Calculation of reaction enthalpy 3 (thermo-15) Glucose (C6H12O6) can be implemented in different ways: aerobic (with O2) and anaerobic (without O2). For both reactions, formulate the reaction equation and calculate the standard reaction enthalpy. Note I: if the ΔHof is unknown, then generally calculate x, y, etc. Note II: state of matter state

  Exercise: thermo-15    

ⓘ    🔄 Reload     🖨️ print           ◀  –   ▶     ?! Question/Answer ?!

Glucose (C₆H₁₂O₆) can be implemented in different ways: aerobic (with O₂) and anaerobic (without O₂).
For both reactions, formulate the reaction equation and calculate the standard reaction enthalpy.
Note I: if the ΔHof is unknown, then generally calculate x, y, etc.
Note II: state of matter state

---

Solution:

ΔHof(C₆H₁₂O₆) = -1268 kJ/mol
ΔHof(O₂)(g) = 0 kJ/mol
ΔHof(C₂H₅OH)(l) = -278 kJ/mol
ΔHof(H₂O)(g) = -286 kJ/mol
ΔHof(CO₂) = -393 kJ/mol

(a) Combustion reaction:
C₆H₁₂O₆ + 6·O₂(g) ⇄ 6·CO₂(g) + 6·H₂O(l)
ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
ΔHoR = (6·-393 + 6·-286)-(-1268+6·0) = – 2806 kJ/mol

(b) Fermentation:
C₆H₁₂O₆ ⇄ 2·C₂H₅OH(l) + 2·CO₂(g)
Note: C₂H₅OH is called alcohol
ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
ΔHoR = (2·-278 + 2·-393)-(-1268) = – 74 kJ/mol

[Author]

Posts