AJC N2014_H2_P1 solns

©2020ASRJC/CHEM 1 H2 Chemistry 9647 2014 A level Paper 1 Suggested Solutions 1 n(Be) present = 9 09.0 = 0.01 mol no. of neutrons present per 9 4 Be atom = 5  no. of neutrons present in 0.09 g of 9Be = 5(0.01)L = 0.05L B 2 ClO2 + e– ClO2– [R] Since no. of electrons transferred must be equal  1 mol of ClO2 gained 1 mol of e  another 1 mol of ClO2 lost 1 mol of e to form Q  O.N. of chlorine in Q changes from +4 (original) to +5 (final) FYI ClO2 + 2OH– ClO3– + H2O + e– [O] balanced equation: 2ClO2 + 2OH– ClO3– + ClO2– + H2O C 3 molecular formula of G = C7H12O3 C7H12O3 + 2 17 O2 7CO2 + 6H2O A 4 Presence of H covalently bonded to O and available lone pairs of e lectrons on O in both water and methanol, hence hydrogen bonds will be the strongest IMF. Methoxymethane does not have H covalently bonded to O , hence only permanent dipoles. A 5 no. of bp no. of lp shape should be A,B 2 2 bent (non–linear) C,D 3 1 trigonal pyramidal A 6 The behaviour of real gas deviates from ideality because volume of gas particles is not insignificant compared to the overall volume occupied by the gas (option D). There are also significant intermolecular forces of attraction between the gas particles (option C), resulting in inelastic collisions (option A). B 7 Lattice energy is proportional to )rr( qq −+ −+ + . Thus the bigger the charge, and the smaller the inter–ionic distance, the more exothermic the lattice energy is. Size of cation: Cs+ > Na+ ; size of anion: Cl– > F– Since Cs+ and Cl– are the larger cation and anion respectively, CsCl will have the least exothermic lattice energy (–661 kJ mol–1). A 8 Hr = Hf (products) – Hf (reactants) = [–1273 + 6(0)] – [6(–394) + 6(–286)] = +2807 kJ mol–1 > 0 As the forward reaction results in a formation of solid (with no change in no. of gaseous molecules), there will be less ways of arranging the particles and S is < 0. C

©2020ASRJC/CHEM 2 9 Cl2 + 2e 2Cl– +1.36 V Br2 + 2e 2Br– +1.07 V I2 + 2e 2I– +0.54 V (SCN)2 + 2e 2SCN– x V Since both aq. C l2 and aq. Br 2 are able to oxidise SCN – ion to (SCN) 2 while aq. I2 is not able to, the Eo(SCN)2/SCN– must be less positive than +1.07 but more positive than +0.54. C 10 rate = k [P] [Q]2 If [P] is doubled while [Q] is halved, rate is halved. Hence only half of the volume collected in first experiment will be produced in the first minute of the 2nd experiment. B 11 To determine the order of reaction wrt H + ions, we have to monitor how the rate of reaction changes as the concentration of H + ions is changed , while keeping all other factors constant. C 12 [H+] = [HNO3] after dilution = )1090( 01.0x10 + = 0.001 mol dm–3 pH = 3 C 13 2H2(g) + CO(g) CH3OH(g) initial amount 2.0 1.0 0 change in amount –x – 1 2 x + 1 2 x amount at eqm 2.0 – x 1.0 – 𝟏 𝟐x 𝟏 𝟐x concentrati