Question: The standard enthalpies of formation of CO2 (g), H2O
(l) and glucose (s) at 25oC are -400 kJ/mol, -300kJ/mol and -1300kJ/mol,
respectively. What will be the standard enthalpy of combustion per gram glucose
at 25oC?
Solution: First let us know the meaning of standard enthalpy of combustion.
The standard enthalpy change of
combustion is the enthalpy change which occurs when
one mole of an element or compound is burned completely in oxygen under
standard conditions.
These standard conditions are:
· A
pressure of 1 atm or 100kilopascals (102kPa)
· A
temperature of 298K (25oC)
· Reactants
and products in physical states, normal for the above conditions.
· A
concentration of 1.0mol dm-3 for solutions.
The enthalpy change of combustion will always have a negative value, of couse, because burning always releases heat.
The
standard enthalpy of combustion can be denoted as ΔHoc where
the o sign indicates standard conditions
When glucose burn in the oxygen it form carbon dioxide and
water. So the proper equation of combustion of
glucose is:
So the
standard enthalpy of combustion will be:
∆Hcombustion
= (6 x ∆Hf CO2 + 6 x ∆Hf H2O) - ∆Hf
C6H12O6
= (6 x -400 + 6 x -300) –
(-1300)
= -2900 kJ /mol
The molar
mass of glucose is 180 grams/mole. So
∆Hcombustion
= -2900/180 kJ/g
= -16.11 kJ/g
Thus, the
standard enthalpy of combustion of per gram glucose at 25oC is
-16.11 kJ/g.