Enthalpy of Formation. DHf.



Definition:



The enthalpy of formation for a compound is defined as the change in enthalpy when exactly 1 mole of a compound is formed at 298 K and 100 kPa from its elements in their standard states.





Examples:



Na(s) + O2(g) + H2(g) NaOH(s) DHf = -426.7 kJ/mol



H2(g) + O2(g) H2O(l) DHf = -285.9 kJ/mol









Notes:

1. DHf is tabulated for 1 mole of the compound.

2. the indicates standard pressure, 100 kPa.

3. the values are tabulated at 298 K.












Using Tabulated Values of DHf to calculate DH298 for a reaction.


















For the above reaction:



DH298 = DHf(N2O4) - 2 DHf(NO2)







In general:



DH298 = niDHf(product)i - niDHf(reactant)i







I like to shorten this even more:



DH298 = miDHf(i)



where mi is the reaction coefficient of substance i, and is positive for product, negative for reactant.














Use of Enthalpy of Formation Tables.


1. Calculate DH for the reaction:



CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)







DH = miDHf(i)



DH298 = DHf(CO2) + 2 DHf(H2O(l)) - DHf(CH4) - 2 DHf(O2)



DH298 = (-393.5) + 2 (-285.9) - (-74.9) - 2 ( 0 )



DH298 = -890.4 kJ



Note

DH is an extensive property, depending on the amount of material.



DH must always be associated with a balanced chemical equation.



DHf is defined for 1 mole of the compound.



DHf for an element in its standard state is 0.














Example 2.


DH298 for the complete combustion of ethyne (acetylene) is -1299.6 kJ / mol. Use the previously used values of DHf(CO2) and DHf(H20(l)) to find DHf(C2H2).







C2H2(g) + 2 O2(g) 2 CO2(g) + H2O(l)



DH = miDHf(i)



DH = 2 DHf(CO2) + DHf(H2O(l)) - DHf(C2H2)



-1299.6 = 2 (-393.5) + (-285.9) - DHf(C2H2)



DHf(C2H2) = 226.7 kJ / mol.