Question

27.4 Anaerobic bacteria use the glycolysis reaction, CAH,Os(aq) + 2 CH2CH(OH)COOH(aq), as a source of en- ergy. In the reaction, glucose, CoH12O6, is first converted to pyruvic acid, CH3COCOOH, which is then converted to lactic acid, CH2CH(OH)COOH, through the action of the enzyme lactate dehydrogenase. Use bond enthalpies and mean bond enthalpies to estimate (a) the enthalpy of formation and (b) the enthalpy of combustion of glucose. Ignore the contributions of enthalpies of fusion and vaporization.

Answer 1

molt combtestion de glucose: C6H₂O6 + 60₂ 600 + 6H₂O _AH.comb. : [GAH (CO2) +GAH (H20)]- [AH ICC His Og )+6 Aurel A H comb: [6*1(-393:5) + 6x (-285:52]- (-1260+0] __ :(-2361-1713+1260) kJ/mol : - 2814 kJ/mol Formation de glucose: LGC (S. graphite) + 6 42 (g) + 30, () CGHED OG (S) for Alpañequired come ore : Lọc + O + CO, AH = -39 3.5 kJ l mol 2 H + 102 H₂O AH = -285.5 kJ/mol 3) (4 Haz O4+602 +6C02 + 6H2O DH=-2814 kolmol Multiply equ - and by G and add 2 60+ Go + 6H₂ +30, 500, + 6H₂O 2) 6C + GH+902 -> GCO2+ 6H2O — Now, switract equ. bom equ.@ 6 6+644 + 30,- C5 Hug 06 – 60, +6CO2 + 6H20 - GCO, - GH, 6C+ 6H2 + 302 -> C6H12 Or — ☺ Equ. ③ is the required equ. Ant- [6X (-193.5) + 6 6285.5) - 6-28149) I kJ/mol Ang [-2361-1713 + 2814] kJ Imol A48 - 1260 kJ/mol