Question

alculate the standard enthalpy of formation of ethanoic acid given that the standard enthalpy of combustion for carbon is –394 kJ mol-1 , hydrogen is –286 kJ mol-1 and ethanoic acid is –876 kJ mol-

Answer 1

Answer:

The standard enthalpy of formation of ethanoic acid is -484 kJ/mol.

Explanation:

$C(g)+O_2(g)\rightarrow CO_2(g),\Delta H_{1, comb}=-394 kJ/mol$...[1]

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l),\Delta H_{2, comb}=-286 kJ/mol$..[2]

$CH_3COOH(l)+2O_2(g)\rightarrow 2CO_2(g)+2H_2O(l),\Delta H_{3, comb}=-876 kJ/mol$..[3]

The standard enthalpy of formation of ethanoic acid :

$2C(g)+2H_2(g)+O_2(g)\rightarrow CH_3COOH, \Delta H_{4}=?$..[4]

Using Hess's law to calculate :

2 × [1] + 2 × [2] - [3] = [4]

$\Delta H_4=2\times (-394 kJ/mol)+2\times (-286 kJ/mol) - (-876 kJ/mol)$

$=\Delta H_4=-484 kJ/mol$

The standard enthalpy of formation of ethanoic acid is -484 kJ/mol.