DHS_H2_CHEM_P1_SOLUTIONS Prelim

2011 DHS H2 Chemistry Paper 1 (Solutions) © DHS 2011 1 9647/01(Solutions) 1 Answer: D Topic: mole concept Solution: CnH2n+2 + ((3n+1)/2)O 2 nCO2 + (n+1)H 2O Residual gas, V = CO2 and unreacted O2 at rtp. For 10 cm3 CH4  CO2 = 10 cm3 O2 unreacted = 70 – 20 = 50 cm3   V = 60 cm3 Only option D starts with 60 cm3. 2 Answer: A Topic: Mole Concept Solution: Ammonium cyanate  CO(NH2)2 Ammonium, NH 4 + To balance the atoms and charge: Cyanate should have 1 C, 1N, 1O; overall –1 charge. Hence, CNO–. 3 Answer: D Topic: Energetics Solution: A : O2 is a covalent molecule. B : Atomising O2 requires breaking strong O=O bond. C: Irrelevant to explai n existence of oxides. D: Refer to Born–Haber cycle. Lattice enthalpy is highly exothermic and it counterbalances the endothermic process including 1 st electron affinity. 4 Answer: D Topic: Energetics Solution: C(s) + O 2(g)  CO2(g) (1) Hc  (C(graphite)) CO 2(g)  CO(g) + ½O2(g) (2) –Hc  (CO) (1) + (2) (3) C(s) + ½O 2(g)  CO(g) (3) Hf  (CO) 5 Answer: B Topic: Electrochem Solution: Analysis: Ni oxidised, cation reduced. From data booklet, E ox Ni 2+ /Ni = – 0.25 V E Co 2+ /Co = – 0.28 V E Fe 3+ /Fe 2+ = + 0.77 V E Mn 2+ /Mn = – 1.18 V E V 2+ /V = – 1.20 V E cell is >0 in reaction with Fe3+. 6 Answer: A Topic: Kinetics, Ionic Eqm Solution: pH 3 [H+] = 10–3 mol dm–3 pH 1 [H+] = 10–1 mol dm–3 Given, reaction is 1st order wrt [H+], ratio of rate = 0.001/0.1 = 0.1 7 Answer: C Topic: Ionic Eqm Solution: When iodide is added to Hg 2+, mass of ppt (HgI2) increases as volume of iodide increases. As more iodide is added, mass of precipitate decreases due to formation of soluble complex, HgI4 2–. Decrease in [iodide] due to formation of complex in (2) causes eqm in (1) to shift left to produce more iodide. Hence ppt dissolves. [Concept is similar to PbI 2 dissolving in excess KI.] Hg2+ + 2I– HgI2 (1) HgI2 + 2I– HgI4 2– (2) 8 Answer: C Topic: Group II Solution: Decomposition of group II nitrate forms an oxide. Hence only A and C are possible answers: BaO is formed. However, question states that an unreactive gas is produced with heat evolved, N2 (high bond energy released N N) is the gas and not NO. Also, NO is rapidly oxidised by air to produce NO 2, hence unlikely to be the product in this reaction. 9 Answer: D Topic: Periodic Table Solution: A : Atomic size decreases across period  N > O > F B, C : Ionic radius across isoelectronic series decreases across period  N3– > O2– >F– Na+ > Mg 2+ > Al 3+ (ionic radius in Data Booklet) D: Na+, Ne and F – are isoelectronic (10 electrons) but nuclear charge is highest for Na+, followed by Ne than F– (refer to proton no.). Hence, electrostatic attraction for the valence electrons decreases fr