Answer:
The standard enthalpy of formation of ethanoic acid is -484 kJ/mol.
Explanation:
[tex]C(g)+O_2(g)\rightarrow CO_2(g),\Delta H_{1, comb}=-394 kJ/mol[/tex]...[1]
[tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l),\Delta H_{2, comb}=-286 kJ/mol[/tex]..[2]
[tex]CH_3COOH(l)+2O_2(g)\rightarrow 2CO_2(g)+2H_2O(l),\Delta H_{3, comb}=-876 kJ/mol[/tex]..[3]
The standard enthalpy of formation of ethanoic acid :
[tex]2C(g)+2H_2(g)+O_2(g)\rightarrow CH_3COOH, \Delta H_{4}=?[/tex]..[4]
Using Hess's law to calculate :
2 × [1] + 2 × [2] - [3] = [4]
[tex]\Delta H_4=2\times (-394 kJ/mol)+2\times (-286 kJ/mol) - (-876 kJ/mol)[/tex]
[tex]=\Delta H_4=-484 kJ/mol[/tex]
The standard enthalpy of formation of ethanoic acid is -484 kJ/mol.
