2019 JPJC Prelim H2 Chem P3 ANS

 Jurong Pioneer Junior College [Turn Over JURONG PIONEER JUNIOR COLLEGE 2019 JC2 H2 CHEMISTRY (9729) Preliminary Examination Paper 3 (Suggested Answers) 1 (a) For HI, violet fumes is observed. E(HCl) = +431 kJ mol1; E(HI) = +299 kJ mol1 Since E(HI) < E(HCl), H‒ I bond is weaker and is more easily broken than H‒C l bond. Hence, H I is less thermally stable and decomposes more readily ( i.e. HI decomposes upon contact with the hot wire). OR radius of Cl = 0.099 nm; radius of I = 0.133 nm Since I has a larger radius than C l, H‒I bond is longer and weaker so it is more easily broken than H‒C l bond. Hence, H I is less thermally stable and decomposes more readily (i.e. HI decomposes upon contact with the hot wire). (b) (i) MnO2(s) + 4H+(aq) + 2e  Mn2+(aq) + 2H2O(l) (ii) E cell = E red  E ox (+1.23) – (+1.36) = 0.13 V < 0 (not energetically feasible) Overall eqn: MnO2 + 4H+ + 2Cl  Mn2+ + 2H2O G = zFE cell where z = total number of electrons transferred per overall eqn = (2)(96500)(0.13) = +25090 J ol1 = +25.1 kJ mol1 Since G > 0, the reaction is not energetically feasible (or not spontaneous) and thus unlikely to occur. (iii) When concentrated HC l is used, [H +] is high and this causes the position of equilibrium of MnO2 + 4H+ + 2e  Mn2+ + 2H2O to shift right so as to react away some H + ions. This makes E(MnO2/Mn2+) becomes more positive than +1.23 V such that the Ecell becomes positive and  energetically feasible. When concentrated HC l is used, [C l] is high and this causes the position of equilibrium of Cl2 + 2e  2Cl to shift left so as to react away some C l ions. This makes E(Cl2/Cl) becomes more negative (or less positive) than +1.36 V such that the Ecell becomes positive and  energetically feasible. The continuous removal of Cl2 gas from the reaction mixture as it is evolved shifts the position of equilibrium of C l2 + 2e  2Clto the left to form back some C l2. This makes E(Cl2/Cl) becomes more negative (or less positive) than +1.36 V such that the Ecell becomes positive and  energetically feasible.

2  Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2019 (c) (i) Type of mechanism: electrophilic addition (ii) Propene can use its two  electrons to form a dative bond with H+ (or to accept a H+). (iii) Nucleophilic substitution (iv) p–p orbital overlap results in the delocalisation of lone pair on O of CO into the  electron cloud of C=O, forming a partial double bond character in the C–O bond. This strengthens the C –O bond in –CO2H, making it difficult to break. Hence, carboxylic acid is unreactive towards HCl. (d) (e) (i) valalavalserglyargasnleuglyval or leuglyval valalavalserglyargasn or serglyargasnleuglyval alavalser (ii) The lone pair on N a is in the hybridised sp 2 orbital which is perpendicular to the p–