Question

The enthalpy of the formation of $CO_2$ and $H_2O$ are - $395\, kJ$ and - $285\, kJ$ respectively and the enthalpy of combustion of acetic acid is $869\, kJ.$ The enthalpy of formation of acetic acid is:

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

Answer 1

Answer: (D)

491 kJ

Answer 2

Combustion reaction for acetic acid is :

$CH _{3} COOH +2 O _{2} \longrightarrow 2 CO _{2}+2 H _{2} O$

Thus,

$H _{R}=\sum H _{f}^{\circ}$ (product) $-\sum H _{f}^{\circ}$ (reactant)

$-869=\left[2 \times H _{f\left( CO _{2}\right)}^{\circ}\right.$
$\left.+2 \times H _{f\left( H _{2} O \right)}^{\circ}\right]\left( H _{ f \left( CH _{3} COOH \right)}^{\circ}+0\right)$
$-869=[2 \times(-395)+2 \times(-285)]$
$-\left[ H _{f\left( CH _{3} COOH \right)}^{\circ}\right]$
$-869=-790-570- H _{f\left( CH _{3} COOH \right)}^{\circ}$
$\therefore H _{f\left( CH _{3} COOH \right)}^{\circ}=-1360+869=-491\, kJ$