Question

The enthalpy of the formation of $C{O}_2$ and $H_{2}O$ are $-395kJ$ and $-285kJ$ respectively and the enthalpy of combustion of acetic acid is $869kJ$. The enthalpy of formation of acetic acid is:

• A. $340kJ$
• B. $420kJ$
• C. $491kJ$
• D. $235kJ$

Answer 1

Answer: (C)

$491kJ$

Combustion reaction for acetic acid is:
$C{H}_{3}COOH+2O_2\to 2C{O}_2+2H_{2}O$

Thus,
$H_R=\sum H_{f_{\left(product\right)}}^{\circ }-\sum H_{f_{\left(reactant\right)}}^{\circ }$

$-869=[2\times H_{f\left(C{O}_2\right)}^{\circ }+2\times H_{f\left(H_{2}O\right)}^{\circ }]-(H_{f\left(C{H}_{3}COOH\right)}^{\circ }+0)$

$-869=[2\times (-395)+2\times (-285\left)\right]-\left[H_{f\left(C{H}_{3}COOH\right)}^{\circ }\right]$

$-869=-790-570-H_{f\left(C{H}_{3}COOH\right)}^{\circ }$

$\therefore H_{f\left(C{H}_{3}COOH\right)}^{\circ }=-1360+869=-491kJ$.