Question

Calculate the enthalpy of formation for glucose (C6H12O6) at standard conditions, given the following enthalpies of formation for its constituent elements: C(graphite) = 0 kJ/mol H2(g) = 0 kJ/mol O2(g) = 0 kJ/mol Hint: Use Hess's Law and standard enthalpies of formation to solve the problem.

Answer 1

To calculate the enthalpy of formation for glucose  C6H12O6  at standard conditions, we need to use Hess's Law and the standard enthalpies of formation for its constituent elements. The balanced equation for the formation of glucose from its constituent elements is:6 C graphite  + 6 H2 g  + 3 O2 g   C6H12O6 s Since the enthalpies of formation for C graphite , H2 g , and O2 g  are all 0 kJ/mol, we need to find the enthalpy change for the reaction. Unfortunately, we cannot directly calculate the enthalpy of formation for glucose without additional information such as the enthalpy of combustion or other related reactions.However, if we were given the enthalpy of combustion for glucose, we could use Hess's Law to find the enthalpy of formation. The enthalpy of combustion for glucose is the enthalpy change when one mole of glucose is completely burned in oxygen to form carbon dioxide and water:C6H12O6 s  + 6 O2 g   6 CO2 g  + 6 H2O l If we were given the enthalpy of combustion for glucose  H_combustion , we could use Hess's Law to find the enthalpy of formation  H_formation  by combining the enthalpy of combustion with the enthalpies of formation for CO2 g  and H2O l :H_formation = H_combustion - [6 * H_formation CO2  + 6 * H_formation H2O ]Without the enthalpy of combustion for glucose or other related reactions, we cannot directly calculate the enthalpy of formation for glucose.