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 ×