2024 EJC H1 Chemistry Prelims P1 Worked Solutions

2024 JC2 Preliminary Examination H1 Chemistry 8873 Paper 1 Worked Solution 1 2HO 3.6 0.200 mol16.0 2 1.0n == + 3NH HC 100 1.0 0.100 mol1000nn = =  =l 32N O NH H O: : 0.10 : 0.10 : 0.20 1:1: 2 xy n n n == x = 1, y = 2  B 2 r 14 10 11 10.855A =  +  =  C 3 reaction 1: 12 [R] 22 2H O H O −− ⎯⎯⎯ → reaction 2: 10 [O] 2 22H O O − ⎯⎯⎯ → (H2O from MnO – 4) reaction 3: 12 [R] 22 2H O H O −− ⎯⎯⎯ → ; 10 [O] 2 22H O O − ⎯⎯⎯ →  B 4 particle no. of p no. of e– P– 40–21 = 19 19+1 = 20 Q+ 38–18 = 20 20–1 = 19 R2– 39–19 = 20 20+2 = 22 S2+ 41–20 = 21 21–2 = 19  B 5 angle of deflection, qq kmm   =  For 1H+: 115º 15º 1kk ++ =  = + A: ( ) 115º 7.5º2 += + = + B: ( ) 215º 10º3 += + = + C✓: ( ) 215º 5º 6 += + = + D: ( ) 315º 3.75º12 += + = +  C 6  C 7  B 8 Due to smaller size of the F atom, its orbitals are more compact and hence there is greater effective overlap between the orbitals, leading to shorter F –F bond length. This should result in a stronger F –F bond. However, due to the short F –F bond, the lone pairs of electrons on the two F atoms are very close, resulting in severe repulsion between them, causing the F–F bond to be weaker. Despite F having a smaller nuclear charge than Cl, Cl has more core electrons than F that confers shielding, nullifying the larger nuclear charge of Cl.  A 9 HCl and HBr have a simple molecular structure. Boiling point of simple molecule is dependent on the strength of the intermolecular forces of attraction. HCl and HBr do not possess H bonding, which requires H bonded to F, O or N. Due to Br having more electrons than C l, the electron cloud of HBr is larger and more polarisable than that of HC l. This leads to stronger instantaneous dipole -induced dipole attractions between HBr molecules, imparting higher boiling point.  D 10 A✓: As T increased, amount of A remaining at equilibrium increased. This shows that position of equilibrium shifts left, hence backward reaction is endothermic. Forward reaction is thus exothermic. B✓: Because strong acids and bases are fully ionised in water, the enthalpy change of neutralisation will be almost identical, as it refers to the heat evolved when H+(aq) reacts with OH–(aq) to give 1 mole of H2O(l). C: Third I.E. of Al should be the energy needed to convert 1 mole of A l2+(g) into 1 mole of Al3+(g). D✓: Lattice energy involves the formation of ionic bonds between gaseous cations and gaseous anions, giving 1 mole of the crystalline ionic lattice, and so is always exothermic.  C 11 Using Hess’ Law ( ) 1 1 1 26.8 2 16.5 kJ mol 6.2 kJ mol H − −  = − − − =+  D 12 1✓: Ultraviolet light provides the energy needed to break the O–O bond to generate O free radicals: 2✓: Adding up the second and third eqns: O3 + O → 2O2 NO catalyses the reaction between O3 and O. 3: NO itself is oxid