Calculate the enthalpy of combustion for acetic acid using standard enthalpies of formation.

HC2H3O2 + 2O2 ==> 2H2O + 2CO2

dHrxn = (n*dHf products) - (n*dHf reactants)

To calculate the enthalpy of combustion for acetic acid (CH3COOH) using standard enthalpies of formation, you can follow these steps:

1. Write and balance the chemical equation for the combustion of acetic acid. In this case, acetic acid reacts with oxygen to produce carbon dioxide and water:

CH3COOH + O2 → CO2 + H2O

2. Determine the standard enthalpies of formation (ΔHf°) for each of the compounds involved in the reaction. The standard enthalpy of formation is the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states.

The standard enthalpies of formation for acetic acid (CH3COOH), carbon dioxide (CO2), and water (H2O) can be found in reference tables or online sources. Let's assume the values are:

ΔHf°(CH3COOH) = -393.5 kJ/mol
ΔHf°(CO2) = -393.5 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol

3. Calculate the enthalpy of combustion using the following equation:

ΔHcombustion = ∑(ΔHf°products) - ∑(ΔHf°reactants)

ΔHcombustion = [ΔHf°(CO2) + ΔHf°(H2O)] - [ΔHf°(CH3COOH) + 0.5(ΔHf°(O2))]

Note that the stoichiometric coefficient for O2 in the balanced equation is 0.5 because it is present as O2 gas.

4. Substitute the known values into the equation and calculate:

ΔHcombustion = [-393.5 + (-285.8)] - [-393.5 + 0.5(0)]

ΔHcombustion = -679.3 - (-393.5)

ΔHcombustion = -679.3 + 393.5

ΔHcombustion = -285.8 kJ/mol

The enthalpy of combustion for acetic acid using standard enthalpies of formation is -285.8 kJ/mol.