SRJC H2 Chem 2013 Prelim P1 Soln

SRJC A 1 Th W wa A B C D Answer One 1o Four 2o 5 mol o 2 Cl ex pr de W Answer Numbe Numbe Answer he diagram When glucos ater are form 4 5 6 7 r: B alcohol con alcohols co of H2O forme luster dec a xperimental rotons that ecay of radi Which row in A B C D r: B r of protons r of neutron below show se reacts w med? Assu nverted to c onverted to ed in total. ay is on e discovery, are heavier um-223. the table c Number of 74 82 74 82 s in A = 88 ns in 223-ra ws the struc ith hot aci d me the cyc carboxylic a ketones: R e rare ins t where a pa r than an  - correctly des protons – 6 = 82 dium = 223 1 ctural formu dified potass lic ring rem cid: RCH2O RR’CHOH + tance of arent atomi -particle. O Ra଼଼ ଶଶଷ  ଵ scribes the Numb 3 – 88 = 135 ula of glucos sium dichro ains intact a OH + 2[O]  [O]  RR’C scientific p ic nucleus e ne of the fi C଺ ଵସ +A nuclear ma ber of neutro 119 127 209 217 5 se. omate(VI), after the rea  RCOOH + C=O + H2O phenomena emits a clus rst predictio ake-up of ele ons Tur how many action. + H2O O a predicte d ster of neut ons was the ement A ? rn Over] moles of d before trons and e nuclear 1222

2   Turn Over] Number of neutrons in 14-carbon = 14 – 6 = 8 Number of neutrons in X = 135 – 8 = 127 3 A given mass of ideal gas occupies a volume V and exerts a pressure p at 30 oC. At which temperature will the same mass of the ideal gas occupy a volume ௏ ଷ and exert a pressure 2p? A 20 oC B 20 K C 202 oC D 202 K Answer: D pV = nRT (2p)( ௏ ଷ) = nR(xT) ଶ ଷ(pV) = x(nRT) x = ଶ ଷ Original T = 30 + 273 = 303 K New T = ଶ ଷ × 303 = 202 K 4 The boiling point of water (100 oC) is greater than that of ammonia (–33 oC). Which statement is a correct explanation of this? A Ammonia has intramolecular hydrogen bonds, which water does not have. B The Mr of water is greater than that in ammonia, so van der Waals’ forces are stronger in water. C There are, on average, more hydrogen bonds between water molecules than there are between ammonia molecules. D The O–H bond requires 460 kJ mol –1 to overcome, while the N–H bond only requires 390 kJ mol–1 to overcome. Answer: C 5 The standard enthalpy change of formation of hydrazine, N2H4(g), is x kJ mol–1. The bond energy of the NN bond is y kJ mol–1. The bond energy of the H–H bond is z kJ mol–1. 1223

3   Turn Over] What is the standard enthalpy change of atomisation of hydrazine? A (x + y + 2z) kJ mol–1 B (y + 2z – x) kJ mol–1 C (x + 2y + 4z) kJ mol–1 D (2y + 4z – x) kJ mol–1 Answer: B By Hess’ Law, Ha = –Hf + BE(NN) + 2BE(H–H) = (y + 2z – x) kJ mol–1 6 The Gsolution Ɵ and Ssolution Ɵ for silver chloride are +55.6 kJ mol –1 and +33.2 J mol –1 K–1 respectively. What is the enthalpy change when 287 g of silver chloride is precipitated under the same conditions? A +65.5 kJ B –65.5 kJ C +131 kJ D –131 kJ Answer: D G = H – TS 55.6 = Hsolution Ɵ – (298)(0.0332) Hsolution Ɵ = +65.49 kJ mol–1 Hppt Ɵ = –65.49 kJ mol–1 For 2 mol of AgCl precipitated, enthalpy change is –131 kJ. 1224

4   Turn Over] 7 The diagram shows the reaction pathway diagram for an uncatalysed reversible reaction. The reaction was then catalysed. What are the changes in the rate constant, equilibrium constant and the reaction pathway diagram? Rate constant, k Equilibrium constant, Kc Energy profile A Unchanged Increase B Increase Unchanged C Increase Increase D Increase Unchanged Answer: D Introduction of catalyst affected only the rate and rate constant will increase. Kc however, is not affected by catalyst. Graph is correct as catalyst merely lower the Ea and does not change the H. 1225

5   Turn Over]    8 Steam dissociates at an initial pressure of 1 atm at T K to form hydrogen gas and oxygen gas. 2H2O(g) 2H2(g) + O2(g) If the total pressure at equilibrium is 1.3 atm, what is the numerical value of the equilibrium constant, Kp, of the reaction at T K? A 0.028 B 0.135 C 0.450 D 0.675 Answer: D 2H2O(g) 2H2(g) + O2(g) Initial pressure 1 0 0 Change -x +x + x/2 Eqm pressure 1-x x x/2 1.3 = 1 -x + x + (x/2) 0.3 = x/2 x = 0.6 Kp = [(0.6)2(0.3)]/(0.4)2 = 0.675 1226

6   Turn Over]    9 The following graphs show the change in pH when four different pairs of acid and base were titrated against each other. In each titration, a 1.0 mol dm −3 solution of an acid is gradually added to 20 cm 3 of a 1.0 mol dm −3 solution of a base. Which pair of solutions could not have given any of the graphs above? A HNO3 and NH3 B HCl and Ca(OH)2 C H2SO4 and NaOH D CH3COOH and NH3 Answer B HNO3 and NH3 represented by top right graph H2SO4 and NaOH represented by bottom right graph (note the endpt for this graph) CH3COOH and NH3 represented by bottom left graph HCl and Ca(OH)2 not represented as more volume of acid (40cm3) is required. 1227

7   Turn Over] 10 A solution contains 1 x 10-3 mol dm-3 of bromide, fluoride, iodide and sulfate ions. Which lead ( II) compound will be precipitated first when 0.01 mol dm -3 of lead (II) nitrate is added dropwise to the solution at 25oC ? Compound Solubility product at 25oC A Lead(II) bromide 4.0 x 10-5 B Lead(II) sulphate 1.6 x 10 −8 C Lead(II) fluoride 2.7 x 10 −8 D Lead(II) iodide 7.1 x 10 −9 Answer: B For lead halide: IP = [Pb2+][X-]2 = (0.01)(1x10-3)2 = 1 x 10-8 mol3 dm-9 IP of Lead (II) bromide < Ksp (no ppt) IP of Lead(II) fluoride < Ksp (no ppt) IP of Lead(II) iodide > Ksp (ppt) For lead sulfate: IP = [Pb2+][SO4 2-] = (0.01)(1x10-3) = 1 x 10-5 mol2 dm-6 IP >> than the Ksp, precipitation will occur readily To determine the critical concentration of Pb2+ to bring about precipitation for lead(II) iodide and lead(II) sulfate. For lead (II) sulfate: 1.6 x 10-8 = [Pb2+](1 x 10-3) Min [Pb2+] = 1.6 x 10-5 mol dm-3 For lead (II) iodide: 7.1 x 10-9 = [Pb2+](1 x 10-3)2 Min [Pb2+] = 7.1 x 10-3 mol dm-3 Thus lead (II) sulfate requires lesser Pb2+ to bring about precipitation. 1228

11 Us Th (S Th Tw B: C: W A B C D Answe E(Sn2+/ E(H+/H2 Reactio Ecell = E = If it is > [Sn2+] c [H+] can more po se of the Da he diagra m Sn2+(aq)/Sn( he e.m.f of t wo students : [H+ (aq)] w : [Sn2+(aq)] Which of thei Both B an B only C only Neither B r: B Sn) = -0.14 2 ) = 0 (red) on occurring Ered – Eoxd 0 + 0.14 0.14, oxida cannot be gr n be greater ositive  E Stan ata Booklet m represe (s)), the sta the cell was s, B and C, was greater was greate ir suggestio nd C B nor C 4 V (oxd) g: 2H+ + 2e Sn + 2e  ation is favo reater than r than 1 mo Ecell become ndard tin ha ce electro t is relevant nts an e ndard elect s found to b suggested than 1.00 m er than 1.00 ons could be  H2  Sn2+ oured or red 1 mol dm-3 ol dm-3 as it s more pos alf ell ode 8 to this ques experiment trode poten be 0.18 V ra possible ex mol dm-3 0 mol dm-3 e correct? duction was as it will ca will facilitat sitive stion. to deter m tial of tin ather than th xplanation. favoured ause the Ece e reduction mine the he expected ell to decreas (by LCP) a H2 (g) 1 Pt 1 mol d 25oC Tur value of d 0.14 V. se. and Ered will atm dm-3 H+ (aq rn Over] the E  l become ) 1229

9   Turn Over] INORGANIC CHEMISTRY 12 The graph below shows the first thirteen ionisation energies for element D. Wha