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IIT-JAM 20111CHEM ACADEMY

CHEM ACADEMY

IIT-JAM 2011Note: Attempt ALL the 44 questions. Question 1-30 (Objective questions) carry three marks each and

question 31-44 (Subjective Questions) carry fifteen marks each.

1. The pair of semimetals in the following is:(a) Al, Si (b) Ge, As (c) Sb, Te (d) Ca, B

2. The most probable oxidation states for both Cr and Mo are(a) +2, +3, +4 (b) +2, +3, +5 (c) +2, +3, +6 (d) +3, +4, +5

3. The correct order of acidic character is:(a) Al2O3 > MgO > SiO2 > P4O10 (b) P4O10 > Al2O3 > MgO > SiO2(c) P4O10 > SiO2 > Al2O3 > MgO (d) SiO2 > P4O10 > Al2O3 > MgO

4. The pair of amphoteric oxides is:(a) VO, Cr2O3 (b) V2O3, Cr2O3 (c) VO2, Cr2O3 (d) V2O5, CrO3

5. In the structure of B4O5 (OH)42–

(a) All four B atoms are trigonal planar(b) One B atom is tetrahedral and the other three are trigonal planar(c) Three B atoms are tetrahedral and one is trigonal planar.(d) Two B atoms are tetrahedral and the other two are trigonal planar

6. The pH of an aqueous solution of Al3+ is likely to be(a) Neutral (b) Acidic (c) Slightly basic (d) Highly basic

7. Hydrolysis of (CH3)2 SiCl2 and CH3SiCl3 leads to(a) Linear chain and cross-linked silicones, respectively(b) Cross-linked and linear chain silicones, respectively(c) Linear chain silicones only(d) Cross-linked silicones only

8. The oxide that has the inverse spinel structure is(a) FeCr2O4 (b) MnCr2O4 (c) CoAl2O4 (d) Fe2CoO4

9. The transition metal monoxide that shows metallic conductivity is:(a) NiO (b) MnO (c) TiO (d) CoO

10. The metal that is extracted by the reduction method is:(a) Al (b) Au (c) Hg (d) Mg

11. The most viscous liquid is(a) Water (b) Methanol (c) Ethylene glycol (d) Glycerol



CHEM ACADEMY

12. In ammonical buffer, oxine (8-hydroxyquinoline) forms yellow precipitate with(a) Mg(II) (b) Ca(II) (c) Ba(II) (d) Sr(II)

13. Addition of an aqueous solution of Fe(II) to potassium hexacyanochromate (III) produces a brick-red colured complex, which turns dark green at 100ºC. The dark green complex is(a) Fe4[Cr(CN)6]3 (b) KFe[Cr(CN)6] (d) KCr[Fe(CN)6] (d) Fe[Cr(CN)6]

14. In the following equation X is:2432419795 Am Bk X

(a) 102 n (b) 1

0n (c) 112 n (d) 4

2He

15. Based on the principle of equipartition of energy, the molar heat capacity of CO2 at constant volumeCv,m is:(a) 3.5 R (b) 6R (c) 6.5 R (d) 9R

16. One mole of a van der waals gas undergoes reversible isothermal transformation from an initialvolume V1 to a final volume V2. The expression for the work done is:

(a) 2

2 11

VRT ln a(V V )V

(b) 2

1 1 2

V b 1 1RT ln aV a V V

(c) 2

1

PRT lnP (d)

2

1 1 2

V b 1 1RT ln aV a V V

17. The scalar product of two vectors u and v, where u = ˆ ˆ ˆ2i 3j 5k and v = ˆ ˆ ˆi j 3k , is:

(a) – 10 (b) ˆ ˆ ˆ2i 3j 15k (c) ˆ ˆ ˆ3i 4 j 2k (d) 10

18. The minimum concentration of silver ions that is required that is required to start the precipitationof Ag2S(Ksp = 1 × 10–51) in a 0.1 M solution of S2– is(a) 1 × 10–49 M (b) 1 × 10–50 M (c) 1 × 10–26 M (d) 1 × 10–25 M

19. Identify the correct statement regarding Einstein’s photoelectic effect(a) The number of electrons ejected depends on the wavelength of incident radiation.(b) Electron ejection can occur at any wavelength of incident radiation.(c) The number of electrons ejected at a given incident wavelength depends on the intensity of theradiation.(d) The kinetic energy of the ejected electrons is independent of the wavelength of incident radiation.

20. The hydrolysis constant (Kh) of NH4Cl is 5.6 × 10–10. The concentration of H3O+ in a 0.1 Msolution of NH4Cl at equilibrium is

(a) 115.6 10 (b) 105.6 10 (c) 5.6 × 10–10 (d) 2.8 × 10–5

21. The acid dissociation constant (Ka) for HCOOH, CH3COOH, CH2ClCOOH and HCN at 25ºC are1.8 × 10–4, 1.8 × 10–5, 1.4 × 10–3 and 4.8 × 10–10, respectively. The acid that gives highest pH at theequivalence point when 0.2 M solution of each acid is titrated with a 0.2 M solution of sodiumhydroxide is:(a) HCOOH (b) CH3COOH (c) CH3ClCOOH (d) HCN



CHEM ACADEMY

22. For an ideal gas undergoing reversible Carnot Cycle, the plot of enthalpy (H) versus entropy (S) is:

(a) (b) (c) (d)

23. Hybridization of the atoms indicated with the asterisk (*) in the following compounds sequantlyare

(a) sp2, sp2, sp3, sp2 (b) sp2, sp3, sp3, sp2 (c) sp3, sp3, sp3, sp2 (d) sp2, sp2, sp3, sp3

24. The Cahn-Ingold-Prelog (CIP) priorities of the group and the absolute configuration (R/S) of thefollowing compounds are

(a) CH2OH > CH(CH3)2 > CH = CH2 > CH3 and S(b) CH2OH > CH = CH2 > CH(CH3)2 > CH3 and S(c) CH2OH > CH = CH2 > CH(CH3)2 > CH3 and R(d) CH2OH > CH(CH3)2 > CH = CH2 > CH3 and R

25. The optically active stereoisomer of the following compound is

(a) (b)

(c) (d)

26. The correct relationship within each pair of the natural products is:(a) Camphor – terpene; insulin – protein; nicotine – alkaloids; streptomycin – carbohydrate(b) Camphor – terpene; insulin – carbohydrate; nicotine – alkaloid; streptomycin – lipid(c) Camphor – alkaloid; insulin – protein; nicotine – terpene; streptomycin – carbohydrate.(d) Camphor – carbohydrate; insulin – protein; nicotine – alkaloid; streptomycin – terpene.



CHEM ACADEMY

27. The correct sequence of relationship between the compounds of the following pairs i-iv is

(a) Identical, enantiomers, diastereomers and structural isomers.(b) Enantiomers, identical, structural isomers and diastereomers.(c) Enantiomers, identical, diasteromers and structural isomers.(d) Identical, identical, diastereomers and structural isomers.

28. The INCORRECT statement in the following is:(a) The nucleobase pairs are aligned perpendicular to the helical axis in DNA.(b) RNA contains uracil and thymine, but DNA contains only thymine.(c) All naturally occuring amino acids with the exception of glycine are chiral(d) All enzymes are proteins, but all proteins are not necessarily enzymes.

29. The product P and Q in the following reactions, respectively, are

(a) (b)

(c) (d)

30. The major product in the following reaction is

(a) (b)



CHEM ACADEMY

(c) (d)

31. (a) In the following reactions, identify X, Y and Z.boiling water

2 3Na SO S X(colorless complex) excess XAgBr Y(soluble complex)

boiling water2 2X Cl H O Z HCl

(b) Draw the structure of S4N4H4 and N4S4F4.32. (a) The magnetic moment of [Fe(phen)2(NCS)2] varies with temperature. The magnetic moments at

200 K and 50 K are 4.9 B.M. and 0 B.M., respectively. Write the d-electron configurations of Fe atboth temperatures and give reason for the observed change in the magnetic moment.(phen = 1, 10-phenanthroline)(b) PCl5 exists as a discrete covalent molecule in the gaseous state, but is ionic in the solid state.Draw the structures of PCl5 in gaseous and solid states.

33. In the following equilibrium and reactions, identify species B to E.Write the balanced chemical equation for the conversion of C to E.

A pH 6 B dil. HCl Coxide of Cr Yellow colour strong oxidizing agentsolid tetrahedralno d-electronsB + diphenylcarbazide D (violet colour)

C + HCl E (greenish yellow gas)34. (a) Identify species A and C in the following:

Write the balanced chemical equation for the conversion of A and A3+.

A + aqua regia AA3+ + NO

A3+ + I– B (black precipitate)

B + I– (excess) C (orange colour)

Hint: C on the dilution with water gives B(b) Draw the structures of X and Y in the following reactions.

(i) Borazine + HCl X

(ii) Borazine + Br2 YY

35. (a) The molar conductances at infinite dilution for BaCl2, KCl, K2SO4 and Cl– are 280, 150, 300 and76 W–1 m2 mol–1, respectively. Calculate the transport number of Ba2+ in BaSO4 solution at infinitedilution.



CHEM ACADEMY

(b) If 4 moles of a MX2 salt in 1 kg of water raises the boiling point of water by 3.2 K. Calculate thedegree of dissociation of MX2 in the solution.

36. (a) For the reaction R P, the plot of ln[R] versus time (t) gives a straight line with a negativeslope. The half life for the reaction is 3 minutes.(ln 2 = 0.693, ln 0.1 = –2.303)(i) Derivative the expression for t1/2.(ii) Calculate the slope of the straight line(iii) Calculate the time required for the concentration of R to decrease to 10% of its initial value.(b) Shown below is the Jablonski diagram that describes various photophysical processes. Thesolid arrows () represent radiative transitions and the wave arrow ( ) represents a non-radiative transition.

(i) Name the photophysical pathways X, Y and Z.(ii) Which of the radiative decays is faster?

37. (a) (i) Given that G = –nFE, derive the expression for the temperature dependence of the cellpotential (E) in terms of the change in entropy (S).(ii) For a cell reaction, E(at 25ºC) = 1.26 V, n = 2 and S = –96.5 J K–1 mol–1. Calculate E at 85ºCby assuming S to be independent of temperature. (F = 96500 C mol–1).(b) The phase diagram for the lead-antimony system at a certain pressure is given below.

(i) Identify the phases and components in region I and region II.(ii) Calculate the number of degrees of freedom (variance) at point M.

38. (a) One mole of an ideal gas initially at 300 K and at a pressure of 10 atm undergoes adiabaticexpansion.(i) Reversibly and(ii) Irreversibly against a constant external pressure of 2 atm until the final pressure becomes equalto the external pressure.Calculate Ssystem for (i) and (ii). For (ii), express the final answer in terms of R. Given: Molar heatcapacity at constant volume Cv,m = 3R/2.(b) For the following equilibrium at 300ºC.



CHEM ACADEMY

2 4 2N O (g) 2NO (g)Calculate Kp when N2O2 is 30% dissociated and the total pressure is 2 bar.

39. (a) The Maxwell probability distribution of molecular speeds for a gas is:3/ 2 2

2 m mvF(v)dv 4 v exp dv2 kT 2kT

where ‘v’ is the speed, ‘m’ the mass of the gas molecule and k the Boltzmann constant.(i) Use F(v) to show that the most probable speed vmp is given by the expression.

1/ 2

mp2RTvM

(ii) Use R = 8 J K–1 mol–1 in the above expression to calculate the vmp for CH4(g) at 127ºC.(b) The wavefunction of a quantum state of hydrogen atom with principal quantum number n = 2 is:

3/ 2

2 m0 0 0

1 1 r r(r, , ) 2 expa a 2a32

(i) Identify the values of quantum numbers l and m and hence the atomic orbital.(ii) Find where the radial node of the wavefunction occurs.

40. (a) Write the possible substitution products in the following reactions. Indicate the types of mecha-nisms (SN1/SN2) that is/are operative in each reaction.

(i) Br CN , DMF ?

(ii) Br

3CH OH ?

(b) Write the elimination products A to C in the following reaction. Identify the major product

H

OH

3 4H PO , A B C

41. (a) Write the structures of A to C in the following reaction sequence.

+

3 2 2

3 3 3 3

1. CF COOOH, CH Cl2. NaOH,

HCl, AlCl CH COCl, AlCl 3. H O3 2H C HC CH A B(Major product) C

(b) Write the structures of D and E in the reactions given below:

H3C CH3

3 2 4 2 32 23 2 2

3 2

1. HNO , H SO 1. Br , FeBr2. SnCl , HCl 2. KOH, H O,3. (CH CO) O 3. NaNO , HCl(major product)

4. H PO

D E



CHEM ACADEMY

42. (a) Write the structures of A to C in the following reaction sequence:CH3

CH3

2

6 4 32

1. NaNHm ClC H COOOH, benzene 2. H O

3. NaNO , HCl(A) B C

(b) Write the structures of D and E in the following reaction.

PhNH2

O

OH

22

Br , OHH O, (intermediate) (stable product)

D E

43. Write the structures of products A to E in the following reaction sequence

N(CH3)2

CH3

3 3Li, NH (liq.) H O isomerizationt BuOH

(A) (B) C

(H3C)3SiO

CH3

D 3H O E

44. Oxanamide O, a tranquilizer, is synthesized according to the following reaction scheme. Write themissing structures and reagents K to O.

C H8 16K: Reagents 2L (C H O and gives positive test with Tollen’s reagent)4 6

NaOH, (–H O)2

(Z)-MAg O, NaOH2

(oxidation) N1. SOCl2. NH

2

3 3. CF COOOH3

O(an epoxy amide)



CHEM ACADEMY

SOLUTIONOBJECTIVE QUESTIONS

1. Al MetalSi Non-metalGe Semimetal, but more like Si.As SemimetalSb SemimetalTe SemimetalCa MetalB NonmetalCorrect option is (c)

2. +2 oxidation state, CrX2, MoX2 (where X = F, Cl, Br, I)+3 oxidation state, CrX3, MoX3, CrO3

+6 oxidation state, CrO3, CrO42–, MoO3

Correct option is (c)3. • Oxides of metal are basic in character and also higher the electropositive character of metal,

higher will be basic character of the oxide, so MgO is more basic than Al2O3.• Oxides of non metals are acidic character.

•54 10P O 4

2SiO 3

2 3Al O 3

MgO

Oxidation state acidic strength.• Correct option is (c)4. Amphoteric oxides get dissoved in both acids as well as bases.

VO dissolves in acids but not in bases.V2O3 dissolves in acids but not in bases.VO3 dissolves in acids as well as bases, so it is an amphoteric oxide.V2O5 dissolves in acids as well as bases, so it an amphoteric oxide.Cr2O3 dissolves in acids as well as bases, so it is an amphoteric oxide.CrO3 dissolves only in bases, but not in acids.Correct option is (c)

5. [B4O5(OH)4]2–



CHEM ACADEMY

Boron with –ve charges, are tetrahedral and Boron with no charge are trigonal planar.Correct option is (d)

6. 3 32 2 6Al 6H O [Al(H O) ]

3 22 6 2 2 5 3[Al(H O) ] H O [Al(OH)(H O) ] H O

As above equilibrium reaction generates H3O+, So aq. solution of Al+3 is acidic.

Correct option is (b)7. Hydrolysis of (CH3)2SiCl2 and CH3SiCl3 can be illustrated by following chemical reactions.

3CH|

Cl Si Cl|

Cl

+ H2O 3CH

|HO Si OH

|OH

Polymerisation

3CH|

O Si O |O|

nCross linked polymers

Correct option is (a)8. A compound AB2O4 (where A is in +2 and B in +3 o.s.), has inverse spinel structure if ccp lattice is

made by oxide ions and

(i) A is occupying 14 th of octahedral site.

(ii) Half of B is occupying 14 th of octahedral site.

(iii) Rest half of of ‘B’ is occupying octahedral site.Tetra hedral site is occupied by metal ion (B+3) which has lower C.F.S.E.In Fe2CoO4, Fe+3 has zero C.F.S.E. So, one of Fe+3 occupies tetrahedral site and it acquires inversespinel structure.Correct option is (d)

9. 3d metal mono-oxide like TiO and VO have high electrical conductivities that decrease with in-creasing temperature.As in these compound conduction band is formed by the overlap of t2g orbitals of metal ions inneighbouring octahedral sites that are oriented towards each other. The radial extension of the d-orbitals of these early d-block elements is greater than for element later in period, hence a bandresult from their overlap this band is only partly filled. Hence these oxide so metallic conductivity.However in MnO, FeO, CoO and NiO, no such type of band formation result due to smaller radialextension of d-orbital, and they behaves as semiconductor.



CHEM ACADEMY

[overlap of the dzx orbitals in TiO to give a t2g band. In perpendicular direction the dyx and dzx orbitalsoverlap in an identical manner]

Correct option is (c)10. Metal Extraction method

Al Electrolysis of Al2O3 in moltel Na3AlF6.Au Leading with CN– first, so soluble complex [Au(CN)2]

– is formed then reductionwith Zn.

Hg Direct reduction of HgS by heat alone (HgS + O2 Hg + SO2)Mg Electrolysis of fused MgCl2 with added KCl.Correct option is (c)

11. Due to 3 O – H groups there will be large number of intermolecular H–bonding possible, which inturn increases the viscosity, as cross linked polymer like network is formed.

Correct option is (d)

12. Mg+2 + oxine reagent in ammonicalbuffer

yellow ppt.

Correct option is (a)



CHEM ACADEMY

13.(II) (III) III II100ºC2 3

6 6 6 6Brick red dark green

Fe K [Cr(CN) ] K Fe[Cr (CN) ]; KFe[Cr(CN) ] K Cr[Fe(CN) ]

Correct option is (c)

14. 243241 4 19795 2 0Am He Bk 2 n

Correct option is (a)15. For CO2, translational degree of freedom = 3

rotational degree of freedom = 2 (because molecule is linear)and vibrational degree of freedom = 3N – 5 = 3 × 3 – 5 = 4 (because CO2 is linear)

Therefore, total internal energy of CO2 molecule, is1 1 1U 3 kT 2 kT 4 kT kT2 2 32

v,mU 13C kT 2

= 6.5 k (per molecule)

= 6.5 k × NA = 6.5 R (per mole)Correct option is (c)

16.2 2

1 1

V V

2V V

RT aw dW Pdv dv(V b) V

2 2a RT aP (V b) RT P

(V b)V V

For isothermal proces, T = constant2

1

V2

1 1 2V

V ba 1 1RT n(V b) RT n aV V b V V

Correct option is (d)

17. Given two vectors, ˆ ˆ ˆu 2i 3j 5k and ˆ ˆ ˆv i j 3k

Scalar product, u . v = (2.1) + (3.1) + (–5.3) = 2 + 3 – 15 = –10Correct option is (a)

18. 22Ag S 2Ag S

2 2spK [Ag ] [S ]

1/ 2 1/ 251sp 25

2

K 10[Ag ] 10 M0.1S

So, for the precipitation of Ag2S, [Ag+] must be greater or equal to 10–25 M.Correct option is (d)



CHEM ACADEMY

19.

0 = threshold wavelength (wavelength above which electrons cannot be ejected)

K.E. of ejected electron = incident 0

hc hc

K.E. is dependent on lincident.Number of ejected electrons depends on intensity of incident radiation only.Correct option is (c)

20. 4 2 3 30.1 x x xNH H O NH H O

3 3h

4

[NH ][H O ]k[NH ]

hx.xk

0.1 x

x2 = kh (0.1 – x)

x2 + kh – kh (0.1) = 0

2 2

h h h h h hk k 4(1)( 0.1 k ) k k 0.4kx

2 1 2

As kh is very small. So, neglecting 2hk .

h hk 0.4k2

(Removing negative sign, as concentration can not be – ve)

10 10115.6 10 2.24 10 5.6 10

2

[neglecting the first term of numerator].

Correct option is (a)21. For any acid base titration, HA + NaOH NaA + H2O

Now, NaA can undergo hydrolysis and will give, pH = 7 + a1 1Pk log c2 2

As Conc. of NaA is same in all cases, so higher pka (lower ka) will have higher pH.For HCN pH will be highest.Correct option is (d).

22. (b) In adiabatic process, qrev = 0 S = 0In isothermal process, T = 0 dH = nCvdT + nRdT ( dT = 0) dH = 0



CHEM ACADEMY

Correct option is (b)23. Whenever there is possibility of resonance, then l.p. of oxygen align itself in a orbital such that it

can participate in resonance (i.e. sidewide overlap of orbitals) is possible, which causes change inhybridisation of oxygen from sp3 to sp2.Correct option is (a)

24. Since 4th group (CH3) is present at above the plane. So, absolute configuration is inverted.

Correct option is (b)25. Options given in (a), (c) and (d) has plane of symmetry, so they are optically inactive

Correct option is (b)26. Correct option is (a)27. Correct option is (c)28. RNA contains adenine guanine, uracil and cytosine, but DNA contains adenine, guanine, thymine

and cytosine.Correct option is (b)

29. Chemical transformation involved in above chemical reaction can be illustrated as

Correct option is (c)



CHEM ACADEMY


Correct option is (c)DESCRIPTIVE QUESTIONS

31. (a) Complete chemical reaction occuring in above chemical transformate can be shown asboiling water

2 3 2 2 3Sodium thiosulfate

(X)

Na SO S Na S O (colourless solid)

2 2 3 3 2 3 2soluble complex

(Y)

AgBr 2Na S O Na [Ag(S O ) ] NaBr

2 2 3 2 2 4(Z)(X)

Na S O 4Cl 5H O 2NaHSO 8HCl

32. (a) Magnetic moment of [Fe(phen)2(NCS)2] varies with temperature from 4.9 BM to 0 BM due tospin crossover from high spin to low spin state of Fe+2 in octahedral field.

At 200 K, (High spin) At 50 K (low spin)

Fe+2, d6 Fe2+, d6

Spin only magnetic moment Spin only magnetic moment = 0 B.M.



CHEM ACADEMY

n(n 2) B.M.

4(4 2) B.M.

24 4.9B.M.

Vibrational entropy change (Svib) is one of the factor for spin crossover.

(b) PCl5 (gas) Trigonal bipyramidal structure.

PCl5 (solid) exists in ionic form as [PCl4]+[PCl6]

– 2PCl5 [PCl4]+[PCl5]

–

A combination with tetrahedral and octahedral structural units.

33. Complete chemical reaction occurs in above chemical transformation can be shown as

dil. HClpH 6 2 23 4 2 7

(B)(A) (C)CrO CrO Cr O

2 32 7 2 2

(E)(C)greenish yellow gas

Cr O 6HCl 8H Cl 2Cr 7H O

34.

3 4(C) orange colour (soluble complex)(C)

BiI I [BiI ]



CHEM ACADEMY

(b)

(ii) (Y) HBr

N

BN

B

NB

Br

H

Br

H

Br

H

(B3N3H3Br3)35. (a) Transport number at infinite dilution for cation or anion is given by

00m

t

here, = stoichiometric coefficient of ion = Ionic conductivity.

2 2

24 2 2 4

0 0 0 0m,BaCl m,BaClm,Cl m,Cl0

0 0 0 0Bam,BaSO m,BaCl m,K SO m,KCl

1 ( 2 ) 2t

2

= 280 2 76 280 152 128 0.457

280 300 2 150 580 300 280

(b) From Van’t hoff equation, for boiling point elevation,

b solute bT i m K

mole k kg3.2K i 4 0.5kg mole

i 1.6

Considering the dissociation of MX2,MX2 M+2 + 2X–

Initially 1 0 0finally 1 – 2here = degree of dissociation (i.e. dissociation from 1 mole) Total number of particles (i) - (1 – ) + + 2



CHEM ACADEMY

i = 1 + 2 1.6 = 1 + 2

0.3 degree of dissociation = 30%

36. (a) (i) For the reaction,kR P where k = rate constant.

The plot of ln[R] vs t is straight line with –ve slope.Above reaction is a first order reaction

d[R] k[R]

dt

On rearranging and then integrating we get,

0[R]n kt

[R] ... (1)

Where [R]0 = initial concentration of reactant.[R] = Concentration of R at time t

At t = t1/2,0[R][R]

2

0

1/ 20

[R]n kt[R] / 2

1/ 2n2 tk

(i) From equation (1),

0ln[R] n[R] kt

0n[R] n[R] kt

Slope = 1

1/ 2

n2 0.693k 0.231mint 3

(iii) t = ? When [R] = [R]0 × 1

10

0

0

[R]n kt([R] /10)

1 2.303t n10 9.97 mink 0.231

(b) (i) X FluorescenceY PhosphorescenceZ Inter system crossing

(ii) Among radiative decays (Fluorescence and (ISC) phosphorescence), Fluorescence is faster be-cause this transition is spin allowed, while Phosphorescence is spin forbidden.



CHEM ACADEMY

37. (a) (i) Given G = –nFE ... (1)We have,

dG = – SdT + VdP

P

G ST

OrP

G ST

OrP

GST

... (2)

Substituting G value from equation (1) in equation (2).Temperature dependence of cell potential in terms of entropy

P

ES nFT

(ii) From relation derived, in part (ii), we have

P

ES nFT

SdE dT

nF

(at constant pressure)

On integration2 2

1 1

E T

E T

SdE dTnF

[ Given S is independent of temperature]

2 1 2 1SE E (T T )

nF

85 25SE E [(273 85) (273 25)]

nF

8596.5E 1.26 60

2 96500

85E –1.26 0.03

E85 = 1.23 V Ans.(b)

(i) Region I has solid (Sb) and melt (Sb + Pb). Components(c) = 2 (Sb and Pb)

Phases (p) = 2 (Solid and melt)



CHEM ACADEMY

Region II has two solid phases having composition either Sb(solid) + Eutectic (solid)Or Pb(solid) + Eutectic (solid), depending upon the mole fraction of Pb less than or more thaneutectic composition composition mole fraction of Pb.[i.e. (i) When XPb < XPb, euectic then Sb(s) + eutectic (s) phase.

(ii) When XPb > XPb, eutectic then Pb(s) + Eutectic (s) phase]So, in region II,

C = 2 P = 2(ii) From phase rule, F = C – P + 2but at constant pressure,

F = C – P + 1At point M, C = 2 [Sb and Pb]

P = 2 [Solid (Pb) and Liquid (Pb + Sb)] F = 2 – 2 + 1

F = 1Therefore, variance or degree of freedom at M = 1

38. (a) From Clausius inequalityqST

; equality holds for reversible process i.e. when q = qrev, then revqST

(i) For reversible adiabatic processq = qrev = 0

S 0 0.R (ii) For adiabatic irreversible process,

q = 0 S > 0

From 1st law of thermodynamics, q = U – WOr, 0 = CV (T2 – T1) + Pex (V2 – V1) (n = 1 mole)

Or, 0 = CV (T2 – T1) + Pex 2 1

2 1

RT RTP P

Given, P1 = 10 atm, P2 = 2 atm = Pex, Cv = 2

R, T1 = 300 K

22 2 2

RT3 R 300 30 R(T 300) 2 0 T 450 T 602 2 10 2

2T 204K

As entropy is a state function, so it is independent of path followed (i.e. reversible or irreversible)if initial and final states are same.



CHEM ACADEMY

Net entropy change can be written as,

2 2v

1 1

T VS nC n nR nT V

2 2

1 1

RT / P3 2041 R n 1 R n2 300 RT / P

3 204 204 / 2R n R n2 300 300 /10

3 204 102R n R n 0.645R2 300 30

(b) N2O(g) 2NO2(g)t = 0 1 mole 0at eqm (1–0.3) mole 0.6 mole Ptotal = 2 bar

Mole fraction of N2O4(g) = 2 4

2 4 2

N O

N O NO

n 0.7 7n n 0.7 0.6 13

Mole fraction of NO2(g) = 2

2 4 2

NO

N O NO

n 0.6 6n n 0.7 0.6 13

2 2 2

2 4 2 4 2 4

2 2 2NO NO total NO total

PN O N O total N O

P (X . P ) X .Pk

P (X .P ) X Or,

2

p p

6 213k ; k 0.79

713

39. (a) From Maxwell probability distribution, the number of molecules having velocity between v andv + dv is,

3/ 2 22 m mvF(v)dv 4 v exp dv

2 kT 2kT

So, velocity (v) at which F(v) will be maximum is said to be most probable velocity, at v = vm,p, F(v) is maximumFor F(v) to be maximum,

mpV V

dF(v) 0dv

andmp

2

2V V

d F(v) 0dv

3/ 2 22 m mvF(v) 4 v exp

2 kT 2kT

3/ 2 2

2dF(v) m mv 2mv4 exp 2v vdv 2 kT 2kT 2kT



CHEM ACADEMY

For,dF(v) 0

dv

2 2mv2v v 02kT

Or, 2 2kTvm

[ neglecting, v = 0, value because at v = 0, F(v) = 0]

Or,2kTvm

and also at 2kTvm

, 2

2d F(v) 0

dv

mp2kT 2RTv vm M

(ii) For CH4(g)M = 16 × 10–3 kg/molT = 127º = 400 KR = 8 J k–1 mol–1.

5

m,p 32 8 400V 4 10 200 10 632.45 m / s16 10

(b) Given3/ 2

2 m0 0 0

1 1 r r(r, , ) 2 expa a 2a32

(i) As the above function is independent of angle (i.e. and ), So, []2 (probility density) is alsoindependent of angle. This is only possible for s-orbital (which is spherical) = 0, m = 0

(ii) For radial node; node2, ,m r r(r, ) 0

and must have different sign (i.e. + ve or –ve for r < rnode and r > rnode)

2, ,m (r, , ) 0 only when node

0

r2 0a

[because exponial term 0

rexp2a

is zero only when r ].

rnode = 2a0

40.

Only SN2 mechanism is operating here because.

(a) C N is a strong nucleophilic.

(b) DMF is an aprotic polar solvent in which nucleophilic does not get solvated.(c) Alkyl halide is primary, so steric hinderence is less.



CHEM ACADEMY

(d) Allylic group increases the electropositive character (because, sp2 carbon is more electronega-tive than sp3 carbon), so nucleophile can attach faster.

(ii)

here SN1 mechanism is operative because:(a) CH3OH is a weak nucleophile.(b) CH3OH is a protic polar solvent which stablises the carbocation formed in SN1 mechanism.(c) Alkyl halide is tertiary,

(d) In SN1 mechanism, carbocation formed is stablished by resonance with double bond.

41. (a)



CHEM ACADEMY

42.



CHEM ACADEMY


NMe2

CH3

2

HH O

(A)

NMe2

CH3

44. K:reagents8 16C H 2L (C4H8O and gives positive Tollen’s reagent test)

Double bond equivalent = 1 So, L is an aldehyde.(Symmetrical alkene) K is an ozonolysis reagent,i.e. (1) O3 (2) Zn/H2O or (CH3)2S/H2O

But is not possible because it can not give alkene as the aldol condensationproduct. L(C4H8O) is CH3 – CH2 – CH2 – CHO



CHEM ACADEMY

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