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Reaction carried out under constant volume. Use a bomb calorimeter. Usually study combustion.

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

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Reaction carried out under constant volume. Use a bomb calorimeter. Usually study combustion.

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

  2. Example A 0.5269 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 0.5269 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. 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. 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. 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 = -191.2 kJ H2 (g) + ½ O2 (g)  H2O (g) ΔH = -241.8 kJ H2O (g)  H2O (l) ΔH = -43.8 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 = -191.2 kJ H2 (g) + ½ O2 (g)  H2O (g) ΔH = -241.8 kJ H2O (g)  H2O (l) ΔH = -43.8 kJ C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2 H2O (l) ΔH = ??? -202.6 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: • H2SO4 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

  15. Using Enthalpies of Formation of Calculate Enthalpies of Reaction • For a reaction

  16. 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)

  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) 283 kJ • 2. C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) -2220kJ • NH3 (g) + HCl (g)  NH4Cl (s) -175.9 kJ

  18. 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) Born-Haber CycleApplication of Hess’s Law

  19. 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.

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