CV =2 1

2 1

E E

T T

Q 7. State and explain Hess's law of constant heat summation.

Ans 7. Statement : The total amount of heat absorbed or evolved in a chemical reaction is the same whether the reaction takes place

in one step or in a number of steps.

OR

The change in enthalpy accompying a chemical reaction is independent of the pathway between initial and final states.

Explanation : Let us consider a chemical reaction in which substance A is converted into substance D in a one step asfollows,

A D ; H = -Q

Now, suppose the same reaction is carried out in three steps as follows,

A B ; H1 = -Q1

B C ; H2 = - Q2

C D ; H3 = - Q3

Then according to Hess's law, H = H1 + H2 + H3

-Q=(-Q1)+(-Q2)+(-Q3) C*-Q=-Q'

Since the enthalpy change is the same in both methods, this can be explained diagrammatically as shown below

Example :The formation of CO2 gas from solid carbon and oxygen gas can be carried by two methods.

1. Method - I :

Carbon dioxide is prepared by combination of solid carbon and oxygen gas directly as,

C(s) + O2{g) CO2(g), H = -395.4KJ

2. Method - II :The same reaction is brought about in two steps as fallows

i) C(s)+

1

2O2(g) CO(g) ; H1 = - 111 KJ


7/20


7

ii) CO(g)+1

2O2(g) CO2(g); H2 = -284.4 KJ

According to Hess's Law,

H = H1 + H2

- 395.4 KJ = -111 KJ + (-284.4 KJ)- 395.4 KJ = -395.4 KJ

Thus total heat evolved in both methods is same, therefore, Hess's law is proved.

Chapter 3_Ionic Equilibrium

Q 1. What are conjugate acid-base pairs ? Explain with suitable example.

Ans 1. Definition : "The pairs of acids and bases formed from each other by mutual gain or loss of proton are called conjugate acid-

base pairs".

ORA pair of an acid and base which differs from one another by a proton is called conjugate acid-base pair.

Explanation :

Consider the following reaction,

21

2Base

Acid

HCl H O12

3BaseAcid

H O Cl

In the above reaction, HCl donates a proton and acts as an acid while water accepts that proton and acts as a base. In the

reverse reaction, hydronium ion donates a proton and therefore it acts as an acid. The chloride ion accepts that proton andbehaves as base.

Thus HCl is a conjugate acid of the base chloride ion ( Cl ) and chloride ion is a conjugate base of the acid HCl. Similarly,

water is the conjugate base of hydronium ion (H3O+) and hydronium ion is a conjugate acid of water.

Thus, we can say that every acid has its (self generated) conjugate base and every base has its (self generated) conjugate acid.

Q 2. Explain Lewis concept of acid and base with suitable examples.Ans 2. According to Lewis concept, acids and bases are defined as follows :

i) Acid : The species (molecule, atom or ion) which accepts a lone pair of electrons to form a co-ordinate bond is called an

acid e.g. Cu++, BF3 etc

ii) Base : .The species (molecule, atom or ion) which donates a lone pair of electrons to form a co-ordinate bond is called

base. e.g. NH3, OH , CN etc.

In short, acid is electron pair acceptor and base is electron pair donor.

Acid-Base reaction :

For example, consider the reaction between boron tri-fluoride and ammonia.

i.e.

In this reaction, NH3 donates lone pair of electron and acts as a base and BF3 accepts it and hence acts as an acid. Thus,

formation of co-ordinate bond between reacting acid and base is neutralisation reaction.

Q 3. State Ostwald's dilution law. Derive an expression for Ostwald's dilution law for weak electrolyte.

OR

Show that degree of dissociation of a weak electrolyte is inversely proportional to square root of its concentration.

Ans 3. Statement : The degree of dissociation of weak electrolyte is inversely proportional to the square root of its concentration or

directly proportional to square root of its dilution.

OR

Whatever may be the concentration, the degree of dissociation of a weak electrolyte varies with its concentration in such away that the value of its dissociation constant remains constant at a given temperature.

Derivation : Let one mole of weak electrolyte BA be present in V dm3 of the solution and 'a' be its degree of dissociation.

Then, BA dissociates as

BA B+ + A- ...(1)

Initial moles 1 0 0

Moles at eqm. (1 - )

conc. at eqm. in (moles/dm3) (1 )V

V

V

By applying law of mass action to eqn(1)


8/20


8

K =[B ][A ]

[BA]

putting the value of concentrations, we have,

K = V V1

V

2

2

V

1V

2

(1 )V

But for weak electrolyte, a is very small1 1

K =2

Vbut C =

1

V

C = concentration in moles/dm3.

K = 2.C

2 =K

Cor 2 = K.V

K

Cor K.V

1

Cor V

This shows that, degree of dissociation of a weak electrolyte is inversely proportional to square root of its concentration or

directly proportional to the square root of its dilution.

Q 4. Explain common ion effect

Ans 4. Definition: "The phenomenon in which the degree of dissociation of weak electrolyte is supressed by adding another strong

electrolyte containing a common ion is called common ion effect."

Explanation :i) Consider the dissociation of weak base NH4OH. The following equilibrium exists in its aqueous solution,

NH4OH NH4+ + OH-

By applying law of mass action

Kb =4

4

NH OH

NH OH

ii) Suppose, a strong electrolyte NH4Cl which contains 4NH ion common to that of NH4OH, is added into above solution,

then NH4C1 being a strong electrolyte is dissociated completely as,

NH4Cl 4NH + ClAs a result, concentration of 4NH ions increases in the solution considerably.

iii) According to Le - Chatelier's principle and to keep K b value constant, the concentration of NH4OH must be increased in

the solution. Hence some 4NH ions combine with OH ions to give more undissociated NH4OH molecules shifting

equilibrium to the left hand side.

Thus, the degree of dissociation of ammonium hydroxide decreases by adding NH 4Cl containing 4NH common ions. This is

called common ion effect.

Q 5. What is buffer solution ? What are its types?

Ans 5. Definition : It is a solution which has definite pH and which resists change in its pH value even wheni) It is diluted.

ii) It is kept for longer time andiii) Small amount of strong acid or strong base is added to it.

There are three types of buffer solutions.

i) Acidic buffer : Itis a mixture of weak acid and its salt with a strong base. e.g.CH3COOH + CH3COONa

It is prepared by dissolving weak acid and its salt with strong base in water.

The pH of acidic buffer solution is always less than 7.

ii) Basic buffer : It is a mixture of weak base and its salt with strong acid. e.g. NH4OH + NH4Cl.It is prepared by dissolving weak base and its salt with strong acid in water.

The pH of basic buffer solution is always more than 7.

iii) Single salt buffer or neutral buffer. It is a solution of a salt of weak acid and weak base in water.

It is prepared by dissolving simple salt of weak acid and weak base in water such as CH3COONH4 (ammonium acetate).

The pH of neutral buffer solution is around 7.

Q 6. Explain "solubility product".

Ans 6. Definition : "It is the product of the molar concentrations of cation and anion of a sparingly soluble salt raised to appropriate

power in its saturated solution at a given temperature."It is denoted by KSp.Explanation :

i) Consider the saturated solution of sparingly soluble salt BA in water.


9/20


9

ii) Then, an equilibrium will exist between the dissolved salt forming saturated solution and excess of undissolved BA

molecules in solid state and its ions.This heterogeneous equilibrium is represented as,

(solid)BA

n(salt sol )BA B A

Applying law of mass action, the equilibrium constant K is given by,

K =B A

BA

Since the salt BA in saturated solution, is in contact with its solid, the concentration of unionised BA molecules is taken

constant.

K =B A

cons tant

K constant = [B+ ][A-]

But, K constant = a new constant called solubility product (Ksp).Ksp = [B

+][A-]

In general, for a sparingly soluble salt like BxAy, dissociating in its saturated solution as,

BxAy xB+y + yA-x

KSP = [B+y]x[A-x]y

e.g. Ca3(PO4)2 3Ca++ + 2PO4

3-

KSP =3 2

34Ca PO

Chapter 4_Electrochemistry

Q 1. State and explain Faraday's first law of electrolysis.

Ans 1. Statement : "The weight of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the

quantity of electricity passed through the electrolyte". Explanation :

Let 'W' kg of a substance be deposited or liberated at an electrode during electrolysis and 'Q' coulomb is the quantity ofelectricity passed through the electrolyte.

Then according to the Faraday's first law of electrolysis,

W Q

W = Z Q

But from Faraday's 1st law W = Z I t

WA = ZA I t .(ii)

WB = ZB I t .(iii)

where ZA and ZB are the electrochemical equivalents of A and B,

Dividing equation (ii) by (iii) we get

A A

B B

W Z I t

W Z I t

A A

B B

W Z

W Z

.(iv)

From equation (i) and (iv) we get

A A

B B

E Z

E Z

E Z

E = F Z

Where, F = constant But, 1 F = 96500 coulomb.

E = 96500 Z or Z =E

96500

Q 2. What is a electrochemical cell? How it is constructed?Ans 2. "A device which converts chemical energy into electrical energy is known as electrochemical cell or voltaic or Galvanic cell".

OR

"The device which can generate electrical energy at the expense of spontaneous redox reaction is called electrochemical cell". Construction of electrochemical cell :

i) Every electrochemical cell is made up of two parts called electrodes or half cells.

ii) When any metal is in contact with its own ions it is called electrode or half cell.

iii) Two half cells or electrodes are required for the construction of an electrochemical cell as follows:a)Anode or Oxidation half cell : It is the electrode where oxidation reaction or liberation of electrons takes place. It has

negative charge in electrochemical cell.

b) Cathode or Reduction half cell: It is the electrode where reduction reaction or gain of electrons takes place. It has

positive charge in electrochemical cell.

iv) In electrochemical cell, these two half cells are connected internally by a porous pot or salt bridge or porous partition andconnected externally by a metallic wire.

v) Due to the spontaneous redox reaction electric current is continuously generated and flows from cathode to anode while

electrons flow from anode to cathode. e.g. Daniell cell.

Q 3. What is salt bridge? Give its functions.

Ans 3. "It is an inverted U-shaped glass tube containing saturated solution of KCl, KNO3 NH4O3in agar-agar gel with glass wool

plugs at the two ends".


10/20


10

Functions of salt bridge :

i) A salt bridge connects the two half cells of an electrochemical cell.,ii) It prevents direct mixing of solutions of two half cells.

iii) It helps in making the electrical contact between the two electrolyte solutions by-means of diffusion of ions (ionic

conductance).

iv) It minimises the liquid junction potential between the two electrolytes.

v) It maintains electrical neutrality in the solution by transport of ions.

Q 4. What is standard hydrogen electrode (SHE)? Describe construction and working of SHE.

Ans 4. "It is an electrode in which pure and dry hydrogen gas is bubbled at 1 atmospheric pressure at 298K about a platinised

platinum plate through a solution containing H+ ions at unit activity." Priniciple : It is a primary reversible reference electrode.

Diagram

Construction:i) The electrode consists of a glass jacket with an inlet at the top and small outlets at the bottom for removal of excess

hydrogen gas.

ii) There is a glass tube fitted inside the glass jacket in which platinum wire is sealed at the bottom whose other end is usedfor electrical contact with other electrode.

iii) A small amount of mercury is placed at the bottom of glass tube for good electrical contact.

iv) The lower end of platinum wire is connected to platinised platinum plate (coated with platinum black).

v) The whole glass assembly is dipped into 1M HCl solution(i.e. H+ ions at unit activity). Working:

i) When pure and dry H2 gas is bubbled at 1 atm and 298K into HCl solution, a small amount of H2 gas gets adsorbed onplatinum plate.

ii) Here, H2 oxidises to H+ ions and passes into solution leaving behind electrons on platinium plate.

iii) Excess of H2 gas escapes through the holes in the outer jacket.iv) An equilibrium is established between H2 gas and H

+ ions in the solution as,

2(g)

1H

2(aq)H e

iv) Electrode potential : The dissolution of H2 gas in the form of H+ ions in the surrounding solution due to oxidation takes

place to a small extent and again, H+ ions from solution accept electrons to form H2 gas back, very few electrons remain on

the plate at equilibrium. Hence, very small oxidation potential is developed which is arbitrarily considered as zero volt for thesake of convenience. Representation :

2(g)(1M)

(Lava)

H H , Pt

Q 5. Give Nernst equation for single electrode potential or cell potential and explain the terms involved in it

Ans 5. i) The potential of single electrode or of a cell, measured under standard conditions (i.e. latm and 298K) is called standardpotential.

ii) But if the concentration of solution is other than 1M or temperature other than 298K, then, the potential of an electrode or


11/20


11

cell can be calculated by Nernst equation. It can be given as,

010

Products2.303RTE E log

nF Reactants

where

E single electrode potential or potential of cell.E standard potential.R gas constant (8.314 JK-1 mol-1)

n no. of electrons involved in reaction.

T temperature in Kelvin.F Faraday (96500 C)

But at 298 K, the factor is2.303RT

F= 0.0591

Hence the equation becomes

o10

[Products]0.0591E E log

n [Reactants]

Q 6. Define E.M.F. series or electrochemical series. What are its applications?Ans 6. "The series of the elements arranged in the decreasing order of their standard reduction potential is known as E.M.F. series or

electrochemical series".

The values of reduction potential are given with respect to SHE scale taking reduction potential of S.H.E. zero.It is also called electrode potential series or activity series.

Applications :

(a) Selection of reducing agents : The elements at the bottom of the series have standard negative reduction potential and

hence they have a tendency to lose electrons readily. Hence, they are strong reducing agents.As, the standard reduction potential value decreases from top to bottom in e.m.f. series, the reducing strength increases. e.g.

The reducing strength of the element decreases in the order, Ii>K>Na>Zn(b) Selection of oxidising agents : The elements at the top of the series have high value of positive standard reduction

potentials. Therefore, they have a tendency to accept electrons readily. Hence they are strong oxidising agents. As standard

reduction potential decreases from top to bottom in the series, oxidising strength decreases in the same order. e.g. F2 > Cl2 >

Br2 > I2.

(c)To study the spontaneity of redox reactions : If the standard E.M.F. of cell is positive, then the cell reaction takes placespontaneously in the forward direction. For e.g. in case of Daniell cell,

2 2(S) (aq) (aq) (S)Zn |Zn ||Cu |Cu

The standard emf of cell is0 0 0ce ll r ed (Cu ) r ed (Zn )

(Cathode) (Anode)

E E E

= 0.34-(-0.76) = 1.1 VSince, Ecell for the above cell is 1.1V i.e. positive, the following cell reaction is spontaneous and Zn reduces Cu

2+ ions to solid

copper.2 2

(S) (aq) (aq) (S)

Zn Cu Zn Cu

Hence, the spontaneity of redox reaction can be studied with the help of emf series.

(d)To study metal displacement reaction: Metals having lower standard reduction potential can displace the metals havinghigher standard reduction potential from their salt solutions.

Hence, the metal in lower part of E.M.F.

series can displace metal placed above it from its salt solution. e.g. Zn is placed below Cu in E.M.F. series.Hence, Zn can displace Cu from its salt solution,

Zn(S) + CuSO4(aq) ZnSO4(aq) +Cu(S)i.e. (S) (aq) (aq) (S)Zn Cu Zn Cu

(e) In calculation of standard E.M.F. of a cell : From the standard reduction potential values of the electrodes from E.M.F.

series, the standard e.m.f. of a cell can be calculated. e.g. Daniell cell0 0 0ce ll r ed (Cu ) r ed (Zn )

(Cathode) (Anode)

E E E

-0.34 -(-0.76) = 1.1 V

f) E.M.F. series also gives an idea whether the metal will be dissolved in non-oxidising acids or not.

Metals with negative 0(red)E value which are placed below hydrogen in e.m.f. series dissolve in conc. HCl liberating hydrogen

gas.Metals with positive 0(red)E values which are placed above hydrogen in e.m.f. series will not dissolve in conc. HCl.

Q 7. Describe the construction and working of lead accumulator.

Ans 7. Principle:Lead accumulator is a secondary, reversible electrochemical cell.


12/20


12

Construction:

i) It consists of acid-proof, non-conducting vessel containing 38 % H2SO4 solution of specific gravity 1.28.

ii) Two lead electrodes are dipped in this sulphuric acid solution which acts as an electrolyte.

iii) The negative electrode or anode consists of series of lead plates connected together.

iv) The positive electrode or cathode consists of series of lead plates coated with PbO2 which are connected together.v) The negative and positive plates are arranged alternatively in the electrolyte, i.e. H2SO4 solution

Working:

Lead accumulator works in two parts.

a) Discharging : It is a process in which chemical energy is converted into electrical energy. Hence it is process of supplying

electrical energy. Here chemical energy is converted into electrical energy. Therefore it act as an electrochemical cell during

discharging process.

i) At negative electrode (anode) :

Lead is oxidised to form Pb2+ ions which combines with SO42- ions of H2SO4 to form insoluble PbSO4.

2(S) (aq)

2 24(aq) 4(S)

2(S) 4(aq) 4(S)

Pb Pb 2e

Pb SO PbSO

.......oxidationPb SO PbSO 2e

ii) At positive electrode (cathode) :

PbO2 get reduced by accepting electron and forms PbSO4(s) .In PbO2 lead has oxidation state +4 and it is reduced to +2oxidation state by accepting two electrons.

PbO2(s) + (aq)4H +24(aq)SO + 2e

- PbSO4(s) + 2H2O(l)

The overall cell reaction during discharing is, the sum of anode and cathode reactions as

Pb(s) + PbO2(s) + 2H2SO4(aq) 2PbSO4(s) + 2H2O{1)During discharging H2SO4 is converted into PbSO4 and H2O. Hence the concentration of H2SO4 falls and specific gravity

decreases upto 1.17, at this stage cell should be charged.

b) Charging :i) It is the process in which electrical energy is converted into chemical enegy. Here electrical energy is stored in the form of

chemical energy. Hence it is also called storage cell.

ii) During charging fresh H2SO4 is added and external e.m.f. slightly greater than 2 volt is applied.iii) Due to applied e.m.f. electrolysis of PbSO4 takes place. Pb is deposited on negative electrode and PbO2 is deposited on

positive electrode and H2SO4 is regenerated. Hence exactly reverse reaction of discharging takes place.

4(s) 2 (l)2PbSO 2H O

Charging

Dischar in(s)Pb + 2(s)PbO + 2 42H SO

E.M.F. of cell is about 2.0 V when 38 % H2SO4 is used but it depends upon concentration of H2SO4.

Representation of cell :

Q 8. What is a fuel cell? Explain the construction and working of H2O2 fuel cell.Ans 8. Fuel cell : "It is an electrochemical or galvanic cell in which chemical energy of a fuel is directly converted into electrical

energy ".

Definition: "It is defined as an electrochemical cell which converts the heat energy obtained from the combustion of fuel

directly into electrical energy".Hydrogen - Oxygen fuel cell: It is also called Bacon cell after the name of scientist who invented it. It is a very common and

successful fuel cell.

Principle : It is based on the combustion of hydrogen i.e. reaction of hydrogen with oxygen to form water.

2H2(g) + O2(g) 2H2O(l) Thus, it is an electrochemical cell in which chemical energy obtained through combustion of

hydrogen is directly converted into electrical energy.


13/20


13

Construction:1. The hydrogen-oxygen fuel cell consists of two porous graphite electrodes impregnated with catalyst such as pt, silver or

palladium.

2. A concentrated solution of aq. KOH or NaOH is placed between the two electrodes which acts as an electrolyte.3. Pure hydrogen and oxygen gases are bubbled under a pressure of 50 atm and temperature 525K through the porous

electrodes into the electrolyte, KOH or NaOH solution. The cell works continuously so long as the gases hydrogen and

oxygen are supplied at the temperature 525K and 50 atm pressure. Working of the cell: Following reactions occur at anode

and cathode during working of the cell.

a) Oxidation half reaction at anode: The pure hydrogen gas bubbled at anode is oxidised to form H+ ions by liberating

electron. These H+ ions are further neutralised by OH- ions furnished from the dissociation of electrolyte, NaOH or KOH.

Reaction at anode :

2(g) (aq)

(aq) (aq) 2 (l)

2(g) (aq) 2 (l)

H 2H 2e

2H 2HO 2H O

H 2HO 2H O 2eoxidation

b) Reduction half reaction at cathode: The oxygen gas bubbled at cathode is reduced by accepting electrons to form OH -ions as follows

Reaction at cathode is,O2(g) + 2H2O(l) + 4e

- 4OH-reductionThe net cell reaction : It is the sum of oxidation half reaction at anode and reduction half reaction at cathode as,

2(g) (aq) 2 (l)

2(g) 2 (l) (aq)

2(g) 2(g) 2 (l)

2H 4OH 4H O 4e oxidation

reductionO 2H O 4e 4OH

2H O 2H O

EMF of cell: The emf of the cell is found to be 1.0 volt.

Water produced in the working of the cell vaporises off due to high temperature (525K). The water vapour leaves the cellform the top which can be condensed and used again.

Q 9. Describe electrochemical theory of corrosion or Explain mechanism of corrosion. Ans. 1. According to this theory corrosion is basically an electrochemical phenomenon.

2. Rusting of iron is a very common example of corrosion. The theory of corrosion can be explained using rusting of iron as

an example as follows. Mechanism of rusting of iron : According to electrochemical theory, the impure iron surface behaves like a small electro

chemical cell in presence of water containing dissolved oxygen or CO2. Such a cell is called corrosion cell or corrosion

couple.In these small corrosion cells on the surface of iron, pure iron acts as an anode and impure surface of iron acts as a cathode.

The water vapour or moisture on the surface of metal dissolves CO2 or O2 from air and the surface of metal sets covered with

solution of CO2 in water i.e. carbonic acid (H2CO3).

H2O + CO2 H2CO3 This H2CO3 acts as an electrolyte, it dissociates in the solution as,

H2CO3 H+ + HCO3

-Following chemical reactions take place at anode and cathode in these small electrochemical cells on the

surface of iron.

a)Reaction at anode : Pure Fe acts as anode and Fe atoms are oxidised to Fe++ ions leaving behind electrons on the surface

which are then pushed into cathodic area.

Fe Fe++ + 2e- E = - 0.44 V Thus the sites or spots on the surface of iron where the above oxidation reaction takes place act

as anodes.

b)Reactions at cathode : Impure iron surface acts as cathode. The electrons liberated at anode are picked up or accepted by

H+ ions which are produced either from H2O or from dissociation of electrolyte, H2CO3The H+ ions thus formed, reduce dissolved oxygen to form H2O. Thus at cathode oxygen is reduced to H 2O.

2H+ +1

2O2 + 2e H2O, 0(red)E = 1.23 V.

The spots where this reaction takes place, act as cathodes.

The electrochemical reaction taking place in the corrosion cell is the sum of the above anode and cathode reaction as follows,

At anode : Fe Fe++ + 2e-, 0oxE = - 0.44V At cathode :

2H+ +1

2O2+ 2e H2O:,

0redE = 1.23V

Net reaction in the cell :


14/20


14

Fe + 2H+ +1

2O2 Fe

++ + H2O, E = 1.67 V

The Fe++ ions so formed move through water and reach on the surface of iron object where they are further oxidised to Fe 3+

ions by the action of atmospheric oxygen and produce the non-sticking material on the surface called rust which is hydrated

ferric oxide. (Fe2O3.nH2O).

2Fe++ +1

2O2 + 2H2O Fe2O3 + 4H

+ Fe2O3 + nFLO 2 3 2(Rust)

Fe O .nH O

Thus rust is a non-sticking compound (Fe2O3.nH2O) Which does not stick to the surface.

Impurities present in iron favour rusting process by setting up of a number of very small corrosion cells.

Mechanism of corrosion or rusting of iron

Chapter 5_Nuclear and RadiochemistryQ 1. Distinguish between isotopes and isobars.

Ans 1.

Isotopes Isobars

1. Atoms of same element having same atomic

number but different mass numbers are calledisotopes:

1. Atoms of different elements having same mass number but

different atomic numbers are called isobars.

2. They occupy same position in periodic table. 2. They have different positions in periodic table.

3. They have similar chemical properties. 3. They have different chemical and physical properties.

4. They contain same number of protons but

different number of neutrons in their nuclei.

4. They contain different number of protons & different number of

neutrons in their nuclei.

5. They contain same number of electrons. e.g.16

8O17

8O18

8O

5. They contain different number of electrons. e.g.40

18Ar40

19Ar40

20Ar

Q 2. Define mass defect. How is it calculated?

Ans 2. Definition : It is the difference between theoretically calculated mass and isotopic (observed) mass of nucleus.

It is denoted by Am.

It is generally expressed in a.m.u.

Calculation of mass defect,1) Let 'X' be the nucleus having atomic number, Z and mass number, A.

2) Let mp be the mass of a proton and mn be the mass of a neutron.

3) Let the observed mass of nucleus i.e. isotapic mass be M..

4) Then mass defect, Am of the nucleus is given as,

Mass Theoretical observed/isotopic

defect massof nucleus massof nucleus

sumof masses observed

of protonsand massof

neutrons nucleus

m = [Zmp+ (A - Z)mn - Mi] a.m.u. OR m = [ZmH + (A- Z)mn- Mi] a.m.u.

were mH = mass of a hydrogen atom, 11H .

Q 3. What is nuclear binding energy ? How is it related to mass defect ?

Ans 3. Energy equivalent to mass defect which is released in binding the nucleons together in the nucleus of an atom is known asnuclear binding energy.

OR

The amount of energy required to break the nucleus of an atom into its isolated nucleons is called nuclear binding energy.Binding energy and mass defect are related to each other by the equation.

B. E. = m 931 MeV Where,

B. E. = Nuclear binding energy in MeV. m = mass defect in a. m. u.Binding energy is directly proportional to mass defect.

Q 4. Define Decay constant

Ans 4. Definition : The fraction of total number of atoms of radioactive element disintegrating per unit time is called decay constant

or disintegration constant or rate constant.


15/20


15

It is denoted by X. Its S.I. unit is sec -1. Other units are per second, per minute, per hour, per day, per year.

Q 5. Explain artificial transmutation with suitable example.

Ans 5. The process of conversion of stable isotope of one element into a stable isotope of another element by bombarding it with

suitable high speed particles or projectiles is called artificial transmutation.

Rutherford in 1919, bombarded high speed -particles on the nucleus of nitrogen then protons were produced and an isotope

of oxygen was obtained.14 4 17 1

7 2 8 1 T arg et EmissionRecoilProjectile

Nucleus

N He O H

In the above example, stable isotope of N 14 is converted into stable isotope of oxygen O17, by artificial means, hence it isartificial transmutation.

Q 6. Give the uses of radioisotopes in medicine.

ORWhat is radiotherapy? Give its applications.

Ans 6. Radiotherapy : The use of radioisotope in cure of disease or to diagnose disease is called radiotherapy.

Applications:

Radioisotopes are used in medicine to diagnose and in treatment of various diseases.

1. Radio cobalt Co60 is used in treatment of cancer.2. I131 is used to detect and cure disorders of thyroid glands.

3. Na24 in the form of NaCl is used to study blood circulation.

4. P32 is used in treatment of leukemia and tumors.

5. Radioactive isotope Fe59 is used to detect the cases of anaemia.

Chapter 6_Chemical Kinetics

Q 1. Define and explain the Instantaneous rate of reaction ? How it is determined graphically?

Ans 1. Definition: "The rate of change of concentration of any one of the reactant or product over a very small interval of time is

called instantaneous rate of reaction".

OR"The rate of reaction measured, for an infinitesimally small time interval is called Instantaneous rate of reaction".

Determination of instantaneous rate of reaction : Consider the reaction, A B

In order to determine the instantaneons rate of reaction, a graph is plotted between the different concentration of reactant, A

against the corresponding values of time.

Determination of instantaneous reaction rateIn order to find out, the rate of reaction at very small interval of time say, dt a tangent is drawn to the curve at point y

corresponding to that time, as shown in graph.

The slope of tangent (tan6) gives the instantaneous rate at that instant of time.

Thus instantaneous ratedx OQ

dy OP

Instantaneous rate is the actual rate of reaction.

Q 2. What is Rate Law expression? Explain Rate law expression with suitable example.

Ans 2. "A mathematical expression, which denotes the true rate of reaction in terms of concentration of reactant, which actually

influences the rate, is called rate law expression". Consider the general reaction, aA + bB cC + dD According to the law of

mass action,

Rate = k[A]a[B]b ----(i)

But according to the rate law, expression can be written as

Rate = k [A]P[B]q(ii) In the above rate law equation, the values of p and q are determined experimentally and they may or

may not be equal to the coefficients of a and b in the reaction.

For example : Consider the reaction between NO2, and 2NO(g) + 2H2(g) N2(g) + 2H2O(g)According to law of mass- action,

Rate = k [NO]2 [H2]2 But the actual mechanism of above reaction is,

2NO(g) + H2(g) N2(g) + H2O2(g) (slow step)H2O2(g) + H2(g) 2H2O(g) (fast step)

Since the slow step is the rate determining step, therefore rate law expression is written according to slow step.

Rate = k [NO]2 [H2] Therefore, it was studied from above example that, the rate law expression must be determinedexperimentally. It can not be written by merely looking at the reaction.

Q 3. Define and explain the term orderof Reaction?Ans 3. Definition: "The sum of concentration terms on which the rate of reaction actually depends, as observed experimentally is


16/20


16

called the order of reaction".

OR"The sum of power of concentration terms, invoked in the rate law expression is called order of reaction".

Consider a general reaction

aA + bB productsAccording to rate Law expression, Rates of Reaction = k [A]p [B]q Therefore, the order of reaction (n) is written as,

n = p + q

The order of reaction with respect to A is P and that of with respect to B is q. The overall order of reaction is p + q. If n = 1,

reaction is first order n = 2, reaction is second order.

Q 4. What is zero order reaction? Explain with suitable example.Ans 4. Definition : "The reaction whose rate is independent of the concentration of reactants is called zero order reaction".

OR"The reaction which proceeds with a constant rate is called a zero order reaction". Consider a general reaction,A products

The rate of disappearances of reactant, A is given by the relation,

0d[A] K[A]dt

d[A]k

dt

Thus the rate of this reaction is proportional to zero01 power of the concentration of the reactant.

For zero order reaction, if a graph of concentration of the reactant against time is plotted, we get a straight line with slope = -k

and the intercept equal to 'a'. Example:The photochemical union of H2 and Cl2 gas in presence of sunlight is an example of zero order reaction.

sunlight2(g) 2(g) (g)H Cl 2HCl

It is found that the rate of such reaction is independent of concentration ofH2 and Cl2.

Rate of reaction =0 0

2 2k H Cl Rate of reaction = k Order of reaction = 0

The unit of rate constant for zero order reaction is mol dm -3 S-1.

Q 5. What is first order reaction? Explain it with suitable example.Ans 5. "The reaction whose rate depends upon the concentration of single reactant only is called the first order reaction ".

OR"The reaction whose rate depends on one concentration term only, is called the reaction of first order". Consider the first order

reaction

A Products

Here, Rate [A]

Rate = k[A]

For example :i) Decomposition of ammonium nitrite is a first order reaction.

NH4NO2 N2 + 2H2O

Rate of reaction = k [NH4NO2]1 Order of reaction = 1

ii) Decomposition of nitrogen pentoxide is also an example of First order reaction

N2O5(g) 2NO2(g) +1

2O2(g)

Rate = k[N2O5]1 Order of reaction1

The units of Rate constant for First order reaction is time-1 i.e. sec-1, min-1, hour-1 etc.

Q 6. Give the units of Rate constant (k) for first order reaction.

Ans 6. We know that, for first order reaction, rate of reaction is dependent upon the concentration of only one reactant

Rate = k[A]

The unit of rate =3Mol.dm

Sec

and [A] = mole.dnr33mol.dm

S= k mole.dm-3

k = S-1


17/20


17

The units of rate constant of first order reaction is time"1.

Q 7. Define and explain the term molecularity of chemical reaction.Ans 7. "The number of molecules (atoms or ions) taking part in a chemical reaction is known as molecularity of chemical reaction ".

OR"The number of reacting species (atoms, molecules or ions) which colloides simultaneously to bring about a chemical reaction

is called molecularity of chemical reaction".

Molecularity of reaction is always a positive integer (i.e. whole number like 1, 2 or 3) and does not depends upon the

experimental conditions. On the basis of molecularity, chemical reactions are classified as follows.i) Unimolecular reaction :

It is the reaction, in which only one molecule of reactant is involved.for e.g. Br2(g) 2Br(g)

ii) Bimolecular reaction : It is the reaction, in which two molecules of reactants are involved.

for e.g. H2 + I2 2HI

(iii) Termolecular reaction : It is the reaction, in which three molecules of reactants are involved.

for e.g. 2NO(g) + O2(g)2NO2(g)

Q 8. Distinguish between molecularity and order of chemical reaction.

Ans 8.

Molecularity Order of reaction

1. It is the total number of reacting species whichbrings about the, chemical change.

1. It is the sum of powers of the concentration terms in the ratelaw expression.

2. It is theoretical concept. 2. It is an experiment ally determined quantity.

3. It is always a whole number. 3. It may be whole number, zero, fractional, positive or negative.

4. It is generally not more than three and never zero. 4. It possesses all sorts of values including zero.5. It can be obtained from balanced chemical equation. 5. It can not be obtained from balanced chemical equation.

6. It does not change with change in temperature andpressure.

6. It changes with change in temperature and pressure.

7. It does not tell anything about the mechanism of the

reaction.

7. It tells us about the slowest step in the mechanism & hence

gives some clue about the reaction mechanism.

Chapter 7_d-block Element

Q 1. "Transition elements show variable oxidation states". Explain.Ans 1. 1. Oxidation state is the apparent or actual charge along with sign present on an atom in a molecule of its compound.

2. All the transition metals except the last member of each series show different oxidation states which are positive due to

their metallic nature.

3. In the case of transition elements, electrons from (n - 1)d orbitals and ns orbitals of the atoms can be used in bonding i.e.

valence electrons belong to last shell (ns orbitals) and last but one shell (n-1) d orbitals.

4. The difference of energy between ns orbitals and (n - 1) d orbitals is very small. Hence, electrons from both the shells canbe used for bonding.

5. Hence, (n - 1) d electrons requires very little energy to promote to valence shell and hence can take part in bonding.

6. Thus, depending upon the number of electrons used from (n - 1)d orbitals and ns orbitals, transition elements can showvariable or different oxidation states.

Q 2. "The transition metals form coloured compound". Explain. Ans 2. i) The compounds of transition metals are coloured in solid as well as in aqueous solution.

ii) A substance appears coloured when it absorbs specific wavelength of incident visible (white) light (400 nm to 780nm) and

reflects or transmits the light of remaining wavelength. Hence, the colour of transmi tted light is the colour of the substance. Itis the complimentary of the colour which is absorbed by the substance.

iii) The transition metal ions containing incompletely filled d orbitals i.e. unpaired electrons in 'd' orbitals or some vacant d

orbitals are coloured.iv) When a beam of light falls on such a transition metal compound, some of the wavelengths of incident light corresponding

to a certain colour are absorbed. This energy is just sufficient to promote or excite 'd' electrons from d orbitals of lower energy

to 'd' orbitals of higher energy. This type of promotion of electrons is called d-d transitions or d-d excitations. These d-dtransitions are possible due to small difference of energy between the two sets of d-orbitals. If energy difference between the

two energy levels is large, then high energy will be required for transition of electrons. Hence, ultraviolet light is absorbed

instead of visible light and such compounds will be, therefore, colourless. Naturally, the transition metal ions which havecompletely filled 'd' orbitals i.e. containing no unpaired electrons and completely vacant 'd' orbitals are obviously colourless

because there are no vaccant 'd' orbitals to permit d-d electron transitions.

The number of 'd' electrons and corresponding/ colours of ions of first transition series are shown in following table :

Q 3. "Cupric salts are blue in colour while cuprous salts are colourless." Explain.

Ans 3. Cu (Z = 29) has electronic configurationThe outer electronic configuration of Cu++ ion will be 3d9, 4s0.

The 3d orbital contains one unpairad electron. Therefore, cupric salts absorb red colour of incident light causing d-d transition

and wavelength of reflected or transmitted light corresponds to blue colour. Hence cupric salts or aqueous solutions of CuSO4

are blue in colour.

In the case of Cu+ ion, outer electronic configuration is 3d10 4s. Here, 3d orbital is completely filled and no unpaired electron


18/20


18

is present.

Therefore d-d excitatious are not possible. For the promotion of d electrons to higher energy 4s orbital, greater amount ofenergy will be required which is not available from visible light.

Therefore, light of visible range is absorbed and the entire light is transmitted and as a result cuprous salts appear white or

colourless.

Q 4. Write a short note on tendency of transition metals to form complexes.

OR

"Transition metals can form complexes." Give reasons.Ans 4. i) A compound in which the central metal ion is linked to a number of negative ions (anions) or neutral molecules having lone

pairs of electrons i.e. Lewis bases is called complex or co-ordination compound.ii) These negative ions or neutral molecules are called Ligands.

iii) The ligands donate lone pairs of electrons to the central transition metal ion forming co-ordinate bonds.

iv) The tendency of transition metal ions to form complexes is due to :a) Small size of cations.

b) High positive charge on the cations.

c) Vacant d orbitals of comparatively low energy.d) Availability of several oxidation states of the cations.

v) Due to the small size and high positive charge, the transition metal cations have a high positive electron density. Hence,

they can readily accept electron pairs donated by the ligands.vi) The transition metal cations possess vacant 'd' orbitals of appropriate energy to accomodate lone pairs of electrons

accepted from ligands.

vii) Due to these reasons, the transition elements have a tendency to form a large number of complexes.For ex:

a) K4[Fe(CN)6], potassium ferrocyanide in which complex ion is [Fe(CN)6]4-

b) [Cu(NH3)4]SO4, copper tetraamine sulphate.

Q 5. Explain catalytic properties of transition metals.

OR

"Transition metals act as good catalysts". Explain.

Ans 5. i) The transition elements and their compounds also show good catalytic properties.ii) The catalytic properties of transition elements are most likely due to the presence of unpaired electrons in incompletely

filled (n-1) d orbitals of their atoms.

iii) Transition elements show several oxidation states and tendency towards complex formation.iv) The catalytic properties of transition elements are mainly due to following two reasons.

a) According to modern theory of catalysis, the transition element working as a catalyst with varying oxidation states may

form intermediate unstable compounds with the reacting substances.

These intermediates provide a new path of reaction having lower activation energy.

The intermediate compound is decomposed readily in reaction with other reactant to give the product and the catalyst i.e.

transition metal is regenerated at the end of reaction.The formation of intermediate unstable compound takes place readily due to variable oxidation states of transition elements.

b) In some cases, transition metals provide a suitable large surface area having free valencies for the adsorption of reactants

on their surface.

Therefore the concentration of the reactant molecules on the surface of the catalyst i.e. transition metal increases. Hence,

according to law of mass action, rate of reaction also increases -

Q 6. Describe the method of extraction of zinc from zinc blende by carbon reduction process. Give the reactions involved.

Ans. Following steps are involved in the extration of Zn from Zinc blende.

A)Concentration of ore

B)Roasting

C)Reduction ( smelting) by vertical retort process

D)Purification A)Concentration of ore : The process of removal of unwanted material (gangue) from the ore is known as concentration or

dressing or benefaction of the ore. Froth floation process :

The ore, zinc blende is first finely ground and then introduced in a big tank.The powdered ore is agitated by compressed air with water and a small quantity of pine oil, fuel oil or eucalyptus oil. The

sulphide ore particles get preferentially wetted by oil and rise to the surface of tank in the form of froth or foam. The stony

matter or impurities get preferentially wetted by water and settle down at the bottom of water layer and are called gangue.

The froth which rises up the tank contains sulphide ore particles is made to overflow and then collected and squeezed throughthe convas bags to remove water and oil. The gangue particles settled at the bottom are then rejected.

Thus earthy and silicious impurities present in the ore are removed by this process.If iron impurities are also present, then they are removed by electromagnetic separation- process.

B)Roasting or calcination : The concentrated sulphide ore is roasted i.e. heated in presence of excess of air (oxygen) at

about 900-950 C in a reverberatory furnace or muffle furnace for the oxidation of ZnS to ZnO.

Following reactions take place during this process.

i) Zinc sulphide (ZnS) is oxidised to zinc oxide (ZnO).

2ZnS + 3O2 2ZnO + 2SO2

ii) Some part of the ore ZnS that may be oxidised to ZnSO4 in this process.


19/20


19

This ZnSO4 is further decomposed to give ZnO at about 900-950c temperature.

ZnS+ 2O2 ZnSO4

2ZnSO4 2ZnO + 2SO2 O2

In this step, SO2 obtained is an important byproduct which can be used for manufacture of H2SO4.

C) Reduction (Smelting) by modern vertical retort process : In this step, zinc oxide is reduced to zinc metal by heating itin presence of reducing agent, coke at 1400-1450C in vertical fire clay retort

The main reaction taking place is,

ZnO + C 1400 1450 Zn + CO

This is a continuous process and is also economical. Hence, it is largely used now a days for the reduction of ZnO.

Construction : The retorts used in this process are made up of highly refractory silicon carbide bricks which can withstand at

very high temperature. Each retort is rectangular in shape having about 6.75 m height, 1.85 m long and 0.3 m broad. Theseretorts are arranged in vertical position.

The retort is enclosed in a heating jacket in which producer gas is burnt for heating the retort.

The heating jacket is provided with an inlet for introducing a mixture of producer gas and air and an outlet for the escape ofburnt gases.

The retort has a charging door at the top for introduction of charge i.e. roasted ore and coke.

There is an inlet near bottom of retort for the entry of producer gas.The bottom of retort is also provided with an extension consisting of a fan and a screw conveyer for the removal of ash.

Working :

i) The roasted ore i.e. ZnO obtained from the previous step is mixed with 40 to 60% of powdered coke and pressed into smallblocks or briquettes .

ii) These briquettes are then introduced periodically in to retort through the charging door at the top.

iii) This charge (ZnO + C) is heated in the retort to about 1400 - 1450c. Now, ZnO is reduced by carbon to zinc and CO isformed. The main reaction taking place is,

ZnO + C Zn + CO

iv) Zinc so formed evaporates at this temperature to form Zn vapours.

v) The vapours of zinc and CO are led to the condenser by applying a suction and by the mild current of producer gas forcedfrom the bottom.

vi) Here, Zinc vapours are condensed to liquid and the vapours of CO are again led to furnace which is used as fuel for

heating the furnace.vii) The crude zinc obtained in liquid form is called zinc spelter.

The molten zinc is taken out from the condenser through the tapping hole from time to time into the receiver.

viii) The fresh charge is introduced through the charging door at the top of retort and the waste residue is taken out of theretort with the help of automatic dishcarge mechanism at the bottom of retort without discontinuing the process. Thus, the

process is made continuous.

D) Purification : The crude zinc (zinc spelter) obtained in the above step is about 97-98% pure and is further purified by

electrolytic method or sometimes by fractional distillation.

Q 7. Explain the electrolytic process for purification of zinc.

Ans 7. i) The zinc metal obtained in the vertical retort process is crude (i.e. zinc spelter) and contains about 97-98/o zinc and rest

impurities of lead, iron, cadmium, antimony, copper, arsenic etc.ii) Impure zinc is purified by electrolytic process as follows :

iii) The electrolyte taken, for the electrolysis is zinc sulphate (ZnSO4) solution acidified with little amount of H2SO4 in an

electrolytic cell.iv) Anode consists of a thick rod or place of impure zinc and cathode is a thin plate of pure zinc metal.

v) These two electrodes are dipped in the electrolyte, ZnSO4 and electric current is passed through the solution.

vi) During electrolysis, anode consisting of impure zinc undergoes oxidation, i.e. Zn atoms pass into solution as Zn++ ions andelectrons are liberated. Due to dissociation, anode decreases in size.

Therefore, reaction at anode,

Zn(s) Znaq++

+ 2e-.....oxidation.

vii) The impurities set free at anode, settle down at the bottom of the electrolytic bath below the anode are called as anode

mud.viii) At cathode, Zn++ ions from the ZnSO4 solution accept electrons and are deposited on cathode as pure zinc metal. The

reaction is,

Znaq++

+ 2e- Zn(s)-----reduction.


20/20


20

ix) In some cases, the aluminium plate is also used as cathode.

Zn is deposited on this plate which is then scrapped off.x) This process gives extremely pure zinc, about 99.99% pure.

Diagram :

raoult law question

Documents