Chem P2 Skema

8/3/2019 Chem P2 Skema

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MARKING SCHEME

CHEMISTRY PAPER 2

TWO AND HALF HOURS (2 )

Section A

No. Answer Section A

Mark

1 (a) (i) [Ar]3d9 1(ii) Period 4, Group 10 2(iii) - The size (or radius) increases

- The nuclear charge remains the same but repulsion resultingfrom the additional electron(s) enlarges the domain of theelectron cloud

3

(b) Before the reaction : B = sp2, N = sp3;After the reaction: B = sp3, N = sp3

4

2 (a) (i)Line I II III IV

R.M.M 2 4 81 83Formula 21H+ 21H+2 8135Br+ 21H8135Br+

2

(ii) The hydrogen & bromine reacted together before ionization 1(b) Order of reaction with respect to BrO3- = x

[ ][ ]

x

33

13

3

1

BrO

BrO

=

r

r

x

3

3

0.20

0.10

2.4x10

1.2x10

=

, x=1

Order of reaction with respect to Br- = y

[ ][ ]

y

Br

Br

r

r

=

2

1

2

1

y

x

x

=

30.0

10.0

106.3

102.13

3

, y=1Order of reaction with respect to H+ = z

[ ][ ]

z

H

H

r

r

= +

+

4

3

4

3 ,z

x

x

=

20.0

10.0

106.9

104.23

3

, z = 2

1

1

1


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(i)

2(b) (ii)

(iii)

rate of reaction = k[BrO3-]1[Br-]1[H+]2 moldm-3s-1

1+1+2= 4

from rate = k[BrO3-]1[Br-]1[H+]2 moldm-3s-1

1.2 x 10-3

= k(0.10)(0.10)(0.10)2

k = 1.2 x 10-3/ 1.0 x 10-4 = 12 mol-3dm9s-1

1

1

2

3 (a) (i) UVCl2 2Cl

1

(ii) Propagation stepsCH4 + Cl CH3 + HCl CH3 + Cl2 CH3Cl + Cl

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(iii) The methyl radical, CH3Cl is the intermediate organic substance 1(b) (i) H

CH4 + Cl2 CH3Cl + HCl

H = Hf (products) - Hf (reactants)

= [Hf (CH3Cl) + Hf (HCl)] [Hf (CH4) + Hf (Cl2)]

= [-82 + (-92)] [-75 + 0]

= -99 kJ mol-1

2

1(ii)

EnergyCl

Ea = 16 kJmol-1

CH3

CH4 + Cl2H= -99kJmol-1

CH3Cl + HCl

R.P.

2

(c) The function of the energy from the UV light is to overcome the Ea.After the reaction has commenced, the energy released by theexothermic process (-99 kJmol-1) is enough for the reaction toproceed to completion without any further assistance from the UVlight.

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4 (a) (i) Boiling point of oxides 1


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SiO2Al2O3

MgO

NaO P4O10 Cl2O7SO3

Na Mg Al Si P S Cl

(ii) - Ionic bonds are found in the oxides of Na, Mg and Al, theelectrostatic force between the positive metallic ions and thenegative oxide ion is very strong. Therefore the boilingpoint of the metallic oxides are high.

- SiO2 has a giant molecular structure where the Si and O

atoms are held by strong covalent bonds. Hence SiO2 has avery high boiling point- P4O10, SO3 and Cl2O7 are simple covalent molecules. The

force of attraction between the molecules are Van der Waalsforces. Therefore, these molecules have low boiling point.As the strength of the V.D.W force depends on the size ofthe molecule (P4O10 > Cl2O7 > SO3) the boiling point of P4O10> Cl2O7 > SO3.

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1

3

(b) (i) From Al3+ + 3e Al E = - 1.66 VSO2-4 + 2e + 4H+ SO2 + H2O E = + 0.17 V

E

cell = + 0.17 - (-1.66) = + 1.83 VAs the Ecell is positivethe reaction will occur spontaneously.

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(ii) 3SO2-4(aq) + 2Al(s) + 12H+(aq) 3SO2(g) + 6H2O(l) + 2Al3+ (aq) 2Section B

5 (a) To halt the reaction abruptly. 2(b) - Le Chateliers principle states that if the conditions of a reversible

reaction is altered and the equilibrium of the system is disturbed, theposition of equilibrium will change to minimize the effects ofaltering the conditions.- Addition of C2H5OH will increase [C2H5OH], thus according to L.C. P., in order to reduce [C2H5OH], the equation position will shift

to the right causing the [CH3CO2C2H5] to increase.

2

(c)

Kc =]3][52[

]2][523[

COOHCHOHHC

OHHCOOCCH 1


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C2H5OH + CH3COOHCH3CO2C2H5 + H2OInitial (mol) 0.05 0.05 - -Change 0.0333 0.0333 0.0333 0.0333Final (mol) 0.0167 0.0167 0.0333 0.0333

Or

n (ethanol)o = 2.3/46 = 0.05 moln (ethanoic acid)o = 3/60 = 0.05 mol

n (NaOH)o = MV/1000 = 1.0 (50)/1000 = 0.05 moln (NaOH) at equilibrium = n (HCl) required

= MV/1000= 1.0 (33.3)/1000= 0.0333 mol

n (NaOH) reacted = n (NaOH) o n (NaOH) at equilibrium

= 0.05 0.0333 = 0.0167 moln (NaOH) reacted = n (ethanoic acid) at equilibriumn (CH3COOH) eq = 0.0167 mol

Assume the total volume is V

Kc = (0.0333/V) (0.0333/V) = 3.98

(0.0167/V) (0.0167/V)

4

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(d) Ethylethanoate 1(e) (i)

O H H H

H C O C C C H

H H H

1

(ii) H H O H

H C C C O C HH H H

1

6 (a) (i) Lowry Bronsted base = proton acceptor 1(b) (i) [H+] = KaC

= 6.3x10-5 (0.020)= 1.1 x 10-3 mol dm-3

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pH = -lg [H+] = - lg 1.1 x10-3 = 2.9 1(ii) pH

13 Shape 110.5 Axis - 1

Point - 16.5

3

25 V of NaOH/cm3

3

(iii) Thymolphthalein because the pKa value of it (9.7) falls within thepH range of 6.5 10.5 thus it will change colour at this pH range toindicate the end-point.

1 + 1

(c) HA + NaOH NaA + H2OWeak acidFrom pH = pKa + lg [NaA]/[HA]4.30 = pKa + lg 1pKa = - lg Ka = 4.30Ka = 5.01 x 10-5 mol dm-3

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pH = 2.50 = -lg [H+] [H+] = 10-2.50 = 3.16x10-3 moldm-3

From [H+] = KaC(3.16x10-3)2 = 5.01 x 10-5 x C

C = 0.199 mol dm-3

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1

7a(i) Enthalpy of solution = the enthalpy change that takes place when onemole of a substance dissolves in a solvent togive an infinitely dilute solution under standardconditions

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(ii) Hsolution = +ve 1AgNO3 (s)Ag2+(aq) + NO3-(aq) , Hsolution = +veAs temperature increases, according to Le Chateliers Principle, thesystem will absorb the extra heat energy, equilibrium position shifts toright, indicating the solubility of AgNO3 increases

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7b(i) Vapour pressure/ Mixture of H2O C6H5ClkPa H2O

C6H5Cl101

98 100 132 Temp/oC

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(ii) - water & C6H5Cl are immiscible- the total v.p of the mixture is the sum of the partial pressure of

the two individual liquids at the same temperature [Ptotal mixture =PH2O + PC6H5Cl]

- when the total v.p equals the atmospheric pressure, the mixturewill boil- the boiling point of the mixture is lower than either of the

individual liquid- the boiling mixture will yield a distillate of a constant

composition as below:

mass of C6H5Cl = v.p of C6H5Cl x Mr (C6H5Cl)mass of H2O v.p of H2O x Mr (H2O)

The distillate will contain a large portion of C6H5Cl due to itshigh molar mass

(the steam distillation is useful for the extraction of organic compoundthat decomposes at a temperature near its B.P since in steamdistillation, this organic compound can be distilled at a temperaturebelow its normal B.P)

5max

(iii) mass of C6H5Cl = v.p of C6H5Cl x Mr (C6H5Cl)mass of H2O v.p of H2O x Mr (H2O)

10.0 = (101-97) x Mr (C6H5Cl)38.8 97.0 x 18

Mr (C6H5Cl) = 112.5

1

18 - structural isomerism exists in compounds that have the same

molecular formula but different structural formula4

max


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- structural formula of C4H9Br

1-bromobutane 2-bromobutane1-bromo-2 -methylpropane

2-bromo-2-methylpropane8a D = 2-bromobutane

- if a haloalkane compound is hydrolysed with aqueous NaOH,alcohol is formed- when alcohol is oxidised, a carbonyl compound C=O is

formed- as this carbonyl compound give positive iodoform test, it is a

ketone that contains group- C CH3

O

+ NaOH + NaBr

+ [O] + H2O

1

1

1

H H H H

H-C-C-C-C-H

H H H Br

H H H H

H-C-C-C-C-H

H H Br H

H

-C-

H-C-C-C-H

H H Br

H

-C-

H-C-C-C-H

H Br H

H H H H

H-C-C-C-C-H

H H Br H

- C C C C -

OH

- C C C C -

Br

- C C C C -

OH

- C C C C -

O

hydrolysis

oxidation


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CH3CH2CCH3 + 3I2 + NaOH CHI3 + CH3CH2COONaO

1

8b Hydrolysis of a haloalkane compound produces a tertiary alcohol thatcannot be oxidised by acifified KMnO4

Thus E = 2- bromo-2-methylpropane

+ NaOH

3 alcohol

1

1

1

8c F = 1-bromobutane

The elimination of HBr from F,

NaOH + HBrethanol

oxidation reaction:

- C-C-C == C- + 2[O] +

propanal methanal

1

1

1

yellow precipitate(iodoform)

C

- C C C

Br

C

- C C C

Br

C

- C C C

OH

- C C C C -

Br

- C C C C -

Br

- C C C = C -

1-butene

- C C C-H

O

C

O


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or - C-C-C == C- + 5[O] - C C COOH + CO2 + 2H2O

Propanal gives +ve silver mirror test,

CH3CH2CH +2[Ag(NH3)2]+ + 2OH- 2Ag(s) +3NH3 +C2H5COONH4 +H2O silver mirrorO

1

9a(i)

+ HNO2 + N2(g) + H2O

2

(ii)

+ Na2CO3(aq) + H2O +CO2(g)2

(iii)

+ C2H5OH + H2O

2

(iv)

+ +HCl(g)2

9b The PE polymer formed by the addition polymerisation process usingthe Ziegler-Natta catalyst is

- a linear polymer

- high density & high melting point

3

9cCH2=CH2 ClCH2-CH2Cl CH2=CHClstep 1, reagent : Cl2 in CCl4

condition: room temperature

step 2, reagent: NaOH in ethanolcondition: reflux

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1

H2N- C COOH

HO- C COOH

nitric(III) acid ornitrous acid

amino ethanoic acid

H2N- C COOH

2 H2N- C COONa

H2N- C COOH

O

H2N- C COC2H5

conc.H2SO4

H2N- C COOH

CH3- C Cl

O

CH3- C N CCOOH

O

step 1 step 2


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110a(i) 2Al(s) + 2OH-(aq) + 6H2O(l) 2[Al(OH)4]-(aq) + 3H2(g) 2(ii) PV = nRT

(1.01x105) x 10-3 = n(8.31)(298)n = 0.0408 mol H2(g)

From equation, 3 mol H2(g) 2 mol Al(s)0.0408 mol H2?

0.0408 x = 0.0272 mol Al(s)mass of Al(s) = 0.0272 x 27

= 0.734g% of Al in alloy = 0.734 x 100%

1.00= 73.4%

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10b N2(g) + 3H2(g)2NH3(g) H = -veA high temperature will lower the production of NH3

a lower temperature is chosen to increase the yield of NH3 but thiswill slow down the rate of reaction a catalyst is used to increase the rate the equation

N2(g) + O2(g)2NO(g) H = +veA high temperature will increases the rate of reaction & at the sametime increases the yield of NO, therefore there is no need for a catalyst

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10c - there is hydrogen bonding between NH3 molecules- the N atom in NH3 is very electronegative & makes use of its

lone pair of electrons to attract a H atom of another molecule- therefore the BP of NH3 is high

- PH3 & AsH3 molecules cannot form H bonds. Only weak Vander Waals forces are formed. Therefore, they have low BP

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