AJC N2014_H2_P3 soln

©2020ASRJC/CHEM 1 H2 Chemistry 9647 2014 ‘A’ Level P3 Suggested Solutions 1 (a) (i) Na2O(s) + H2O(l) 2NaOH(aq); pH = 13 (accept 14) [1] (ii) P4O6(s) + 6H2O(l) 4H3PO3(aq); pH  2 [1] (b) (i) amount of Na2S2O3 reacted = 1000 70.16 x 0.200 = 3.34 x 10–3 mol 1 2 )ONa(n )OSNa(n 22 322 = amount of Na2O2 present = 1.67 x 10–3 mol mass of Na2O2 present = 1.67 x 10–3 x [2(23.0) + 2(16.0)] = 0.131 g mass of Na2O = 0.5 – 0.130 = 0.370 g amount of Na2O present = ]0.16)0.23(2[ 370.0 + = 5.97 x10–3 mol )ONa(n )ONa(n 22 2 = 3 3 10x67.1 10x97.5 − − = 3.57 [1] [1] [1] Comments: (a) Students are expected to include state symbols in the equation. Some thought that P 4O6 was a liquid or a gas, or that H 3PO3 was gaseous. Several gave H 2 as one of the products with Na 2O, and H3PO4 or O2 as popular incorrect products with P4O6. (c) (i) Compound with a lower pKa is the stronger acid. H2O HO– + H+ H2O2 HOO– + H+ In the 2nd equilibrium, the highly electron –withdrawing oxygen attached to the –O– will disperse the negative charge on the O atom, stabilising it. The position of equilibrium will lie more to the right as compared to the 1 st eqm, making H2O2 a stronger acid with lower pKa compared to water. [1] (ii) CH3 C OH O CH3 C O O + H+ CH3 C O O OH CH3 C O O O + H+

©2020ASRJC/CHEM 2 In the first equilibrium, The p–orbital of the C atom overlaps sideways with the p–orbitals of the 2 neighbouring O atoms . The negative charge on the carboxylate ion is distributed equally between the two O atoms. This will stabilise the CH3CO2–, hence making CH3CO2H a stronger acid. In the second equilibrium, this stabilisation by resonance of the –CO2– structure is impossible due to an intervening O atom , hence making CH3CO3H a weaker acid. (or the negative charge on the O atom is locali sed as there is absence of p orbital on the adjacent sp3 O atom, hence making CH3CO3H a weaker acid.) [1] [1] Comments: Those who had successfully argued the H2O2 case often incorrectly used the same arguments with CH3CO3H. Some thought that the more oxygen atoms there were , the more stabilisation by delocalisation. Other students wrongly suggested that CH 3CO3H ionised to give OH – (or OH +), rather than H+. (d) (i) B.E.(C–C) + 2B.E.(C–O) = 350 + 2(360) = +1070 kJ mol–1 [1] (ii) Hr = B.E.(C=C) + 4B.E.(C–H) – [4B.E.(C–H) + B.E.(C–C) + 2B.E.(C–O)] –354 = +610 – [B.E.(C–C) + 2B.E.(C–O)] sum of B.E.(C–C) and 2B.E.(C–O) = +964 kJ mol–1 In a 3 –membered ring where each sp 3 hybridised C atom has only an angle of 60o (<<109.5o), the ring in epoxyethane experiences severe bond strain, resulting in bond weakening. This explains the much lower sum of the C –C and 2 x C –O actual bond energies than the theoretical value.