G.C.E. (A. L.) Support seminar - 2014 Chemistry - Paper I Answer Guide

QuestionNumber

Answer QuestionNumber

Answer

(1) 3 (26) 2

(2) 5 (27) 5

(3) 1 (28) 4

(4) 1 (29) 2

(5) 3 (30) 3

(6) 3 (31) 5

(7) 4 (32) 3

(8) 5 (33) 2

(9) 1 (34) 5

(10) 5 (35) 2

(11) 3 (36) 3

(12) 3 (37) 5

(13) 2 (38) 1

(14) 3 (39) 1

(15) 4 (40) 2

(16) 4 (41) 2

(17) 3 (42) 5

(18) 2 (43) 1

(19) 5 (44) 4

(20) 1 (45) 5

(21) 2 (46) 3

(22) 4 (47) 4

(23) 2 (48) 1

(24) 5 (49) 4

(25) 1 (50) 2

[see page 2

[see page 3

1. (a) (i) He < O < Cl < Mg

(ii) Li

(iii) chlorine

(iv) CH4 < CO < CO2

(v) NO2

+ < NO2

− < NO3

−

(4×5 = 20 marks)

(b) (i) (+) (+)

(−)

COa Ob

Oc ≡ N (10 marks)

(ii) (+) (+)

(−)

CO ≡O

O N

(+) (+) CO ≡

O

O N

(−)

(1) (2)

As the charge distribution is equal, stabilities are similar to each other. But, stability decreases due to the presence of minus charges on more electronegative N

and O atoms.

(5×3 = 15 marks)

(iii) CO/ Carbon monoxide (02 marks)

(c) Atom Hybridization Electron geometry around the atom

Oa SP3Tetrahedral

Na SP2Trigonal planar

C SP Linear

(3×6 = 18 marks)

(d) (i) Electron configuration

Zn − 4s2 3d10 Co − 4s2 3d7

Configuration of Zn is relatively more stable. Less tendency for the de-localization of

electrons. Strength of the metallic bond is low.

(2×5 = 10 marks)

(ii) C

O

OHCH3 has intermolecularH bonds. But C

O

HCH3 has only permanent

dipole-permanent dipole attractions, which is relatively less stronger. (3×3 = 9 marks)

(Total marks92)

G.C.E. (A.L.) Support seminar - 2014 Chemistry - Paper II Answer Guide

- 2 -

[see page 4

- 3 -

2. (a) (i) A - Na B - Mg

(ii) C - NaOH E - MgO (16 marks)

(iii) H2 $Hydrogen (04 marks)

(iv) 4 Na + O2 2Na2O

2Na + Excess O2 Na2O2

(v) Mg(s) + dil. H2SO4 (aq) MgSO4 (aq) + H2(g)

Mg(s) + conc. 2 H2SO4 (aq MgSO4(aq)+ SO2 + 2H2O(l) (04 marks)

(vi) NaCl(aq), NaOCl (aq) , H2O(l) (2×3 = 6 marks)

(vii) NaOH - strongly basic Mg(OH)2 - basic Al (OH)3 - amphoteric H2 SiO3 - very weakly acidic H3 PO4 - weakly acidic H2 SO4 - strongly acidic HClO4 - very strongly acidic (2×3 = 6 marks) (6×6) + (7×2) marks

(b) (i) Co (05 marks)

(ii) Is2 2s2 2p6 3s2 3p6 3d7 4s2 (10 marks)

(iii) + II CoO (02 marks) + III Co2O3 (02 marks)

(iv) [Co(H2O)6]2+ (06 marks)

(v) (a) Co(OH)2 , [Co(OH)2(H2O)4 ] - blue precipitate (05 marks) [Co(NH3)6]

2+ - yellow brown solution (05 marks)

(b) [Co(H2O)6]2+ + 2OH− Co(OH)2 + 6H2O /

[Co(H2O)6]2+ + 2OH− [Co(OH2) (H2O)4] + 2H2O (05 marks)

Co(OH)2 + 6 NH3 [Co(NH3 )6]

2+ + 2OH− (05 marks)

(vi) Colour intensity increases. / Yellow brown turns to orange brown. (05 marks) [Co(NH3)6]

2+ complex oxidizes into [Co(NH3)6]3+

' (05 marks)

3. (a) (i) By PV = nRT .

V = lA . P1lA = nTRT ; nT = P1lA RT

(ii) For the first instance, P1lA = nTRT nT = P1lA / RT

For the second instance, P2 lA = nTR2T nT = P2lA / 2RT As the total number of gas moles does not change,

P1lA RT

P2lA 2RT

=

P2 = 2P1

[see page 5

- 4 -

(iii)

Fraction of molecules with a given energy

Speed / ms-1

T1

2T

^4.0marks&

(b) Vapour pressure

XA

10

Vapour pressure

Vapour pressure

XA

Vapour pressure

PB°

PA° PA

°

PC°

XBXc0 0

1 1 01

PAB

PAC

PA

PCPA

PB

diagram (1) diagram (2)

(v) Mixture AB is negatively deviated from the ideal condition whereas the mixture AC is positively deviated from the ideal condition.

^6.0marks&

4. (a) (i)

CCH3

CH3 NH2

CH3

C

CH3

CH3 OH

CH3

C F (ii)

C H

C2H5 NH2

CH3

C

OC2H5 CH3

C

H

C3H5 OH

CH3

B E Y (iii)

CH3CH2CH2CH2NH2 C H

CH3 NH2

CH3

CH2

(iv)

C H

CH2 NH2

CH3

CH3 C H

CH2 OH

CH3

CH3 C C H

O

H

CH3

CH3

A D X (8 × 3 = 24) marks

[see page 6

- 5 -

(b) (i) CH3 OH (4 marks)

(ii) (4 × 4 = 16) marks

(iii)

Reaction Reaction type

1

2

3

O E

AE

(2 × 3 = 6) marks 30 marks

(c) (i)

H3C +8

CH3

CH3

BrC−8

H3C+ : Br−

CH3

CH3

C+

First step

H3C

CH3

CH3

OHCH3C

H3C

CH3

C+ : OH

−

(7 × 2 = 14) marks

(ii) CH3− CH2 - Br is a primary alkyl halide' It is formed by the carbocation CH3

CH2

+ '

As it is a primary carbocation, it is unstable. ∴ Breaking of the C - Br bond, as well as the formation of the bond with the

nucleophyle (OH− takes place at the same time, it is a single step reaction. (10)

H

H

++ Br−

CH3

+8BrC H3C

H

H

OHC : OH−

−8

(05)

But, as (CH3)3C+ is a tertiary carbocation, it is stable. (01)

30 marks

5. (a) (i) PV = nRT P × 2.5 × 10−3m3 = 0.15mol × 8.314JK−1mol−1 × 700K P = 3.49 × 105Nm−2

(ii) PV = nRT 8.5 × 105Nm−2 × 2.5 × 10−3m3 = n × 8.314JK−1mol−1 × 700K n = 0.365 mol

[see page 7

- 6 -

(iii) 2H2(g) + X2(g) 2H2X(g)

(x − 0.08) + (0.15 − 0.04) + (0.08) = 0.365 x = 0.255 mol

(iv) 2

[H2X(g)]

[X(2g)] [H2(g)]

2 Kp =

P2

H2X(g)

P P2

H2(g) × X2(g)

[H2X(g)] =

[X2(g)] =

[H2(g)] =

0.08 2.5

0.11 2.5 0.175 2.5

mol dm−3

mol dm−3

mol dm−3

(v)

=

2 0.08 2.5 )(

0.11 2.5 )( 0.175

2.5 )( mol−1

dm3

4.7 mol−1 dm3

2 Kc =

2)( Kp =

=

0.080.365 × 8.5 × 105

2)( 0.1750.365 × 8.5 × 105 )( 0.11

0.365 × 8.5 × 105

Pa−1

8 × 10−7 Pa−1

=

= PH2X(g)

2)( 0.080.365 × 8.5 × 105

)( 0.1750.365 × 8.5 × 105

)( 0.110.365 × 8.5 × 105

= PH2(g)

PX2(g)

Pa

Pa

Pa

5a - ^60marks&

(b) (i) Kw = [H+

(aq)] [OH−

(aq)]

(ii) −log Kw = −log {[H+

(aq)] × [OH−

(aq)]}

= −log [H+

(aq)] −log [OH−

(aq)]

pKw = pH + pOH

(iii) BOH(aq) B(aq) + OH(aq)-+

Kb =[B+

(aq)][OH−

(aq)][BOH(aq)]

Kb =[OH−

(aq)]C

[OH−

(aq)] = (KbC)

12

12

12

12

12

12

pOH = pKb − log C

14 − pH = pKb − log C

pH = 14 − pKb + log C

− log[OH−

(aq)] = log Kb − log C

12

12

2

(iv)

= 11

= 14 − − 0.552

PH = 14 + log (1× 10−5 mol dm−3) + log 0.1 mol dm-312

12

^30)

1/2

[see page 8

- 7 -

(v) BCl + NaOH BOH + NaCl

NaOH + HCl NaCl + H2O

BOH + HCl BCl + H2O (3 × 4 = 12) marks

(vi) (I)

pH

phn

V HCl

Me.O

15 35

}

}

(2 × 6 = 12) marks

(II)

= 1.5 × 10−3 mol

} = (35 − 5) 0.11000

Number of HCl moles reacted with BOH

} 0.11000

Number of HCl moles reacted with NaOH

= × 15

= 2 × 10−3 mol

= (1.5 × 10−3 + 2 × 10−3) × 50 25

NaOH moles in X

= 7 × 10−3 mol 7 × 10−3

25C2 = × 1000 = 0.28 moldm−3

2 × 10−3

25C1 = × 100 = 0.08 moldm−3

(8 × 2 = 16) marks 5b - ^90marks&

6. (a) (i) i. Substances which are completely dissociated into ions in the aqueous solution. or

Substances in which ions exist in its solid, liquid and in the aqueous solution.

ii. Ion concentration is very high in the saturated solutions of such compounds. There are attractive forces between the ions. (Ions are not independent on each other).

- 8 -

(ii) Ú First, Ag2CrO4(s) dissociates. Ú At a particular instance, the rate of the forward and backward reactions becomes equal to each other.

Ag2CrO4(s) 2Ag+(aq) + CrO4 (aq) −2 ^01)

Ú From that instant, concentration of ions remain constant, the ionic product,

2−[Ag+(aq)]2 [CrO4- (aq)] is equal to KSP of the compound at the given temperature.

When NH3 is added,

Ú As the equilibrium,Ag+(aq)+2NH3(aq) [Ag(NH3)2]

+(aq)

(2) reaches,

Ag+ ion concentration decreases. There, according to Le Chatterlier's principle,

as the reaction (1) above, gets shifted towards the products, [CrO42−(aq)] increases.

(2 × 10 = 20) marks (iii) Ag2CrO4(s) 2Ag+

(aq) + CrO4(aq)2−

2 × 10−4

1 × 10−4equlibrium

concentration mol dm−3

KSP = [Ag+(aq)]

2 [CrO4(aq)] = (2 ×10

−4mol dm−3)2 (1 ×10−4mol dm−3)

= 4 ×10−12mol3

dm−9

2−

^20marks&

(iv) New [CrO4 ] = (5 ×10− 4 + 1) mol dm−3

= 1 mol dm−3

[Ag+(aq)] =

Ksp 2−

[CrO4 (aq)

1/2

=

= 2 ×10−6moldm−3

Ag2CrO4 moles precipitated = (2 ×10

−4 −

2 ×10−6 ) × 500

1000mol

= 1 ×10−4mol

2−

1/24 ×10−12 mol3

dm−9

1 moldm−3 )(

^20marks&

(v) KC = [Ag(NH3)+

(aq)] [Ag+

(aq)] [NH3 (aq)]2

1 × 108dm6mol−2

= (2 ×10

−4) mol dm−3

(2 ×10−6 mold m−6) [NH3 (aq)]

2

[NH3 (aq) ] = 1 × 10−3 mol dm−3

Total NH3 moles = (1 × 10−3) + (2 × 2 × 10

−4) mol 2 = 7 × 10

−4mol ^20marks&[see page 9

- 9 - (b) (i)

I. ∆H° = (2 × 105 + 0) − (0 + − 89) kJ mol−1

= 299 kJ mol−1 (12)II. ∆S° = (2 × 72 + 27) − (42 × 2 − 137)JK−1 mol−1

= 224 JK−1 mol−1 (12)

III. ∆G° = ∆H°− ∆S°T = 299 kJ mol−1 − 224 JK−1 mol−1 × 298 K

= 299 kJ mol−1 − 66.752 kJmol−1

= (299 − 66.752)kkJmol−1

= 232.248 kJmol−1 (12)

(ii) (+) ve value obtained for ∆G°. The reaction given, is not spontaneous. According

to that, X should act as the oxidizing agent and Y should act as the reducing agent. Therefore, Y is above X in the electro-chemical series. (10)

(iii) I. Anode - Y Cathode − X II. Anode Y(s) Y2+

(aq) + 2e III. Cathode X2+

(aq) + 2e X(s) IV. Y(s) Y2+(aq, 1moldm−3) X2+(aq, 1moldm−3) / X(s) ^14marks&

6b - (60 marks)

7' (a) (i)

C ≡ C − H C ≡ C − NNa+

+ 2Na 22 + H2 (10)

(ii) C ≡ C − NNa+

+ CH3 C H O

CH3 C C ≡ CO− NNa+

H

(iii) C ≡ C − HH

C ≡ C− NNa+

Hg2+

Na

dil. H2SO4

CO

CH3

CH3 C CC ≡ O− NNa+

CH3 C CC ≡ OH

CH3 C CC ≡ Br

CH3 C CC Br

H2O

PBr3

Br2 / CCl4 Br

Br Br

Br

(5 × 11 = 55) marks

7a - (75 marks)

[see page 10

(vi) i. W;afm%arlhla Ndú; ls¯fï jdish

l=ula o @

ii. W;afm%arlhla m%;sl%shdj

- 10 -

[see page 11

(b) (i) a, h and f (10)

(ii) a - acid base h - acid base f - nucleophylic substitution (15) ^25marks&

7b - (25 marks)

(c) (i)

CUCNKCN

0 - 5 °C A

NH2 NN2Cl

OH COOHOOH

CNNaNO2(i)

dil. H2SO4

(ii) dil. HCl

Room temperature

conc.HH2SO4

O

O

C

B

(9 × 4 = 36) marks

(ii)

+

+ NaCl

+ dil. HCl

+ dil.HCl

NaOH(aq)NN

H O

CC O− NNa+

O

C O− NNa+

O

NH2

NH2

he resultant solution is treated with dil.HCl solution.

The resultant white precipitate is filtered and separated. It is benzoic acid. (14)

Alternative answer :

NH3

+Cl

−

(aq)

COOH

dil. HCl

The resultant white precipitate is filtered and separated. It is benzoic acid.

N

H O

C

NH3

+Cl

−

(aq) COOH

+

The compound given below is heated with aqueous NaOH.

7c - ^50marks&

8' (a) (i) Presence of PbI2 is confirmed as the dark yellow precipitate is dissolved in water when heated and precipitated back when cooled. As the pink compound dissolves in a acid, it is a basic compound. The species which forms blue with conc.HCl, is Co2+.

∴ Contain PbI2 and Co(OH)2 . (2 × 8) + (4 + 4) = 24 marks

(ii) [CoCl4]2−

Tetrachlorocobaltate(II) ion (3 × 2 = 6) marks

8a - 30 marks (b) (1) The white precipitate of lead, which turns black when heated, is PbS2O3 . (3 × 3 = 9) marks (2) ÷ Filtrate Y contains a reducing agent. ÷ That reducing agent is removed with Ca2+ ions by forming a precipitate. ÷ This precipitate should be CaC2O4 . (3 × 3 = 9) marks

(3) There are Ca2+ ions remain in the filtrate Z. As a precipitate was formed on heating, HCO3 ions have dissociated to CO3

2− ions. CaCO3 precipitates.

(3 × 3 = 9) marks

(4) Possible to be NO3

−

or NO2

− . As an oxidation is not possible in the filtrate Z, it can not be NO2

− . (2 × 4 = 8) marks

∴ Four anions are,

C2O4

2−,S2O3

2− , HCO3 −

and NO3

−

. (4 × 4 = 16) marks

8b - ^50maximummarks& (c) (A) Determination of the concentration of Cl

−

AgCl mol = Ag+ mol

= 0.002mol Concentration of Cl

− = 0.08moldm−3

= 0.287g

143.5gmol−1

^7marks&

(B) 2Fe3+ + 3H2S 2FeS + S(4)+ 6H+

4Fes + 7O2 2Fe2O3 + 4SO2(4) (01)S + O2 SO2

H2O + 5SO2 + 2MnO4 5S2O4 +

2Mn2+(4)+ 4H+

6H+ + 5C2O4 + 2MnO4 10CO2 +

2Mn2+(4)+ 8H2O

C2O4 mol = 0.12 × 25 1000

= 3 × 10−3mol

moles MnO4 remain = 3 × 10−3 × 2

5 = 1.2 × 10−3mol

MnO4 moles reacted with SO2 = 0.048 × 50 1000

− 1.2 × 10−3

= 1.2 × 10−3mol

− −

2−

2−

−

−

(4) ^3 × 5 = 15marks&

Fe3+ mol = Fes mol = SO2 mol = 1.2 × 10−3mol × 5

2× 2

3 Fe3+ concentration =

2 × 10−3

× 1000 25

= 0.08 mol dm−3 (3 × 8 = 24 marks)

- 11 -

[see page 12

(C)Determination of the concentration of H+ :

NaOH mol = overall H+ mol in the filtrate (B)

= 0.6

× 201000

= 12 × 10−3mol

H+ mol from H2S = 3 × Fe3+ mol

= 3 × 2 × 10−3mol = 6 × 10−3mol

H+ mol in the initial solution = (12 × 10−3

− 6 × 10−3

)

= 6 × 10−3mol concentration of H

+ in the =

6 × 10−3

× 1000

25initial solution

= 0.24 moldm−3

(3 × 8 = 24) marks

9' (a) (i) (i) Ca(PO4)/ Ca(PO4)2 CaX2 (05) (ii) As its water solubility is low. (05)

(iii) 3Ca3(PO4)2 CaX2 + 14HNO3 3Ca(H2 PO4)2 + 7Ca(NO3)2 + 2HX (10) (iv) Due to high water solubility, can be applied for short term crops, too. Those ions are

easily absorbed by the roots of plants. (05)

(v) Petroliem

air

water

Superphosphate

Y

XH2(g)

P = 250 atmT = 450°Ccatalyst Fe

NH3(g)

N2(g) fractional distillation

P = 1 - 9 atmT = 850 - 1225°Cpt/ Rd catalyst

NO(g)

NO2(g)

HNO3 (aq)

Apatite

partial

electrolysis

- 12 -

[see page 13

(I) A - H2((g) B - N2(g) C - NH3(g) D - NO(g) E - NO(g) F - Apatite (6 × 4 = 24) marks

(II) P - Fractional distillation

Q - Electrolysis

R - Partial acidulation (3 × 3 = 09) marks

(III) X - 250 atm, 450°C, Fe - catalyst Y - 850°C - 1225°C ,1 - 9 atm, Catalyst Platinum / Rhodium (6 × 2 = 12) marks

9a - ^70marks&

(b) (i) Anode - Titanium

Cathode - Nickel (3 × 2 = 06) marks

(ii) 2Cl(aq)

−

Cl2 (g) + 2e

( At the anode ) (04)

2H2O (l) + 2e

−2OH(aq) + H2 (g) (At the cathode) (04)

(iii) As charge removes at the anode and charge is supplied at the cathode / By moving Na+ ions from anode to cathodic compartment. (04)

(iv) As anions are not exchanged through the membrane, it is prevented the moving of OH−

ions, by that prevents the collision with Cl2. (04)

(v) For the absorption of CO2 and other acidic gases ; production of soap, paper, dyes ; refining of water ; removing heavy metals ; removing the acidity of water. (4×2 = 08)

9b - ^30marks&

(c) (i) In the photosynthesis, breaking of water molecules by sunlight (photo-lysis) (3×2= 06) (ii) NH4 , NO2 , NO3

+ − − (3×2= 06)

(iii) Acceptable answers such as, Ú Bringing down the human health Ú Growth retarding in plants

Ú Destroying the structures made of marble and metals.

Ú Weathering of rocks. Ú Weather changing etc.

(3×4= 12) (iv) 1. Acids such as sulphuric, nitric etc. dissolve the the substance of alumino-silicates

in the soil, and release Al3+ ions into water. This disturbs the function of the gills in fish.

2. Acidic rain water which flows through soil, removes nutrients and releases Al3+ ions. Therefore instead of the essential nutrients such as Ca2+ and Mg2+ , plants absorb Al3+ ions. (4×2= 08)

- 13 -

[see page 14

Cl2, ClO2, O3 laf,dfrd weók

(v) Ozone, aldehyde, peroxy acetyl natrates (PAN), peroxy benzyl natrates (PBN)

(3×2= 06) (vi) Reforestation and conservation of forests.

Minimizing the combustion of fuels. (3×2= 06)

(vii) Tuning the engine Using catalytic converters.

(3×2= 06)

9c- ^50marks&

10. (a) Xn+

+

XO2 + (5 − n) × 10 (4)

12H+ + 2IO3

−+ 10e I2 + 6H2O × (5 − n) (4)

12(5 − n) H+ + 2(5 − n) IO3

−+ 10Xn+ 10 XO2 + (5 − n) I2 + 6(5 − n)H2O (4)

IO3

−: Xn+ = (10 − 2n) : 10

(5 − n) : 5 (2)

Moles of Xn+ consumed = 0.2 × 101000 = 2 × 10−3 mol (2)

5 × 1.2 × 10−3 = (5 − n) 2 × 10−3 (2)

6.0 × 10−3 = 10 × 10−3 − 2 × 10−3 n

− 4 × 10−3 = −2 × 10−3 n

n = 2 (3)

IO3

−

Xn+ = (5 − n) 5

1.2 × 10−3 mol 2 × 10−3 mol

= (4)

10a - ^25marks&

(b) 1. P - H2S Q - S R - Cr2(SO4)3 X - H2O Y - Na2S2O3 Z - SO2 (6 × 4 = 24) marks

2. (i) 3H2S(g) + 4H2SO4(aq) + K2 Cr2O7(aq) K2 SO4 (aq) + Cr2 (SO4)3 (aq) + 3S(s) + 7H2O(l)

(ii) H2S(g) + excess 2NaOH(aq) Na2 S (s) + 2H2 O(l)

(iii) 4S(s) + 6NaOH(aq) 2Na2 S (s) + Na2 S2O3(aq)+ 3H2 O

(iv) Na2 S2O3(aq) + 2 HCl 2NaCl(aq)+ H2 O(l) + S(s) + SO2(g)

(v) S(s)+ hot conc.' 2H2SO4 (aq) 3SO2 (g) + 2H2 O(l)(5 × 4 = 20) marks

- 14 -

[see page 15

3. SO2(g) + 2H2O(l) H2SO4 (aq) + 2H+ + 2e

X + nH+ + ne XHnflower petals/coloured dyes colourless

compounds

(04)

10b - ^50marks&

(c) (i) I R = K [A]x [B]Y [C]Z

(10)

II 1 × 10−3 mol dm−3s−1 = K (0.05mol dm−3)x (0.05 mol dm−3)Y (0.05 mol dm−3)Z ......... (1) 3 × 10−3 mol dm−3s−1 = K (0.15mol dm−3)x (0.05 mol dm−3)Y (0.05 mol dm−3)Z ......... (2) 9 × 10−3 mol dm−3s−1 = K (0.15mol dm−3)x (0.15 mol dm−3)Y (0.05 mol dm−3)Z ......... (3) 9 × 10−3 mol dm−3s−1 = K (0.15mol dm−3)x (0.15 mol dm−3)Y (0.05 mol dm−3)Z ......... (4)

^5 × 4 = 20)marks

1

2

1 × 10−3

3 × 10−3 = (0.05)(0.15)

X

X

x = 13

2

9 × 10−3

3 × 10−3 =

y = 1

0.150.05

Y( )

4

3

9 × 10−3

9 × 10−3 =

z = 0

11

Z( )

(02)

(02)

(02)

(02)

(02)

III R = K [A] [B] (10)

(ii) I Amount of A consumed = 0.2 mol (02) Amount of B formed = 0.2 mol (02) ∴ n = 2 (02) As, 0.2 mol of A is consumed, when 0.2 mol of B is formed. (02)

II Kc = [P] [A]

n

= (0.4 ) mol dm−3

(0.4) mol dm−3= 1

(02) (02) (02) III R = K [P] [B]

(02)

IV Kc = [P] [A] [P] = Kc [A]R = K Kc [A] [B]

(02) (02)

(02) 10c - ^75marks&

* * *

- 15 -