the enthalpy of combustion of C, H , and sucrose are -393.5 , -286.2, and -5644.2 kJ/mol
calculate the enthalpy of formation of sucrose .
ans:-2226kJ
The combustion reaction of sucrose can be written as:
C12H22O11 + 12 O2 → 12 CO2 + 11 H2O (∆H = -5644.2 kJ/mol)
The combustion of C and H can be written as:
H2 + 1/2 O2 H2O (∆H = -286.2 kJ/mol)
C + O2 CO2 (∆H = -393.5 kJ/mol)
Since ∆H = Σ∆Hf(products) - Σ∆Hf(reactants), then
∆H = [(12) ∆Hf CO2 + (11) ∆Hf H2O] - [∆Hf C12H22O11 + (12)O2]...........(1)
∆Hf(O2) = 0 kJ/mole
Putting all the values in eq (1) we get,
-5644.2 kJ = [(12)-393.5 kJ + (11)-286.2] - [∆Hf C12H22O11 + (12) 0]
-5644.2 kJ = -4722.0 kJ -3148.2 -[∆Hf C12H22O11]
∆Hf C12H22O11 = 5644.2 kJ - 7870.2 kJ
= 2226 kJ
C12H22O11 + 12 O2 → 12 CO2 + 11 H2O (∆H = -5644.2 kJ/mol)
The combustion of C and H can be written as:
H2 + 1/2 O2 H2O (∆H = -286.2 kJ/mol)
C + O2 CO2 (∆H = -393.5 kJ/mol)
Since ∆H = Σ∆Hf(products) - Σ∆Hf(reactants), then
∆H = [(12) ∆Hf CO2 + (11) ∆Hf H2O] - [∆Hf C12H22O11 + (12)O2]...........(1)
∆Hf(O2) = 0 kJ/mole
Putting all the values in eq (1) we get,
-5644.2 kJ = [(12)-393.5 kJ + (11)-286.2] - [∆Hf C12H22O11 + (12) 0]
-5644.2 kJ = -4722.0 kJ -3148.2 -[∆Hf C12H22O11]
∆Hf C12H22O11 = 5644.2 kJ - 7870.2 kJ
= 2226 kJ