Question

Calculate 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-1

Answer 1

`\Delta H^0_f(CH_3COOH) = -484 kJ/mol`

Step-by-step 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

`CH_3COOH + 2O_2\rightarrow 2C0_2 + 2 H_20`

entalpy of formation

`2C_2+ 2H_2 + O_2 \rightarrow CH_3COOH`

`C+ O_2\rightarrow CO_2`

`H_2+(1)/(2)O_2\rightarrow H_2O`

now,

`\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)`

`\Delta H^0_f(CH_3COOH) = 2(-394)+ 2(-286)-(-876)`

`\Delta H^0_f(CH_3COOH) = -484 kJ/mol`