RI 05a Energetics Part 1 Tut(ans to self-check)

© Raffles Institution 2024 7 Suggested Answers to Tutorial 5a Chemical Energetics 1: SelfCheck Questions 1 Refer to lecture notes. Please give specific definition when asked for the enthalpy change for a specific substance. Example: (a) Standard enthalpy change of formation of CH3OH(l) is the energy change when 1 mole of CH3OH(l) is formed from its constituents elements, C(s), O2(g) and H2(g) at 298 K and 1 bar. C(s) + ½ O2(g) + 2H2(g)  CH3OH(l) Hf [CH3OH(l)] 2 No. of moles of butane = 1.2/24.0 = 0.0500 mol Heat produced by combustion = −Hc(butane) x nbutane = −(−3000x103) x 0.0500 = + 1.50x 105 J Given 80% efficiency, heat absorbed by water = 80/100 x 1.50 x 105 = +1.20 x 105 J Applying q = mcT +120000 = m x 4.18 x (100 – 20) Mass of water heated, m = 359 g 3 2 1N2(g) + 2 3H2O(g)  NH3(g) + 4 3O2(g) ∆Hr = [2 1BE(NN) + 3BE(O-H)] – [3BE(N-H) + 4 3BE(O=O)] = [ 2 1(944) + 3(460)] – [3(390) + 4 3(496)] = +310 kJ mol1 [Note: whenever only bond energies are used, the species must be in the gaseous states.] 4 Note: This question is different from example in Section 3.7 of lecture notes. The heat absorbed by the copper calorimeter was not ignored here since the specific heat capacity of copper was given in the question. Method 1: Heat capacity of apparatus, C = 120 x 0.387 + 150 x 4.18 = 673.44 J K−1 Heat absorbed by the apparatus = C∆T = 673.44 x (85−20) = +43774 J Method 2: Total heat absorbed = q(calorimeter) + q(water) = 120 x 0.387 x (85−20) + 150 x 4.18 x (85−20) = +43774 J Given that efficiency is 75%, heat producted by combustion = +43774/0.75 = +58365 J Mass of ethanol burnt = 43.56 – 41.36 = 2.20 g Amount of ethanol burnt = 2.20 / 46.1 = 0.047722 mol Enthalpy change of combustion of ethanol = −58365 / 0.047722 = −1220 kJ mol–1 Note: ∆T in K is the same as ∆T in oC

© Raffles Institution 2024 8 5 Hsolno(CaBr2) CaBr2(s) Ca 2+(aq) + 2Br (aq) LE(CaBr2) Hhydo(Ca2+) 2 Hhydo(Br) Ca 2+ (g) + 2Br (g) Hsolno(CaBr2) = Hhydo(Ca2+) + 2Hhydo(Br) – LE(CaBr2) = −1591 + 2(−284) −(−2132) = −27 kJ mol–1 8 D PCl5(g)  PCl3(g) +Cl2(g) ∆ H ∆H = BE of bonds broken in reactants  BE of bonds formed in products = 2BE(P−Cl) − BE(Cl−Cl) = 2 x 330 – 240 = +420 kJ mol−1 9 C + + q qLE r r     The magnitude of lattice energy is proportional to product of ionic charges and inversely proportional to the sum of ionic radii (or interionic distance). 6 C Recall: Enthalpy change of formation of carbon monoxide is the energy change when one mole of gaseous carbon monoxide is formed from its constituent elements, C(s) and O2(g), in their standard states.  Option A is incorrect. C(g) should be C(s). C in its standard state is C(s). O(g) should be ½O2(g). C in its standard state is O2(g).  Option B is incorrect. C(g) should be C(s). C in