Question

Calculate the enthalpy of formation of acetic acid $\left(C{H}_{3}COOH\right)$ if its enthalpy of combustion is $-867kJmo{l}^{-1}$. The enthalpies of formation of $C{O}_2\left(g\right)$ and $H_{2}O\left(l\right)$ are $-393.5$ and $-285.9kJmo{l}^{-1}$ respectively.

Answer 1

Solution:-

Let enthalpy of formation of acetic acid be $xkJ/mol$.

Combustion of acetic acid-

$C{H}_{3}COOH+3{O_2}_{\left(g\right)}⟶2{C{O}_2}_{\left(g\right)}+2{H_{2}O}_{\left(l\right)}\Delta H=-867kJ/mol$

$\Delta H^0={\sum \Delta {H^0}_f}_{\left(product\right)}-{\sum \Delta {H^0}_f}_{\left(reactant\right)}$

$\therefore -867=(2\times (-393.5)+2\times (-285.9))-(x+0)$

$\Rightarrow x=-1358.8+867=-491.8kJ/mol$

Hence the enthalpy of formation of acetic acid will be $-491.8kJ/mol$.