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Given the balanced chemical equation: C6H12O6 + 6O2 -> 6CO2 + 6H2O The enthalpy of formation of glucose is -1273 kJ/mol and the enthalpy of formation of water is -286 kJ/mol. Calculate the enthalpy change of the oxidation of glucose in kJ/mol.

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To calculate the enthalpy change of the oxidation of glucose, we need to use Hess's Law, which states that the total enthalpy change of a reaction is the sum of the enthalpy changes of formation of the products minus the sum of the enthalpy changes of formation of the reactants.In this case, the balanced chemical equation is:C6H12O6 + 6O2 -> 6CO2 + 6H2OWe are given the enthalpy of formation of glucose  C6H12O6  as -1273 kJ/mol and the enthalpy of formation of water  H2O  as -286 kJ/mol. We also need the enthalpy of formation of carbon dioxide  CO2 , which is -393.5 kJ/mol.Now we can apply Hess's Law:H = [6 -393.5  + 6 -286 ] - [-1273]H = [-2361 +  -1716 ] -  -1273 H = -4077 + 1273H = -2804 kJ/molSo, the enthalpy change of the oxidation of glucose is -2804 kJ/mol.

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