2019 ASRJC Prelim H2 Chem P3 ANS

ASRJC JC2 PRELIMS 2019 9729/03/H2 1 ANDERSON SERANGOON JUNIOR COLLEGE H2 Chemistry 9729 2019 JC2 Prelim Exam Paper 3 Solutions 1 (a) (i) Kc = [Fe(SCN)2+] [Fe3+][SCN–] [1] units = mol–1 dm3 [1] (ii) Kc = [Fe(SCN)2+] [Fe3+][SCN–] 4.99 x 102 = [Fe(SCN)2+] [Fe3+][5.00×10–3] [Fe(SCN)2+] [Fe3+] = 2.495 [1] Let % of Fe(SCN)2+ = x %, then % of Fe3+ = 100 – x, x 100 - x = 2.495 x = 249.5 – 2.495x 3.495x = 249.5 x = 249.5 3.495 = 71.4% [1] (iii) When temperature is increased (to 500 K), backward endothermic reaction is favoured, shifting the equilibrium position to the left to decrease the temperature by absorbing the extra heat. [1] The red colour of the solution becomes less intense/fade. [1] (b) Na, Mg and Al have giant metallic lattice structure with high melting points, as a large amount of energy is required to overcome the strong electrostatic forces of attraction between cations and sea of delocalised electrons. [1] Melting point increases from Na to Al as the number of valence electrons for metallic bonding increases, and cationic radii decreases. Hence, metallic bond strength increases. [1] Si has a giant molecular structure with high melting point as very large amount of energy is required to overcome the extensive covalent bonds between atoms in the three dimensional network structure. [1] P4, S 8 and Cl2 have simple molecular structure with low melting points as small amount of energy is required to overcome the weak instantaneous dipole –induced dipole (id –id) attractions between the molecules. [1] Melting point: S8 > P4 > Cl2 o Number of electrons of S8 > P4 > Cl2 o Hence, the strength of id–id attractions between the molecules decreases across the period from S8 > P4 > Cl2 Amount of energy required to overcome the id–id attractions between the molecules decreases across the period from S8 > P4 > Cl2 [1]

ASRJC JC2 PRELIMS 2019 9729/03/H2 2 (c) (B and C could be metal oxides or chlorides due to the high melting points while D and E could be non–metal oxides or chlorides due to the low melting points.) B is SiO2 (as it has a high melting point, insoluble in water and only reacts with hot concentrated NaOH.) Reaction of B with hot NaOH: SiO2 (s) + 2OH– (aq)  SiO32– (aq) + H2O (l) C is NaCl (as it has a high melting point and dissolves in water to form a neutral solution. Due to low charge density of Na+, NaCl will not hydrolyse in water.) D has a low melting point and dissolves in water to give an acidic solution. In addition, the ratio of D to precipitated Cl– = 1: 5. D is PCl5 PCl5 (s) + 4H2O(l)  H3PO4(aq) + 5HCl(aq) E is SO3 It has low melting point and dissolves readily in water to give a strongly acidic solution. SO3 (g) + H2O (l)  H2SO4 (aq) (It is an acidic oxide that reacts with aqueous sodium hydroxide in the ratio of 1:2.) SO3 (g) + 2OH– (aq)  SO42– (aq) + H2O (l) [6] 2 (a) (i) pV = nRT n1 = 125 x 101325 x 60 x 10–3 8.31 x (273 + 27) = 304.8 [1] 125 n1 = 50 n2 n1 n2 = 2.5 n2 = 121.9 (number of moles of He remaining in the tank) Number of moles of helium used to fill up the balloons = 304.8 –121.9 = 183 mol [1] (ii) At constant temperature P1V1 = P2V2. (Boyle's law) 50 x 60 x 10–3 = 1.2 x V2 V2 = 2.5 m3 [1] Let n = number of balloons and Vb = the volume of each blown–up balloon. Vb = 4πr3 3 = 4π(0.15)3 3 = 1.414 x 10–2 m3 n = V2/Vb = 176.8 The remaining gas in the tank can fill up 176 balloons. [1]

ASRJC JC2 PRELIMS 2019 9729/03/H2 3 (iii) Helium behaves more ideally in balloons. [1] Since the pressure is lower in the balloon (1.2 atm) as compared to in the tank (50 atm), the helium gas molecules will be further apart in the balloon. The volume of the gas molecules therefore becomes insignificant compared to the volume of the balloon. OR The intermolecular forces of attraction are negligible. [1] (b) (i) CH3CH(OH)CH3(l) + 9/2O2(g) 3CO2(g) + 4H2O(l) 3C(s) + 4H2(g) + 5O2(g) -318 3 x (-394) 4 x (-286) ? [1] for cycle Hoc = –(–318) + 3(–394) + 4(–286) = –2008 or –2010 kJ mol–1 [1] (ii) Energy taken in during Bond breaking Energy released during Bond forming 2 x C–C = 2 x 350 7 x C–H = 7 x 410 1 x C–O = 1 x 360 1 x O–H = 1 x 460 9/2 x O=O = 4.5 x 496 3 x 2 x C=O = 6 x 805 4 x 2 x O–H = 8 x 460 6622 8510 Hoc = 6622 – 8510 = –1888 kJ mol–1 [1] CH3CH(OH)CH3(g) + 9/2 O2(g) 3CO2(g) + 4H2O(g) CH3CH(OH)CH3(l) + 9/2 O2(g) 3CO2(g) + 4H2O(l) Hovap = –2008 + 4 x (41) –(–1888) = +44.0 kJ mol–1 [1] (iii)  H for combustion is negative  S for combustion is positive as the number of gaseous particles increases. [1]  –TS is negative  Since G = H – TS  G will always be negative and the reaction is spontaneous at all temperatures. [1] –1888 –2008 4 x 41 Hovap

ASRJC JC2 PRELIMS 2019 9729/03/H2 4 (c) (i)  (least acidic) 2 –methylpropan–2–ol, 2,2 –dimethylpropanoic acid, 2–chloro–2–methylpropanoic acid, (most acidic) [1] Compare alcohol with acids:  Both 2–chloro–2–methylpropanoic acid and 2,2 –dimethylpropanoic acid are more acidic than 2–methylpropan–2–ol as the negative charge on the carboxylate ion is distributed equally [1] between the two O atoms and hence stabilises the ion to a greater extent. The dissociation of the acid to release H+ is highly favoured. Compare 2,2–dimethylpropanoic acid with 2–chloro–2–methylpropanoic acid  2–chloro–2–methylpropanoic acid is the most acidic due to the presence of electron–withdrawing C l atom, which disperses the negative charge on the carboxylate ion to an even greater extent [1] and stabilises it more. OR  2,2–dimethylpropanoic acid is less acidic due to the presence of electron–donating alkyl (methyl) group, which intensifies the negative charge on the anion (carboxylate ion) and destabilises it. Therefore, the dissociation of this acid to release H + is less favourable. (ii) CH3 C Cl CH3 CH3 K CN in ethan ol heat CH3 C CN CH3 CH3 CH3 C COOH CH3 CH3 dilu te HCl heat [1] [1] [1] (iii) Step 1: aq. NaOH, heat [1] Step 2: acidified K2Cr2O7 or KMnO4, heat [1] Step 3: CH3MgBr, followed by H3O+ [1] G: CH3 C OH CH3 H [1] H: C O CH3 CH3 [1] 3 (a) (i) [1] each NO3– : trigonal planar PO43–: tetrahedral both correct shapes [1] (ii) sp2 [1] N O O O x x  xx xx x x x   x  xx  x x x x x x  ¯ P O O O x x  xx x x x x   x  xx  x x x x x x  3– O x x x x x x x

ASRJC JC2 PRELIMS 2019 9729/03/H2 5 (iii) The continuous overlap of p orbitals between the N and O atoms allow the lone pair /