HCI Sec 4 EOY Examination Chemistry Marking Scheme 2008 (updated on 26 Oct 08)

Paper 1

1. D 2. C 3. A 4. C 5. B 6. D 7. B 8. D 9. A 10. B 11. A 12. B 13. C 14. D 15. A 16. D 17. D 18. C 19. B 20. A 21. C 22. C 23. B 24. D 25. B 26. A 27. B 28. D 29. A 30. B

Paper 2 / Section A

A1. a, b

Description of substance a) Structure b) Name

(i) conducts electricity as a liquid at room temperature B mercury

(ii) is used to make drill bits D diamond

(iii) is a black solid which undergoes sublimation E iodine

(iv) liberates a yellowish-green gas during electrolysis C Any chloride compound

c. Structure A, silicon(IV) oxide, has a much higher melting point than structure E, iodine.

In A, the silicon and oxygen *atoms are joined together by (strong) *covalent bonds to form a *macromolecule / macromolecular structure / giant molecular structure / giant covalent structure.

Iodine is made of *(simple) molecules held by van der Waals forces / intermolecular forces.

*A lot more energy is required to break all the covalent bonds in diamond than to overcome the weak van der Waals’ forces between iodine molecules.

A2a. Dot and cross diagram for CH2Cl2

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H

Cl C H

Cl

x

x

- ½ mark for using symbols other than dot and cross

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b. *Hydrogen bonds exist between water molecules while *van der Waals’ forces / dispersion forces exist between dichloromethane molecules.

Deduct ½ m if ‘molecules’ is not mentioned.

Hydrogen bonds between water molecules are *stronger than van der Waals’ forces between dichloromethane molecules; *more energy is required to separate the molecules in ammonia than in dichloromethane.

Award ½ m if answer is ‘the intermolecular forces between water molecules are stronger than between dichloromethane molecules.

c.

Award ½ m if δ+ and δ- are missing.

A3a. Na2CrO4 +6 Cr2O7

2- +6[No marks if + is missing. Penalise only once.]

b. (NH4)2Cr2O7 is reduced to Cr2O3 as the oxidation state of chromium decreases from +6 (in (NH4)2Cr2O7 ) to +3 (in Cr2O3 ).

(NH4)2Cr2O7 is oxidised to N2 as the oxidation state of nitrogen increases from -3 (in (NH4)2Cr2O7 ) to 0 (in N2).

[If product is not mentioned, ½ for each answer.]

c. Aqueous solution of Cr2O72- ions turned from *orange to *green.

d. Hydrogen peroxide is a reducing agent / is used to reduce Cr2O72- to Cr3+.

A4a. The *power stations, located at the west of Singapore, burn carbon fuels *containing sulfur (as impurity).

b. Sulfur dioxide is a *pollutant / poisonous gas/ toxic gas/ harmful gas (½) which *irritates the eye, or attacks the lungs or causes breathing difficulties (½). [State any one effect.]

OR Sulfur dioxide causes acid rain (½) which can damage limestone buildings or corrodes metal structures (½).

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H H H | |H – C – C – O – H ------- O | | H H H

δ+

δ-

Or H H | |H – C – C – O – H | | H H H H

O δ+

δ-

c. NO2 is formed when *nitrogen reacts with oxygen (in the air) *at high temperatures in the car engines.

d. All the cars in Singapore are fitted with *catalytic converters which will reduce/convert nitrogen dioxide to *nitrogen.

A5. Need to know:

(i) MgCO3 (ii) H2SO4 (iii) HCl

No. of moles 10/84 = 0.119 0.025 X 2.00 = 0.0500

0.025 X 2.00 = 0.0500

Acid is the limiting reactant. Magnesium carbonate is in excess.

Sulfuric acid is dibasic. It has twice the concentration of hydrogen ions.1 mol of H2SO4 gives 1 mol of CO2.Volume of carbon dioxide obtained = 0.05 x 24000 = 1200 cm3

Hydrochloric acid is monobasic.2 mol of HCl gives 1 mol of CO2.Volume of carbon dioxide obtained = ½ x 0.05 x 24000 = 600 cm3

a. MgCO3 + 2H+ Mg2+ + H2O + CO2 or CO32- + 2H+ H2O + CO2

b. (i)

(ii) *Increase concentration of HCl *to 4.00 mol/dm3

Or Use HCl of concentration 4.00 mol/dm3 / Use 4.00 mol/dm3 HCl

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- Label 1200 cm3 1 m Correct end-time for rxn

½ m

- Label 600 cm3 1 m - Gradient of H2SO4 graph steeper than HCl ½ m (Only one label/wrong label – deduct 0.5 m once

No label at all, only mark for volume on the y-axis.

Volume of gas (cm3)

Time (min)

Reaction I

Reaction II

1200

600

t1 t2

A6.a. (i) To a portion of river water, add aqueous sodium hydroxide and a few pieces of

aluminium foil. Warm the mixture.

A (pungent) gas that turns *moist red litmus blue is produced. *Ammonia is evolved.

[Warming not stated. Can award marks for observation.]

(ii) Ammonium salts / fertilisers can also react with aqueous sodium hydroxide (on warming) to produce ammonia. Hence, can’t confirm presence of nitrate ions.

Ammonium salts / fertilizers can interfere with the results of the reaction.

b. No. of mol of iodine liberated = 0.508/254 = 0.002001 mol of oxygen produces 2 mol of iodine.No. of mol of oxygen present = 0.00100Concentration of dissolved oxygen = 0.001/2 = 0.000 500 mol/dm3

No. of mol of I2 is calculated wrongly and mole ratio is not givenBut, show that no. of mol of oxygen present is [½] no. of mol of I2.

[Minus ½ m if final answer is not corrected to 3 sf. ]

A7.Chemical reagent

(i) Functional group

(ii) Organic product

(1) aqueous bromine

| | – C = C –

(2) acidified potassium dichromate(VI)

– O – H

(3) aqueous sodium hydroxide

O ||

– C – O – H

[6][Drawing of functional group: Minus ½ m ONCE if lines representing bonds are missing.]

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H Br Br O | | | ||H – O – C – C – C – C – O – H | | | H H H

O O || ||H – O – C – C = C – C – O – H | | H H

H O | ||H – O – C – C = C – C – O Na+

| | | H H H

-

b. The aqueous bromine will turn from *reddish-brown to colourless *immediately [2x½] [Any one point missing, deduct ½ m]

c. (i) Addition polymer [1]

(ii) Condensation polymer [1]

[Minus ½ m ONCE if the two end lines are missing. ]

d. Polymers are *non-biodegradable. Thus they lead to land pollution / cause land-fill problems.

OR burning of polymers releases poisonous / toxic / harmful fumes / gases.

Paper 2 / Section B

B8a.(i) and (ii)

element melting point /oC atomic radius colour of vapour

fluorine -200 to -240 [1m] 64 Yellow gas / vapour

(iii) Going down the group, the atoms have more fully filled electron orbitals / quantum shells / inner shells / shells.

b.

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CH2OH COOH CH2OH COOH | | | |––––––– C ––––– C ––––– C ––––– C –––––––– | | | | H H H H

H O H O | || | ||–––––– O – C – C = C – C – O – C – C = C – C –––––– | | | | | | H H H H H H

H H | | H – N – N – H

c. N2H4 (g) + 2Cl2 (g) N2 (g) + 4HCl (g)

Energy required for bond breaking = 167 + (4 x 386) + (2 x 240)= 167 + 1544 + 480= 2191 kJ

Energy released during bond forming = 942 + (4 x 428)= 942 + 1712= 2654 kJ

H = 2191 - 2654 (subtraction is consequential)= - 463 kJ

Alternatively:Enthalpy change for bond breaking = +2191 kJ Enthalpy change for bond making = - 2654 kJ Enthalpy change for the reaction = 2191 - 2654 (subtraction is consequential)

= -463 kJ

(Deduct 0.5 m once for no unit/wrong unit like kJ/mol or putting the sign if statement is energy taken in to break bonds …; no penalty if student writes everything in one line provided working is clear and correct)

d. 1. For the same number of mole of oxygen / same mass of oxygen used, the energygiven out for hydrazine is greater than for ethane.

2. *Hydrazine is a liquid and ethane is a gas at room temperature. *Liquids are easier to transport / store than gases.

3. Burning of ethane produces *CO2 as a greenhouse gas.*Hydrazine is environment-friendly / burning of hydrazine produces harmless nitrogen/ harmless products.

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Correct curve: 1 m Label activation energy with

correct arrow: ½ m Label enthalpy change with

correct arrowed: ½ m Reactants and products not

labelled, deduct ½ m Axis not labelled, deduct ½ m

Consequential marking, mark for endo diagram if answer in (ii) is +ve.

N2H4 + 2Cl2

N2 + 4HCl

EA

ΔH = - 463 kJ

Energy

Progress of reaction

Or +2191kJ

Or time

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B9.a. (i) HF exists as (simple) molecules which are not charged / cannot conduct

electricity. [Award ½ m for ‘no mobile ions/free electrons’.]

(ii) Hydroxide ion is lower than fluoride ion in the electrochemical series. Hence, *OH- is preferentially discharged / OH- is discharged instead of F- to produce *oxygen.

b. (i) The litmus indicator is bleached.(ii) Chloride ions are discharged / *oxidised / lose electrons to form chlorine, which

bleaches the litmus indicator.

Or(i) The solution turns blue.(ii) *Hydrogen ions are discharged, leaving behind a *higher concentration of OH-/

leaving behind NaOH / resulting in the electrolyte becoming more alkaline. indicator.

c. (i) The forward reaction *releases 197kJ of energy for *every mole of oxygen that reacts with sulfur dioxide.

(ii) 1. *Cooling / lowering the temperatures favours the forward reaction / shiftsthe equilibrium position to the right. *Heat energy is released in the reaction *to offset the effect of lowering the temperature. *More sulfur trioxide is formed / yield of sulfur trioxide / yield of product increases.

2. To ‘recycle’ the energy released in the reaction / to tap the energy released / to use energy to maintain the reaction at 450 o C.

(iii) Vanadium(V) oxide *catalyses or speeds up the rate of forward and backward reaction to the *same extent.Or, vanadium(V) oxide speeds up the attainment of equilibrium.Or, vanadium(V) oxide shortens time to attain equilibrium.

B10. EITHERa. (i) Aspirin

*Accept a value from pH 3 – 6. * Aspirin is a weak acid; * it dissociates/ionises partially in water to give hydrogen ions.*Low concentration of hydrogen ions gives it pH 4 – 6.

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(ii) Full structural formula of the acid and alcohol that are used to produce aspirin.

Acid: Alcohol:

(iii) *Boil / warm / reflux the mixture of acid and alcohol with *concentrated sulfuric acid as a catalyst.

b. (i) titration(ii) *Aqueous ammonia and dilute sulfuric acid are soluble reactants and *product is

a soluble salt.

Bonus ½ m *An indicator can be used during titration to determine exact volumes of acid and alkali that neutralises each other, so as to obtain pure salt.

(iii) 2NH3.H2O + H2SO4 (NH4)2SO4 + 2H2OOr, 2NH3 (aq) + H2SO4 (aq) (NH4)2SO4 (aq)

c. No. of mol of ZnCO3 = 1.25/125 = 0.0100 molNo. of mol of HCl = (30/1000) x 1.0 = 0.0300 mol

1 mol of ZnCO3 reacts with 2 mol of HCl0.01 mol of CaCO3 react with 0.02 mol of HClBonus ½ m: Zinc carbonate is the limiting reactant.

1mol of ZnCO3 produces 1 mol of ZnCl2

Mass of ZnCl2 produced = 0.01 x 136 g = 1.36 g

B10. OR

a. (i)Carbon Oxygen Hydrogen

% by mass 40.0 53.3 6.7Ar 12 16 1No of mol 40.0/12

= 3.33353.3/16 = 3.331

6.7/1 = 6.7

Molar ratio 3.333/3.331 = 1 3.331/3.331 = 1 6.7/3.331 = 2

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H O | || H – C – C – O – H | H

| C H-C C – C | ||H-C C – H

C | H

O|| – O – H

H – O

Award 0.5 m for each correct calculation of no. of mol. That is students can score up to max of 1.5 m here.

Hence empirical formula is COH2

Let the molecular formula be (COH2)n

Mr of (COH2)n = 9030n = 90 n = 3

Hence molecular formula is C3O3H6

(ii) X (alcohol), oxidized to Y (acid)

Full structural formula of X:

(iii) 1. Full structural formula of Z:

Draw at least 1 –OCH3 group at the correct position Draw at least 2 –OCH3 group at the correct position

2. *Boil / warm / reflux the mixture of Y and methanol with *concentrated sulfuric acid as a catalyst.

b. (i) Wash the residue with deionised water.(ii) The filtrate may contain unreacted *aqueous barium nitrate (or aqueous potassium

sulfate). *The crystals formed may not be pure / may be contaminated.

(iii) Titration(iv) KOH + HNO3 KNO3 + H2O

Or K2CO3 + 2HNO3 2KNO3 + CO2 + H2O

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H H | | H – O – C – C – C – O – H | | || H H O

H H H | | | H – C – O – C – C – C – O – C – H | || | || | H O H O H