1. Thermochemistry Chapter 5

2. Introduction All chemical and physical changes involve energy Thermodynamics is the study of energy and its transformations Thermochemistry is the study of the energy changes involved in chemical reactions

3. 5.1 The Nature of Energy Key Concepts Thermodynamics Thermochemistry Kinetic energy Potential energy System Surroundings Work Force Heat

4. 5.1 The Nature of Energy Three fundamental forms of energy Kinetic energy Potential energy Mass energy E = mc 2 discussed elsewhere

5. Kinetic Energy Any object in motion possesses KE or E k KE = ½ mv2 Examples A cow of mass 2000 kg, walking at a speed of 0.5 m/s has…. KE = 0.5 x 2000 kg x (0.5 m/s) 2 = 250 Joules KE1 J = 1 kg∙m 2 /s 2 How fast would a 2.0 g fly have to fly to have the same KE? = 500 m/s =1118 mph

6. Potential Energy Potential energy is “stored” energy Energy possessed by an object by virtue of its position relative to other objects Gravity imparts PEPE = mgh Electrostatic and chemical energy are forms of PE

7. Energy Content Total energy of an object is the sum of KE and all forms of PE E tot = KE + Σ PE Example: A biker at the top of a hill has PE relative to the bottom of the hill A biker riding down the hill has both KE & PE As a biker reaches the bottom of the hill, all PE has been converted to KE

8. Conservation of Energy

9. Units of Energy Joule 1 J = 1 N∙m = 1kg∙m 2 /s 2 1 cal = 4.184 J calorie The energy required to increase the temperature of 1 g of water by 1°C Nutritional calories … Are actually kilocalories 1 Cal = 1000 cal = 1 kcal

10. System & Surroundings Thermodynamic studies are conducted within a defined system System: portion of the universe defined for study Surroundings: everything else Open system can exchange energy with surroundings Close system cannot exchange energy and matter with surroundings self-contained –self-contained –Isolated from surroundings

11. Transferring Energy: Work and Heat What is work? Force: a push or pull exerted on an object Work is the energy used to cause an object to move Work is the product of Force x distance w = FdUnits? Energy is transferred as work

12. Heat Energy is also transferred as heat. What is heat? Heat is energy transferred from an object of higher temperature to an object of lower temperature Heat is not temperature!

13. Energy So energy can be defined as… Capacity to do work or to transfer heat ΔE = w + q

14. Sample Problem A bowler lifts a 5.4-kg bowling ball from ground level to a height of 1.6 m, and then drops it back to the ground a)What happens to PE as it is raised? b)How much work was done to raise the ball? c)What is the speed of the ball at impact? d)What is the weight of the ball in newtons?

15. More sample problems Calculate the KE in J for An Ar atom moving with a speed of 650 m/s A mole of Ar atoms moving at the speed of 650 m/s

16. 5.2 Key Concepts Internal energy First law of thermodynamics Endothermic Exothermic State functions

17. 5.2 The First Law of Thermodynamics Energy is conserved Energy is neither created nor destroyed, but can be transformed. The total energy of the universe is a constant

18. First Law of Thermodnamics If the energy of a system decreases, it must be transferred to the surroundings If energy is gained by a system, it must be transferred from the surroundings

19. Internal Energy Sum of all KE and PE of all components of a system E = ΣKE + ΣPE Change in internal energy is … ∆E = E f - E i E cannot be known, but ∆E can be measured

20. Internal Energy E = the sum of all kinetic and potential energies of all components of the system

21. Internal Energy Thermodynamic quantities must be described with a number, a unit, and a sign When the system has absorbed energy from its surroundings… +∆E….. E f > E i, therefore ∆E is positive When the system loses energy to its surroundings… -∆E ….. E f < E i, therefore ∆E is negative

22. Changing Internal Energy A system of H 2 and O 2 gas has more internal energy than one composed of H 2 O (l) E is lost to surroundings when H 2 O (l)is formed E is gained when H 2 O (l) is decomposed

23. Increasing Internal Energy Heat & work are added to the system E f > E i Internal energy increases ∆E = E f -E i ∆E > 0 ∆E = q + w

24. Sample Problem Relating ∆E to heat & work H 2 (g) & O 2 (g) ignite System loses 1150J of heat to surroundings As gases expand, piston rises 480 J work done on the piston. What ∆E? ∆E = q + w ∆E = -1150J + (-)480J ∆E = -1630 J 1630 J was transferred from system to the surroundings

25. Endothermic & Exothermic Processes Endothermic…. System absorbs heat from surroundings Exothermic…. System releases heat into surroundings

26. Energy Diagram Exothermic Reaction

27. State – specific characteristics of a system (physical or chemical) Examples: Pressure Temperature Composition State Function – a property of a system that characterizes a state of a system and is independent of how the system got to that state. Examples: Pressure Volume Composition www70.homepage.villanova.edu/kathleen.thrush/chapter_6_powerpoint_le.ppt

28. Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. energy, pressure, volume, temperature  E = E final - E initial  P = P final - P initial  V = V final - V initial  T = T final - T initial Energy it took each hiker to get to the top is different !! www70.homepage.villanova.edu/kathleen.thrush/chapter_6_powerpoint_le.ppt

29. State Functions In the first law of thermodynamics ΔE = q + w ΔE is a state function q and w are not See fig 5.8 p. 152

30. 5.3 Enthalpy Symbol H Like internal energy, enthalpy cannot be measured, but ∆H can ∆H equals the heat, q p, gained or lost by a system under constant pressure Exothermic processes: ∆H < 0 Endothermic processes: ∆H > 0 ∆H = H final – H initial = q p

31. 5.4 Enthalpies of Reaction Thermochemical Equations ∆H = H final – H initial ∆H rxn = H products – H reactants 2H 2 (g) + O 2 (g)  2H 2 O (g) ∆H = -486.3 kJ When 2 moles of H 2 gas combusts to form 2 moles of H 2 O under conditions of constant pressure, 486.3 kJ of heat are released by the system. Coefficients indicate molar amounts associated with ∆H rxn

32. Enthalpy Diagrams

33. Guidelines for Thermochemical Equations 1.Enthalpy is an extensive property CH 4 (g) + O 2 (g)  CO 2 (g) + 2H 2 O(l) ∆H = -890 kJ 2.∆H for reaction is equal in magnitude but opposite in sign to ∆H for the reverse rxn CO 2 (g) + 2H 2 O(l)  CH 4 (g) + O 2 (g) ∆H = +890 kJ 3. ∆H depends on the state of the reactants & products CH 4 (g) + O 2 (g)  CO 2 (g) + 2H 2 O(g) ∆H = -802 kJ 2H 2 O(l)  2H 2 O(g) ∆H = +88 kJ

34. 5.5 Calorimetry Calorimetry is the measure of heat flow involved in a chemical reaction Heat capacity: amount of heat energy required to raise the temperature of an object by 1K Molar heat capacity: amount of heat energy required to raise the temperature of 1 mole of a pure substance by 1K Specific heat: amount of heat energy required to raise the temperature of 1 g of a substance by 1K = heat capacity of 1 gram

35. Determining specific heat During a thermochemical process, –Measure mass of substance –Measure change in temperature –Measure heat energy transferred Specific heat = q/mΔT Units: J/g∙K q = mc(T f -T i )

36. Sample Problem 5.5 How much heat is needed to warm 250g of water from 22C to 98C. (Specific heat of water is 4.18 J/gK) Determine the molar heat capacity of water

37. Constant Pressure Calorimetry Calorimeter An insulated container in which aqueous chemical reactions are conducted If the mass and specific heat of the solvent (water) is known, q can be calculated if ΔT of the water is measured q soln = - q rxn and q rxn = - q soln q rxn = - (m soln )(c soln )ΔT

38. Sample 5.6 50 mL of 1.0 M HCl mixed with 50 mL of 1.0 M NaOH. Given: T i = 21.0C T f = 27.5C; density of soln = 1.0 g/mL; c soln = 4.18 J/gK Calculate enthalpy change for the neutralization reaction m soln = V x d = 100mL x 1.0 g/mL = 100 g ΔT soln = T f - T i = 27.5 – 21.0C = 6.5C = 6.5K

39. Sample 5.6 q rxn = - q soln = - (m soln )(c soln )ΔT soln = - (100 g)(4.18J/gK)(6.5K) = - 2700 J = - 2.7 kJ How would you express this change on a molar basis?

40. Sample 5.6 At constant pressure ΔH = q Express ΔH on a molar basis HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Molar quantities of reacants = (0.050 L)(1 mol/L) = 0.050 mol ΔH = (-2.7 kJ/0.050 mol) = -54 kJ/mol

41. 5.6 Hess’s Law If a rxn is carried out in steps, ΔH rxn = sum of ΔH of each step. ΔH rxn = Σ ΔH 1 + ΔH 2 + ΔH 3 + …step. A  D ΔH rxn = -850 kJnet rxn A  B ΔH 1 = -175 kJstep 1 B  C ΔH 2 = -400 kJstep 2 C  D ΔH 3 = -275kJstep 3

42. Hess’s Law Eq 1 CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) ΔH = -802 kJ Eq 2 2H 2 O (g)  2H 2 O (l) ΔH = -88 kJ Add Eq 1 + Eq 2 CH 4 (g)+2O 2 (g)+2H 2 O (g)  CO 2 (g)+2H 2 O(g)+2H 2 O (l) ΔH = -890 kJ Net Eq CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (l) ΔH = -890 kJ

43. Using Hess’s Law to Calculate ΔH Sample 5.8 Given: (1)C(s) + O 2 (g) → CO 2 (g) ΔH 1 = -393.5 kJ (2)CO(g) + ½ O 2 (g) → CO 2 (g) ΔH 2 = -283.0 kJ Determine the ΔH for the reaction (3) C(s) + ½ O 2 (g) → CO(g) ΔH 3 = ? kJ

44. Using Hess’s Law to Calculate ΔH Sample 5.8 The basic idea: rearrange equations 1 and 2, if necessary, in such a way that, when added, produce equation 3

45. 5.7 Enthalpies of Formation Thermodynamic quantities: ΔHºStandard Enthalpy Standard conditions: P = 1 atm, T = 298K ΔH vap Enthalpy of vaporization Vaporize liquids to gases ΔH f Enthalpy of fusion Melting solids to liquids ΔH c Enthalpy of combustion ΔH f º Enthalpy of formation

46. Enthalpy of Formation ΔH f º for the formation of 1 mol of a substance from its elements 2 C (graphite) + 3H 2 (g) + ½ O 2 (g)  C 2 H 5 OH (l) ΔH f º = -277.7 kJ By definition, ΔH f º = 0 for any element

47. Calculation of  H using  H f ° We can use Hess’s law in this way:  H =   n  H f ° products –   m  H f ° reactants where n and m are the stoichiometric coefficients.

48.  H= [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)] = [(-1180.5 kJ) + (-1143.2 kJ)] – [(-103.85 kJ) + (0 kJ)] = (-2323.7 kJ) – (-103.85 kJ) = -2219.9 kJ C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) Calculation of  H using  H f °

49. Energy in Foods Most of the fuel in the food we eat comes from carbohydrates and fats.

50. Energy in Fuels The vast majority of the energy consumed in this country comes from fossil fuels.