Question
Calculate the enthalpy of Combustion of ethanol using appendix 1 and then again using mean bond dissociation energies. Compare the results.
Appendix 1: thermochemic data This brief list has been constructed to allow students given in Appendix 2, and to allow instructors to setfuet Physical chemistry, 5th edn, P. W. Atkins, oUP 1995 to tackle problems and H chemisry and physies, 6lst Edh, R. C. Weast (Ed), CRC P All data refer to 298.15 K and l bar pressure. Units of kJ mol-: Units of S and op are J K-1 mol-1 Ho and .e Compound 0 57 8.5 +1.9 +2.9 24 61 -110.5-1372 197.7 291 -393.5 -3944 2137 37 -74.8-50.7 1863 35.3 -84.7328 2296 526 +52.3 +68.2 219.6 436 +226.7 +209.2 200.9 43.9 -103.9 -23.5 269.9 73.5 126.2-170 310.2 975 -146.4 -8.2 3484 120.2 0.2 204.3 1421 -224.4 +1.0 3286 224.3 156.5 +82.9 +129.7 269.3 817 +49.0 +124.3 173.3 136.1 +50.0 +1220 320.7 103.6 -2387-166.3 126.8 81 -200.7-161.9 2398 439 -2777 -1748 1607 1115 -235.1-168.5 282.7 654 C(graphite) C(diamond) Carbon Carbon monoxide CO) Carbon dioxide CH4) C2Hs 6(0) C2H(0) C2H2(a) C3He(a) G4H100) CsH12(0) Ethyne Butane Pentane -198.7 C7H1 CeH12 Heptane 156.0 +26.8 Benzene Toluene Methanol Methanol Ethanol Ethanol CH3OHo) CHgOHo) C2HsOH() C8H6OH(0) CH3COOHo) -165.0 -50.9 146.0 Acetic acid Benzoic acid -484.5 -389.9 159.8 1243 -3851 2453 1676 1468 -2222 -1543 360. 1268 -910 212 C12H22011() CeH1206(8) CH2(NH2)C02H) CO(NH2)20) Br20) Br B-o-glucose Glycine o) 532.9 -373.4 103.5 992 -333.5 -1973 1046 93 Bromine 0 1522 75.7 +30.9 +31 2455 36 -364 53.5 1987 29 ne 2(0) Hydrogen bromide HBra) Calcium 0 414253
Explanation / Answer
combustion reaction is
C2H5OH(g) + 3 O2 (g)....................................> 2 CO2(g) + 3 H2O (g)
considering heat of formation values:
Hrxno = [Hfo (products) - Hfo (reactants)]
Hrxno = [3 *Hfo (H2O (g)) + 2 *Hfo (CO2(g)] – [1*Hfo (C2H5OH(g)) – 3*Hfo (O2(g)]
Hrxno = [3 * (-241.8) + 2 * (-393.5) ] - [1 * (-235.1) + 3*0.00]
Hrxno = -1277.2 KJ/mole.
considering the bond dissociation energy:
Hrxno = E (reactants) - E (products)
C2H5OH(g) + 3 O2 (g)....................................> 2 CO2(g) + 3 H2O (g)
Hrxno = [5* E(C-H) + 1 * E (C-C) + 1 * E(C-O) + 1 * E (O-H)] - [ 4 * E(C=O) + 6 * E(O-H) ]
Hrxno = [5 * 413 + 1 * 347 + 1 * 358 ] - [4 * 799 + 5 * 467] = -2761 KJ /mole.
