# 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.

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