RI Tutorial 5a Energetics I (suggested solutions)

© Raffles Institution 2024 1 Tutorial 5a Chemical Energetics 1: Suggested Solutions to Practice Questions 11 (a) Hess’ Law states that the enthalpy change of a reaction is determined by the initial and final states of the system and is independent of the pathways taken. It allows one to calculate Hr of a reaction that is difficult or impossible to perform in the laboratory. (b) Hfo 3C(s) + 4H 2(g) C 3H8(g) 3Hfo[CO2(g)] 4Hfo[H2O(l)] Hco[C3H8(g)] 3 CO 2(g) + 4 H2O (l) By Hess’ Law, Hfo [C3H8(g)] = 3 Hfo [CO2(g)] + 4 Hfo [H2O(l)] – Hco [C3H8(g)] = 104 kJ mol1 (c) Total amount of propane and butane present = 5.60 / 22.7 = 0.2467 mol Let the amt of propane be x mol, then amt of butane is (0.2467 – x) mol. (Note: Use H = – ୯ ୬ఽ × a , q = n(H) / a) total q = x ( Hco [C3H8(g)]) + (0.2467 x)(Hco [C4H10(g)]) = +654 kJ ⇒ x (2220) + (0.2467– x)(2877) = +654 kJ ⇒ x = amt of propane = 0.08486 mol amt of butane = 0.2467 – 0.08486 = 0.1618 mol By Avogadro’s law, % composition of propane = (0.08486 / 0.2467) x 100 = 34.4% % composition of butane = (0.1618 / 0.2467) x 100 = 65.6% 12 (a) (i) K 2CO3(s) + 2HCl(aq)  2KCl(aq) + CO2(g) + H2O(l) (ii) q = 30.0 x 4.18 x 5.20 = +652.1 J (or heat produced by rxn = 652.1 J) Molar mass of K2CO3 = 2(39.1) + 12.0 + 3(16.0) = 138.2 g mol1 Amt of K2CO3 = 2.76 / 138.2 = 0.01997 mol Amt of HC l = 30.0/1000 x 2 = 0.0600 mol; hence, HCl is in excess. Enthalpy change of reaction, H1 = 652.1/0.01997 =32650 J mol1 = 32.7 kJ mol1 (iii) Amt of HCl = 30.0/1000 x 2 = 0.0600 mol. Since HC l is in excess, its concentration need not be specified accurately. (b) (i) KHCO 3(s) + HCl(aq)  KCl(aq) + CO2(g) + H2O(l) (ii) q = 30.0 x 4.18 x (−3.70) = −464 J (or heat absorbed by rxn = 464 J) Molar mass of KHCO3 = 39.1 + 1.0 + 12.0 + 3(16.0) = 100.1 g mol1 Amt of KHCO 3 = 2.00 / 100.1 = 0.01998 mol Amt of HC l = 30.0 / 1000 x 2 = 0.0600 mol; hence, HCl is in excess. Enthalpy change of reaction, H2 = −(−464.0 /0.01998) = +23220 J mol1 = +23.2 kJ mol1 + 5 O2 (g) + 3 O2 (g) + 2 O 2 (g)

© Raffles Institution 2024 2 (c) 2KHCO3(s) K 2CO3(s) + H2O(l) + CO2(g) heat +2HC l(aq) +2HCl(aq) 2H2 H1 2KC l(aq) + 2H2O(l) + 2CO2(g) By Hess’ Law, 2Hr = 2H2  H1 = +79.1 kJ mol1 Enthalpy change for the decomposition of 1 mole KHCO3 = +39.6 kJ mol1 (to 3 sf) 13 (a) Standard enthalpy change of neutralisation is the energy change when an acid and a base react to form one mole of water at 298 K and 1 bar. (b) Final temperature as extrapolated on graph to time of mixing at 2 min = 10 oC Maximum temperature fall by extrapolation = 10.0 oC Temperature change, ∆T = (10.0 – 20.0) ºC = −10.0 ºC q = 50 × 4.18 × –10.0 = −2090 J (or heat absorbed by reaction = 2090 J) Amt of NaHCO3 = 3.50 / 84.0 = 0.04167 mol Limiting reagent = NaHCO3 since amt of HCl = 0.05 mol Enthalpy change of reaction = –(–2.090 kJ) / 0.04167 = + 50.2 kJ mol1 Assumption made  Heat gain from the surroundings is accounted for from the extrapolation of the graph.  Heat capacity of the cup/calorimeter is negligible. 2Hr Note: ∆T in K is the same as ∆T in oC

© Raffles Institution 2024 3 (c) CO2(g) + NaOH(aq) NaHCO3(aq) Hr = 57.3  (50.2) + (17.6) =  89.9 kJ mol1 (d) HCN is a weak acid and energy is required to ionise it to produce one mole of H+ ions before neutralisation. NH3 is a weak base and energy is required to ionise it to produce one mole of OH ions before neutralisation can take place. Enthalpy of ionisation of the weak acid and weak base are endothermic and that resulted in a less exothermic enthalpy change of neutralisation of the weak acidweak base as compared to that of a strong acidstrong base. 14 (a) The bond energy of a CH bond is the average energy absorbed when 1 mole of CH bonds of methane are broken in the gaseous state. CH4(g)  C(g) + 4H(g) H = 4 BE(CH) (b)(i) Method 1: Constructing energy cycles and using Hess’ Law BE(C C) + 6BE(CH) C 2H6(g) 2C(g) + 6H(g) Hfo = 84.7 kJ mol1 2 Hatomo (C) + 3 BE(H-H) 2C(s) + 3H2(g) BE(CC) + 6BE(CH) = Hfo + 2 Hatomo (C)+ 3 BE(HH) BE(C C) = (84.7) + 2(+715) + 3(436)  6(410) BE(C C) in C2H6 = + 363 kJ mol1 Method 2: Hfo = energy consumed in bond breaking + energy released in bond formation 84.7 = 2 Hatomo (C)+ 3 BE(HH)  [BE(CC) + 6 BE(CH)] BE(C C) in C2H6 = + 363 kJ mol1 Hydrocarbon C2H6 C2H4 C2H2 BE(carbon-carbon) / kJmol-1 +363 +610 +819 Comments:  Since bond energies are a reflection of the bond strengths, Bond strength: CC > C=C > CC  Bond energy of C=C is not twice that of CC  -bond is weaker than a -bond. (b)(ii) The C–C bond energy found in the Data Booklet is an average value and not specific for the C–C bond in ethane. Draw similar energy cycles for C2H4 and C2H2. NaCl(aq) + H2O(l) + CO2(g) +HCl(aq) 57.3 kJ mol1 Hr NaHCO3(s) +HCl(aq) +17.6 kJ mol1 +50.2 kJ mol1 + aq Note that this is also ΔHatom [CH4(g)]

© Raffles Institution 2024 4 15 (a) When an ionic solid dissolves in water, two enthalpy terms are involved. 1. The ions must be separated from the ionic lattice. The energy required is the lattice enthalpy (LE). It is also called the lattice dissociation energy. 2. The specific gaseous ions interact with water molecules. Energy is released (Hhyd) due to the formation of ion-dipole interactions between the gaseous ions and water. Hsoln = LE + Hhyd energy/ kJ mol1 (b) LiCl(s) + aq  Li+(aq) + Cl(aq) Hsoln (LiCl) Hsoln (LiCl) = Hhyd (Li+) + Hhyd (Cl)  LE (LiCl) = 40 kJ mol1 (c) Li+ has a smaller radius and a higher charge density than K+. Since LE is inversely proportional to interionic distance, the lattice energy of LiCl is slightly more exothermic than that of KCl. However, since the magnitude of hydration energy is proportional to charge density, Li+ will have a much more exothermic hydration enthalpy compared to K+. Therefore, the enthalpy change of solution of LiCl is more exothermic than that of KCl. 16 (a) 2Cu+(g) + O2(g) 2Cu+(g) + O(g) + 2e ½ BE [O2] = ½(+496) kJ mol1 1st EA [O] =  141 kJ mol1 2Cu+(g) + O(g) + e 2nd EA [O] = +791 kJ mol1 2Cu+(g) + ½ O2(g) + 2e 1st IE [Cu] x 2 = 2(+745) kJ mol1 2Cu(g) + ½ O2(g) LE [Cu2O] Hatom [Cu] x 2 = 2(+339) kJ mol1 2Cu(s) + ½ O2(g) Hf [Cu2O] =  166 kJ mol1 Cu2O(s) Lattice energy of Cu2O = 3232 = 3230 kJ mol1 (b) The experimental lattice energy of CuO is expected to deviate more from the theoretical value compared to that of Cu2O. Cu2+ has a higher charge but smaller size than Cu+, hence Cu2+ has a higher charge density and greater polarizing power than Cu+, leading to greater covalent character in CuO than in Cu2O. The greater covalent character in CuO led to the greater deviation. LiCl(s) Li +(aq) + Cl(aq) Li+(g) + Cl(g) LE (LiCl) = 843 Hhyd (Li+) + Hhyd (Cl) = 883 Hsoln (LiCl) Energy / kJ mol1 0 +aq

© Raffles Institution 2024 5 (c) Cu2O(s) CuO(s) + Cu(s) LE (Cu2O) = 3232 kJ mol1 LE (CuO) =  4143 kJ mol1 2Cu+(g) + O2(g) Cu 2+(g) + O2(g) + Cu(s) 2nd IE (Cu) 1 st IE (Cu)   Hatom = +339 kJ mol1 = +1960 kJ mol1 = +745 kJ mol1 Cu+(g) + Cu2+(g) + O2(g) +e Cu 2+(g) + O2(g) + Cu(g)  H = 35 kJ mol1 (d) q = 60.0 x 4.18 x 8.9 = +223