Answer:
[tex]\Delta H^0_f(CH_3COOH) = -484 kJ/mol[/tex]
Explanation:
given,
enthalpy of combustion of carbon = –394 kJ mol-1
enthalpy of combustion of hydrogen = –286 kJ mol-1
enthalpy of combustion of ethanoic acid = –876 kJ mol-1
combusion of ethanoic acid
[tex]CH_3COOH + 2O_2\rightarrow 2C0_2 + 2 H_20[/tex]
entalpy of formation
[tex]2C_2+ 2H_2 + O_2 \rightarrow CH_3COOH[/tex]
[tex]C+ O_2\rightarrow CO_2[/tex]
[tex]H_2+\dfrac{1}{2}O_2\rightarrow H_2O[/tex]
now,
[tex]\Delta H^0_f(CH_3COOH) = 2\Delta H^0_c(C)+ 2\Delta H^0_c(H_2)-\Delta H^0_c(CH_3COOH)[/tex]
[tex]\Delta H^0_f(CH_3COOH) = 2(-394)+ 2(-286)-(-876)[/tex]
[tex]\Delta H^0_f(CH_3COOH) = -484 kJ/mol[/tex]
