JPJC 2026 Chemical Energetics Revision Answer

2026 JPJC JC1 H2 Chemistry (9476) 1 Topic 5: Chemical Energetics JURONG PIONEER JUNIOR COLLEGE Topical Revision for Physical Chemistry (Topic 5: Chemical Energetics) FOUNDATION KIT 1 [PJC MYE 2011/P1/Q13] Ans: B NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) Hneutralisation From the mole ratio, 25 cm 3 of 1.0 mol dm –3 NaOH(aq) is completely neutralised by 25 cm3 of 1.0 mol dm–3 HCl(aq). n(H2O) formed = n(NaOH) reacted = 25 1.0 1000 = 0.0250 mol For 0.025 mol of water formed, q = mcT = (25 + 25)  4.18  6.8 J = 1421. 2 J From the mole ratio, 50 cm 3 of 0.5 mol dm –3 NaOH(aq) is completely neutralised by 50 cm3 of 0.5 mol dm–3 HCl(aq). n(H2O) formed = n(NaOH) reacted = 50 0.5 1000 = 0.0250 mol Since same amount of water is formed, q is constant.  q = mcT = 1421. 2 J (50 + 50)  4.18  T = 1421.2  T = 3.4 oC 2 [PJC Promo 2011/P1/Q12] Ans: A Given Hsoln(NaF) = +71 kJ mol–1 = + qn(NaF) q = 71  n(NaF) kJ = 71  4.2 (23.0 +19.0) = 7.1 kJ = 7100 J Given c = 4.2 J cm–3 K–1  q = vcT 7100 = 200  4.2  T  T = 8.5 oC Since H is positive, heat is absorbed and temperature falls.  the final temperature should be lower than the initial temperature.  Final temperature = 25 – 8.5 = 16.5 oC

2026 JPJC JC1 H2 Chemistry (9476) 2 Topic 5: Chemical Energetics 3 [PJC Promo 2013/P1/Q11] Ans: D By Hess’ Law, Hf(CO) = Hc(C) – Hc(CO) 4 [PJC Promo 2012/P1/Q8] Ans: B By Hess’ Law, Hc(CH3OH) = – (–239) + (–394) + 2(–286) = –727 kJ mol–1 5 [PJC MYE 2010/P1/Q14] Ans: B By Hess’ Law, Hr = – (–20.5) + (–243.0) = –222.5 kJ mol–1 Or H2S(g) + ½ O2(g)  H2O(g) + S(s) Hf : (–20.5) ½ (0.0) (–243.0) (0.0) By formula, Hr = mHf(products) – nHf(reactants) = (–243.0) – (–20.5) = –222.5 kJ mol–1 CO(g) C(s) + ½ O2(g) Hf CO2(g) + ½O2(g) Hc(CO) + ½ O2(g) Hc(C) CH3OH(l) + O 2(g) CO2(g) + 2H2O(l) Hc C(s) + 2H2(g) + 2O2(g) (–239) (–394) 2(–286) H2S(g) + ½O2(g) H2O(g) + S(s) Hr H2(g) + ½ O2(g) + S(s) (–20.5) (–243.0)

2026 JPJC JC1 H2 Chemistry (9476) 3 Topic 5: Chemical Energetics 6 [AJC Promo 2016/P1/Q20] Ans: C 5N=N=O(g) + H–CC–H(g)  H–O–H(g) + 2O=C=O(g) + 5NN(g), Hr = –1668 kJ mol–1 Bonds broken Bonds formed 5 N=N 5(+418) 2 O–H 2(+460) 5 N=O 5 E(N=O) ? 4 C=O 4(+805) 2 C–H 2(+410) 5 NN 5(+944) 1CC 1(+840) Total 5E(N=O) +3750 Total +8860 Hr = E(all bonds in reactants) – E(all bonds in products) = –1668 5E(N=O) +3750 – (+8860) = –1668  E(N=O) = +688.4 kJ mol–1 7 [PJC MYE 2011/P1/Q20] Ans: D Ca(s) + 2H2O(l)  Ca(OH)2(s) + H2(g) Hf : (0.0) 2Hf(H2O(l)) Hf(Ca(OH)2(s)) (0.0) By formula, Hr = mHf(products) – nHf(reactants) Hr = Hf(Ca(OH)2(s)) – 2Hf(H2O(l)) 8 [PJC Promo 2011/P1/Q27] Ans: A – LE (MgCl2) = +2526  LE (MgCl2) = – 2526 kJ mol–1 (Option 1 is correct.) Hsoln(MgCl2) = – LE (MgCl2) + Hhyd(Mg2+) + 2Hhyd(Cl–) = +2526 –1890 + 2(–384) = – 132 kJ mol–1 (Option 3 is correct.) MgCl2(aq) MgCl2(s) –1890 + 2(–384) kJ mol–1 = Hhyd(Mg2+) + 2Hhyd(Cl–) +2526 kJ mol–1 = – LE (MgCl2) enthalpy Mg2+(g) + 2Cl – (g) can be determined in the lab to be calculated will need this information Hsoln(MgCl2) (Option 2 is correct.) ∆Hr

2026 JPJC JC1 H2 Chemistry (9476) 4 Topic 5: Chemical Energetics 9 [AJC Promo 2016/P1/Q21] Ans: B The enthalpy diagram indicates that the enthalpy change is exothermic. 1. Enthalpy change of solution involves dissolving of an ionic compound completely in water to give aqueous ions such that no further heat change occurs upon adding more water. Hsoln = – LE + Hhyd(ions) This process can be exothermic or endothermic , depending on whether the hydration energy released is sufficient to supply the energy needed to break the ionic bonds between ions in the ionic compound. 2. Enthalpy change of hydration involves formation of aqueous ions from its gaseous ions via iondipole interactions which release energy, such that no further heat change occurs upon adding more water. Hence this process is always exothermic. 3. Enthalpy change of neutralisation involves formation of H2O from H+(aq) & OH(aq). This process is always exothermic due to bond formation between H+ and OH–. 10 [PJC Promo 2013/P1/Q28] Ans: A From the Born-Haber cycle, These are the information needed to estimate the lattice energy of CaF3:  enthalpy change of formation of CaF3(s)  enthalpy change of atomisation of calcium  enthalpy change of atomisation of fluorine or bond energy of F–F bond  first and second ionisation of calcium  first electron affinity of fluorine Ca(s) + F 2(g) CaF3(s) Ca(g) 3F(g) Ca2+(g) + 3F–(g) LE 1st + 2nd IE (Ca) 31st EA Ha Hf 3Ha or E(F–F) ve ve ve

2026 JPJC JC1 H2 Chemistry (9476) 5 Topic 5: Chemical Energetics 11 [PJC MYE 2012/P1/Q20] Ans: B 1 () P(red) has a higher energy content than P(black). 2 () Lower energy  more stable Since energy level of P(black) < P(red) < P(white), relative stability of P(black) > P(red) > P(white) 3 () P(black)  P(white) is an endothermic process. 12 [JPJC Promo 2020/P1/Q14] Ans: A 1 () Since graphite has a lower energy content, graphite is more energetically stable than diamond. 2 () Hc of diamond is more exothermic than Hc of graphite. energy / kJ mol–1 combustion product −2941 P(red) −2971 −2925 energy / kJ mol1 C(diamond) H > 0 C(graphite) energy / kJ mol1 C(diamond) H > 0 C(graphite) CO2(g) Hc (graphite) Hc (diamond) P(white) P(black) H > 0 (endothermic)

2026 JPJC JC1 H2 Chemistry (9476) 6 Topic 5: Chemical Energetics 3 () Ha of diamond is less endothermic than Ha of graphite. 13 [PJC MYE 2011/P1/Q15] Ans: D During phase change (at dynamic equilibrium), G = 0. H2O(s)  H2O(l) H = +6.0 kJ mol–1 Ice melts at 0 C. G = H – TS 0 = +6.0 – (0 + 273) S S = +0.0220 kJ mol–1 K–1 = +22.0 J mol–1 K–1 (for melting of 1 mole of ice) amount of ice melted = 5418.0 = 3.00 mol  S = (+22.0)  3.00 = +66 J K–1 14 [JPJC Promo 2019/P1/Q10] Ans: B During phase change (at dynamic equilibrium), G = 0. G(l)  G(g) H = +10.0 kJ mol–1 G boils at 266 K. G = H – TS 0 = +10.0 – (266) S Svapouration = +0.0376 kJ mol–1 K–1 = +37.6 J mol–1 K–1 Condensation is the reverse process of vapourisation. Hence, Scondensation = –37.6 J mol–1 K–1 15 Ans: D Mg(s) + 2H2O(l)  Mg(OH)2(s) + H2(g)  H is negative (< 0) since heat is released upon mixing which warms up the food.  S is positive (> 0) since there is an increase in the disorderliness of the system as the reaction proceeds with an increase in number/amount of gas particles from 0 to 1 mol.  G is negative ( < 0 ) since the reaction is energetically feasible as heat is spontaneously released upon mixing. energy / kJ mol1 C(diamond) H > 0 C(graphite) C(g) Ha (graphite) Ha (diamond)

2026 JPJC JC1 H2 Chemistry (9476) 7 Topic 5: Chemical Energetics 16 Ans: D Reactions with negative H and positive S are spontaneous at all temperatures since G < 0 at all T. G = H – T S –ve +ve –ve Options A and C are rejected since these reactions have positive H. For Option B, S < 0 since there is a decrease in disorderliness of the system as the reaction proceeds with a decrease in the number/amount of gaseous particles from 9 mol to 4 mol. F