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. [1] [1] Comments: Fewer students were able to explain the difference in the values. Most students wrote about either heat being lost to the surroundings, non–standard conditions or average bond energies were used. Some students appreciated that the difference was significantly la rger than one might expect from using average bond energies. (e) (i) CH3 C C CH3 H CH2 OH H H C CH3 C C CH3 H CH3 E CH3 C CH3 C H H CH2 F [1]: correct structure for C [1]: correct structures for E and F (structures can be swopped) Thinking process Given that CH 3COCH3 and CH3CO2H are formed when one of the alkenes, E or F are treated with hot/MnO4-, you can deduce the identity of one of the alkenes to be (CH3)2C=CHCH3 (labelled as E, in this case). Since the alkene is obtained from dehydration of the alcohol C, you can add back the –H and –OH across the C=C of the alkene deduced to form C. Here, you need to also take note that the alcohol C formed must have the structure CH3CH(OH)R so that it can give a positive iodoform test. After which, you can then deduce the structure of alkene F. [2]

©2020ASRJC/CHEM 3 Comments: Although most gave structures with five carbon atoms, there were a number who suggested straight chain compounds rather than branched chain ones. The idea that E and F must have the same overall carbon skeleton was not appreciated by some students. Many ga ve straight chain alkenes thus only incorporating part of their evidence and some gave C as a primary alcohol, which could not have given a positive tri–iodomethane reaction. (ii) Limiting Cl2 and uv light [1] (iii) CH2CN Observe the gain of a C and N atoms with loss of Cl atom. [1] (iv) CH2CO2H [1] (v) Reaction II:KCN in ethanol, heat (under reflux) (accept NaCN in ethanol) Reaction III: dilute H2SO4, heat [1] [1] (vi) CH2 C O CH CH(CH3)2 CH3 An easier way to deduce A is to derive the ester B first and then remove the –O- atom to get ketone A. [1] 2 (a) [2] EAH = −73 Li(s) ∆Hato = +159.5 Li (g) 1st IELi= +519 Li+ (g) + + ½H2(g) ½ B.E.= ½ (+436) H (g) H– (g) ∆Hfo = −90.5 Li+H–(s) Lattice energy

©2020ASRJC/CHEM 4 By Hess’ Law, LE = (−90.5) – [159.5 + 519 + ½ (436) + (−73)] = –914 kJ mol−1 [1]: correct substitution of relevant values from question and from Data Booklet [1]: final answer with correct units Comments: For this question, students are expected to extract relevant data from the question and Data Booklet. Students can first tackle the question by writing the equation that represents lattice energy of LiH(s). Next, identify the enthalpy changes from the question and data booklet that can be linked to the reactants [Li+(g) & H –(g)] and product [LiH(s)] . The determination of L.E. can subsequently be done by completing the energy cycle. Note: ∆H atm (H) = ½ B.E.(H –H), all the energy terms should be quoted and t he signs of the enthalpy changes should be correct. (b) 4LiH(s) + AlCl3(s) LiAlH4(s) + 3LiCl(s) Hr = Hfo (products) – Hfo (reactants) –276 = Hfo (LiAlH4) + 3(–408.5) – [4(–90.5) + (–704)] Hf (LiAlH4) = –116.5 kJ mol–1 [1]: balanced equation [1]: correct substitution into mathematical expression [1]: final answer with units [3] Comments: It is important to be able to first extract the reactants and products correctly from the question, before balancing the equation for the reaction. Since enthalpy change of reaction is given in the question, and all data available from the question are enthalpy change of formation, students can use the mathematical expression “ Hro = Hfo(products) – Hfo(reactants)” to determine Hfo(LiAlH4) without drawing any energy cycle. Students should take note of both the signs and multipliers in the calculation of the enthalpy change. (c) (i) Go = Ho– TSo, (–27.68) = (+3.46) – (298)So So = +0.104 kJ mol–1 K–1 or +104 J mol–1 K–1 S is positive as the reaction results in an increase in number of moles of gaseous particles which increases disorder / brings about more ways of arranging / distributing the particles in the system. [1]: final answer with units [1]: correct explanation [2] (ii) O.N. of H in LiAlH4 = –1 O.N. of Al in Li3AlH6 = +3 [1] [1] (iii) octahedral [1]

©2020ASRJC/CHEM 5 Comments: (i) Students should state the units for So correctly. If the substitution of values for Ho and Go are in kJ mol–1, the calculated value should be in kJ mol–1 K–1, not J mol–1K–1. Do not omit the K–1. As this question required students to comment on why So is positive “with respect to the equation”, students should link the explanation to the equation given in the question. (ii) Note: O.N. of Li = +1 and O.N. of H in hydrides = –1 (not +1). Hence, O.N. of Al is +3 (not –3, –9) Calculatio