1. Bomb Calorimetry (Constant Volume Calorimetry)
Reaction carried out under constant volume.
Use a bomb calorimeter.
Usually study combustion.

2. Example
A g sample of octane is placed in a bomb calorimeter with a heat capacity of 11.3 kJ °C-1. The original temperature of the water in the calorimeter is 21.93°C. After combustion, the temperature increases to 24.21°C. Determine the enthalpy of combustion per mole and per gram of octane.

3. Example
A g sample of octane is placed in a bomb calorimeter with a heat capacity of 11.3 kJ °C-1. The original temperature of the water in the calorimeter is 21.93°C. After combustion, the temperature increases to 24.21°C. Determine the enthalpy of combustion per mole and per gram of octane.
-48.9 kJ g-1
-5590 kJ mol-1

4. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -802 kJ
Hess’s law: if a reaction is carried out in a number of steps, 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 = -88 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ

5. Note that:
H1 = H2 + H3

6. What is ΔH for the following? FeO (s) + CO (g)  Fe (s) + CO2 (g)
Example
Given:
Fe2O3 (s) + 3 CO (g)  2 Fe (s) + 3 CO2 (g) ΔH = -23 kJ
3 Fe2O3 (s) + CO (g)  2 Fe3O4 (s) + CO2 (g) ΔH = -39 kJ
Fe3O4 (s) + CO (g)  3 FeO (s) + CO2 (g) ΔH =+18 kJ
What is ΔH for the following?
FeO (s) + CO (g)  Fe (s) + CO2 (g)

7. What is ΔH for the following? FeO (s) + CO (g)  Fe (s) + CO2 (g)
Example
Given:
Fe2O3 (s) + 3 CO (g)  2 Fe (s) + 3 CO2 (g) ΔH = -23 kJ
3 Fe2O3 (s) + CO (g)  2 Fe3O4 (s) + CO2 (g) ΔH = -39 kJ
Fe3O4 (s) + CO (g)  3 FeO (s) + CO2 (g) ΔH =+18 kJ
What is ΔH for the following?
FeO (s) + CO (g)  Fe (s) + CO2 (g)
-11 kJ

8. Example
The bombardier beetle uses an explosive discharge as a defensive measure. The chemical reaction involved is the oxidation of hydroquinone by hydrogen peroxide to produce quinone and water:
C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2 H2O (l)

9. C6H4(OH)2 (aq)  C6H4O2 (aq) + H2 (g) ΔH = +177.4 kJ
H2 (g) + O2 (g)  H2O2 (aq) ΔH = kJ
H2 (g) + ½ O2 (g)  H2O (g) ΔH = kJ
H2O (g)  H2O (l) ΔH = kJ
C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2 H2O (l)
ΔH = ???

10. C6H4(OH)2 (aq)  C6H4O2 (aq) + H2 (g) ΔH = +177.4 kJ
H2 (g) + O2 (g)  H2O2 (aq) ΔH = kJ
H2 (g) + ½ O2 (g)  H2O (g) ΔH = kJ
H2O (g)  H2O (l) ΔH = kJ
C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2 H2O (l)
ΔH = ??? kJ

11. 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): 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.
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.

12. Examples – write formation reactions for each:
H2SO4
C2H5OH
NH4NO3

13. Examples – write formation reactions for each:
H2SO H2 (g) + S (s) + 2O2 (g) → H2SO4
C2H5OH 2C (s) + 3H2 (g) + 1/2O2 (g) → C2H5OH
NH4NO3 N2 (g) + 2H2 (g) + 3/2O2 (g) → NH4NO3

14. Enthalpies of Formation

16. Using Enthalpies of Formation of Calculate Enthalpies of Reaction
For a reaction

17. Examples
Use the tables in the back of your book to calculate ΔH for the following reactions:
4 CuO (s)  2 Cu2O (s) + O2 (g)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
NH3 (g) + HCl (g)  NH4Cl (s)

18. Examples
Use the tables in the back of your book to calculate ΔH for the following reactions:
4 CuO (s)  2 Cu2O (s) + O2 (g) kJ
2. C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) kJ
NH3 (g) + HCl (g)  NH4Cl (s) kJ

19. Born-Haber Cycle Application of Hess’s Law
Vaporize the metal (enthalpy of vaporization)
Na (s)  Na (g)
Break diatomic nonmetal molecules (if applicable) (bond enthalpy)
½ Cl2 (g)  Cl-
Remove electron(s) from metal (ionization energy)
Na (g)  Na+ (g) + e-
Add electron(s) to nonmetal (electron affinity)
Cl (g) + e-  Cl- (g)
Put ions together to form compound (lattice energy)
Na+ + Cl-  NaCl (s)

20. Overall Reaction:
Na (s) + ½ Cl2 (g)  NaCl (s)
This is useful because all quantities are directly measurable except lattice energy. The Born-Haber cycle can be used to calculate lattice energy from the other values.

21. Step
Energetics 1 2 3 4 5

22. Step
Energetics 1
endothermic 2 3 4
Exothermic 5
EXOTHERMIC (highly)