To calculate the standard enthalpy of formation of glucose C6H12O6 , we can use the following equation:Hf = Hf products - Hf reactants The balanced equation for the formation of glucose is:6C + 6H2 + 3O2 C6H12O6The standard enthalpies of combustion for glucose, carbon, and hydrogen are given as -2803.6 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively. We can use these values to find the standard enthalpies of formation for the reactants and products.For carbon C , the standard enthalpy of formation is 0 kJ/mol since it is in its elemental form.For hydrogen H2 , the standard enthalpy of formation is also 0 kJ/mol since it is in its elemental form.For oxygen O2 , the standard enthalpy of formation is 0 kJ/mol since it is in its elemental form.Now, we can use the standard enthalpies of combustion to find the standard enthalpy of formation for glucose C6H12O6 . The balanced equation for the combustion of glucose is:C6H12O6 + 6O2 6CO2 + 6H2OUsing the given standard enthalpies of combustion:-2803.6 kJ/mol = 6 -393.5 kJ/mol + 6 -285.8 kJ/mol - Hf C6H12O6 - 6 0 kJ/mol -2803.6 kJ/mol = -2361 kJ/mol - 1714.8 kJ/mol - Hf C6H12O6 Hf C6H12O6 = -2361 kJ/mol - 1714.8 kJ/mol + 2803.6 kJ/molHf C6H12O6 = -272.2 kJ/molTherefore, the standard enthalpy of formation of glucose C6H12O6 is -272.2 kJ/mol.