The standard enthalpy of combustion of sucrose is `-5645 kJ mol^(-1)`. What is the advantage (in `kJ mol^(-1)` of energy released as heat) of complete

Question

The standard enthalpy of combustion of sucrose is `-5645 kJ mol^(-1)`. What is the advantage (in `kJ mol^(-1)` of energy released as heat) of complete

The standard enthalpy of combustion of sucrose is `-5645 kJ mol^(-1)`. What is the advantage (in `kJ mol^(-1)` of energy released as heat) of complete aerobic oxidation compared to anaerobic hydrolysis of sucrose to lactic acid ? `Delta H_(f)^(@)` for lactic acid, `CO_(2)` and `H_(2)O` is `-694, -395.0` and `-286.0` respectively.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - `5396" kJ. mol"^(-1)`
`C_(12)H_(22)O_(11)+12O_(2) rarr 12CO_(2) +11H_(2)O" "` [A erobic oxidation]
`DeltaH=-5645" kJ/mole"`
`C_(12)H_(22)O_(11)+H_(2)O rarr 4 CH_(3)-overset(OH)overset(|)(CH)-COOH`
`DeltaH=-249`
Advantage `=5645-249=5396" kJ mol"^(-1)`

5 views Answered 2023-02-05 15:29