2019 NJC Prelim P3 ANS

[Turn over Suggested Answer for SH2 H2 Chemistry 2019 Prelim Paper 3 Section A Answer all the questions from this section. 1 Potassium is an element found in Period 4 of the Periodic Table. (a) (i) Write an equation for the second ionisation energy of potassium. [1] K+(g) → K2+(g) + e− Must show state symbol (ii) Fig 1.1 shows the second ionisation energy for the consecutive elements from Period 3 and Period 4. Fig 1.1 Suggest which of the above elements, A to H, is potassium. [2] The second electron of potassium is removed from the 3p subshell (inner quantum shell) as compared to the second electron of calcium which is removed from the 4s subshell (outer quantum shell). Much more energy is required to remove the second electron from K as compared to Ca. Hence element G is potassium. Note: It is insufficient to mention that 2nd I.E. of K is much higher than 1st I.E. as the comparison with 1st I.E. is not reflected in Fig 1.1 (b) When 0.400 g of potassium dichromate(VI), K2Cr2O7, was heated using a strong flame, a yellowish green solid mixture was obtained and a gas was evolved. The gas collected ignited a glowing splint. Upon adding water to the yellowish green solid, part of the solid dissolved to give a yellow solution of K2CrO4, leaving a green solid. (i) State the identity of the gas. [1] Oxygen gas.

NJC/H2 Chem Prelim/03/2019 2 (ii) The green solid that remained was collected and dried. It weighed 0.052 g and was found to be 68.4% chromium by mass and contains only chromium and oxygen. Determine the chemical formula of the green solid. [2] Cr O Mass 68.4 100 × 0.052 = 0.03557 31.6 100 × 0.052 = 0.01643 Ar 52.0 16.0 Moles 0.000684 0.001027 ÷ by smallest no. 1 1.5 Mol ratio 2 3 Hence chemical formula is Cr2O3 (iii) Using your answers to b(i) and b(ii), write the equation for the decomposition of potassium dichromate(VI). [1] 4K2Cr2O7  4K2CrO4 + 2Cr2O3 + 3O2 (c) State two physical properties of chromium that differs from calcium. Explain the reasons for those differences. [2] TM has higher density due to larger atomic mass and smaller radius giving rise to dense close-packed structure (allowing more small size but heavy atoms to be packed within a specific volume.) TM has higher melting point due to more delocalised electrons contributed from 3d and 4s subshells, resulting in stronger metallic bonding. TM has higher electrical/heat conductivities due to more delocalised electrons contributed from 3d and 4s subshells. Note: Variable oxidation states and reactions to give coloured complexes are chemical properties. (d) Vanadate(V) ion, VO3−, changes into pale yellow oxovanadium(V) ions, VO2+, when acid is added. Upon addition of excess zinc, the pale yellow solution changes to blue, to green and finally to violet. (i) With the aid of an equation, explain the ty pe of reaction that occurs when vanadate(V) ion changes into oxovanadium(V) ions. [2] VO3– + 2H+ → VO2+ + H2O Acid-base reaction (ii) Explain why vanadium is able to exhibit variable oxidation states. [1] Because of the close proximity in energy of the 4s and 3d electron in V, it is able to lose electrons from both the 3d and 4s subshells, hence it can exhibit variable oxidation states.

NJC/H2 Chem Prelim/03/2019 3 [Turn over (iii) With the aid of the Data Booklet, deduce the final oxidation state of vanadium in the violet solution. [3] Zn2+ + 2e− Zn Eo = – 0.76 V VO2+ + 2H+ + e− VO2+ + H2O Eo = + 1.00 V Reaction between VO2+ and Zn, Eo = + 1.00 – (–0.76) = +1.76 V > 0 (reaction is feasible) VO2+ + 2H+ + e− V3+ + H2O Eo = + 0.34 V Reaction between VO2+ and Zn, Eo = + 0.34 – (–0.76) = +1.10 V > 0 (reaction is feasible) V3+ + e− V2+ Eo = − 0.26 V Reaction between V3+ and Zn, Eo = − 0.26 – (–0.76) = +0.50 V > 0 (reaction is feasible) V2+ + 2e V Eo = − 1.20 V Reaction between V2+ and Zn, Eo = − 1.20 – (–0.76) = −0.44 V < 0 (reaction is NOT feasible) Hence final oxidation state of V is +2 Note: It is important to show that V2+ do not undergo further reduction when excess Zn is present. Solid vanadium(V) oxide, V2O5, is used as a catalyst for reaction 1. reaction 1: SO2(g) + 1 2O2(g) → SO3(g) (iv) State the type of catalysis and outline the mode of action when V2O5 is used for reaction 1. [3] Heterogeneous catalysis is involved in the reaction: Reactants are adsorbed on the surface of the catalyst; the reactant molecules are brought closer together with correct orientation and the bonds to be broken are weakened, thus Ea for the reaction is lowered. The products formed desorb and leave the surface of the catalyst. (v) NO2 can also act as a catalyst for reaction 1. Write two equations to show how NO2 acts as a catalyst in reaction 1. [2] NO2 + SO2 → SO3 + NO NO + ½ O2 → NO2 Note: Eqns can be found under Org Book 1 > Alkane > Environmental pollutants. NO2 is acting as a homogeneous catalyst and it should produce a gaseous intermediate during catalysis. [Total : 20]

NJC/H2 Chem Prelim/03/2019 4 2 Lithium oxide, Li 2O, is used in traditional ceramic glazing to create a blue hue with copper on ceramics. (a) Using the data below and any appropriate data from the Data Booklet, construct a labelled energy level diagram to calculate the standard enthalpy change of formation of Li2O(s). / kJ mol–1 first electron affinity of oxygen –141 second electron affinity of oxygen +798 lattice energy of lithium oxide –2863 standard enthalpy change of atomisation of lithium +159 [5]  State symbol  Balanced equation  Correct enthalpy change By Hess’ Law, Hof (Li2O) = 2[Hoatm (Li)] + 2[1st IE (Li)] +Hatm (O2) + 1st Ea (O) + 2nd Ea (O) + L.E. = 2(+159) + 2(+519) + ½ (+496) + (–141) + (+798) + (–2863) = –602 kJ mol–1 0 2 Li (s) + ½ O2 (g) Li2O (s) 2 Li (g) + ½ O2 (g) 2 Li+ (g) + 2 e– + ½ O2 (g) 2 Li+ (g) + 2 e– + O (g) Hof (Li2O) Hoatm (Li)  2 1st IE (Li)  2 ½ BE(O=O) 2 Li+ (g) + e– + O– (g) 2 Li+ (g) + O2– (g) 2nd Ea 1st Ea L.E. of Li2O Enthalpy / kJ mol–1

NJC/H2 Chem Prelim/03/2019 5 [Turn over (b) Lithium oxide is produced during the thermal decomposition of lithium peroxide, Li 2O2, at 450 oC. Li2O2(l) → Li2O(s) + 1 2O2(g) (i) Given that the standard enthalpy change of formation of Li 2O2(l) is –606 kJ mol –1 and using your answer in (a), calculate the enthalpy change for the thermal decomposition of Li2O2. [1] enthalpy change for the thermal decomposition of Li2O2 = Hof (products) – Hof (reactants) = Hof (Li2O) + Hf (O2) – Hof (Li2O) = (– 602) + 0 – (– 606) = + 4 kJ mol–1 (ii) Explain why the thermal decomposition of Li2O2 is feasible. [2] G = H - TS The increase in the entropy (S >0) from the production of O2 gas and the elevated temperature readily offset the slightly endothermic enthalpy change causing G to be negative, hence the reaction is feasible. (c) Table 2.1 gives the melting points of two lithium ionic compounds. compound melting point / oC Li2O 1400 Li2O2 195 Table 2.1 Explain the difference in the melting points. [2] |L.E| ∝ 𝑞+ × 𝑞– 𝑟+ + 𝑟– Both Li2O and Li2O2 have the same Li+ cation, but different anion. Since O22–in Li2O2 has a larger ionic radius as compared to the O2– in Li2O with a smaller ionic radius, the magnitude of the lattice energy of Li2O2 is smaller (or lattice energy of Li2O2 is less exothermic) than