1. STANDARD MOLAR ENTHALPY OF FORMATION
Enthalpy change when 1 mol of species is formed in its Standard State at a Specified Temperature from the most stable forms of its constituent elements in their standard forms (at the same temperature).

2. MOST STABLE FORM OF ELEMENT
= Form favored in Equilibrium at 1 Atmosphere and specified temp. (usually K)
e. g. for C at K and 1 atmos., most stable form is GRAPHITE (not diamond!)

3. STANDARD ENTHALPY OF FORMATION FOR CO2
Hf0 CO2(gas) = Standard Enthalpy of Reaction for:
C (s, graphite) + O2 (g) CO2(g)
Nomenclature:
Hf Std. State at 25 0C understood
Formation

4. ENTHALPY OF FORMATION
N.B. Hf0 for an ELEMENT in its Standard State = 0
If not in its Standard State = 0
e.g. For C (s, graphite) C (s, diamond)
H0 = kJ mol –1

5. CALCULATION OF H FROM TABLE OF H VALUES
1. Break down steps of reaction into:
(a) Decomposition of Reactants into Elements in Standard Forms
(b) Formation of Products from Elements in Standard States
2. Apply HESS’S LAW

6. HESS’S LAW
Add 2 (or more) Reactions to give New Reaction, then Add Enthalpies in same manner to give Enthalpy of New Reaction

7. HESS’S LAW

8. CALCULATE ENTHALPY OF FORMATION FOR CO
Need Reaction:
C(s, graphite) + ½ O CO2 (g) H= ??

9. CALCULATE ENTHALPY OF FORMATION FOR CO (cont.)
I. C(s, graphite) + O2 (g) CO2 (g) ,
H0 = kJ mol –1
II. CO2 (g) – ½ O2 (g) CO (g) ,
H0 = kJ mol –1

10. APPLICATION OF HESS’S LAW
ADD I + II:
C(s, graphite) + O2 (g) CO2 (g) ,
H0 = kJ mol –1
II. CO2 (g) – ½ O2 (g) CO (g) ,
H = kJ mol –1
C(s, graphite) + ½ O2 (g) CO (g)
H0 = kJ kJ = kJ mol-1

11. HESS’S LAW

12. DIFFERENT ALLOTROPIC & PHYSICAL FORMS OF ELEMENTS
e.g. red, white and black P
different forms of S
C (graphite and diamond)
Spacing and arrangement of atoms is different in Graphite and Diamond, and requires energy input to effect the transition from one form to the other.

13. BOND ENTHALPIES
Energy used to BREAK Specific Bond in Gas Phase Reaction
N.B. Bond Enthalpies are ALWAYS +.