CCC HEEP WOH COLLEGE 2011 – 2012 MOCK...

Marking Scheme / 1

2011 - 2012 / Form Seven / AL Chemistry / Mock Examination / Marking Scheme / C.K.Lau

CCC HEEP WOH COLLEGE

2011 – 2012 MOCK EXAMINATION

F7 AL CHEMISTRY

Marking Scheme

Paper 1

Section A

1. (a) (i) N2O N2O3 1,1

(ii) N2O: O.S. of central N= +2; O.S. of peripheral N= 0 ½,½

(+1; ½ marks only)

N2O3: O.S. of N in N=O = +2; O.S. of N in O=N-O = +4 ½,½

(+3; ½ marks only)

(b) (i) 1

(ii) CoCl2(H2NCH2CH2NH2)2 + CN- → CoClCN(H2NCH2CH2NH2)2 + Cl- ½

CoClCN(H2NCH2CH2NH2)2 + CN- → Co(CN)2(H2NCH2CH2NH2)2 + Cl- ½

(c) (i) Since the intermolecular force between NH3 molecules is intermolecular hydrogen

bonding, the association of these molecules by strong, extensive intermolecular bonds

means that they require an extra amount of energy for breaking hydrogen bond in

order to enter the vapour phase. 1

Whereas the intermolecular force between PH3 molecules is only weak van der Waal’s

force. 1

(ii) Fluorine is the most electronegative element, it always shows an oxidation state of –1

in its compounds. It does not exhibit any positive oxidation state. 1

No low lying empty (vacant) d orbital to expand its octet. (1)

Moreover, fluorine is a strong oxidizing agent which simply oxidizes water.

2F2(g) + 2H2O(l) → 4HF(g) + O2(g) 1

2. (a) NO2 has a bent shape / V-shape 1

1

(b) either structure 1

(c) Kc = [N2O4(g) / [NO2(g)]2 1

= [(0.2 x 0.46 / 2) / 5] / [0.2 x 0.54] / 5]2 1

= 19.72 mol-1 dm3 1

Marking Scheme / 2


(d) Since the reaction is exothermic, 1

Kc decreases with increasing temperature and ½

causes the equilibrium shift to the left. ½

(e) It is favoured by high pressure. ½

By Le Chatelier’s Principle, the system opposes the change by favouring the side with less

gaseous molecules. ½

3. (a) (i) Structure of X Structure of Y

½,½

(ii) X: butanal Y: methylpropanal 1,1

(iii) Butanal has a straight chain which is long and thin ½

while methylpropanal has a branched chain which is more rounded. ½

Therefore, butanal has a larger surface area than methylpropanal. ½

The van der Waals’ forces between butanal molecules are stronger. ½

Thus, butanal has a higher boiling point than methylpropanal.

(iv) Butanal and methylpropanal can be separated by fractional distillation 1

because they are two miscible liquids which have very close boiling points. 1

(b) (i) The higher colour intensity of the riboflavin solution, the larger amount of light is

absorbed, i.e. higher absorbance. 1

(ii) Correct drawing [1]; Correct labeling [1] 2

(iii) From the calibration curve, the concentration of the riboflavin solution is equal to 3.7 ×

10−5 M. 1

4. (a) (i) Step 1: (NH2)2CO → HCNO + NH3 1

Step 2: 6HCNO → C3H6N6 + 3CO2 1

(ii) Endothermic reaction 1

Bond breaking of NC bonds involved in decomposition of urea. 1

Absorbance

Concentration of riboflavin (× 10−5 M)

Marking Scheme / 3


(iii) 1

(iv) 1

(v) Ethanoyl chloride / ethanoyl anhydride / ethanoic acid 1

(b) Absorption peak at 2250 cm-1 showing the presence of C=N. 1

Broad absorption peak at 3200 to 3500 cm-1 showing the presence of alcohol –OH. 1

Alternate structure: 1

5. (a) (i) Because the attraction between the ions and the polar water molecules liberate

heat. 1

(ii) 1

M(OH)2(s)∆Hsolution M2+(aq) + 2OH-(aq)

M2+(g) + 2OH-(aq)

-∆Hlattice

∆Hhydration[M2+(aq)] +

2∆Hhydration[OH-(aq)]

∆Hsolution[Mg(OH)2(s)] = –(–2995) + [(–1890) + 2(–550)]

= +5 (kJ mol-1) ½

∆Hsolution[Ba(OH)2(s)] = –(–2320) + [(–1275) + 2(–550)]

= –55 (kJ mol-1) ½

(iii) For salts of small anions, lattice energy is the dominant factor.

Since lattice energy of Mg(OH)2 is more negative, 1

more energy used to dissociate lattice of Mg(OH)2.

Therefore Ba(OH)2 is more soluble. 1

Marking Scheme / 4


(b) (i) Electronegativity is the ability of an atom to attract electrons in a covalent bond

towards itself. 1

(ii) H−F. 1

It is because H and F have the largest electronegativity difference among the other

combinations. ½

The greater the difference, the more separation of charge occurs, and hence the

higher the polarity it has. ½

6. (a) Add 0.02 moles of FeSO4(NH4)2SO4•6H2O crystal to excess known volume (V1 dm3) ½

and known concentration (M1) of silver nitrate solution in a conical flask. ½

Stopper and shake the conical flask well. ½

Stand the flask overnight to allow the equilibrium to be attained and the ½

sedimentation of solid residue.

Pipette 25cm3 of the liquid ½

Titrate the liquid with known concentration (M2) of KSCN solution ½

Data treatment:

Let V2 dm3 be the volume of KSCN solution used in titration. ½

Initial concentration of [Fe2+] = 0.02 / V1 ½

No. of moles of Ag+ in 25cm3 = M2V2

[Ag+]eqm = M2V2 / 0.025 ½

[Fe3+]eqm = M1 – (M2V2 / 0.025) ½

[Fe2+]eqm = (0.02/V1) –[Fe3+]eqm ½

Kc = [Fe3+]eqm / [Ag+]eqm [Fe2+]eqm ½

(b) (i) Fluorine shows only one oxidation state while others show various oxidation states.

Formation of extensive hydrogen bonds in hydride. (any ONE) 1

(ii) Fluorine is the most electronegative atom. ½

It has no low-lying vacant orbitals in its valence shell. ½

(c) (i) The color of solution changes from blue to brown. ½

White / red / brown / grey / black precipitates formed (depends on the purity) ½

4I-(aq) + 2Cu2+(aq) → 2CuI(s) + I2(aq) 1

(ii) The solution color changes from brown to colorless. ½

I2(aq) + 2S2O32-(aq) → 2I-(aq) + S4O62-(aq) ½

(iii) Add excess KI solution and then thiosulphate solution to it. 1

7. (a) (i) C 1

(ii) D 1

(iii) E 1

(iv) B 1

(b) (i) A: B:

1,1

CH2CCH2NH2

CH3

OH

CH=CCH2NH2

CH3

Marking Scheme / 5


(ii) Unimolecular nucleophilic substitution forms A ½

Elimination forms B ½

(iii) 3o halide stabilizes the carbocation 1

(iv) 2

8. (a) (i) Approximate pH = 8 – 9 1

Due to the presence of the salts formed from the neutralization of weak acid and

strong base. 1

(ii) Phenolphthalein 1

(iii)

1

(iv) No. of moles of HCl used = 8.35 x 10-3 x 0.2 = 1.67 x 10-3 mole

No. of moles of NaOH left and reacted with HCl = 1.67 x 10-3 1

No. of moles of NaOH reacted reacted with aspirin

= 0.1 x 0.1 – 1.67 x 10-3 mol = 0.833 x 10-3 mole

No. of moles of aspirin presented in the solution

= 0.833 x 10-3 / 2 = 4.165 x 10-3 mole 1

% by mass of aspirin = [(4.165 x 10-3 x 180) / (0.5 + 0.5)] x 100% = 75.0% 1

(b) H2S(g) � 2H+(aq) + S2-(aq)

At eqm, 1.2 x 10-20 = [H+(aq)]2 [S2-(aq)] / [H2S(g)] ½

When pH = 3, 1.2 x 10-20 = [10-3]2 [S2-(aq)] / [10-1]

[S2-] = 1.2 x 10-15M 1

Product of Zn ion and sulphide ion

= 1.2 x 10-15 x 10-1 = 1.2 x 10-16 mole ½

It is far more than the solubility product of ZnS ½

ZnS will be precipitated at pH 3 ½

½

½

1

COOH

OCOCH3

COO-Na+

OH + 2NaOH + CH3COO-Na+ + H2O

Marking Scheme / 6


9. A. Introduction (Max. 2)

(i) Electronic configuration (max. 1 mark)

- Form ions with incomplete d-orbital [except Zn and Sc]

- e.g. V2+ [Ar]3d3

Mn2+ [Ar]3d5

Fe2+ [Ar] 3d6

(ii) Atomic properties (max. 1 mark)

- Small variation of atomic radius, I.E., electronegativity

(iii) Physical properties (max. 1 mark)

- close packed structure

e.g. V: b.c.c.

Fe: b.c.c. / f.c.c.(above 1000°C)

- higher m.p., b.p., densities, heat of vaporization and fusion.

- high mechanical strength

(iv) Chemical properties (max. 1 mark)

- reacts slowly with acid

- highly polarizing (high charge/ radius ratio)

B. Characteristics (Max. 8)

(i) Variable oxidation states (max. 3 marks)

- because energy difference between 3d and 4s orbitals is small.

- common oxidation states: +2 and +3

- V and Mn never form simple ions with higher oxidation states.

- highest oxidation states only occurs in complex ions and covalently bonded

oxides. e.g. V2O5, VO3-, MnO4-

- maximum oxidation number corresponds to the maximum number of electrons

available for bonding

- V and Mn: electrons from 3d to 4s subshells used for bonding

- Fe: only unpaired electron in the subshells can be used for bonding.

- Variable O.S.:

V: +2 in V2+; +3 in V3+; +4 in VO2+; +5 in VO2+

Mn: +2 in Mn2+; +3 in Mn(OH)3; +4 in MnO2; +6 in MnO42-; +7 in MnO4-

Fe: +2 in Fe2+; +3 in Fe3+

(ii) Complex formation (max. 3 marks)

- vacant, low energy orbitals in central metal atom accepting lone pair electrons

of ligands

- formation of dative covalent bond

- High charge density of central metal atom

- coordination numbers are usually 4 and 6

- C.N. is 6, octahedral arrangement e.g. [Fe(H2O)6]2+

C.N. is 4, tetrahedral arrangement/ square planar e.g. CoCl42-

- Stability constant

Marking Scheme / 7


(iii) Catalytic properties (max. 2 marks)

- ability to vary their oxidation states

- presence of empty orbitals and lone pairs of electrons in 3d, 4s and 4p subshells.

- Heterogeneous catalysis

- by adsorption theory

- involves 2 different phases

- e.g. MnO2 used in catalysis of decomposition of H2O2.

2H2O2 → 2H2O + O2

- Homogeneous catalysis

- same phase

- provides alternate reaction pathway with lower activation energy

- e.g. Use of Fe3+

2I- +2Fe3+ → I2 + 2Fe2+

2Fe2+ + S2O82- → 2Fe3+ + 2SO42-

(iv) Colored compounds (max. 1 mark)

- d-orbitals split into 2 sets of orbitals with small energy difference correspond to

visible light region of EM wave

- d-d transition causes emission of visible light upon absorption of energy.

- e.g. V2+ is violet

Mn2+ is pink

Fe2+ is pale green

C. Comparison: (max. 2 marks)

s-block:

- low charge to radius ratio

- low polarizing power

- no low lying vacant d-orbitals

d-block:

- higher charge to radius ratio

- high polarizing power

- low lying vacant d-orbitals

10. A. Introduction: (max. 1 mark)

- 1o, 2o, 3o haloalkanes and halobenzene

- favors nucleophilic addition and substitution due to polar C-X bond.

- structures

- use of halogeno compounds

B. Chemical properties: (max. 6 marks)

- Monomolecular nucleophilic substitution:

- reaction mechanism

- rate equation

- order of reaction

- factors affecting the rate of reaction

- energy level diagram

- formation of racemic mixture

- Bimolecular nucleophilic substitution:


- rate equation

- order of reaction

- factors affecting the rate of reaction

- energy level diagram

- inversion of product

Marking Scheme / 8


- Examples:

- Substitution with hydroxides and alkoxides

- Substitution with cyanide

- Substitution with ammonia

- Elimination:


- reaction conditions

- formation of alkenes/ alkynes

C. Comparison: (max. 5 marks)

Differences:

- Halogeno compounds

- nucleophilic substitution

- due to electron deficient carbon atom directly bonded to halogen

- Trigonal planar (SN1) and bipyrimidal shaped(SN2) intermediates are formed.

- Carbonyl compounds

- nucleophilic addition

- due to electron deficient site in carbon of the carbonyl group

- tetrahedral intermediate is formed

Similarities:

- Both attacked by nucleophilies

Paper II

1. (a) (i) EDTA + 4NaOH → Na4[EDTA] + 4H2O 1

(ii) EDTA possesses four negatively charged carboxylate group in their salts while free acid

is neutral. 1

The oxygen atom on the carboxylate group can form a stronger bond to metals and is

thermodynamically favourable. 1

With this negative charge on the salt of EDTA, rate of binding to metal ion would be

increased and is kinetically favourable. 1

(iii) It can remove hard ions of Ca2+ and Mg2+ in water. 1

Na4[EDTA] + Ca2+ → Ca[EDTA]2- + 4Na+ 1

Or Na4[EDTA] + 2Mg2+ → Mg2[EDTA] + 4Na+

(iv) The calcium ions in the salt of EDTA are replaced by lead ions,

Ca[EDTA]2- + 2Pb2+ � Pb[EDTA]2- + 2Ca2+ 1

The equilibrium lies on the right as the lead-EDTA complex formed is more stable.

½

The lead-EDTA complex is soluble and can be excreted from human body. ½

(v) Chromatography followed by mass spectrometry. 1,1

Or + EDTA, Cu2+ + EDTA → purple, colourimetry (1,1)

(b) (i)

Trigonal pyramidal T-shaped

(1 mark each for the structure and 1 mark each for the shape) 4

Marking Scheme / 9


(ii) 2

(c) The sample is first vapourized into gaseous atoms. ½

In the ionisation chamber, they are turned into positive ions by bombarding these neutral

atoms with a stream of high energy electrons. 1

These positive ions are accelerated in an electric field produced by a high voltage applied

between the plates in the acceleration chamber. 1

The ions then enter the magnetic field where they are deflected. ½

Ions with small masses are deflected most and ions with large masses are deflected least.

½

The magnetic field is varied so that ions of different masses are detected and traced out

on the recorder. ½

2. (a) (i) 2N2H4 → N2 + 2NH3 + H2 1

3N2H4 → 2N2 + 2NH3 + 3H2 (1)

(ii) ∆H = 4E(NH) + E(NN) – 3E(NH) – ½E(N≡N) – ½E(HH) 1

= 4(388) + 163 – 3(388) – ½(944) – ½(436)

= -139 kJ mol-1 1

∆H = -126.3 kJ mol-1 (1)

(iii) 1,1-dimethylhydrazine is energetically more stable to store 1

It is because CH bond is stronger than NH bond. 1

(iv) no. of more of gas = (4 / 2) (15 / 32) = 0.938 mole 1

PV = nRT

(1.92 x 103)(V) = (0.938) (8.31) (600+273)

⇒ V = 3.54 m3 1

V = 4.13 m3 (1)

(b) (i) HF has the highest b.p. because of the intermolecular hydrogen bond ½,½

b.p. increase from HCl to HI as the molecular size increase, 1

the chance of polarizability increase. ½

therefore, the van der Waals’ forces between molecules increase ½

(ii) The molecular size of BrF3 is larger than ClF3 ½

Therefore the dipole-dipole attraction between molecules increase ½

(iii) ∠F-Br-F < ∠F-Cl-F ½

electron cloud of Br > Cl and more diffused 1

the lone pair – bond pair repulsion of BrF3 > that of ClF3 ½

(c) (i) It is very difficult to break the strong N≡N ½

P4 is bonded by single bonds ½

∠P-P-P = 60o, ½

PP is weak as the strong repulsion between bond pair electrons ½

(ii) N has high electronegativity 1

Hydrogen bond exist between NH3 and water molecules ½

Wile no hydrogen bond exist between PH3 and water molecules ½

(iii) Conc. H2SO4 is a strong oxidizing agent which can displace HX from halides ½

Conc. H3PO4 is a typical acid only. ½

(iv) Cl2 + H2O � HCl + HclO ½

HCl decompose under sunlight, ½

Equilibrium shift right and [Cl2] decrease ½,½

Marking Scheme / 10


3. (a) (i) (I) Let the rate equation be

rate = k[A]m 1

where m is the order of reaction with respect to A.

By substituting the given information into the rate equation, we have

3.0 × 10−4 = k (1.0)m………(1)

12.0 × 10−4 = k (2.0)m……..(2)

Dividing (2) by (1)

4 = (2)m

m = 2

∴ the order of reaction with respect to A is 2. 1

(II) At 273 K,

3.0 × 10−4 = k (1.0)m

3.0 × 10−4 = k (1.0)2

k = 3.0 × 10−4 mol−1 dm3 s−1 1

At 298 K

3.0 × 10−4 = k (0.5)m

3.0 × 10−4 = k (0.5)2

k = 12.0 × 10−4 mol−1 dm3 s−1 1

(ii) (I) By substituting values of k1, k2, T1 and T2 into the Arrhenius equation and then

solving for Ea,

k1 = 3.0 × 10−4 T1 = 273 K

k2 = 12.0 × 10−4 T2 = 298 K

ln k1 – ln k2 = Ea / R [T2-1 – T1-1] 1

ln (3.0 x 10-4) – ln (12.0 x 10-4) = Ea / 8.314 [298-1 – 273-1] 1

Solving for Ea = 37.46 kJ mol−1 1

(II) Let k be the rate constant at 345 K.

ln k – ln (12.0 x 10-4) = 37464 / 8.314 [298-1 – 345-1]

k = 9.4 x 10-3 mol−1 dm3 s−1 1

(b) (i) When the temperature is increased, the kinetic energy of the reactant particles

increases. 1

Then, the number of reactant particles with energy equal to or greater than the

activation energy increases. 1

Therefore, the number of effective collisions increases and the reaction rate is thus

increased. 1

(ii) At a higher temperature, the number of reactant particles with energy equal to or

greater than the activation energy increases. 1

Therefore, the reaction rate increases with the temperature. However, enzymes are

proteins which are easily denatured at a higher temperature. 1

Therefore, the reaction rate decreases at a certain high temperature.

Marking Scheme / 11


(c) (i) Phosphoric acid acts as a catalyst in the reaction. 1

(ii) Assume that one mole of ethanol is produced.

Mass of atoms in CH2=CH2 = (12.0 × 2 + 1.0 × 4) g = 28.0 g

Mass of atoms in H2O = (1.0 × 2 + 16.0) g = 18.0 g

Mass of atoms in CH3CH2OH = (12.0 × 2 + 1.0 × 6 + 16.0) g = 46.0 g

Atom economy = %1000.180.28

0.46×

+ 1

= 100% 1

(iii) Using catalysts 1

The process uses phosphoric acid as the catalyst. The use of catalyst can speed up

the chemical reaction and get more products in a relatively shorter period of time.

1

Maximizing atom economy 1

The atom economy of the process is 100%. 1

4. (a) (i) W is butanoic acid. 1

Z is ethyl ethanoate. 1

(ii) They are functional group isomers. 1

(iii) Butanoic acid contains the carboxyl group which can participate in hydrogen bond

formation. 1

The butanoic acid molecules are held together by strong hydrogen bonds 1

while the ethyl ethanoate molecules are held together by weak van der Waals’ forces

only. 1

Therefore, butanoic acid has a higher boiling point than ethyl ethanoate.

(iv) Simple distillation 1

can be used because they have a large difference in boiling points. 1

(b) (i) Let the mass of compound X be 100 g. Thus,

the mass of carbon in the compound = 48.7 g

the mass of hydrogen in the compound = 8.1 g

the mass of oxygen in the compound = 43.2 g

Carbon Hydrogen Oxygen

Mass (g) 48.7 8.1 43.2

Number of moles (mol) 1.40.12

7.48= 1.8

0.1

1.8= 7.2

0.16

2.43=

Mole ratio 5.17.2

1.4= 3

7.2

1.8= 1

7.2

7.2=

Simplest mole ratio 3 6 2

1

∴ the empirical formula of compound X is C3H6O2. 1

(ii) From the mass spectrum, the peak at m/e = 74 corresponds to the molecular ion of

compound X. 1

Therefore, the molecular mass of compound X is 74. 1

Marking Scheme / 12


(iii) Referring to the IR spectrum, the presence of a strong absorption peak at around 1750

cm−1 indicates that compound X contains a C=O group. 1

The absence of a broad absorption peak in the region of 2500−3500 cm−1 indicates that

compound X does not contain a O−H group. 1

Referring to the mass spectrum, the peak at m/e = 74 corresponds to the molecular ion.

Therefore, the molecular mass of compound X is 74. The molecular formula of compound X

is C3H6O2. 1

For saturated hydrocarbon with 3 carbon atoms, there will be 8 hydrogen atoms. Degree

of unsaturation = 12

68=

− (in C=O). As a result, compound X should be an ester. 1

The peaks at m/e = 59, 43 and 15 should correspond to CH3COO+, CH3CO+ and CH3+

respectively. 3

Therefore, the possible structure of compound X is CH3COOCH3. 1

5. (a) (i) The set-up is applied to determine the boiling point of propan-2-ol. 1

(ii) A: anti-bumping granule 1

B: Liebig condenser 1

(iii) Propan-2-ol is flammable. ½

It may catch fire to heat it in a naked flame. ½

(iv) If propan-2-ol is pure, its boiling point should be quite sharp. 1

(b) (i) 2

glucose [1] fructose [1]

(ii) Glucose belongs to aldehydes. ½

Fructose belongs to ketones. ½

(iii) Glucose and fructose are functional group isomers. 1

(iv) Glucose and fructose are added to acidified potassium dichromate solution

respectively. 1

The aldehyde group of glucose is oxidized by acidified potassium dichromate solution

and the solution changes from orange to green. 1

However, there is no reaction between fructose and acidified potassium dichromate

solution and thus no observable changes are made. 1

H

Marking Scheme / 13


(c) (i)

3

(ii)

2

(iii)

3

The End

CCC HEEP WOH COLLEGE 2011 – 2012 MOCK...

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