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As per your syllabus, the standard enthalpy of combustion of GLUCOSE (C₆H₁₂O₆) is which?
A−2802 kJ/mol
B+2802 kJ/mol
C−74 kJ/mol
D−180 kJ/mol
Answer & Solution
Correct answer: A. −2802 kJ/mol
As per your syllabus, C₆H₁₂O₆ + 6 O₂
→ 6 CO₂ + 6 H₂O; Δ_c H° = −2802 kJ/mol.
Body extracts this energy through metabolism.
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