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? △H∫∘​ for lactic acid,CO_{2} and H2​O is -694,-395 and -286.0 respectively

A

advantage=4356 kJ.mol−1

B

advantage=5396 kJ.mol−1

C

advantage=4756 kJ.mol−1

D

advantage=5657 kJ.mol−1

Sucrose=C12​H22​O11​
Lactic acid=C3​H6​O3​=LA

Given, C+O2​(g)→CO2​(g)   Δf​HCO2​​=−395kJ/mol   ...(a)
H2​(g)+21​O2​(g)→H2​O(g)   Δf​HH2​O​=−286kJ/mol   ...(b)
3C+3H2​+23​O2​→C3​H6​O3​   Δf​HLA​=−694kJ/mol   ...(c)

Now, multiply (a) by 12 and (b) by 11 and add both to get
12C+11H2​+235​O2​→12CO2​+11H2​O   Δf​H=−7886kJ/mol
or 12CO2​+11H2​O→12C+11H2​+235​O2​   Δf​H=+7886kJ/mol   ...(d)

Now combustion or aerobic oxidation reaction is 
C12​H22​O11​+12O2​→12CO2​+11H2​O   ΔC​H=−5645kJ/mol   ...(e)
(d)+(e) gives
C12​H22​O11​→12C+11H2​+211​O2​   ΔH=2241kJ/mol
or 12C+11H2​+211​O2​→C12​H22​O11​   Δf​Hsucrose​=−2241kJ/mol   ...(f)

This equation gives heat of formation of sucrose.
Now consider anaerobic hydrolysis of sucrose
 

C12​H22​O11​+H2​O(g)→4C3​H6​O3​   Δh​H=?
Δh​H=Δf​Hproducts​−Δf​Hreactants​
Δf​HLA​=−694kJ/mol (given)
so Δh​H=4(−694)−[Δf​Hsucrose​+Δf​HH2​O​]
=4(−694)−[−2241−286]=−249kJ/mol

So enthalpy change in anaerobic hydrolysis=−249kJ/mol
enthalpy change in aerobic oxidation=−5646kJ/mol
So advantage of aerobic oxidation to anaerobic hydrolysis =−249−(−5645)=5396kJ/mol