JPJC Prelim 2025 Paper 1 Worked Solutions H2 Chem

© Jurong Pioneer Junior College [Turn Over ANSWER SCHEME 1 C ( ) 5 -3 mpV= RTMr 18.010 ×5.7×10 = (8.31)(273)Mr Mr = 71.6 Let a be the %mass of 35Cl2 ( 𝒂 𝟏𝟎𝟎 × 𝟕𝟎) + [𝟏 − 𝒂 𝟏𝟎𝟎 × 𝟕𝟒] = 𝟕𝟏. 𝟔 a = 60 2 B From the first 2 properties, E must be a main group metal. Hence, option C and D are incorrect. Since it was obtained by removing electrons from the same orbital, E must be either a metal in Group 1 or 2. 3 A Since its shape is trigonal pyramidal, that must mean that around O there are 3 bp and 1lp. The 3 bp are formed between one electron from F and 1 from O and the lp belongs to O. With only 5 valence electrons around O than the usual 6, the entire ion is short of 1 electron and hence n = 1 4 C CH4 + 2O2 → CO2 + 2H2O 10 cm3 CH4  10 cm3 CO2 C2H6 + 3.5O2 → 2CO2 + 3H2O 20 cm3 C2H6  40 cm3 CO2 5 B Amount of MnO4- = 37.5 0.021000 = 7.5 ×10-4 MnO4- + 8H+ + 5e ⇌ Mn2+ + 4H2O Amount of MnO4- : amount of electrons: amount of G ion : 7.5  10−4 : 3.75  10−3 :1.25  10−3 3: 1 Given that ion of G is oxidised to GO3-, O.N. +5, Original O.N of ion of G is +5 –3 = +2 50 cm3 CO2

2 © Jurong Pioneer Junior College 9729/01/J2 PRELIM EXAM/2025 6 B Assuming ideal gas behaviour (i.e. pV = nRT), 1) value of pV of L = nLRTL = 3x value of pV of J = nJRTJ = x 2) J deviates from Ideal gas behaviour more than L, indicating more significant/stronger intermolecular forces of attraction between J molecules than L (i.e. 𝐧𝐋𝐓𝐋 𝐧𝐉𝐓𝐉 = 𝟑) 1: LL JJ nT nT = ( ) ( ) + + 0.5 273 50 0.5 273 25 = 1.08 ✓2: J=NH3 (stronger hydrogen bonds between NH3 molecules) L = CH4 (weaker id-id attractions between CH4 molecules) LL JJ nT nT = ( ) ( ) + + 1.5 273 25 0.5 273 25 = 3 3: J = H2 (weaker id-id attractions between CH4 molecules) L=SO2 (stronger pd-pd bonds between NH3 molecules) LL JJ nT nT = ( ) ( ) + + 0.75 273 25 0.25 273 25 = 3 7 A K: 1s2 2s2 2p6 3s2 3p6 4s1 Cl: 1s2 2s2 2p6 3s2 3p5 ✓1: K has one more quantum shell, valence electrons are further away from nucleus. Increase in shielding effect by inner shell electrons outweighs the increase in nuclear charge hence, decreased attraction between nucleus and valence electrons / decreased n uclear attraction. Thus, the atomic radius of potassium is greater than atomic radius of chlorine atom. 2: Similarly, the 4s electron is further away from the nucleus, decreased nuclear attraction between nucleus and valence electrons, thus, less energy is required to remove an outermost/valence electron. 3 K+: 1s2 2s2 2p6 3s2 3p6 Cl-: 1s2 2s2 2p6 3s2 3p6 Both are isoelectronic, with potassium having a bigger nuclear charge. Hence, there is great nuclear attraction between nucleus and valence electrons. Thus the ionic radius of potassium is smaller than that of chlorine.

3 © Jurong Pioneer Junior College 9729/01/J2 PRELIM EXAM/2025 [Turn Over 8 D reaction 1 Br2(aq) + 2NaOH(aq) → NaBr(aq) + NaOBr(aq) + H2O(l) 0 -1 +1 reaction 2 3NaOBr(aq) → 2NaBr(aq) + NaBrO3(aq) +1 -1 +5 Reaction 1: Br in Br2 is oxidised to OBr- and reduced to Br- Reaction 2: Br in OBr- is oxidised to BrO3- and reduced to Br-. Hence both are disproportionation reactions. 9 D A: Dipole moments cancel out. B: Dipole moment is smaller than that in D since S is less electronegative than O. C: Dipole moment is smaller than that in D as the net dipole moment of C–F bonds reduces the dipole moment of C=O bond D: 10 B sodium magnesium oxygen bromine Charge +1 +2 -2 -1 Ionic size 0.186 0.160 0.140 0.195 MgO MgBr2 Na2O NaBr LE  13.33 5.63 6.13 2.62 Based on the above formula, MgO will be 2nd least exothermic 11 C H = (5(BE(N=N) + BE(N=O) + 2BE(C-H) + BE(CC)) – (2BE(O-H) + 4BE(C=O) + 5(NN)) = (5(418 + 686) + 2(410) + (840)) – (2(460) + 4(805) + 5(944)) = - 1680 kJ mol-1 q +  q− LE  r+ + r−

4 © Jurong Pioneer Junior College 9729/01/J2 PRELIM EXAM/2025 12 C Since average IMF in water are much stronger than the average IMF in steam, and average IMF in ice are slightly stronger than the average IMF in water ✓1: The numerical value of Hb is greater than Hm 2 (H2O(l) = H2O(g)) Hb is positive as the energy required to break stronger IMF in liquid water is more than energy released when forming IMF between steam molecules (H2O(s) = H2O(l)), Hm is positive as the energy required to break stronger IMF in solid water is more than energy released when forming IMF between liquid water molecules. ✓3: The intermolecular forces in ice and water are similar – hydrogen bond 13 A Let the rate equation be rate = k[N]n [P]q. Comparing Experiment 2 and 3, keeping [N] constant, ( ) ( ) ( ) ( ) 3 3 0.006 0.86.4 10 1.6 10 0.006 0.4 nq nq k k − −  =  q = order of reaction w.r.t [P] = 2 (correct; reject Option D) Comparing Experiment 1 and 2, ( ) ( ) ( ) ( ) − −  = 13 31 0.006 0.41.6 10 1.6 10 0.003 0.4 n n k k  n = order of reaction w.r.t [N] = 0 Since the order of reaction w.r.t [N] is not one, the half–life of N is not constant. (correct; reject Option C) The rate equation is rate = k[P]2 and hence, the units of k = mol‒1 dm3 min−1. (correct; reject Option B). The rate equation implies that the rate–determining step involves 2 units of P only. Hence, this also implies that N is involved in other steps and thus, the reaction is not a one-step reaction. 14 D ×A Reaction rate is defined as the speed at which a chemical reaction proceeds, defined as the change in concentration of a reactant or product per unit of time. ×B At higher concentration, particles are closer together per unit volume. ×C At higher concentration, the kinetics energy of the particles remain the same. ✓D With more collisions per second, the rate will increase.

5 © Jurong Pioneer Junior College 9729/01/J2 PRELIM EXAM/2025 [Turn Over 15 A ROH(l) + CH3COOH(l) = CH3COOR(l) + H2O(l) Initial amt 3 2 0 1 change -0.5 -0.5 +0.5 +0.5 Eqm amt 2.5 1.5 0.5 1.5 Eqm [ ] 2.5 V 1.5 V 0.5 V 1.5 V 32 c 3 [CH COOR][H O]K= [CH COOH][ROH] ✓A c 0.5 1.5[ ][ ] K= 1.5 2.5[ ][ ] VV VV = 0.2 ×B Ho < 0 i.e exothermic, When temperature decrease, POE will shift to the right, favouring the exdothermic reaction to release heat. Hence, the [CH3COOH] will reduce below 1.5 mol. ×C KC has no units. ×D The equilibrium amount of ester is the not the same as the equilibrium amount of water. (See working above) 16 B 2 CO (g) + O2 (g)  2CO2 (g) ∆H < 0 At time = t 1, shape of graph shows a gradual increase in CO and decrease in CO 2, indicating POE shifts to the left. Possible changes include • increasing temperature as increasing temperature will favour endothermic reaction, i.e. backward reaction thus POE shifts left • remove O2 as removing O2 will result in POE shifting left. (Cancel A and D) At time = t 2, shape of graph shows a sharp increase in both CO and CO 2, indicates a decrease in the volume of system. When volume of system decrease, pressure of system increase. POE shifts to the right side with lesser mol to reduce the pressure. (Cancel C) Addition of catalyst will not result in any equilibrium shift. Addition of an inert gas at constant volume will not result in any equilibrium shift as partial pressure of the other gases remains the same.

6 © Jurong Pioneer Junior College 9729/01/J2 PRELIM EXAM/2025 17 D Amount of NH4+ = 100 0.101000 = 0.01 mol (excess) Amount of OH- = 40 0.151000 = 0.006 mol In resulting solution, Amount of NH4+ left 0.01 – 0.006 = 0.004 mol Amount of NH3 formed = 0.006 mol    + 4 b 3 [NH ]pOH=pK +lg [NH ]   0.004pOH=4.75+lg 0.006 =4.57 pH = 14 – 4.57 = 9.43 18 B ×A Since solubility of Group 2 hydroxides