1. Standard Enthalpy of Formation
College Chemistry

2. Enthalpies of Formation
If 1 mol of compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation, Hof .
Standard conditions (standard state): Most stable form of the substance at 1 atm and 25 oC (298 K).
Standard enthalpy, Ho, is the enthalpy measured when everything is in its standard state.
Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states.

3. Enthalpies of Formation
If there is more than one state for a substance under standard conditions, the more stable one is used.
Standard enthalpy of formation of the most stable form of an element is zero.

4. Enthalpies of Formation
Substance
DHof (kJ/mol)
C(s, graphite)
O(g)
247.5
O2(g)
N2(g)

5. Enthalpies of Formation C3H8(g) + 5O2(g)  3CO2(g) + 4H2O()
Using Enthalpies of Formation to Calculate Enthalpies of Reaction
For a reaction
Note: n & m are stoichiometric coefficients.
Calculate heat of reaction for the combustion of propane gas giving carbon dioxide and water.
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O()

6. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O()
∆Hof (kJ/mol):

7. Hess’s Law
Hess’s law: if a reaction is carried out in a number of steps or if a product can’t easily be made form its standard state, H for the overall reaction is the sum of H for each individual step.
For example:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -802 kJ
2H2O(g)  2H2O(l) H= kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ

8. Another Example of Hess’s Law
Given:
C(s) + ½ O2(g)  CO(g) DH = kJ
CO2(g)  CO(g) + ½ O2(g) DH = kJ
Calculate DH for: C(s) + O2(g)  CO2(g)