Question

Calculate heat of formation of acetic acid if its heat of combustion is - 867 kJ/mol. The heat of formation of carbon dioxide and water are -393.5 and -285.9 kJ/mol. *

Answer 1

The chemical equation for formation of the products is;

$2CH_3COOH+CaCO_3\to(CH_3COO)_2Ca+CO_2+H_2O$

The equation for calculating the standard enthalpy for formation is;

$\Delta H°=\sum n\Delta H°_f(products)- \sum n \Delta H°_f(reactants)$

$\Delta_f°$ for $CO_2=(-393.5kJ/mol)$

$\Delta_f°$for $H_2O=(-285.9kJ/mol)$

$\Delta _f°$for $CH_3COOH=(-867kJ/mol)$

$\Delta H°=[1molCO_2(-393.5kJ/mol)+1mol H_2O(-285.9kJ/mol)]-[2molCH_3COOH(-867kJ/mol)]$ $=[-679.4kJ/mol]-[-1734kJ/mol]$

$=1054.6kJ/mol$

Since the reaction is exothermic, the standard heat of formation is thus;

$-1054.6kJ/mol$