VJC Prelim P1

Detailed Ans wers for H1 Chemistry Prelim Paper 2009 1 C 6 B 11 C 16 A 21 D 26 B 2 A 7 A 12 B 17 D 22 B 27 D 3 A 8 B 13 D 18 B 23 C 28 C 4 B 9 D 14 B 19 D 24 A 29 A 5 C 10 C 15 C 20 D 25 C 30 A Page 1 of 3 1 A Correct: 1 mole of methane contains 4 moles of H atoms = 4 x 6.02 x 1023 = 2.41 x 1024 H atoms. B Correct: Ar of 12C is 12.0. C Wrong: One mole of hydrogen gas contains 2 moles of hydrogen atoms = 2 x 6.02 x 1023 atoms of hydrogen. D Correct: One mole of methane (16.0g) contains one mole of carbon (12.0g) 12.0 100%16.0 = 75% Answer: C 2 This is a disproportionation reaction where Cu 2O is reduced to Cu and is oxidised to CuSO 4 simultaneously. Hence, it can act as both an oxidising and reducing agent. H 2SO4 is not reduced or oxidised in this reaction. Answer: A 3 Let the oxidation state of Fe in [Fe(CN) 6]4- be x. x + 6(-1) = - 4 x = +2 Electronic configuration of Fe is [Ar]3d 64s2. Electronic configuration of Fe2+ is [Ar]3d6. Answer: A 4 The state of a compound depends on its bonding. CO2 gas is made up of simple molecule with weak dispersion forces (a type of van der Waals’ forces) of attraction between the molecules. SiO 2 has strong covalent bonds throughout the giant molecular structure. Therefore, more energy is required to overcome the strong covalent bonds in SiO 2 compared to the weak dispersion forces in CO 2 gas. The boiling point of SiO 2 is much higher than CO 2, therefore it is a solid at r.t. Answer: B 5 A CO2 is linear (0 lp, 2 bp); SO2 is bent (1 lp, 2 bp) B NH3 is trigonal pyramidal (1 lp, 3 bp); NO 3 - is trigonal planar (0 lp, 3 bp) C CO2 is linear (0 lp, 2 bp); I3 - is linear (3 lp, 2 bp).The linear shape of I3 - is due to 3 lp occupying the equatorial positions in a trigonal pyramid. D BCl 3 is trigonal planar (0 lp, 3 bp); PCl3 is trigonal pyramidal (1 lp, 3 bp) Answer: C 6 The standard enthalpy change of formation is when one mole of a pure compound is formed from its constituent elements in their standard states , under standard conditions of 298K and 1 atm. Answer: B 7 H c (C3H4) = Hf (product) - Hf (reactant) – 1938 = (-394 X 3) + (-286 X 2) - Hf (C3H4) Hf (C3H4) = + 184 kJ mol-1 Answer: A 8 For C2H6(g) ⇌ C2H4(g) + H2(g), ∆Hrxn = Total bond energy for reactants – Total bond energy for products = [BE(C-C) + 6BE(C-H)] – [BE(H-H) + BE(C=C) + 4BE(C-H)] = [350 + 6(410)] – [436 +610 + 4(410)] = +124 kJ mol-1 > 0 When temp increases, position of eqm shifts left to counteract the increase in temp, hence, eqm yield of ethene decreases. Rate of formation of ethene will decrease at lower temperature due to smaller frequency of collision between reactants molecules to form the products. Answer: B 9 A 2 c 2 [H ][CO]K Unit: mol dm [H O] -3 B  2 3 4 c 24 3 [Cu NH ]K  [Cu ][NH ] Unit: (mol dm-3)-4 C 2 2 c 24 [NO ]K Unit: mol dm [N O ] -3 D 32 3 2 c