Chapter 6 Thermochemistry

1. Chapter 6 Thermochemistry

2. Final Exam .

3. Study of energy and its transformation ― thermodynamics Thermodynamics of chemical reactions ― thermochemistry Energy ― capacity to do work (W) and transfer heat (Q) unit: J W and Q ― two ways to transfer energy

4. Work = Force x Distance W = F • d SI unit: W ― J, F ― N, d ― m J = N • m

5. Kinetic energy Two forms of energy Potential energy Kinetic energy: energy from motion potential energy: energy from the interaction between objects. depends on the objects and the relative distance.

10. BrNO + BrNO  2NO +Br2 exothermic reaction

11. CH4 + 2O2  2H2O + CO2 Exothermic Process

12. system surroundings universe Energy of the universe is constant. Or say energy of the universe is conserved. First law of thermodynamics

13. ∆E = Q + W E: energy of the system, including Ek and Ep. Or called internal energy of the system. ∆E: change of internal energy of the system. ∆E = Ef − Ei System absorbs heat from surroundings, Q > 0. System releases heat to surroundings, Q < 0. System does work on surroundings, W < 0. Surroundings do work on system, W > 0.

14. Calculate ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system.

15. State Function A property of a system whose value is determined only by specifying the system’s state, not on how the system arrived at that state. The state of a system is specified by parameters such as temperature, pressure, concentration and physical states (solid, liquid, or gas).

17. Ef Ei ∆E = Ef − Ei

18. System: a battery E: State Functions

19. State functions: E, P, V, T, do not depend on path. W, Q: not state functions, depend on process.

20. PV work: W = −P∆V

21. Example 6.4, page 245 To inflate a balloon you must do pressure–volume work on the surroundings. If you inflate a balloon from a volume of 0.100 L to 1.85 L against an external pressure of 1.00 atm, how much work is done (in joules)?

23. Enthalpy: H = E + PV Only PV work, constant P ∆H = Qp

24. ∆H = Qp ∆H > 0 ↔ endothermic ∆H < 0 ↔ exothermic Recall ….

25. Enthalpy

26. Example 6.6, page 250 Identify each process as endothermic or exothermic and indicate the sign of H. (a) sweat evaporating from skin (b) water freezing in a freezer (c) wood burning in a fire

27. Q = c m ∆T c: specific heat (capacity) J·°C−1·g−1 g m: mass ∆T: change of temperature °C same sign as Q

29. Q = cm ∆T C = c m C: heat capacity, J ·°C−1 Q = C ∆T

30. Example 6.2, page 242 Suppose you find a copper penny (minted before 1982, when pennies were almost entirely copper) in the snow. How much heat is absorbed by the penny as it warms from the temperature of the snow, which is –8.0 C, to the temperature of your body, 37.0 C? Assume the penny is pure copper and has a mass of 3.10 g.

32. Example 6.3, page 243 A 32.5-g cube of aluminum initially at 45.8 C is submerged into 105.3 g of water at 15.4 C. What is the final temperature of both substances at thermal equilibrium? (Assume that the aluminum and the water are thermally isolated from everything else.)

33. A constant-pressure coffee-cup Calorimeter ΔHrxn = − Qsoln = − c m ΔT Read example 6.8, page 253

34. ∆E = Q + W PV work: W = − P∆V PV work only, constant P: ∆H = Qp PV work only, constant V: ∆E = Qv

35. A constant-volume Calorimeter ∆Erxn = −Qcal = −Ccal ΔT Read example 6.5, page 248

36. Characteristics of enthalpy change of a reaction ∆H = Hproducts − Hreactants 2H2(g) + O2(g)  2H2O(g) ∆H = −483.6 kJ ∆H: enthalpy (change) of reaction, heat of reaction • It is important to specify the state of each species in a thermochemical reaction. • ∆H of the reverse reaction is the negative of the original reaction. 3. ∆H depends on how the reaction is written.

37. An LP gas tank in a home barbeque contains 13.2 kg of propane, C3H8. Calculate the heat (in kJ) associated with the complete combustion of all of the propane in the tank. Example 6.7, page 252

38. HB HC HA ∆HAB = HB − HA ∆HAC = HC − HA ∆HCB = HB − HC ∆HAB = ∆HAC + ∆HCB

39. H1 = H2 + H3 Hess’s Law: ∆H for the overall reaction is equal to the sum of the enthalpy changes of each individual step.

40. The enthalpy of reaction for the combustion of C to CO2 is −393.5 kJ/mol C, and the enthalpy for the combustion of CO to CO2 is −283.0 kJ/mol CO: C(s) + O2(g)  CO2(g) ∆H1 = −393.5 kJ CO(g) + ½ O2(g)  CO2(g) ∆H2 = −283.0 kJ Using these data, calculate the enthalpy for the combustion of C to CO: C(s) + ½ O2(g)  CO(g) ∆H = ?

41. Calculate the ∆H for the reaction 2C(s) + H2(g)  C2H2(g) Given the following chemical equations and their respective enthalpy changes: C2H2(g) + 5/2 O2(g)  2CO2(g) + H2O(l) ∆H1 = −1299.6 kJ C(s) + O2(g)  CO2(g) ∆H2 = −393.5 kJ H2(g) + ½ O2(g)  H2O(l) ∆H3 = −285.8 kJ Try example 6.9 and for practice 6.9, page 256-257

42. Reno, NV 3500 ft 4500 ft Atlanta, GA 1000 ft Sea level Standard 0 ft

43. H = H (P, T, phase), phase = s, l, g Standard state P = 1 atm T = temperature of interest, often 25 °C state = most stable form The standard enthalpy of formation of a pure substance, ∆H°f, is the change in enthalpy for the reaction that forms one mole of the substance from its elements, with all substances in their standard states.

44. “Sea level” : free elements at standard state

45. Example 6.11, page 261 Calculate the standard enthalpy change for the following reaction 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)

46. ΔHrxn° = Σnp ΔHf°(products) − Σnr ΔHf° (reactants)

47. For practice 6.11, page 261 Calculate the standard enthalpy change for the following reaction 2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(s) Al(s) + ½ Fe2O3(s)  ½ Al2O3(s) + Fe(s)