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 oxidised by O3 to NO2 in the second equation, hence acting as a reducing agent.  B 13  C 14 A catalyst decreases the activation energy of the forward and backward reaction by the same amount, C. Hence rate constant, for the forward and backward reaction increases by the same factor, C RTe : a a aE C E E CC RT RT RT RT RTk Ae Ae e Ae −− − + − = = =     B 15 1✓: As reaction is exothermic, the eqm shifts left to absorb heat when temperature increases, causing the yield to decrease. Rate of both forward and backward reaction increases (but to different extent) since the rate constant of both forward and backward reaction increases (by different factor). 2: Increase in [H2S] causes the eqm to shift right by LCP, thereby leading to an increase in yield. The rate of forward reaction will increase with [H2S]. 3✓: Pressure  concentration. Rate of reaction decreases with pressure. When pressure decreases, eqm shifts to the left with more gaseous particle, leading to a decrease in yield.  A 16 A✓:       4 4 22 c 2 2 42 CO H 0.1 0.1 0.010.1 0.1CH H O K = = =  B:      2 2 3 c 3 3 22 NH 0.1 1000.1 0.1NH K = = =  C:      2 2 c 22 H 0.1 1H 0.1 0.1K = = =  I I D:     24 c 2 2 2 NO 0.1 100.1NO K = = =  A 17 A: Na2CO3 is a Brønsted base as it accepts H+ from HCl. B: NH3 is a Brønsted base as it accepts H+ from HNO3. C: H2O is a Brønsted base as it accepts H+ from HCl. D✓: KOH is an Arrhenius base as it contains a OH group and undergoes dissociation to give OH – ions, which then reacts with the H+ from HNO3.  D 18 24H SO 20 0.1 0.00200 mol1000n =  = NaOH 80 0.1 0.00800 mol1000n =  = H2SO4 + 2NaOH → Na2SO4 + 2H2O NaOH remaining 0.00800 2 0.00200 0.00400 mol n = −  = 30.00400OH 0.0400 mol dm20 80 1000 −− == + ( ) pH 14 pOH 14 lg OH 14 lg 0.0400 12.6 −= − = +  = + =  B

19 The major buffer in human blood is the H2CO3/HCO – 3 buffer. Since lactic acid is produced during exercise, to maintain the pH, H 3O+ from lactic acid must be removed by the HCO – 3.  C 20 A: Atomic radius generally decreases across Period 3 as the shielding remains constant while nuclear charge increases, resulting to an increase in effective nuclear charge. B: Na, Mg and A l are metals with high electrical conductivity, while Si is a metalloid with low conductivity. C: Although first I.E. generally increase across Period 3. However, there is a dip in first I.E. at Group 13 (as electrons are removed from 3p rather than 3s orbital) and Group 16 (due to interelectronic repulsion between pair 3p electrons). D✓: Melting point increases from Na to Al as the number of delocalised electrons increase leading to stronger metallic bonds. Si has a giant molecular structure with strong Si–Si bond in a lattice structure, imparting the highest melting point.  D 21 1: AlCl3 but not Al2O3 hydrolyses in water. 2: Al2O3 being an amphoteric oxide reacts with HC l. However, A lCl3 does not react with HCl. 3✓: Al2O3 is ionic but AlCl3 is simple covalent.  D 22 The stronger the halide ion as a reducing agent, the larger the number of halogen it can reduce. X– can reduce both Y2 and Z2, while Z– can only reduce Y2, and Y– cannot reduce any of the halogen. Hence X– is the strongest, while Y– is the weakest reducing agent.  B 23 Butanone and butanal are constitutional isomers, so they must have the same empirical and molecular formula, only differing in their structures , i.e. structural and skeletal formula, which shows that they are different compounds.  B 24 Alkanes’ ‘lack of affinity’ is due to the molecules being non -polar, hence do not attract negative nucleophiles nor positive electrophiles, and is unreactive.  C 25 A✓: Combustion 2CH3CH2CH2OH + 9O2 → 6CO2 + 8H2O B✓: Elimination C: D✓: Oxidation  C 26  D 27  C 28  C 29 A: Hydrogen bonds required H bonded to F, O or N, which is not present in the polyester chain. B✓: Softening of a thermoplastic involves weakening of the intermolecular forces of attraction between the chains, which in these case are instantaneous dipole-induced dipole and permanent dipole -permanent dipole attractions (due to polar ester –CO2– group). C: Softening of a thermoplastic does not involve breaking the covalent bonds. D: Softening of a thermoplastic does not involve breaking the covalent bonds.  B 30 Nanoparticles are materials which have structured components with at least one dimension in the size range between 1 and 100 nm. 1: 8.7 m = 8700 nm; 2.2 m = 2200 nm 2✓: 50 nm is between 1 and 100 nm 3: 2.5 m = 2500 nm 4: 0.5 m = 500 nm  A Answer Key Qn Ans Qn A