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What would you do in order to solve this question: "The enthalpy of combustion for benzoic acid is -3223.6kj/mol. Use the values of the standard enthalpies of formation of liquid water and CO2 to calculate standard enthalpy of formation of benzoic acid."

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  1. The combustion of benzoic acid is
    C7H6O2 + 15/2 O2 ==> 7CO2 + 3H2O and since you want the enthalpy of FORMATION (and not combustion), reverse the equation (and the sign for dH combustion) to obtain
    7CO2 + 3H2O ==> C7H6O2 dH = +3223.6 kJ
    Then dHrxn = (n*dHf benzoic acid)- (n*dHf products). Substitute 3223.6 kJ for dHrxn and substitute values for dHf H2O and dHf CO2 and calculate dHf benzoic acid.

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