1. Standard Heat of Formation
Enthalpies

2. Standard Heat of Formation
Standard heat of formation is defined as the change in enthalpy that accompanies the formation of one mole of the compound at its standard state from its elements in their standard state.
Standard state is 1 atm & 25°C (298K)
Every free element in its standard state is assigned a ∆Hf of 0.0 kJ.
Free elements include single elements and any diatomic molecule (H2O2N2Cl2Br2I2F2)

3. Heat of formation equation
Ex: Ca(s) + C(s) /2 O2(g)  CaCO3(s)
Practice: Write a formation equation for 1 mole of Na2CO3(s)
2Na(s) + C(s) /2 O2(g)  Na2CO3(s)

4. Hrxn = Σ Hf°(products) - Σ Hf° (reactants)
USE TABLE OF VALUES

5. Example 1
CH4(g) +2O2(g)  CO2(g) H2O(l)
∆Hf(CO2(g)) = -394kJ
∆Hf(H2O(l)) = -286 kJ
∆Hf(CH4(g)) = -75kJ
∆Hf(O2(g)) = 0 kJ
Σ  H°f(products) - Σ  H°f(reactants) [-394kJ + 2(-286kJ)] – [-75kJ + 2(0)] -966kJ + 75 kJ -891 kJ, exothermic

6. Example 2: Use table C-13 2HF(g)  H2(g) + F2(g) Hrxn= 41.26 kJ
endothermic

7. 2H2S(g) + 3O2(g)  2H2O(l) + 2SO2(g)
Example 3: Use table C-13
2H2S(g) O2(g)  2H2O(l) SO2(g)
Hrxn= kJ
exothermic

8. Example 4: Use table C-13 4Fe(s) + 3O2(g)  2Fe2O3(s)
Hrxn= kJ
exothermic

9. Example 5: Use table C-13 2NO(g) + O2(g)  2NO2(g) Hrxn= -114.14 kJ
exothermic

10. Example 6: Use table C-13 2CO(g) + O2(g)  2CO2(g) Hrxn= -565.968 kJ
exothermic

11. Answers to Examples -891 kJ, exothermic1 542.2 kJ, endothermic