nvs2 numerical made easy

CHEMISTRY NUMERICALS MADE EASY…

My dear students,

In CBSE XII class Chemistry, Numericals asked from Unit -1,2,3 & 4 it

carries Eight to Ten marks in the Board Examination. Most of the time in the Board

examination, one numerical problem for Three (3) or Two (2) mark question are usually

asked from each unit. The following important points to remember to carry numericals

and some solved numericals also given for your ready reference.

SOLID STATE

In the Numerical problem, usually calculating efficiency of packing in

FCC and BCC, Density of the Crystal, calculating the radius ratios and finding out the

Co-Ordination number abd also the type of the crystal to be identified may be asked.

Before starting, let us remember the following points:

1. NUMBER OF ATOMS PRESENT IN A UNIT CELL:

(a) If it is a simple cubic unit cell, only ONE (1) atom will be present.

(b) If it is a body centred cubic unit cell, only TWO (2) atoms will be present.

(c) If it is a face centred cubic unit cell, only FOUR (4) atoms will be present.

2. RELATION BETWEEN THE LENGTH OF THE UNIT CELL, RADIUS OF

THE ATOM AND NEAREST NEIGHBOUR DISTANCE:

(a). For a Simple cube : a = 2r = d

(b) For B.C.C. : a = 3

4r =

3

2d

( c) For F.C.C : a = 2 2 r = 2 d

Let us identify each term in it :

a = Length of the Unit Cell, r = Radius of the Atom,

K.G.MALLIKARJUN, VICE PRINCIPAL, JNV, ANANTHAPUR(A.P.)

1


d = Nearest neighbour distance

HOW TO FIND OUT THE CO – ORDINATION NUMBER :

When r+ and r – values were given, we can find out the Co – Ordination Number.

a). If −

+

r

r ratio is between 0.225 – 0.414, the Co – ordination number is four (4)

and the shape is Tetrahedral

b) If −

+

r

r ratio is between 0.414 – 0.732, the Co – ordination number is six(6)

and the shape is Octahedral.

c). If −

+

r

r rafio is between 0.732 – 1.000, the Co –ordination number is Eight(8)

and the shape is B.C.C.

NOTE : -

Some time , Co –ordination number and one of the radius will be given and asked

to calculate the other radius. In this case, −

+

r

r ration should be equated to the

MINIMUM value of the radius ratio. For example. The Co – ordination number of

crystal is sic(6). Then, you have to equate the radius ratio value to 0.414 and calculate the

other radius.

HOW TO CALCULATE THE EFFICIENCY OF PACKING:

(a) For a simple cubic

a = 2r


2


∴ Volume occupied by all the spheres in a unit cell = 1 x 3

4 πr3 ( 1 sphere per unit

cell)

Volume of the unit cell = (2r)3 (volume = a3)

Volume occupied by all the sphere in unit cell Hence, packing efficiency = x 100

Volume of unit cell

= 3

3

)2(

1003

41

r

r ×× π

= 67

22

× = 52%

(b) In B.C.C

There are two atoms per unit cell.

Packing efficiency = 3

3

)3

4(

1003

42

r

r ×× π ( aa

3

4= )

= 2 x 3

4 x

444

33

xx x 100 = 68 %

(c) In F.C.C .

Number of atoms per unit cell = 4

a = 2

4r = 2 2 r

Packing efficiency = 3

3

)2

4(

1003

44

r

r ×× π = 74 %


3


HOW TO CALCULATE THE DENSITY OF A CRYSTAL:

d = ANa

zM3


d = Density of the crystal

z = Number of atoms in the Unit cell

M = Mass of the Unit cell

a = Length of the Edge of the Unit cell

NA = Avogadro’s Number.

NOTE : -

(a). Some times, the distance between the two(2) ions will be given. You can

calculate the length of the edge of the Unit cell (a) by multiplying the distance

between the Two(2) ions with 2.

2 x Distance between the two ions = Length of the edge of the Unit cell(a).

(b). To calculate the distance between the two ions, you divide the length of the

edge of the unit cell with 2.

2

a will give the distance between the two ions.


4


SOLVED PROBLEMS

1.Calculate the efficiency of packing in case of metal crystal for :-(a) simple cubic (b) Body centred cubic (c) face centred cubic (with the assumption that atoms are touching each other)

(a) Simple cubic:-

Let cell edgelength be a and radius of an atom be r

a = 2rVol. of cube = a3 = 8r3

z = 1; Volume of one sphere = (4/3) πr3

Efficiency of packing= vol. of all spheres in one cube x 100/ Vol. of cube= [(4/3) πr3 x 100 ]/ 8r3

= 52.3%

(b)Body centred cubic(BCC)

Body diagonal = 4r(4r)2=a2+(√2a)2=a2+2a2=3a2

16r2=3a2

3

4ra =

Vol. of cube = 33

64 33 r

a =

z=2; Vol. of two spheres = 23

4 3 ×rπ

Efficiency of packing= vol. of two spheres x 100/ Vol. of cube

=

33

64

1003

42

3

3

r

r ××× π

= 68%

(c) Face centred cubic (FCC)

Face diagonal = √2a = 4r (or) 2

4ra =

Vol. of cube = a3 =22

64 3r

z = 4 ; Vol. of 4 spheres = 3

3

44 rπ×


5


Efficiency of packing = \\3

3

22

64

1003

16

r

r ×π

= 74%

2. Niobium crystallizes in body centred cubic structure.. If Density is 8.55 g/cm3. Calculate atomic radius of niobium (At. Mass of Niobium=93).

Ans:- ANa

Mz3

×=ρ

8.55= 233 10023.6

932

×××

a

a3= 2310023.655.8

932

×××

a=[3.6x10-23 ]1/3 = 3.302 x 10-8 cm = 3.302 x 10-10 m

For BCC unit cell √3a = 4r : r = nmma

29.144

103.3732.1

4

3 10

=××=−

3. Formula mass of NaCl is 58.45 g/mol and density of it’s pure form is 2.167 g/cm3. Average distance between adjascent sodium and chloride ions in the crystal is 2.814x10-8 cm.Calculate Avogadro constant.

M =5 8.45 g/molρ = 2.167 g/cm3

a = 2 x distance between Na+ and Cl- = 2 x 2.814x10-8 cm = 5.628 x10-8 cm

ANa

Mz3

×=ρ

NA = ( )23

38100523.6

167.210628.5

45.584 ×=××

×−

4. Analysis shows that nickel oxide has formula Ni0.98 O1.00. What fractions of Nickel exist as Ni2+ and Ni3+ ion?

Let no of Ni2+ be ‘x’ and the no. of Ni3+ be ‘y’.


6


x + y = 0.98 (1)

The total positive charge per formula unit = 2+, since charge of one oxide ion is 2-. Therefore

2x + 3y = 2 (2) Multiply (1) with 2 => 2x + 2y = 1.96 (3)

From (3) and (2) y = 0.04 & x = 0.94

No. of Ni 3+ = 0.04, No of Ni 2+ = 0.94

% of Ni3+ = %08.498.0

10004.0 =×

% of Ni2+ = 95.92%

5. The compound CuCl has ZnS structure. Its density is 3.4 g/cm3 . What is the length of the edge of the unit cell?

ANa

Mz3

×=ρ

ie, a3 = 231002.64.3

994

×××

= 96.736 x 10-23cm3

a = 5.75 x 10 -8 cm

6. Thallium chloride, TlCl crystallizes in either a simple cubic lattice or a face centred cubic lattice of Cl- ion with Tl+ ion in the holes. If the density of the solid is 9 gcm-3 and edge of the unit cell is 3.85 x 10-8 cm, what is the unit cell geometry?

ANa

Mz3

×=ρ ,

( ) 2338 10023.61085.3

2409

××××= z

240

10023.61085.385.39 2324 ×××××=−

z

z =1.289 ≈ 1

So TlCl is simple cubic.


7


7. Gold (atomic radis =0.144 nm). Crystallises in face centred Unit cell. What is the length of a side of a cell?

For FCC, a = 2√2r = 2 x 1.414 x 0.144nm = 0.4072 nm

8. KF has NaCl structure. What is the distance between K+ and F- in KF. If density is 2.48 g/cm3.

ANa

Mz3

×=ρ ,

( ) 3/1231002.648.2

584

×××=a

= 5.374 x 10-8cm

Distance b/w K+ & F- = ½ a = ½ (5.374 x 10-8 cm) = 2.687x10-8 cm = 269 pm

9..A metal ( atomic mass = 50) has a body centred cubic crystal structure. The density of the metal is 5.96 g /cm3. Find the volume of its unit cell ( NA

= 6.023 x 1023 atoms / mol.)

Ans:- Mass per unit cell = 2310023.6

502

××

Density of unit cell = 5.96 g /cm3

Volume = ?

Density = volume

Mass

Volume = 3233

23108.2

96.5

1

10023.6

502cmcm

Density

Mass −×=××

×=


8


SOLUTIONS

My Dear Students,

Let us see the various formulas involved in this chapter.

HOW TO CALCULATE THE CONCENTRATION OF A SOLUTION:

1. MOLARITY CAN BE CALCULATED AS :

No. of moles of the soluteM =

Volume of solution in liters

2. MOLALITY CAN BE CALCULATED AS :

No. of moles of the solutem =

Volume of the solvent in Kilo grams.

3. NORMALITY CAN BE CALCULATED AS :

No. of gram equivalents of soluteN =

Volume of solution in liters

4. MOLE FRACTION OF EACH COMPONENT :

No. of moles of the componentsXA (or) XB =

Total Number of moles

HOW TO CALCULATE THE NUMBER OF MOLES:

Weight of the componentNumber of moles =

Molecular weight of the component

HOW TO CALCULATE THE WEIGHT OF THE SOLVENT:

Weight of the solvent = Weight of the solution - Weight of the soluteNOTE :

1. Students, while calculating Molality (m), you will have to convert the weight of

the solvent in Kgs.


9


2. While calculating the Molarity(M), you will have to convert the volume of the

solution in Liters.

3. While calculating the mole fraction, normality Component A is taken as Solvent

where as Component B is taken as Solute.

XA = BA

A

nn

n

+ XB = BA

B

nn

n

+

Where XA and XB are the mole fractions of the components of A and B

4. XA + XB = 1 . If you calculate XA, you can calculate XB = 1 - XA

HOW TO CALCULATE PARTIAL PRESSURE OF THE GAS (HENRY’S LAW):

Partial pressure of the gas in solution = KH x Mole fraction of the gas in solution

(or)

P = KH . X

KH = Henry’s law constant.

PROBLEMS ON COLLIGATIVE PROPERTIES:

1. Relative Lowering of Vapour pressure :

Xsolute = solvent

osolutionsolvent

o

P

PP −(or)

WM

Mw

solute

solvent

××

= solvent

osolutionsolvent

o

P

PP − (Xsolute =

WM

Mw

solute

solvent

××

)


Posolvent = Vapour pressure of the pure Solvent


10


Psolution = Vapour pressure of the solution.

XB = Mole fraction of the solute

w = Weight of Solute

W = Weight of Solvent

Msolvent = Molecular mass of solvent

Msolute = Molecular mass of solute

NOTE: -

From the value of mole fraction (XB), you can find out the molecular weight

of the Non – Volatile solute(MB).

XB = BA

B

nn

n

+ where nB= B

B

M

W

Where WB = Weight of the Non – Volatile solute

MB = Molecular weight of the Non –Volatile solute.

2. Problems on Elevation of boiling point:

Msolute = WT

wK

b

b

×∆××1000

Let us identify the each term in it :

Msolute = Molecular weight of the non-volatile solute

Kb = Molal Elevation constant

w = Weight of the non – volatile solute

W = Weight of the solvent


11


bT∆ = Elevation of boiling point.

NOTE : -

Students, it is observed that you are doing a mistake while calculating the

bT∆ value. Its value always should be in Kelvin only.

KTKTTb 12 −=∆

Where T1K = Boiling point of the pure Solvent, T2K = Boiling point of the solution.

3. Problem on Depression in Freezing Point:

Msolute = WT

wK

f

f

×∆××1000


Msolute = Molecular weight of the non-volatile solute

Kf = Molal Elevation constant

w = Weight of the non – volatile solute

W = Weight of the solvent

fT∆ = Depression in Freezing point.

NOTE :-


12


Students it is observed frequently that you are doing mistake while

calculating fT∆ value. Its value always should be in Kelvin only. Some times, fT∆

value is given in the negative sign. But, you should take only the positive value only.

fo

fo

f TTT −=∆

Where Tof = Freezing point of pure solvent, Tf = Freezing point of the solution.

4. Problems on Osmotic Pressure :

MRT=Π

(or)

RTV

n=Π (or) Msolute = V

wRT

Π


Π = Osmatic pressure in atmospheres only.

R = Gas constant = 0.0821 Ltrs. Atms.

V = Volume of the solution in Litres only

T = Temperature in Kelvin only.

n = No. of moles of the Non – Volatile solute.

M = Concentration of the solution in molarity.

NOTE :-

1. Students, the common mistake committed by most of you is substituting

the ‘R’ value. It is observed that the value of ‘R’ is taken as 8.314 instead

of 0.0821.


13


2. As the value of ‘R is in Ltrs Atms, care should be taken that the volume

of the solution should be in Litres and also the value of Osmotic Pressure

in Atmospheres inly.

3. Some times, Osmatic pressure value will be given in mm mercury. Hence,

it is to be converted into atmospheres. We know that 760mm Hg = 1

atmosphere.

4. Isotonic solutions were given to find out the molecular weight of the non

– volatile solute, Then equate both the Osimatic pressure values and use

this formula:

n 1 = n 2

Where n1 , n2 are the number of moles of the two solutes.

HOW TO CALCULATE THE VAN’T HOFF FACTOR (i) , DEGREE OF

ASSOCIATION AND DEGREE OF DISSOCIATION.

Normal Moleculat Massi =

Observed Molecular mass

NOTE :-

In case of dissociation the value of i < 1 and in case of association the value of i > 1.

To calculate degree of dissociation and association:

1.For Dissociation:

i = 1

1 α+ (or) i = 1 + α

Let us Know :

i = Van’t Hoff factor, α = Degree of dissociation


14


% of Dissociation = Degree of dissociation x 100

2.For Association:

i = 1 - 2

α

Let us Know : i = Van’t Hoff factor, α = Degree of association

% of Association = Degree of Association x 100

SOLVED PROBLEMS

1. 18 grams of glucose ( molar mass 180 gram /mol) is present in 500 cm3 of its aq. Solution. What is the molarity of the solution? What additional data is required if the molarity of the solution is also required to be calculated.

Ans ; - The amount of glucose present in one litre of solution

= 18grams x 3

3

500

1000

cm

cm = 36 grams

Number of moles of glucose present in one litre of solution

= 36 grams x grams

mol

180

1 = 0.2 mol.

2. Calculate the molarity and molality of a 15% solution (by weight) of H2SO4 of density 1.020 g / cm3.

Ans ; - Molarity of 15% H2SO4

Weight of H2SO4 in 100 grams of solution = 15 grams

Volume of 100 grams of solution = 02.1100 = 98 ml

( Density of H2SO4 = 1.02 g /cm3)

Moles of H2SO4 present in 98 ml of solution = 9815

Molarity of solution = 9815 x 98

1000 = 1.56M

Molality of 15% H2SO4

Weight of H2SO4 in 100 grams of solution = 15 gramsWeight of water in 100 grams of solution = 85 grams

Moles of H2SO4 present in 85 grams of water = 9815

Hence Molality = 9815 x 85

1000 = 1.8m.


15


3. If N2 gas is bubbled through water at 293K, how many millmoles of N2 gas would dissolve in 1 liter of water. Assume that N2 exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N2 at 293K is 76.48 K. bar.

Ans ; - The solubility of gas is related to its mole fraction in the aq. Solution. The mole fraction of the gas in the solution is calculated by applying Henry,s law. Thus.

xN2 = H

N

K

p2

= bar

bar

76480

987.0 = 1.29x 10-5

As a litre water contains 55.5mol of it (unit 8, class – XI) therefore, if n represents number of moles of N2 in solution.

XN2 = 5.55+n

n =

5.55

n = 1.29 x 10-5

Thus n = 1.29 x 10-5 x 55.5 mol. = 7.16 x 10-4 mol.

= 7.16 x 10-4mol x 1

1000m mol

= 0.716 m mol.4. The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A

non-volatile non electrolyte solid weighing 0.5grams is added to 39.0 grams of benzene(molar mass 78 gram / mol.). The vapour pressure of the solution then is o.845 bar. What is the molecular mass of the solid substance?

Ans :- p0Solvent = 0.850 bar pSolution = 0.845 bar

MSolvent = 78 gram / mol. w = 0.5 gram and W = 39 grams.

solvent

solutionsolvent

p

pp0

0 − =

WM

Mw

solute

solvent

××

bar

barbar

850.0

845.0850.0 − =

gramsMmol

gramgrams

solute 39

785.0

×

×

Msolute = 39005.0

850.0785.0

×××

bar

barmolgramgrams

= 170 grams / mol

5. A solution of 12.5 grams of urea in 170 grams water gave a boling elevation of 0.63K calculate the molecular mass of urea taking Kb = 0.52 K / m.

Ans: - ∆Tb = kb . m (or) Msolute = WT

wk

b

b

×∆××1000


16


W = 170 grams w = 12.5 grams, ∆Tb = 0.63, kb = 0.52K/m

Msolute = 63.0170

5.12100052.0

×××

= 60.7 grams / mol.

6. An aq. Solution of a weak mono basic acid containing 0.1 gram in 21.7 gram of water freezes at 273.813 K. If the value of kf for water is 1.86 degree / mol. What is the molecular mass of the mono basic acid.

Ans:- ∆Tf = kf . m (or) Msolute = WT

wk

f

b

×∆××1000

W = 21.7 grams w = 0.1 grams, ∆Tf = 273.813 – 273 = 0.813K/m, Kf= 1.86K/m

Msolute = 813.07.21

1.0100086.1

×××

= 10.54 grams / mol.

7. The osmatic pressure of blood is 8.21 atms. At 370. How much of glucose should be used per litre for an intravenous injection that is at the same osmatic pressure on blood.

Ans:- We know MSolute = v

RTwsolute

π.

π = 8.21 atms. T = 37 + 273 = 310K v = 1 litreMsolute = 180 wsolute = ?

wsolute = 310082.0

21.8180

××

= 58.06 grams.

8. One litre aqueous solution of sucrose ( molar mass = 342 grams/ mol)weighing 1015 grams is found to record an osmatic pressure of 4.82 atms. At 293K. What is the molarity of the sccrose solution( R = 0.0821 atom / mol./K)

Ans :- weight of solution = 1015 grams

We know π = vn RT

π = 4.82 atms. R= 0.0821 atms/mol./K , T = 293K

4.82 = vn x 0.0821 x 293

4.82 = molarity x 0.021 x 293 ( vn = molarity)


17


Molarity = 0.2 molar.

Mass of solute in one liter of solution = 0.2 x 342 = 68.4 grams

Mass of water = (1015 – 68.4) grams = 946.6 grams = 0.9466 kg

Hence molarity = 0.2 / 0.9466 = 0.211 kg . mol.

9. Two elements ‘A’ and ‘B’ form compounds having molecular formulae AB2

and AB4, when dissolved in 20grams of C6H6, 1 grams of AB2 lowers the freezing point by 2.3K, while 1.0 grams of AB4 lowers it by 1.3K. The molar depression constant for benzene is 5.1kg/mol. calculate atomic mass of A and B.

Ans :- Using the relation ∆Tf = kf . m For compound AB2, we get

2.3K = 5.1 K kg/mol x 1

1M x kg02.0

1

( M1 = Molecular mass of AB2)

M1 = 02.03.21.5

× = 110.8 / mol --------(1)

For compound AB4, we get

1.3K = 5.1K.kg/mol x 2

1M x kg02.0

1 )

( M2 = Molecular mass of AB4)

M2 = 02.03.11.5

× = 196.15 / mol -------- (2)

From (1) and (2)

A + 2B = 110.8 ---------(3)A + 4B = 196.15 ---------(4)

Solving (3) and (4)- 2B = - 85.35

B = 42.675

A = 25.43

10. At 300K, 36 grams of glucose present per litre in its solution has an osmatic pressure 4.98 bars. If the osmotic pressure of solution is 1. 52 bar at the same temperature, what would be its concentration.

Ans:- Using the relation π = CRT , we get

π1 = C1RT and π2 = C2RT


18


2

1π

π =

2

1C

C

52.198.4 = 180

36 x 2

1C

C2 = 0.061 mol./litConcentration of solution will be 0.061 mol./ lit.

ELECTROCHEMISTRY

My dear student,

This chapter involved very few formulas, Hence, you can score

more marks in this chapter.

How to calculate Molar Conductivity ( λm)

Molar Conductivity (λm) = M

1000×κ

Let us identify each term in it :-

κ (kappa) = Specific Conductivity

M = Concentration of the solution in Molarity.

How to calculate Equivalent Conductivity

Equivalent Conductivity = N

1000×κ

Let us identify each term in it :-

κ (kappa) = Specific Conductivity


19


N = Concentration of the solution in Normality

How to Calculate Degree of Dissociation

Degree of Dissociation ∞≈m

m

λλλ

Let us identify each term in it

mλ = Molar Conductivity

∞mλ = Molar Conductivity at infinity dilution.

How to calculate cell constant :

Cell Constant = A

l

Let us identify each term in it:

l = Distance between the electrodes

A = Area of the electrode

Cell Constant can also be calculated as:

Cell Constant = Specific conductivity X Resistance

How to calculate Ionization constant (K):

K = α

α−1

2C



20


K = Ionisation Constant

α = Degree of dissociation

How to calculate cell potential :

Eo Cell = Eocathode - Eo

Anode

NOTE:

Students, you are not able to Identify which is cathode and which is anode by

looking at Electrode potential Values. Let us see, how to decide them on the basis of

Electrode potential values.

1. If Eo Values for both the Electrodes were in the NEGATIVE SIGN, the LEAST

VALUE is taken as ANODE . For example Eo of Metal X = - 3.05 V and Eo of

Metal Y = - 2.92V. Here X is taken as ANODE and Y is taken as CATHODE.

2. Similarly, if E0 values for both Electrodes were in the POSITIVE SIGN, We have

to follow the same as in case of the above.

3. If both are in different signs, the NEGATIVE values is taken as ANODE AND

positive IS TAKEN AS cathode. For example E0 of Metal X = - 2.37V, E0 = of

Metal Y = + 0.34V. Here X is taken as ANODE and Y is taken is CATHODE.

PROBLEMS ON NERNST EQUATION:

This equation is useful to find out the electrode potential of an Electrode in

which concentration of Metal ions is not 1M.

Ecell = E0cell +

n

0592.0 log

( )ellOxidationc

lductioncelRe


21



n = Number of moles of electrons Transferred

(Reduction Cell) = Concentration of the reduction half cell

(Oxidation Cell) = Concentration of the Oxidation half cell

E0cell = Standard cell potential

NOTE:

1. Some times the number of Electrons lost by one Electrode may not be tha

same that of the other Electrode gaining.

For example :

Cr / Cr+3(0.1) // Fe+2(0.01) / Fe

Here chromium is loosing three(30 electrons where as Iron is gaining only

two(2) electrons. Hence, Redox equation is to be balanced.

2 x ( Cr - 3e- → Cr+3) ⇒ 2Cr - 6e- → 2Cr+3

3 x (Fe +2 + 2e- → Fe ) ⇒ 3Fe+2 +6e- → 3 Fe

2Cr + 3 Fe+2 → 2Cr+3 + 3 Fe

Hence , the concentration terms of oxidation and reduction half cells are to be

raised to the respective powers. That is:

Ecell = E0cell +

6

0592.0 log

[ ][ ]2

3

1.0

01.0

Similarly whenever Silver Electrode is connected to other electrodes, you

also should be very careful about the value of ‘n’ and also in raising the concentration

terms.

How to calculate the single electrode potential


22


Ecell = E0cell +

6

0592.0 log +nM

1


E0cell = Standard cell Potential

n = Number of moles of electrons transferred

[ ]+nM = Concentration of Electrolyte.

How to calculate the Equilibrium Constant:

E0 cell = n

0592.0 log Kc



n = Number of moles of electrons Transferred

K = Equilibrium constant

How to calculate the Free energy change (or) Maximum work done by the cell

0cellnFEG −=∆


∆G = Free energy change

n = Number of moles of electrons transferred

F = 96,500 Coloumbes


How to calculate mass of Metal deposited at Cathode:

Charge = Current X Time (or) Q = It


23


NOTE:- One mole of the electron is required 1 Faraday(96,500 C) for the reduction of 1

mole of Metal .

How to calculate Efficiency of a cell

Efficiency of a cell (η) = H

G

∆∆

= Totalwork

Usefulwork

SOLVED PROBLEMS

1. How much charge is required for the following reduction of Ans:-

(i) Al3+ + 3e- Al

Three mole electrons are required for the reduction of 1 mole of Al ions. We

know

that the charge on one electron= 1.6021 x 10-19 C.

Charge on 3 mole = 3x6.023x1023 x 1.6021x10-19 =3F

(ii) Cu2+ + 2e- Cu

Similarly 2F

(iii) MnO4- Mn2+

MnO4- + 8H+ + 5e- Mn2+ + 4H2O

Similarly 5F

2. How much electricity in terms of Faraday is required to produce.Ans:-

(i) CaCl2 Ca

Ca2+ + 2e- Ca (40g)

Therefore one F is required to produce 20g Ca.


24


ii) Al2O3 Al

Al3+ + 3eAl

ie, to get one mole Al(27g) 3F is required. So in order to get 40g of Al.

27

3 x 40 = 4.44 F is required

3. How much electricity is required in coulomb for the oxidation of Ans:-

(i) H2O O2

H2O ½ O2 + 2H+ + 2e-

2F is required.

(ii) FeO Fe2O3

2FeO + H20 Fe2O3 + 2H+ + 2e-

1F is required

4. A solution of Ni(NO3)2 is clectrolysed between platinum electrodes using a current

of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode.?

Ans:-t = 1200s

Charge = current x time

= 5A x 1200s = 6000C

Ni2+ + 2e- Ni

We require 2F or 2x96487 C to deposit 1mol or 59 g of Ni

For 6000C the mass of Ni deposited

=1

1

964872

600059−

−

××cmol

cgmol

=1.834 g

5. Three electrolytic cells A,B,C containing solution of ZnSO4, AgNO3, and CuSO4

respectively are connected in series. A steady current of 1.5 amperes was passed

through them until 1.45 grams of silver deposited at the cathode of cell B. How

long did the current flow? What mass of copper and of zinc were deposited?

Ans:-


25


By Faradays second law,

E1/m1 = E2/1.45g = E3/m3

Mass of Cu (or) m1 = 31.77 x 108

45.1 g = 0.426g

Mass of Zn (or) m3 = 0.439g

108g Ag is deposited by passing 96500 C

1g Ag is deposited by passing 10896500

1.45 g Ag is deposited by passing 10896500 x 1.45

Q=It

t = 5.1108

45.196500

××

= 863.7s =14.39 minutes

6. Using the standard electrode potential given in the table 5.4(page no.95 in Text book) predict if the reaction between the following is feasible.

a) Fe3+ (aq) and I- (aq)

E0 Fe3+/Fe2+ = 0.77 V

E0 I2/I- = 0.54 V

The reaction is feasible. Fe3+ is 0.77 V which is more than that of I2

(0.54V).

2Fe 3+ + 2I - 2Fe 2+ + I2

b) Ag+(aq) and Cu(s)

E0 Ag+/Ag = 0.80 V

E0 Cu2+/Cu = 0.34 V

Reaction b/n Ag+ and Cu is feasible because E0 of Ag, 0.8V is more than

that of Cu, 0.34 V.

2Ag+ (aq) + Cu(s) Cu 2+ (aq) 2Ag(s)

c) Fe3+ (aq) and Br- (aq)

E0 Fe3+/Fe2+ = 0.77 V


26


E0 Br/Br- = 1.09V

The reaction b/w Fe3+ and Br- is not feasible since E0 of Fe3+ is less than

that of Br-.

d) Ag(s) and Fe3+ (aq)

E0 Ag+/Ag 0.80 V

E0 Fe3+/Fe2+ 0.77V

The reaction b/w Ag and Fe3+ is not feasible since E0 of Fe3+ is less than

Ag.

e) Br (aq) and Fe2+ (aq)

E0 Br/Br- =1.09 V

E0 Fe2+ /Fe=-0.44 V

The reaction is feasible

7. Calculated the standard cell potentials of galvanic cell in which the following reactions take place.

(i) 2Cr(S) + 3Cd+2(aq) → 2Cr+3(aq) + 3 Cd, (ii) Fe+2((aq) + Ag+(aq) → Fe+3(aq) + Ag(s) i) Cathode

3Cd2+ +6e 3Cd E0 = -0.40V

Anode

2Cr 2Cr3+ + 6e E0 = -0.74V

ECell = Ecathode - Eanode

= -0.40 - (-0.74)

= 0.34V

We know ∆G0 = -nFE0

∆G0 = -6 x 96487 x 0.34 = -196.86 kJ mol-1

E0cell = 0.0592 x logk / n => K =3.039x1034


27


ii) Cathode => Ag++e Ag, E = 0.80V

Anode => Fe2+ Fe3+ + e, E = 0.77 V

Ecell =Ecathode-Eanode = 0.80-0.77 = 0.03V

We know ∆G0=-nFE0

= -1 x 96487 x 0.03 = -2.89461 kJ mol-1

We know at equilibrium

E0cell = 0.592 x log K/n ; n = 1 => K = 3.211

8. Write the Nerest equation and e.m.f. of the following cells at 298K

(i) Mg(s) /Mg2+ (0.001M) // Cu2+ (0.0001M) /Cu(s)

Mg Mg2+ (0.001M)

Cu2+ + 2e Cu

Mg + Cu2+ Mg2+ +Cu

K=[ ][ ][ ][ ]+

+

2

2

CuMg

CuMg

K=0.001M/0.0001M = 10

Ecell =E0cell +

10

1log

2

0592.0

E0cell = 0.34 – (-2.36) = 2.70V

Ecell = 2.70 - 0.0592/2 x 1 = 2.6704 V

(ii) Fe(s)/Fe2+ (0.001M)// H+/H2(g) (1 bar)/ Pt(s)

Fe Fe2+ + 2e

2H++2eH2

Fe+2H+ Fe2+ + H2

K=[ ][ ]

M

M

H

HFe

1

001.022

=+

+

=0.001


28


Ecell = 0.44 + 1000log2

0592.0

= 0.5288 V

(iii)Sn(s)/Sn2+(0.05M)//H+(0.02M)/H2(g)

SnSn2+ + 2e

2H+ + 2e H2

Sn + 2H+ Sn2+ + H2

K= [ ][ ][ ][ ] ( ) 125

02.0

050.022

22

==+

+

HSn

HSn

Ecell = 0.14 +125

1log

2

0592.0

Ecell = 0.078V

iv) Pt(s)/Br2(l)/Br(0.01M)//H+(0.03M)/H2(g)/H2(g)/bar/Pt(s)

Br2 2Br- + 2e

2H+ + 2e H2

Br2+2H+ 2Br- + H2

K = [ ] [ ]

[ ][ ]( )( ) 9

1

03.0

01.02

2

2

2

22

==+ M

M

HBr

HBr

Ecell =E0cell + 0.0592/2 log 1/k

Ecell = -1.09 + 0.0592/2 log 9 = -1.0618V

9. The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500Ω . What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 x 10-3S cm-1.

Ans : - Conductivity = 0.146 x 10-3 S cm-1

R=1500Ω


29


We know R = ρ l/a ; l / a =cell constant

ρ = 1/(0.146x10-3 Scm-1)

l / a= 1500 x 0.146x10-3 = 0.2190cm-1

10.Conductivity of 0.00241M acetic acid is 7.896 x 10-5 S cm-1 . Calculate its molar conductivity and if λ0 for acetic acid is 390.5 S cm2 mol -1, what is its dissociation constant.

Ans:- Conductivity = 7.896x10-5cm-1

molar conduvctivity= molarity

tyconductivi

= M00241.0

10896.7 5−×

=32.76 Scm2mol-1

α= 00838.05.390

76.320

==λλ

K = α

α−1

2C

= ( )

8467.10838.01

0838.000241.0 2

=−

× x 10-5

11. In the button cell widely used in watches and other devices the following reation takes place

Zn (s) + Ag2O (s) + H2O (l) Zn 2+ (aq) + 2Ag (s) + 2 OH- (aq) Determine Eo and ∆G0 for the reaction.

Ans :-

E 0 cell = 0.344V-(-0.76V) = 1.104V

∆G0 = -nFE0 = - 2 x 96500Cmol-1 x 1.104V = - 2.131 x 105 J mol-1

12. Depict the galvanic cell in which the reaction Zn(s) + 2 Ag+ (aq) → Zn+2 (aq) + 2 Ag( s) takes place. Further show (i) Which of the electrode is negatively charged (ii) The carries of the current in the cell (iii) Individual reaction at each electrode.

Ans:-

(i) Zn electrode is negatively charged

ii) Ag+ and Zn2+ ions are the carriers of current in the cell.


30


iii) At anode (oxidation)

Zn Zn2+ + 2e-. Here electrons are generated. It is rich in electrons. Thus

it is negative electrode.

At Cathode(reduction)

2Ag+ + 2e 2 Ag

CHEMICAL KINETICS

My Dear Students,

Even this chapter contains very less formulas. Hence you can score full

marks in this chapter.

1. How to write the Rate expression for a given equation :

Examples:-

1. 2NO(g) + O2 (g) → 2NO2 (g)

Rate Expression = - ( )dt

NOd

2

1 = -

( )dt

Od 2 = + ( )dt

NOd 2

2

1

2. 5Br-(aq.) + BrO3- (aq.) + 6H+ (aq.) → 3Br2 (aq.) + 3H2O (l)

Rate Expression = - ( )dt

Brd −

5

1 = -

( )dt

BrOd −3 = -

( ) ( )dt

Brd

dt

Hd 2

3

1

6

1 +=+

(Note : - In aq. Solution changes in concentration of water is very small and

hence we do not use change in concentration of water for expressing the rate)

2. Units of Rate constant for different order can be calculated as :


31


For the First order Reaction:

Rate = k [Concentration]1

( )dt

ionconcentratd = k [ concentration]1

ond

litermoles

sec = k [ moles / liter]1

k = litermolesondlitermoles 1

sec ×

k = 1secsec

1 −= ondond

Similarly we can calculated the unitsfor the other orders also.

NOTE:- For :- . Zero Order :- k = mol L-1 s-1

Second Order : - k = L mol-1 s-1

nth Order : - k = (mol / L)(1-n) s-1

3. When one Fast reaction and one Slow reation is given, how to calculate the

order of a reaction?

Examples:-

1. In the reaction 3ClO- → ClO3- + 2 Cl-

Various steps are:

ClO- + ClO- → ClO2- + Cl- (Slow step)

ClO2- + ClO- → ClO3

- + Cl- (Fast step)

Rate determine step is always SLOW STEP

∴ RATE = k1 [ClO-] [ClO-]

= k1 [ClO-]2

ORDER = 2


32


2. In the reaction NO (g) + O2 (g) → 2NO2 (g)

Various steps are :

NO (g) + O2(g) → NO3 (g) (Fast step)

NO3 (g) + NO (g) → 2NO2 (g) (slow step)

Rate = k [NO3] [NO]

But in the actual reaction NO and O2 are the reactants, Hence NO3

intermediate product.

NO3 → NO + O2

Rate = [NO] [O2] [NO]

= [NO]2 [O2]

ORDER = 3

4. Order of a reaction is also calculated by :

a. Integrated Method

b. By initial rate Method

c. Half – life period method

Integrated Method

In this method, we have to make use of the general equation pertaining to

First Order reaction. The equation can be written as

K = R

R

t0log

303.2 (or) k =

2

1

12

log303.2

R

R

tt −

Let us identify the terms in it : -


33


k = Rate constant, R0 = Initial concentration, R = Concentration left after

time ‘t’, t = time , R1 and R2 are the concentration of the reactants at time t1

and t2

If the k value is the same for different values given in the problem, then it is a

First order constant.

By Initial rate method

In this method, a table comprises of concentration values of the reactants in

different experiments and also the rate constant values will be given. If you look at it, it is

clear that the concentration of one of the teactants will be kept constant and the other will

be changed. To begin the problem, first of all you have to write the rate equation for the

given chemical reaction.

Let us take some examples:-

Example:1:-

Consider the following data for the reaction : 2A + B → C + D

Determine the order of the reaction with respect of ‘A’ and with respect to

‘B’ and the overall order of the reaction? What are the units of the rate

constant?


Run [A]

Mol. L-1

[B]

Mol. L-1

Rate of reaction(mol. L-1 sec.-

1)

01 0.1 0.1 6.0x10-3

02 0.3 0.2 7.2x10-2

03 0.3 0.4 2.88 x 10-1

04 0.4 0.1 2.4 x 10-2

34


Solution:- Rate law = k [A]p [B]q

By looking at the table, you can find that, in experiment 1 and 4, the

concentration of reactants B is the same. It is clear that, in experiment 2 and 3 the

concentration of reactants A is the same.

Rate Law is : Rate = k [A]p [B]q

Where ‘p’ and ‘q’ values can be calculated by dividing the given experimental values.

Experiment 3 / 2 : ( )( )q

q

litmol

litmol

..2.0

..4.0

102.7

1088.22

1

=××

−

−

4 = (2)q

22 = 2q q = 2

Experiment 4 /1: Similarly If you find ‘p’ value, we get, p = 1

Order of ‘A’ = 1, Order of “B’ = 2

Rate : k [A]p [B]q , Rate = k [A]1 [B]2

Then order of the reaction = 3

Example: 2:- If the initial rate is given and the concentration are not given for some

Experiments at that time by using following method we can calculate the concentrations


35


The reaction between A and B is first order with respect to A and zero order

with respect to B. fill in the blank in the following table.

Experiment [A]/M [B]/M initial rate / M min-1

1 0.1 0.1 2 x 10-2

2 -- 0.2 4 x 10-2

3 0.4 0.4 --

4 -- 0.2 2 x 10-2

Rate = k [A] 1 [B] 0 = k [A] 1

From (1) 2 x 10-2 = k x 0.11 à (1)

K = (2 x 10-2 / 0.1) = 2 x 10-1 min-1

Form (2) 4 x 10-2 = k [A]

[A] = 4 x 10-2 / 2 x 10-1) = 0.2 M.

Form (3) rate = k x 0.41 = 2 x 10-1 x 0.41 = 8 x 10-2 M min-1

From (4) 2 x 10-2 = k [A]

1

2 x 10-2 = 2 x 10-1 [A]

[A] = (2 x 10-2 / 2 x 10-1) = 0.1M.

Half life method

t1/2 = k

693.0

t1/2 = Half life time, k = Rate constant

For Zero order reaction t1/2 α [R0]

First order reaction t1/2 is independent of [R0]

Second order reaction t1/2 α [ ]0

1

R

n th order reaction t1/2 α [ ] 10

1−nR

[R0] = initial concentration

5. How to calculate the ActivationEnergy:


36


log

−=21

12

1

2

303.2 TT

TT

R

E

k

k a


k1 and k2 are the rate constant at the temperature T1T2

Ea = Activation Energy, R = Gas Constant.

Note :-

ra

far EEH −=∆ −

ΔrH- = Enthalpy of the reaction

Eaf = Activation energy of the forward reaction

Ear = Activation energy of reverse reaction.

For endothermic reaction ( ΔHθ > 0 ) hence Ear < Ea

f and for exothermic

reaction ( ΔHθ < 0 ) hence Ear > Ea

f

6. How to calculate slope of the line:

R

Eslope a

303.2

−=

Where Ea = Activation Energy, R = Gas Constant.

CHEMICAL KINETICS

1. Form the rate expression for the following reactions determine their order of

reaction and dimensions of the rate constants.

a) 3NO(g) à N2O (g) + NO2 (g) Rate = K [NO] 2

b) H2O2 (aq) + 3 I – (aq) + 2H+ à 2H2O (l) +I-

3 Rate =- k [H2O] [I-]

c) CH3 CHO(g) à CH4(g) + CO(g) Rate = k [CH3 CHO]3/2

d) CHCl3 (g) + Cl2 (g) à CCl4 (g) + HCl (g): Rate = [CHCl3] [Cl2] 1/2


37


e) C2 H5 Cl(g) à C2H4(g) = HCl(g) : Rate = k [C2H5Cl]

Ans:-

a) Order = 2 Unit of rate constant is s-1 mol-1 L [M-1s-1]

b) Order = 2; Unit of rate constant is s-1 mol-1 L or M-1 s-1

c) Order = 3/2 ; Unit of rate constant is s-1 ; mol-1/2 L1/2

d) Order = 3/2 ; Unit of rate constant is s-1 mol-1/2 L1/2

e) Order = 1 Unit of rate constant is s-1

2. For the reaction

2A + B + C à A2 B+C

The rate = k[A] [B]2 with k = 2 x 10-6 M-2 S-1, calculate initial rate of the

reaction when [A] = 0.1M [B] = 0.2M and [c] = 0.8 M. If the rate of reverse

reaction is negligible, then calculate the rate of reaction after [A] is reduced to

0.06M.

Ans:-

Rate = k [A] [B] 2 = 2 x 10-6 M-2 s-1 x 0.1 M x (0.2M) 2

= 8 x 10-9 5Ms-1.

When [A] is reduced to0.06m?

[B] = 0.2 – 0.02 = 0.18 M

∴ Rate = 2 x 10-6 x 0.06 x (0.18) 2

= 38.88 x 10-10

= 3.888 x 10-9 Mol L-1 s-1.

3. The rate of decomposition of NH3 on platinum surface is zero order. What are the

rate of production of N2 and H2 if k = 2.5 x 10-4 M-1 s-1.

Ans:-

2NH3 à N2 + 3H2

Rate of decomposition of NH3 = k[NH3]0 = k = 2.5 x 10-4 M s-1

- [ ] [ ] [ ]

dt

Hd

dt

Nd

dt

NHd 223

3

1

2

1 ==


38


Rate of formation of N2 = [ ]

dt

NHd

22

1 3

= ½ x 2.5 x 10-4 M s-1 = 1.25x10-4

Rate of formation of H2 = [ ]dt

NHd 3

2

3

= 14105.22

3 −−×× Ms

= 3.75 x 10-4 M s-1

4. In a pseudo first order hydrolysis of ester in water following results were

obtained.

T/s 0 30 60 90

[Ester]/M 0.55 0.31 0.17 0.085

i) Calculate the average rate of reaction between the time intervals 30 to 60

seconds.

ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.

i) Average rate = ( ) ( )

3060

31.017.0

12

12

−−−=

−−−tt

CC

= 30

14.0 = 4.6 x 10-3 Mol s-1

ii) at t = 30 min. k = R

R

t0log

303.2

= 31.0

55.0log

30

303.2

= 12 min1091.1249.030

303.2 −−×=×

At t = 60 min. k = 17.0

55.0log

60

303.2

= 51.060

303.2 × = 1.95 x 10-2 min-1

Since the value of k [H2O] is constant so it is a pseudo first order reation.

5. In a reaction between A and B, the initial rate of reaction was measured for

different initial concentration of A and B as given below.


39


[A]/M 0.2 0.2 0.4

[B]/M 0.3 0.1 0.05

r0 / Ms-1 5.07 x 10-5 5.07 x 10-5 7.6 x 10-5

What is the order of reaction with respect to A and B?

- [ ]dt

Ad = rate = k [A] x [B]y

5.07 x 10-5 = k x 0 .2x x 0 .3y à (1)

5.07 x 10-5 = k x 0 .2x x 0.1y à (2)

7.6 x 10-5 = k x 0 .4x x 0.05y à (3)

Dividing (1) by (2)

1 = (0.3./0.1)y = 3y

y = 0.

Dividing equation (2) by (3)

( ) xx

22.0

4.0

1007.5

106.145

5

=

=

××

−

−

2.87 = 2x

log2.87= xlog2

x = 1.5

Order with respect of [A] is 1.5

Thus order or reaction w.r.t. A is 1.5 and w.r.t. B is zero. The overall order is 1.5.

6. Reaction between NO2 and F2 to give NO2 F takes place by the following

mechanism.

NO2 (g) + F2(g) slow > NO2 F(g) + F(g)

NO2(g) + F(g) fast > NO2F(g)

Write the rate expression for the reaction.

The rate-determining step is the slow step. So rate law is

Rate = k[NO2] [F2].

7. The rate constant for a first order reaction 60 s-1. How much time will it take to

reduce the initial concentration of the reactant to its 1/16th value?


40


k = 60 s-1 , Initial concentration = R0

Concentration at time t =16

1 R0

t = tR

R0log60

303.2

= 16

log60

303.2

0

0

RR

= 2041.160

303.216log

60

303.2 ×=

= 4.62 x 10-2 s.

8. The rate of most of the reactions double when their temperature is raised from

298k to 308k. Calculate their activation energy.

log 1

2

k

k=

−

21

11

303.2 TTR

Ea

Ea =

12

221 1log303.2

TTk

kTRT

−

×

= 298308

2log298308314.8303.2

−××××

= 52.897 kJ mol -1.

9. The half-life for radioactive decay of 14C is 5730 y. an archeological artefact

contained wood had only 80% of the 14C found in a living tree. Estimate the age

of sample.

λ = 2/1

693.0

t = years5750

693.0

t = tN

N0log303.2

λ

= 80

100log

693.0

5730303.2 years×

= 1845 years.


41


10. During nuclear explosion, one of the product is 90 Sr with half life 28.1 Y. if 1µg

of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how

much of it will remain after 10 years and 60 years if it is not lost metabolically?

λ = yearst 1.28

693.0693.0

2/1

=

t = tN

N0log303.2

λ

10 years = tN

gµ1log

693.0

1.28303.2 ×

1071.02.28303.2

693.0101log =

××=

tN

tN

1 = 1.279

Nt = 279.1

1

= 0.7818 μg

When t = 60 years

6427.02.28303.2

693.0601log =

××=

tN

tN

1= 4.392

Nt= 392.4

1

= 0.228 μg.


42

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