Question

Calculate the calorific value of sugar if its heat of combustion is 5645 kj mol-1​

Answer 1

ANSWER:-

Sucrose=C12 H22 O11

Lactic acid=C3 H6 O3 =LA

Given, C+O2(g)→CO2 (g) Δ f HCO2 =−395kJ/mol ...(a)

H 2(g)+ 21 O2(g)→H2O(g) Δ f H H2O=−286kJ/mol ...(b)

3C+3H2+ 23O2→C3H 6O Δ f H LA=−694kJ/mol ...(c)

Now, multiply (a) by 12 and (b) by 11 and add both to get

12C+11H 2+ 235O 2 →12CO 2+11H 2O Δ f H=−7886kJ/mol

or 12CO 2+11H 2O→12C+11H2 + 235 O2 Δ f H=+7886kJ/mol ...(d)

Now combustion or aerobic oxidation reaction is

C 12 H22 O11 +12O 2 →12CO2 +11H2O Δ CH=−5645kJ/mol ...(e)

(d)+(e) gives

C12 H22 O 11 →12C+11H2 + 211O2 ΔH=2241kJ/mol or 12C+11H2 + 211O2 →C12 H22 O11 Δ f H sucrose =−2241kJ/mol ...(f)

This equation gives heat of formation of sucrose.

Now consider anaerobic hydrolysis of sucroseC12 H22 O11 + H2O(g)→4C 3H6O3 Δ h H=?

Δ h H=Δ f H products−Δ f H reactants

Δf H LA =−694kJ/mol (given)

so Δ h

H=4(−694)−[Δ f H sucrose+Δ f H H2O]

=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

I HOPE THIS HELPS YOU

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