The enthalpy of combustion of ethyl alcohol is 1380.7 KJ/mol. If the enthalpies of formation of CO2 and H20 are 394.5 and 286.6 KJ/mol. Calculate the enthaply of formation of ethyl alcohol.
We have to calculate the enthalpy of formation of ethyl alcohol, so we have to calculate ΔH for the following reaction:
2C + 6H + 1/2 O2 -----------------------> C2H5OH ..........................................................(a)
Combustion of ethyl alcohol is given by:
C2H5OH + 3O2 ---------> 2CO2 + 3H2O ΔH = 1380.7 KJ/mol ....................................(i)
Formation of CO2 is given by:
C + O2 ---------> CO2 ΔH = 394.5 KJ/mol ..........................................................(ii)
Formation of H2O is given by:
H2 + 1/2 O2 ---------> H2O ΔH = 286.6 KJ/mol .....................................................(iii)
To Obtain equation (a), we have to multiply equation (ii) by 2 and equation (iii) by 3 and then add these two equations, we get
2C + 3H2 + 7/2 O2 -------------> 2CO2 + 3H2O ΔH = 1648.8 KJ/mol ..............................(iv)
Now subtract equation (i) from equation (iv), we will get
2C + 3H2 + 1/2 O2 -------------> C2H5OH ΔH = 268.1KJ/mol
So, enthalpy of formation of ethyl alcohol is 268.1KJ/mol.