Question

Calculate the enthalpy of formation of acetic acid (\[C{{H}_{3}}COOH\]) if its enthalpy of
combustion is - 867 kJ/mol. The enthalpies of formation of \[C{{O}_{2}}\] (g) and \[{{H}_{2}}O\] (l)
are - 393.5 kJ/mol and - 285.9 kJ/mol respectively.

Answer 1

Hint: The standard enthalpy of formation is the change of enthalpy during the formation of one mole of the substance from its constituent elements.
Enthalpy of combustion is the amount of heat released during the combustion of the specified amount of substance.

Complete step by step solution:
Given in the question AH of combustion of acetic acid = - 867 kJ/mol
ΔH of formation of $CO_2$ = -393.5 kJ/mol
ΔH of formation of $H_2O$ = -285.9 kJ/mol
The chemical reaction for the combustion of acetic acid (\[C{{H}_{3}}COOH\])
\[C{{H}_{3}}COOH(l)\,+\,3{{O}_{2}}(g)\,\to \,2C{{O}_{2}}(g)\,+{{H}_{2}}O(l)\] (ΔH = -867 kJ/mol)
Let's take the enthalpy of formation of acetic acid as X.
ΔH (reaction) = ΔH (product) - ΔH (reactant)
Thus,
-867 = (2 × (enthalpy of formation of \[C{{O}_{2}}\]) + 2 × (enthalpy of formation of \[{{H}_{2}}O\])) - (enthalpy of formation of acetic acid)
-867 = (2 × (-393.5) + 2 × (-285.9)) - (X)
-867 = -1358.8 + X
 X = -1358.8 + 867
 X = - 491 kJ/mol

Additional Information:

Standard states for the enthalpy of formation are:
For a gas it should obey the ideal gas equation at 1 bar pressure.
For a solute present in an ideal solution, a concentration of 1 mol/liter.
For an element, the form of element which is most stable in standard conditions.

Note: Elements like oxygen gas and solid carbon have standard enthalpy of formation of zero as there is no change involved in their reaction. If the value of enthalpy of formation is negative it is considered more stable.