Enthalpy of Formation. Df.

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) + ½ O2(g) + ½ H2(g) NaOH(s) Df = -426.7 kJ/mol

H2(g) + ½ O2(g) H2O(l) Df = -285.9 kJ/mol

Notes:

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

2. the º indicates standard pressure, 100 kPa.

3. the values are tabulated at 298 K.

Using Tabulated Values of Df to calculate D298 for a reaction. For the above reaction:

D298 = Df(N2O4) - 2 Df(NO2)

In general:

D298 = åniDf(product)i - åniDf(reactant)i

I like to shorten this even more:

D298 = åmiDf(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 DHº for the reaction:

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

DHº = åmiDf(i)

D298 = Df(CO2) + 2 Df(H2O(l)) - Df(CH4) - 2 Df(O2)

D298 = (-393.5) + 2 (-285.9) - (-74.9) - 2 ( 0 )

D298 = -890.4 kJ

Note

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

DHº must always be associated with a balanced chemical equation.

Df is defined for 1 mole of the compound.

Df for an element in its standard state is 0.

Example 2.

D298 for the complete combustion of ethyne (acetylene) is -1299.6 kJ / mol. Use the previously used values of Df(CO2) and Df(H20(l)) to find Df(C2H2).

C2H2(g) + 2½ O2(g) 2 CO2(g) + H2O(l)

DHº = åmiDf(i)

DHº = 2 Df(CO2) + Df(H2O(l)) - Df(C2H2)

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

Df(C2H2) = 226.7 kJ / mol.