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

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

Step-by-step explanation:

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

`H_2(g)+(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* (-394 kJ/mol)+2* (-286 kJ/mol) - (-876 kJ/mol)`

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

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