Thermodynamics and Energy Changes in Reactions

1. Chapter 5 Thermochemistry: Energy Changes in Reactions

2. Particulate Review What is the equation for the titration between hydrochloric acid and sodium hydroxide? • HCl(aq) + KOH(aq) →H2O(ℓ) + KCl(aq) • HCl(aq) + NaOH(aq) →H2O(ℓ) + NaCl(aq) • HClO(aq) + NaOH(aq) →H2O(ℓ) + NaClO(aq) • 2 HCl(aq) + NaOH(aq) →H2(g) + 2 NaOCl(aq)

3. Particulate Review (cont. 1) When performing the titration between hydrochloric acid and sodium hydroxide which solution is in the buret? • Hydrochloric acid • Sodium hydroxide • Water • Sodium chloride

4. Particulate Preview When ozone molecules absorb UV light from the sun, the ozone breaks down into oxygen molecules and oxygen atoms. Does bond breaking occur when energy is absorbed or released? • Absorbed • Released • Neither

5. Chapter Outline 5.1 Sunlight Unwinding 5.2 Forms of Energy 5.3 Systems, Surroundings, and Energy Transfer 5.4 Enthalpy and Enthalpy Change 5.5 Heating Curves, Molar Heat Capacity, and Specific Heat 5.6 Calorimetry: Measuring Heat Capacity and Enthalpies of Reaction 5.7 Hess’s Law 5.8 Standard Enthalpies of Formation and Reaction 5.9 Fuel, Fuel Values, and Food Values

6. Definitions • Thermodynamics – the study of energy and its transformations • Thermochemistry – the study of the relation between chemical reactions and changes in energy • Thermochemical equation: 2 H2(g) + O2(g) → 2 H2O(ℓ) + energy • Thermal equilibrium – a condition in which temperature is uniform throughout a material and no energy flows from one point to another

7. Definitions (cont. 1) • Heat − the energy transferred between objects because of a difference in their temperatures • Work − a form of energy: the energy required to move an object through a given distance w = F × d • w is work. • F is force. • d is distance.

8. Two Types of Energy • Potential energy (PE): the energy stored in an object because of its position PE = m × g × h • m = mass • g = acceleration due to gravity • h = vertical distance • Kinetic energy (KE): the energy due to motion of the object KE = 1/2mu2 • m = mass • u = velocity

9. Work and Energy

10. Potential Energy: A State Function Depends only on the difference between initial and final state of the system and not how it achieved that state

11. Total Energy Total energy = PE + KE = m·g·h+ ½mu2

12. The Nature of Energy • Law of conservation of energy • Energy can be neither created nor destroyed. • Energy can be converted from one form to another. • Potential energy → kinetic energy • Chemical energy → heat • Thermal energy – kinetic energy of atoms, ions, and molecules

13. Energy at the Molecular Level • Kinetic energy at the molecular level: • Mass, velocity of the particle (KE = ½mu2) • Temperature • As T increases, molecular motion and KE increase. • Potential energy at the molecular level: • Electrostatic interactions:

14. Electrostatic Potential Energy

15. Energy of Chemical Reactions

16. Terminology of Energy Transfer • System – the part of the universe that is the focus of a thermochemical study • Isolated – exchanges no energy or matter with surroundings • Closed – exchanges energy but no matter with surroundings • Open – exchanges both energy and matter with surroundings • Surroundings – everything in the universe that is not part of the system • Universe = system + surroundings

17. Examples of Systems

18. Heat Flow • Exothermic process: energy flows out of system to surroundings (q < 0) • Endothermic process: energy flows into system from surroundings (q > 0) q is quantity of energy transferring.

19. Phase Changes and Heat Flow Endothermic Exothermic

20. Energy and Phase Changes • Absorbed heat increases kinetic energy of molecules. • Loss of kinetic energy is caused by release of heat by molecules.

21. Internal Energy • Internal energy (E): • State function ∆E = Efinal – Einitial • Sum of KE and PE of all components of the system • Types of molecular motion: • (a) translational • (b) rotational • (c) vibrational

22. Change in Internal Energy ∆E = q + w • ∆E= change in system’s internal energy • q = heat, w = work • Work • w = –P∆V • Where P = pressure, V = change in volume • Work done by the system = energy lost by the system, hence the negative sign. • ∆E = q + w = q + (–P∆V) = q–P∆V

23. Units of Energy • Calorie (cal) • The amount of heat necessary to raise the temperature of 1 g of water by 1oC • Joule (J) • The SI unit of energy • 4.184 J = 1 cal • Energy = heat and/or work (same units!)

24. First Law of Thermodynamics • First law of thermodynamics = law of conservation of energy • Energy of the universe is constant! • Universe = system + surroundings • Energy gained or lost by a system must equal the energy lost or gained by the surroundings. • ∆Esystem= –∆Esurroundings

25. Energy Flow Diagram

26. Practice: Calculation of Work • Calculate the work in L• atm and joules associated with the expansion of a gas in a cylinder from 54 L to 72 L at a constant external pressure of 18 atm. (Note: 1 L • atm = 101.32 J) • Collect and Organize: • Expansion of gas in a cylinder: Vinitial = 54 L Vfinal= 72 L P = 18 atm • We need to calculate the amount of work performed.

27. Practice: Calculation of Work (cont. 1) • Calculate the work in L• atm and joules associated with the expansion of a gas in a cylinder from 54 L to 72 L at a constant external pressure of 18 atm. (Note: 1 L • atm = 101.32 J) • Analyze: • w= –PV • 1 L·atm = 101.32 J • The sign associated with w will be negative, since work is done by the system.

28. Practice: Calculation of Work (cont. 2) • Calculate the work in L• atm and joules associated with the expansion of a gas in a cylinder from 54 L to 72 L at a constant external pressure of 18 atm. (Note: 1 L • atm = 101.32 J) • Solve:

29. Practice: Calculation of Work (cont. 3) • Calculate the work in L• atm and joules associated with the expansion of a gas in a cylinder from 54 L to 72 L at a constant external pressure of 18 atm. (Note: 1 L • atm = 101.32 J) • Think About It: • The negative sign indicates work done by the system—energy lost from the system to the surroundings.

30. Enthalpy, Change in Enthalpy • Enthalpy (H): The sum of the internal energy and the pressure−volume product of a system H = E + PV • Enthalpy change (∆H): The energy absorbed by reactants (endothermic) or energy given off by products (exothermic) for a reaction carried out at constant pressure: ∆H = ∆E + P∆V

31. Enthalpy, Change in Enthalpy (cont.) • Enthalpy change (∆H) ∆H = ∆E + P∆V ∆H = qP = ∆E + P∆V • ∆H > 0, endothermic; ∆H < 0, exothermic • Subscripts indicate ∆H for specific processes.

32. Enthalpy Change Exothermic Endothermic

33. Heating Curves Heat in → kinetic energy

34. Heating Curves (cont. 1) Heat in → phase change

35. Heat Capacities • Molar heat capacity (cP): • Quantity of energy required to raise the temperature of 1 mole of a substance by 1oC q = ncPTcP= J/(mol • oC) • Specific heat (cs): • The energy required to raise the temperature of 1 gram of a substance by 1oC (at constant P) q = ncsTcs= J/(g • oC) • Heat capacity (CP): • Quantity of energy needed to raise the temperature of an object by 1oC (at constant P)

36. Molar Heat Capacities

37. Enthalpy in Change of State • Molar heat of fusion (Hfus): • The energy required to convert 1 mole of a solid at its melting point into the liquid state q = nHfus • Molar heat of vaporization (Hvap): • The energy required to convert 1 mole of a liquid at its boiling point to the vapor state q = nHvap

38. Phase Changes in Heating Curve

39. Practice: Specific Heat Capacity • During a strenuous workout, a student generates 2,000 kJ of heat energy. What mass of water would have to evaporate from the student’s skin to dissipate this much heat? • Collect and Organize: • We know: q = 2000 kJ • We want to know the mass of water.

40. Practice: Specific Heat Capacity (cont. 1) • During a strenuous workout, a student generates 2,000 kJ of heat energy. What mass of water would have to evaporate from the student’s skin to dissipate this much heat? • Analyze:q = nHvap • q = 2000 kJ • Hvap (H2O) = 40.67 kJ/mol • Solve for moles, then convert moles to mass.

41. Practice: Specific Heat Capacity (cont. 2) • During a strenuous workout, a student generates 2,000 kJ of heat energy. What mass of water would have to evaporate from the student’s skin to dissipate this much heat? • Solve:

42. Practice: Specific Heat Capacity (cont. 3) • During a strenuous workout, a student generates 2,000 kJ of heat energy. What mass of water would have to evaporate from the student’s skin to dissipate this much heat? • Think About It: • 2000 kJ is a significant amount of energy, so we would expect that it would evaporate a significant amount of water. This is consistent with our answer.

43. Cooling Curves • Heat transfer: ice @ –8.0C to water @ 0.0C • Temp change: q1 = ncPT • Phase change: q2 = nHfus

44. Practice: Calculating Energy • Calculate the amount of energy required to convert 237 g of solid ice at 0.0°C to hot water at 80.0°C. Hfus = 6.01 kJ/mol, cP= 75.3 J/(mol • °C) • Collect and Organize: 237 g H2O (ice) final temperature = 80.0°C Initial temp = 0.0°C Hfus = 6.01 kJ/mol cP = 75.3 J/(mol• °C)

45. Practice: Calculating Energy (cont. 1) • Calculate the amount of energy required to convert 237 g of solid ice at 0.0°C to hot water at 80.0°C. Hfus = 6.01 kJ/mol, cP= 75.3 J/(mol • °C) • Analyze: Convert mass of water to moles of water. Calculate energy of ice converting to liquid H2O. Heat the water from 0.0°C to 80.0°C.

46. Practice: Calculating Energy (cont. 2) • Calculate the amount of energy required to convert 237 g of solid ice at 0.0°C to hot water at 80.0°C. Hfus = 6.01 kJ/mol, cP= 75.3 J/(mol • °C) • Solve:

47. Practice: Calculating Energy (cont. 3) • Calculate the amount of energy required to convert 237 g of solid ice at 0.0°C to hot water at 80.0°C. Hfus = 6.01 kJ/mol, cP= 75.3 J/(mol • °C) • Think About It: • We need to first take energy to melt ice into water and then we need to heat the water up. These two processes will require the addition of much energy.

48. Calorimetry • Calorimetry: • The measurement of quantity of heat transferred during a physical or chemical process • Calorimeter: • A device used to measure the absorption or release of heat by a physical or chemical process Closed system: –qsystem= qcalorimeter

49. Measuring Molar Heat Capacity • (a) 23.5 g of Al beads are heated to 100.0C in boiling water. • (Heat transferred from water to Al until the final temp of Al beads = 100.0C)

50. Measuring Molar Heat Capacity (cont. 1) • (b) Al beads (at 100C) are placed in a calorimeter containing 130.0 g of water at 23.0C. • (c) Heat transferred from Al beads to water until thermal equilibrium is achieved at 26.0C qaluminum= –qwater mAlcs,AlTal= –(mwatercs,waterTwater)