Explain why the enthalpy change of combustion per gram of hydrogen (H₂) is significantly greater than that of glucose (C₆H₁₂O₆), even though both compounds contain hydrogen atoms that are ultimately oxidised to water during combustion.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
The standard enthalpy of combustion of hydrogen is −286 kJ mol⁻¹ and that of glucose is −2803 kJ mol⁻¹.
Model answer (4 marks)
1. Convert the standard enthalpy of combustion to a per‑gram basis.
• Hydrogen: ΔH°c = –286 kJ mol⁻¹ ÷ 2 g mol⁻¹ = –143 kJ g⁻¹.
• Glucose: ΔH°c = –2803 kJ mol⁻¹ ÷ 180 g mol⁻¹ = –15.6 kJ g⁻¹.
Thus hydrogen releases roughly nine times more energy per gram.
2. The molar mass of glucose (180 g mol⁻¹) is far larger than that of hydrogen (2 g mol⁻¹). Dividing the large molar enthalpy by a large molar mass gives a much smaller value per gram for glucose.
3. Glucose already contains oxygen atoms bonded to carbon (C–O bonds). During combustion only the remaining C–H bonds are oxidised, so fewer new C–O bonds are formed and less energy is released per gram.
4. Hydrogen consists solely of H–H bonds. In combustion all of these bonds are broken and new H–O bonds are formed in water; every atom contributes to the energy change, with no ‘wasted’ mass from non‑combustible atoms or pre‑oxidised bonds.
• Hydrogen: ΔH°c = –286 kJ mol⁻¹ ÷ 2 g mol⁻¹ = –143 kJ g⁻¹.
• Glucose: ΔH°c = –2803 kJ mol⁻¹ ÷ 180 g mol⁻¹ = –15.6 kJ g⁻¹.
Thus hydrogen releases roughly nine times more energy per gram.
2. The molar mass of glucose (180 g mol⁻¹) is far larger than that of hydrogen (2 g mol⁻¹). Dividing the large molar enthalpy by a large molar mass gives a much smaller value per gram for glucose.
3. Glucose already contains oxygen atoms bonded to carbon (C–O bonds). During combustion only the remaining C–H bonds are oxidised, so fewer new C–O bonds are formed and less energy is released per gram.
4. Hydrogen consists solely of H–H bonds. In combustion all of these bonds are broken and new H–O bonds are formed in water; every atom contributes to the energy change, with no ‘wasted’ mass from non‑combustible atoms or pre‑oxidised bonds.
Examiner tips
- Show the per‑gram calculation explicitly; examiners expect the division by molar mass. Use the word ‘approximately’ when giving the final values. Explain the role of pre‑existing oxygen in glucose. Mention that hydrogen has no other atoms to ‘waste’ energy on.
Common mistakes
- Failing to convert to a per‑gram basis and giving only molar values. Confusing the sign of ΔH°c (students sometimes write +286 kJ). Overlooking the effect of glucose’s pre‑oxidised C–O bonds.
Mark scheme (4 marks)
- Converting to a per-gram basis: hydrogen releases approximately 143 kJ g⁻¹ whereas glucose releases approximately 15.6 kJ g⁻¹, so hydrogen releases roughly 9 times more energy per gram.
- Glucose has a very high molar mass (180 g mol⁻¹) compared to hydrogen (2 g mol⁻¹), so dividing the molar enthalpy by molar mass gives a much lower value per gram for glucose.
- Glucose is already partially oxidised — it contains oxygen atoms bonded to carbon — so fewer bonds to oxygen need to be formed during combustion, releasing less energy per gram.
- Hydrogen consists entirely of H–H bonds, which upon combustion are fully converted to H–O bonds in water; all bonding change contributes to energy release, so there is no 'wasted' mass from non-combustible atoms or pre-existing oxidised bonds.
Key terms in this question
Related
- All IB DP Chemistry Higher Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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