The standard enthalpy of formation of glucose is -1273.3kJ/mol, and for carbon dioxide it is -393.5kJ/mol, and for water -285.8 kJ/mol. What is the standard enthalpy of combustion of glucose, C6H12O6?

The reaction equation has the form C₆H₁₂O₆ + xO₂ -> yCO₂ + zH₂O. Balancing carbons, y=6. Balancing hydrogens, z=6, so by balancing oxygen atoms, x=6. Then we use a Hess's law cycle (to be shown during interview) to find that the enthalpy of combustion is -(-1273.3kJ/mol)+(-6x-393.5kJ/mol+-6x-285.8kJ/mol) = -2802.5kJ/mol.