Thermochemistry

1. Thermochemistry • Outline • Energy • Types • Law of Conservation of Energy • First Law of Thermodynamics • State Functions • Work • Enthalpy • Specific Heat • Calorimetry • Hess’ Law Chapter 6

2. Kinetic & Potential Energy

3. What are the manifestations of energy? Tro: Chemistry: A Molecular Approach, 2/e

4. What is a system?

5. Sign Conventions

6. How does the internal energy change? • ─DEsystem= DEsurroundings

7. How does the internal energy change? • DEsystem= ─ DEsurroundings

8. What is a state function?

9. How does the Law of Conservation of Energy apply to an Ecosystem?

10. How does a chemical reaction do work?

11. How does a chemical reaction do work?

13. Chapter 6: Examples – Work Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm.

15. Chapter 6: Examples – Energy A 100 W electric heater (1 W = 1 J/s) operates for 20 minutes to heat a gas cylinder. The gas expands from 2.04 L to 2.54 L against an atmospheric pressure of 1.0 atm. What is the change in internal energy of the gas?

16. What is the sign of the enthalpy change?

18. What is the sign of the enthalpy change?

19. Where does the heat flow? • Enthalpy transferred out of reactants  • exothermic  • H =  • Enthalpy transferred into products  • endothermic  • H = +

20. How are the enthalpy of a forward and a reverse reaction related? H2O(g)  H2(g) + 1/2 O2(g) H = +241.8 kJ H2(g) + 1/2 O2(g)  H2O(g) H = 241.8 kJ Hforward = Hreverse (For reversible reactions)

21. How does the amount of substance undergoing change affect the enthalpy? H2O(g)  H2(g) + 1/2 O2(g) H = +241.8 kJ 2 H2O(g)  2 H2(g) + 1 O2(g)H = +483.6 kJ

22. Does the physical state of reactions and products affect the enthalpy? H2O(g)  H2(g) + 1/2 O2(g) H = +241.8 kJ H2O(l)  H2(g) + 1/2 O2(g) H = +285.8 kJ

23. Chapter 6: Examples – Energy Stoichiometry If 5,449 kJ of energy is released when gaseous water is decomposed into its elements, how many grams of water was used? H2O (g) H2 (g) + ½ O2 (g) ∆H = 241.8 kJ

24. Chapter 6: Examples – Energy Stoichiometry When 2.00 moles of sulfur dioxide gas reacts completely with 1.00 moles of oxygen gas to form 2.00 moles of sulfur trioxide gas at 25 °C and a constant pressure of 1.00 atm, 198 kJ of energy is released as heat. Calculate ΔH and ΔE for this process.

25. Chapter 6: Examples – Energy Stoichiometry Solid sulfur, S, reacts with carbon dioxide gas to produce sulfur dioxide gas and carbon solid, ΔH = - 75.8 kJ. If 12.9 g of sulfur react with 9.70 g of carbon dioxide, how many kJ of energy are released or absorbed?

27. Calorimetry – constant pressure

28. Chapter 6: Examples – Calorimetry • A reaction known to release 1.78 kJ of heat takes place in a calorimeter containing 0.100 L of solution. The temperature rose by 3.65 ˚C. • Next, 50 mL of hydrochloric acid and 50 mL of aqueous sodium hydroxide were mixed in the same calorimeter and the temperature rose by 1.26 ˚C. What is the heat output of the neutralization reaction?

29. Chapter 6: Examples – Calorimetry A 55.0 g piece of metal was heated in boiling water to 99.8 ˚C and dropped into an insulated beaker with 225 mL of water (ρ = 0.997992 g/mL) at 21.0 ˚C. The final temperature of the metal and the water is 23.1 ˚C and cwater is 4.184 J/g · ˚C. Calculate the specific heat of the metal assuming that no heat was lost to the surroundings.

30. Chapter 6: Examples – Calorimetry A 33.14 g sample of copper (cCu = 0.385 J/g · ˚C) and aluminum (cAl = 0.902 J/g · ˚C) was heated to 119.25 ˚C and dropped into a calorimeter containing 250.0 g of water at 21.00 ˚C. The temperature rose to 23.05 ˚C. Assuming that no heat was lost to the surroundings, what is the mass of copper in the sample?

31. Bomb Calorimeter – constant volume

32. Chapter 6: Examples – Calorimetry Octane, C8H18, a primary constituent of gasoline burns in air. Suppose that a 1.00 g sample of octane is burned in a calorimeter that contains 1.20 kg of water. The temperature of the water and the bomb rises from 25.00 ˚C to 33.20 ˚C. If the heat capacity of the bomb is 837 J/˚C, cwater is 4.184 J/g · ˚C, calculate the molar heat of reaction of octane.

37. Chapter 6: Examples – Hess’ Law Determine the ΔHsublimation of ice to water vapor: H2O (s)→ H2O (g) Given: H2O (s)→ H2O (l)ΔH = 6.02 kJ H2O (g)→ H2O (l)ΔH = -40.7 kJ

38. Chapter 6: Examples – Hess’ Law Two forms of carbon are graphite, the soft, black, slippery material used in “lead” pencils, and as a lubricant for locks and diamond, the beautiful, hard gemstone. Using the enthalpies of combustion for graphite (-394 kJ/mol) and diamond (-396 kJ/mol), calculate ΔH for the conversion of graphite to diamond: C graphite (s)→ C diamond (s) Given: C graphite (s) + O2 (g) → CO2 (g)ΔH = -394 kJ C diamond (s) + O2 (g) → CO2 (g)ΔH = -396 kJ

39. Chapter 6: Examples – Hess’ Law Calculate the enthalpy change of the formation of methane, CH4, from solid carbon as graphite and hydrogen gas: C(s) + 2 H2 (g) → CH4 (g) Given: C(s) + O2 (g) → CO2 (g)ΔH = -393.5 kJ H2 (g) + ½ O2 (g) → H2O (l)ΔH = -285.8 kJ CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2 O (l)ΔH = -890.3 kJ

40. Chapter 6: Examples – Hess’ Law Calculate the enthalpy change for S (s) + O2 (g) → SO2 (g) Given: 2 SO2 (g) + O2 (g) → 2 SO3 (g)ΔH = -196 kJ 2 S (s) + 3 O2 (g) → 2 SO3 (g)ΔH = -790 kJ

41. CH4(g)+ 2 O2(g)→ CO2(g) + H2O(g) DH° = [(DHf° CO2(g) + 2∙DHf°H2O(g))− (DHf° CH4(g) + 2∙DHf°O2(g))] CH4(g) → C(s, graphite) + 2 H2(g) DH° = + 74.6 kJ C(s, graphite) + 2 H2(g) → CH4(g) DHf°= − 74.6 kJ/mol CH4 DH° = [((−393.5 kJ)+ 2(−241.8 kJ)− ((−74.6 kJ)+ 2(0 kJ))] = −802.5 kJ C(s, graphite) + O2(g) → CO2(g) DHf°= −393.5 kJ/mol CO2 2 H2(g) + O2(g) → 2 H2O(g) DH° = −483.6 kJ H2(g) + ½ O2(g) → H2O(g) DHf°= −241.8 kJ/mol H2O CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) DH° = −802.5 kJ Tro: Chemistry: A Molecular Approach, 2/e

42. Chapter 6: Examples – Hess’ Law Calculate ΔHf˚ for the combustion of methane: CH4 (g) + O2 (g) → CO2 (g) + 2 H2O (l)

43. Chapter 6: Examples – Hess’ Law Benzene, C6H6, is an important hydrocarbon. Calculate its enthalpy of combustion (ΔH˚): C6H6 (l) + O2 (g) → 6 CO2 (g) + 3 H2O (l) Given: ΔHf˚ [C6H6 (l)] = +49.0 kJ/mol ΔHf˚ [CO2 (g)] = -393.5 kJ/mol ΔHf˚ [H2O (l)] = -285.8 kJ/mol ΔHf˚ [O2 (g)] = 0 kJ/mol

44. Chapter 6: Examples – Hess’ Law Benzene, C6H6, is an important hydrocarbon. Calculate its enthalpy of combustion (ΔH˚): C6H6 (l) + O2 (g) → 6 CO2 (g) + 3 H2O (l) Given: 6 C (s) + 3 H2 (g)  C6H6 (l) ΔH = 49.0 kJ C (s) + O2 (g)  CO2 (g) ΔH = -393.5 kJ H2 (g) + ½ O2 (g)  H2O (l) ΔH = -285.8 kJ

45. Chapter 6: Examples – Hess’ Law Calculate the enthalpy of formation of the fermentation of glucose to ethanol: C6H12O6 (s) → 2 C2H5OH (l) + 2 CO2 (g) Given: ΔHf˚ [C6H12O6 (s)] = -1,260 kJ/mol ΔHf˚ [CO2 (g)] = -393.5 kJ/mol ΔHf˚ [C2H5OH (l)] = -277.7 kJ/mol

46. Chapter 6: Examples – Hess’ Law Calculate ΔHf˚ (in kilojoules) for the synthesis of lime (calcium oxide) from limestone (calcium carbonate), an important step in the manufacture of cement. CaCO3 (s) → CaO (s) + CO2 (g) Given: ΔHf˚ [CaCO3 (s)] = -1,206.9 kJ/mol ΔHf˚ [CaO (s)] = -635.1 kJ/mol ΔHf˚ [CO2 (g)] = -393.5 kJ/mol

47. Chapter 6: Examples – Hess’ Law Nitroglycerin is a powerful explosive, giving four different gases when detonated: 2 C3H5(NO3)3 (l) 3 N2 (g) + ½ O2 (g) + 6 CO2 (g) + 5 H2O (g) Given: ΔHf˚ [C3H5(NO3)3 (l)] = -364 kJ/mol ΔHf˚ [CO2 (g)] = -393.5 kJ/mol ΔHf˚ [H2O (g)] = -241.8 kJ/mol ΔHf˚ [O2 (g)] = 0 kJ/mol ΔHf˚ [N2 (g)] = 0 kJ/mol Calculate the energy liberated when 7.00 g is detonated.

49. Chapter 9: Examples – Bond Energy Approximate the ΔHrxn for the production of ammonia by the Haber process: N2 (g) + 3 H2 (g)  2 NH3 (g)

50. Chapter 9: Examples – Bond Energy Approximate the ΔHrxn for the combustion of methane: CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)