Question

Calculate the standard enthalpy change for the reaction: C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (l)Given the following standard enthalpies of formation: ΔHf(C6H12O6 (s)) = -1273 kJ/molΔHf(CO2 (g)) = -393.5 kJ/molΔHf(H2O (l)) = -285.8 kJ/mol

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the formula:H_reaction =  Hf products  -  Hf reactants For the products, we have 6 moles of CO2  g  and 6 moles of H2O  l : Hf products  = 6   -393.5 kJ/mol  + 6   -285.8 kJ/mol  = -2361 kJ/mol - 1714.8 kJ/mol = -4075.8 kJ/molFor the reactants, we have 1 mole of C6H12O6  s  and 6 moles of O2  g . The standard enthalpy of formation for O2  g  is 0 kJ/mol since it is in its elemental form: Hf reactants  =  -1273 kJ/mol  + 6  0 kJ/mol = -1273 kJ/molNow, we can calculate the standard enthalpy change for the reaction:H_reaction =  Hf products  -  Hf reactants  = -4075.8 kJ/mol -  -1273 kJ/mol  = -2802.8 kJ/molTherefore, the standard enthalpy change for the reaction is -2802.8 kJ/mol.