Find the standard enthalpy of combustion

Question: The standard enthalpies of formation of CO₂ (g), H₂O (l) and glucose (s) at 25^oC are -400 kJ/mol, -300kJ/mol and -1300kJ/mol, respectively. What will be the standard enthalpy of combustion per gram glucose at 25^oC?

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 (25^oC)

·       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 ΔH^o_c  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:

∆H_combustion = (6 x ∆H_f CO₂ + 6 x ∆H_f H₂O) - ∆H_f C₆H₁₂O₆

                      = (6 x -400 + 6 x -300) – (-1300)

                         = -2900 kJ /mol

 The molar mass of glucose is 180 grams/mole. So

∆H_combustion = -2900/180 kJ/g

                   = -16.11 kJ/g

Thus, the standard enthalpy of combustion of per gram glucose at 25^oC is -16.11 kJ/g.