Stpm Che2 Ans (sbh)

Suggested Answers For Chemistry STPM Section A 1

(a)

(i) nucleus has no (zero) influence over the electron.

1m

(ii) Energy required = 1310 / 6.0 x 1023 = 2.18 x 10-21 kJ or 2.18 x 10-18 J

1m

(iii) Difference in energy (J per electron) between n = 2 and n = 4 = (1230 – 981) x 103 / 6.0 x 1023 J per electron transition. = 4.15 x 10-19 J per electron transition E = hf f = 4.15 x 10-19 / 6.3 x 10-34 = 6.26 x 1014 s-1

1m

(b) (i)

P1: 35Cl+ ion and P3:

37

Cl+

(ii) Peak height P2 : Peak height P4 100 : 33 = 3 : 1 3 : 1 which conforms to the relative abundance of 35Cl and 37Cl (c)

(i) At very high pressures, the volume occupied by the gas molecules cannot be ignored because the volume of the gas is small and the molecules are closer together. (ii) At medium pressure : -negative deviation caused by stronger intermolecular forces of attraction since HCl is a bigger molecule than hydrogen. - HCl is a polar molecule; hence stronger intermolecular forces exist .

1m 1m 1m 1m 1m

1m 1m 10m

2

(a)

(i) 0 = 2(-127) + 2∆Hf (ClO2)- 2 (-25) ∆Hf (ClO2) = + 102 kJ mol-1

(ii) The enthalpy of formation of ClO2 is endothermic, hence ClO2 is unstable compared to chlorine and oxygen. (b) (i) Fe3+(aq) + e Fe3+(aq); EӨ = +0.77 V + ClO2 (aq) + 4H (aq) + 5 Fe3+ Cl- (aq) + 5Fe2+ + 2H2O ; E.m.f of the cell = 1.50 – (+0.77 ) = + 0.73 V Since e.m.f of the cell is positive, ClO2 oxidises Fe2+ to Fe3+ spontaneously. (ii) S + 2H + + 2e H2S (aq); EӨ = +0.14 V ClO2 + 4H2S 2Cl (aq) + 5S + 4H2O + 2H+ E.m.f of the cell = 1.50 – (+0.14 ) = + 1.36V Since e.m.f of the cell is positive, ClO2 oxidises H2S to sulphur spontaneously. (iii)

1m 1m 1m

1m 1m

1m 1m

2m

* Correct arrangement of apparatus as well as functional; 1 mark * correct materials and prperly labelled : 1 mark (c) 3

ClO2-

2ClO2 + 2OH-

+ ClO3- + H2O

The minimum energy required to displace/remove one mole of electron from one mole of atom in the gaseous state (b) 1. atomic size increases 2. screening effect (shielding effect) increases (c) -Valence electron configuration: Mg: 3s2 ; Al: 3s23p1 - more energy is required to remove an electron from a fully occupied s-orbital than from a singly occupied p orbital. (d) -the small boron ion will distort the electron cloud of its neighbouring anions, giving a covalent characteristic to its compounds. - the big Ba2+ ion will not show much polarization; thus barium compounds are more ionic.

1m 10m

(a)

(e)

1m 1m 1m 1m 1m 1m 1m

x

(i)

Mg2+

x

N3-

x

1m

x (ii) 4

Mg3N2

+ 6 H2O

2NHs(g) + 3Mg(OH)2

2m 10m

(a) C No. of moles 35.2/12 = 2.9 Mole ratio 4 Molar mass : 136.9 Molecular formula : C4H9Br (b) (i)

H 6.5/1 = 6.5 9

Br 58.3/80 = 0.73 1

1m 1m

CH3

1m

CH3 - CH - CH2Br (ii) (c)

Nucleophilic substitution SN2 (primary halogen alkanes)

1m

(i) H

H 1m

CH3 - C = C- CH3 or CH3 CH = CHCH3 (ii) CH3

1m

CH3 - C - CH3 Br (iii) Isomer: CH3 CH2 – CH = CH2 or CH3- C = CH2

1m

CH3 Equation: CH3 CH2 – COOH +CO2+ H2O

1m

(d) -Warm with aqueous acidified KMnO4. -Alcohol P does not decolourise the purple solution. -2-methyl-1- propanol decolourises the purple solution. (oxidation of primary alcohols)

1m

CH3 CH2 – CH = CH2 + 5[O] Or C(CH3)2=CH2 + 4[O]

(CH3)2C=O + CO2 + H2O

1m 10m

SECTION B 5

(a)

(i) Relative atomic mass = Mass of one atom of element X 1/12 x Mass of one atom of 12C

(b)

1mark (ii) Isotopes means atoms of an element having the same number of protons but different number of neutrons. 2 marks Ethanol molecule: CH3CH2OH During bombardment by high speed electrons, various bonds can be broken, positive ions which are fragments are formed. m/e 15 28 31 45 46

Ion CH3+ CH3CH+ CH2OH+ CH3CH2O+ CH3CH2OH+ 5marks

(c)

6

(a)

(i)

(ii)

(i) Water molecule undergoes autoionisation. H2O <=== > H+ + OH- ΔH = Endothermic Ionic product of water, Kw = [H3O+ ] [OH-] = 1 x 10-14 mol2dm-6 at 250C When temperature increases, the position of equilibrium shifts to the right forming more H+ and OH- . Hence the value of Kw increases. 4marks Ca(s) EӨ = -2.87V (ii) Ca2+ (aq) + 2e The standard electrode potential of calcium is the potential difference obtained when the half cell of Ca/Ca2+(aq) is connected to a standard hydrogen electrode under standard conditions. 3marks

KNO3(s) + aq K+ (aq) + NO3- (aq) The forward reaction of dissolution of KNO3 is an endothermic 1m process as the temperature drops when it dissolves. The backward reaction of crystallization is an exothermic process. On cooling a saturated solution, heat is given off/ removed from the system. By Le Chatelier’s Principle, processes occur in the system to counteract the lost of heat by favouring the reaction that gives out heat. 1m Since the crystallization is an exothermic process, it is favoured and (2) crystals formed. Ca2+(aq) + SO42- (aq) CaSO4(s) When concentrated sodium sulphate solution is added to a saturated solution of calcium sulphate, concentration of sulphate ions increases. 1m By Le Chatelier’s Principle, processes occur in the system to

counteract this increase of sulphate ions by favouring the process that 1m decreases the concentration of sulphate ions. The equilibrium shifts to 1m the right to favour the precipitation reaction, a white precipitate of (3) calcium sulphate forms. (iii)

(b)

2NO2 (g) N2O4(g) Δ H –ve (brown) (colourless) Since cooling favours the exothermic reaction which gives out heat, 1m the forward reaction is exothermic. On compressing the gaseous mixture, total pressure in the system increases; By Le Chatelier’s Principle, the forward reaction that 1m produces less molecules of gas is favoured, the colour fades as the brown NO2 gas forms the colourless dimer. On warming, heat is supplied to the system. According to Le Chatelier, the process that absorbs heat is favoured. Since backward 1m process is endothermic, it is favourd. The brown colour is restored as colourless N2O4 dissociates to form brown NO2. (3) C(s)

+

H2O(g)

H2(g)

+ CO (g)

1m

1m Initial partial press./atm 1.0 0 0 Equilm partial press./atm 0.1 0.9 0.9 Change in partial pressure of steam = 1.0- 0.1 = 0.9 atm. Since 1 mol of steam produces 1 mol of H2 and 1 mol of CO, partial pressure of H2 = partial pressure of CO = partial pressure of steam used = 0.9 atm 1m Kp = 0.9 x 0.9 / 0.1 = 8.1 (3) (c)

Ksp = [Mg2+ ] [ OH-]2 At pH 9, pOH = 14-9 = 5 and [OH-] = 10-5 = 1 x 10-5 mol dm-3 Max. value of [ Mg2+] = Ksp / [OH-]2 = 1.8 x 10-12 / (1 x 10-5)2 = 1.8 x 10-2 mol dm-3 At pH 6.5, pOH = 14-6.5 = 7.5; [OH-] = 10-7.5 = 3.16 x 10-8 mol dm-3 Max value of [ Mg2+] = Ksp / [OH-]2 = 1.8 x 10-12 / ( 3.16 x 10-8 )2 = 1803 mol dm-3 Peat based soil is preferred.

1m 1m

1m 1m(4) (15m)

7.

(a)

(b)

Silver bromide is used in black and white photography. The surface of 5m the paper is spread with a layer of silver bromide. When exposed to sunlight, the bromide ion will be oxidized to bromine gas, leaving a layer of silver on the surface of the paper. 1m (i) [Co(NH3)3(Cl)3] (ii) NH3 2m Cl NH3 Co Cl

NH3

Cl fac-triamminetrichlorocobalt(III) NH3

2m

Cl

NH3 Co

Cl

Cl

NH3 mer-triamminetrichlorocobalt(III) (c)

(i) Observation: no visible change Explanation: There is no reaction between the solution and silver nitrate because there is no chloride ion present.

2m

(ii) Observation: White precipitate forms when silver nitrate is added 3m + Explanation: Ag will react with the chloride ion from the complex to form AgCl solid which appears as a white precipitate. (15m) 9

(a)

HCl or H2SO4or H+

or

acid

conc(if HCl only)/dilute/aqueous + heat (b) (c) (d)

1m 1m

two rings only (1 ring around the α-C of tyrosine & 1 around the α-C of lysine) + NH3CH2CO2 (or displayed formula)

(i) NH2CH2CO2- (Na+) (either -CO2-Na+or -CO2Na but NOT –CO-O-Na)

1m 1m 1m

(ii) (Na+) –O-C6H4-CH2CH(NH2)CO2- (Na+) (iii) (Cl-)+NH3(CH2)4CH(NH3+)CO2H (Cl-)

1m+1m 1m+1m

(iv) HO-C6H2Br2-CH2CH(NH2)CO2H (if shown, Br at 2,6 to OH group)

1m (6m )

(e) 2m

(f) 3m

15m

10 (a)

(i) AlCl3/FeCl3/Al/Fe/I2 (+ heat) (N.B. NOT AlBr3 etc.) (or names)

1m

(ii) (sun)light/hf/UV

1m

(b)

SOCl2/PCl3/PCl5 (or names)

1m

(c)

(i) C > B > A

1m

(ii) (acyl chloride fastest) highly δ + carbon atom joined to 2 1m electronegative atoms or addition-elimination mechanism is possible (aryl chloride slowest) delocalisation of lone pair over ring → stronger C- 1m Cl bond Or

impossibility of ‘backside’ attack on the C-Cl bond

(i) C6H5-CO2C6H5

1m

(d)

(i) E

1m 1m (3m) 1m

- CH3CH2CH2CO2-(Na+)

[NOT C3H7COO-Na or C3H7COOH] but allow CH3CH2CH2CO2Na] - CHl3 or name

1m

(ii) the alcohol from E has four different groups around a carbon atom

1m

∴ it is chiral/asymmetric or it is produced as a 50:50 mixture of mirror images

1m

(ii) C6H5-CONHCH3 (iii) C6H5-CONHCH3

Or structural formula

(e)

1m

or its mirror images are non-superimposable.

1m

(15m ) END OF MARKING SCHEME