2. Thermodynamics

3. Antoine Lavoisier [1743-94]

5. Julius Robert Meyer

6. James Joule : 1818~1889

7. Hermann von Helmholtz (1821-1894)

8. Rudolf Clausius (1822--1888),

9. THERMOCHEMISTRY The study of heat released or required by chemical reactions Fuel is burnt to produce energy - combustion (e.g. when fossil fuels are burnt) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) + energy

10. What is Energy? Energy Kinetic energy (E K ) Potential energy (E P ) Energy due to motion Energy due to position (stored energy)

11. Total Energy = Kinetic Energy + Potential Energy E = E K + E P Kinetic energy & potential energy are interchangeable Ball thrown upwards slows & loses kinetic energy but gains potential energy The reverse happens as it falls back to the ground

12. Law of Conservation of Energy Law of Conservation of Energy: the total energy of the universe is constant and can neither be created nor destroyed; it can only be transformed. internal energy The internal energy, U, of a sample is the sum of all the kinetic and potential energies of all the atoms and molecules in a sample i.e. it is the total energy of all the atoms and molecules in a sample

13. Systems & Surroundings In thermodynamics, the world is divided into a system and its surroundings A system is the part of the world we want to study (e.g. a reaction mixture in a flask) The surroundings consist of everything else outside the system SYSTEM CLOSED OPEN ISOLATED

14. Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. open mass & energyExchange: closed energy isolated nothing SYSTEM SURROUNDINGS

15. OPEN SYSTEM: can exchange both matter and energy with the surroundings (e.g. open reaction flask, rocket engine) CLOSED SYSTEM: can exchange only energy with the surroundings (matter remains fixed) e.g. a sealed reaction flask ISOLATED SYSTEM: can exchange neither energy nor matter with its surroundings (e.g. a thermos flask)

16. HEAT and WORK HEAT is the energy that transfers from one object to another when the two things are at different temperatures and in some kind of contact e.g. kettle heats on a gas flame cup of tea cools down (loses energy as heat) Thermal motion (random molecular motion) is increased by heat energy i.e. heat stimulates thermal motion

17. Work is the transfer of energy that takes place when an object is moved against an opposing force i.e. a system does work when it expands against an external pressure Car engine: petrol burns & produces gases which push out pistons in the engine and transfer energy to the wheels of car Work stimulates uniform motion Heat and work can be considered as energy in transit

18. UNITS OF ENERGY S.I. unit of energy is the joule (J) Heat and work ( energy in transit) also measured in joules 1 kJ (kilojoule) = 10 3 J Calorie (cal): 1 cal is the energy needed to raise the temperature of 1g of water by 1 o C 1 cal = 4.184 J

19. INTERNAL ENERGY (U) Internal energy changes when energy enters or leaves a system  U = U final - U initial  U change in the internal energy Heat and work are 2 equivalent ways of changing the internal energy of a system

20. += Change in internal energy Energy supplied to system as heat Energy supplied to system as work  U = q (heat) + w (work) q w q w U U like reserves of a bank: bank accepts deposits or withdrawals in two currencies (q & w) but stores them as common fund, U.

21. First Law of Thermodynamics: the internal energy of an isolated system is constant Signs (+/-) will tell you if energy is entering or leaving a system + indicates energy enters a system - indicates energy leaves a system

22. EXPANSION WORKAn important form of work is EXPANSION WORK i.e. the work done when a system changes size and pushes against an external force e.g. the work done by hot gases in an engine as they push back the pistons WORK HEAT In a system that can’t expand, no work is done (w = 0)  U = q + w when w = 0,  U = q (at constant volume)

23. A change in internal energy can be identified with the heat supplied at constant volume ENTHALPY (H) (comes from Greek for “heat inside”) the change in internal energy is not equal to the heat supplied when the system is free to change its volume some of the energy can return to the surroundings as expansion work   U < q

24. The heat supplied is equal to the change in another thermodynamic property called enthalpy (H) i.e.  H = q this relation is only valid at constant pressure As most reactions in chemistry take place at constant pressure we can say that: A change in enthalpy = heat supplied

25. EXOTHERMIC & ENDOTHERMIC REACTIONS Exothermic process: a change (e.g. a chemical reaction) that releases heat. A release of heat corresponds to a decrease in enthalpy Exothermic process:  H < 0 (at constant pressure) Burning fossil fuels is an exothermic reaction

26. Endothermic process: a change (e.g. a chemical reaction) that requires (or absorbs) heat. An input of heat corresponds to an increase in enthalpy Endothermic process:  H > 0 (at constant pressure) Photosynthesis is an endothermic reaction (requires energy input from sun) Forming Na + and Cl - ions from NaCl is an endothermic process

27. Measuring Heat reaction Exothermic reaction, heat given off & temperature of water rises Endothermic reaction, heat taken in & temperature of water drops

28. How do we relate change in temp. to the energy transferred? Heat capacity (J/ o C) = heat supplied (J) temperature ( o C) Heat Capacity = heat required to raise temp. of an object by 1 o C more heat is required to raise the temp. of a large sample of a substance by 1 o C than is needed for a smaller sample

29. b. Specific heat is the amount of heat required to raise the temperature of 1 kg of a material by one degree (C or K). 1) C water = 4184 J / kg C 2) C sand = 664 J / kg C This is why land heats up quickly during the day and cools quickly at night and why water takes longer.

30. Why does water have such a high specific heat? Water molecules form strong bonds with each other; therefore it takes more heat energy to break them. Metals have weak bonds and do not need as much energy to break them.

31. How to calculate changes in thermal energy Q = m x  T x C p Q = change in thermal energy m = mass of substance  T = change in temperature (T f – T i ) C p = specific heat of substance

32. c. A calorimeter is used to help measure the specific heat of a substance. First, mass and temperature of water are measured Then heated sample is put inside and heat flows into water  T is measured for water to help get its heat gain This gives the heat lost by the substance Knowing its Q value, its mass, and its  T, its Cp Cp can be calculated

33. Specific heat capacity is the quantity of energy required to change the temperature of a 1g sample of something by 1 o C Specific Heat Capacity (Cs) Heat capacity Mass = J / o C / gJ / o C g =

34. Vaporisation Energy has to be supplied to a liquid to enable it to overcome forces that hold molecules together endothermic process (  H positive) Melting Energy is supplied to a solid to enable it to vibrate more vigorously until molecules can move past each other and flow as a liquid endothermic process (  H positive) Freezing Liquid releases energy and allows molecules to settle into a lower energy state and form a solid exothermic process (  H negative) (we remove heat from water when making ice in freezer)

35. Reaction Enthalpies All chemical reactions either release or absorb heat Exothermic reactions: Reactants products + energy as heat (  H -ve) Endothermic reactions: Reactants + energy as heat products (  H +ve) e.g. burning fossil fuels e.g. photosynthesis

36. Bond Strengths Bond strengths measured by bond enthalpy  H B (+ve values) bond breaking requires energy (+ve  H) bond making releases energy (-ve  H) Lattice Enthalpy A measure of the attraction between ions (the enthalpy change when a solid is broken up into a gas of its ions) all lattice enthalpies are positive I.e. energy is required o break up solids

37. Enthalpy of hydration  H hyd the enthalpy change accompanying the hydration of gas- phase ions Na + (g) + Cl - (g) Na + (aq) + Cl - (aq) -ve  H values (favourable interaction) WHY DO THINGS DISSOLVE? If dissolves and solution heats up : exothermic If dissolves and solution cools down: endothermic

38. Breaking solid into ions Ions associating with water Dissolving += Lattice Enthalpy + Enthalpy of Hydration = Enthalpy of Solution Substances dissolve because energy and matter tend to disperse (spread out in disorder) 2 nd law of Thermodynamics

39. Making H 2 O from H 2 involves two steps. H 2 (g) + 1/2 O 2 (g) ---> H 2 O(g) + 242 kJ H 2 O(g) ---> H 2 O(liq) + 44 kJ ----------------------------------------------------------------------- H 2 (g) + 1/2 O 2 (g) --> H 2 O(liq) + 286 kJ Example of HESS’S LAW — If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns. Making H 2 O from H 2 involves two steps. H 2 (g) + 1/2 O 2 (g) ---> H 2 O(g) + 242 kJ H 2 O(g) ---> H 2 O(liq) + 44 kJ ----------------------------------------------------------------------- H 2 (g) + 1/2 O 2 (g) --> H 2 O(liq) + 286 kJ Example of HESS’S LAW — If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns. USING ENTHALPY

40. Hess’s Law & Energy Level Diagrams Forming H 2 O can occur in a single step or in a two steps. ∆H total is the same no matter which path is followed.

41. Hess’s Law & Energy Level Diagrams Forming CO 2 can occur in a single step or in a two steps. ∆H total is the same no matter which path is followed. Forming CO 2 can occur in a single step or in a two steps. ∆H total is the same no matter which path is followed.

42. This equation is valid because ∆H is a STATE FUNCTIONThis equation is valid because ∆H is a STATE FUNCTION These depend only on the state of the system and not on how the system got there.These depend only on the state of the system and not on how the system got there. V, T, P, energy — and your bank account!V, T, P, energy — and your bank account! Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H.Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H. This equation is valid because ∆H is a STATE FUNCTIONThis equation is valid because ∆H is a STATE FUNCTION These depend only on the state of the system and not on how the system got there.These depend only on the state of the system and not on how the system got there. V, T, P, energy — and your bank account!V, T, P, energy — and your bank account! Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H.Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H.  ∆H along one path =  ∆H along another path  ∆H along another path  ∆H along one path =  ∆H along another path  ∆H along another path

43. Standard Enthalpy Values Most ∆H values are labeled ∆H o Measured under standard conditions P = 1 bar = 10 5 Pa = 1 atm /1.01325 Concentration = 1 mol/L T = usually 25 o C with all species in standard states e.g., C = graphite and O 2 = gas Most ∆H values are labeled ∆H o Measured under standard conditions P = 1 bar = 10 5 Pa = 1 atm /1.01325 Concentration = 1 mol/L T = usually 25 o C with all species in standard states e.g., C = graphite and O 2 = gas

44. Enthalpy Values H 2 (g) + 1/2 O 2 (g) --> H 2 O(g) ∆H˚ = -242 kJ 2 H 2 (g) + O 2 (g) --> 2 H 2 O(g) ∆H˚ = -484 kJ H 2 O(g) ---> H 2 (g) + 1/2 O 2 (g) ∆H˚ = +242 kJ H 2 (g) + 1/2 O 2 (g) --> H 2 O(liquid) ∆H˚ = -286 kJ H 2 (g) + 1/2 O 2 (g) --> H 2 O(g) ∆H˚ = -242 kJ 2 H 2 (g) + O 2 (g) --> 2 H 2 O(g) ∆H˚ = -484 kJ H 2 O(g) ---> H 2 (g) + 1/2 O 2 (g) ∆H˚ = +242 kJ H 2 (g) + 1/2 O 2 (g) --> H 2 O(liquid) ∆H˚ = -286 kJ Depend on how the reaction is written and on phases of reactants and products

45. Standard Enthalpy Values NIST (Nat’l Institute for Standards and Technology) gives values of ∆H f o = standard molar enthalpy of formation ∆H f o = standard molar enthalpy of formation — the enthalpy change when 1 mol of compound is formed from elements under standard conditions. See Table 6.2 NIST (Nat’l Institute for Standards and Technology) gives values of ∆H f o = standard molar enthalpy of formation ∆H f o = standard molar enthalpy of formation — the enthalpy change when 1 mol of compound is formed from elements under standard conditions. See Table 6.2

46. ∆H f o, standard molar enthalpy of formation Enthalpy change when 1 mol of compound is formed from the corresponding elements under standard conditions H 2 (g) + 1/2 O 2 (g) --> H 2 O(g) ∆H f o (H 2 O, g)= -241.8 kJ/mol By definition, ∆H f o = 0 for elements in their standard states. Enthalpy change when 1 mol of compound is formed from the corresponding elements under standard conditions H 2 (g) + 1/2 O 2 (g) --> H 2 O(g) ∆H f o (H 2 O, g)= -241.8 kJ/mol By definition, ∆H f o = 0 for elements in their standard states.

47. Using Standard Enthalpy Values Use ∆H˚’s to calculate enthalpy change for H 2 O(g) + C(graphite) --> H 2 (g) + CO(g) Use ∆H˚’s to calculate enthalpy change for H 2 O(g) + C(graphite) --> H 2 (g) + CO(g) (product is called “water gas”)

48. Using Standard Enthalpy Values H 2 O(g) + C(graphite) --> H 2 (g) + CO(g) From reference books we find H 2 (g) + 1/2 O 2 (g) --> H 2 O(g) ∆H f ˚ = - 242 kJ/molH 2 (g) + 1/2 O 2 (g) --> H 2 O(g) ∆H f ˚ = - 242 kJ/mol C(s) + 1/2 O 2 (g) --> CO(g) ∆H f ˚ = - 111 kJ/molC(s) + 1/2 O 2 (g) --> CO(g) ∆H f ˚ = - 111 kJ/mol H 2 O(g) + C(graphite) --> H 2 (g) + CO(g) From reference books we find H 2 (g) + 1/2 O 2 (g) --> H 2 O(g) ∆H f ˚ = - 242 kJ/molH 2 (g) + 1/2 O 2 (g) --> H 2 O(g) ∆H f ˚ = - 242 kJ/mol C(s) + 1/2 O 2 (g) --> CO(g) ∆H f ˚ = - 111 kJ/molC(s) + 1/2 O 2 (g) --> CO(g) ∆H f ˚ = - 111 kJ/mol

49. Using Standard Enthalpy Values H 2 O(g) --> H 2 (g) + 1/2 O 2 (g) ∆H o = +242 kJ C(s) + 1/2 O 2 (g) --> CO(g)∆H o = -111 kJ C(s) + 1/2 O 2 (g) --> CO(g) ∆H o = -111 kJ --------------------------------------------------------------------- ----------- H 2 O(g) --> H 2 (g) + 1/2 O 2 (g) ∆H o = +242 kJ C(s) + 1/2 O 2 (g) --> CO(g)∆H o = -111 kJ C(s) + 1/2 O 2 (g) --> CO(g) ∆H o = -111 kJ --------------------------------------------------------------------- ----------- To convert 1 mol of water to 1 mol each of H 2 and CO requires 131 kJ of energy. The “water gas” reaction is ENDOthermic. H 2 O(g) + C(graphite) --> H 2 (g) + CO(g) ∆H o net = +131 kJ ∆H o net = +131 kJ H 2 O(g) + C(graphite) --> H 2 (g) + CO(g) ∆H o net = +131 kJ ∆H o net = +131 kJ

50. Using Standard Enthalpy Values In general, when ALL enthalpies of formation are known: Calculate ∆H of reaction? ∆H o rxn =  ∆H f o (products) -  ∆H f o (reactants) Remember that ∆ always = final – initial

51. Using Standard Enthalpy Values Calculate the heat of combustion of methanol, i.e., ∆H o rxn for CH 3 OH(g) + 3/2 O 2 (g) --> CO 2 (g) + 2 H 2 O(g) ∆H o rxn =  ∆H f o (prod) -  ∆H f o (react) ∆H o rxn =  ∆H f o (prod) -  ∆H f o (react) Calculate the heat of combustion of methanol, i.e., ∆H o rxn for CH 3 OH(g) + 3/2 O 2 (g) --> CO 2 (g) + 2 H 2 O(g) ∆H o rxn =  ∆H f o (prod) -  ∆H f o (react) ∆H o rxn =  ∆H f o (prod) -  ∆H f o (react)

52. Using Standard Enthalpy Values ∆H o rxn = ∆H f o (CO 2 ) + 2 ∆H f o (H 2 O) - {3/2 ∆H f o (O 2 ) + ∆H f o (CH 3 OH)} - {3/2 ∆H f o (O 2 ) + ∆H f o (CH 3 OH)} = (-393.5 kJ) + 2 (-241.8 kJ) = (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-201.5 kJ)} - {0 + (-201.5 kJ)} ∆H o rxn = -675.6 kJ per mol of methanol ∆H o rxn = ∆H f o (CO 2 ) + 2 ∆H f o (H 2 O) - {3/2 ∆H f o (O 2 ) + ∆H f o (CH 3 OH)} - {3/2 ∆H f o (O 2 ) + ∆H f o (CH 3 OH)} = (-393.5 kJ) + 2 (-241.8 kJ) = (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-201.5 kJ)} - {0 + (-201.5 kJ)} ∆H o rxn = -675.6 kJ per mol of methanol CH 3 OH(g) + 3/2 O 2 (g) --> CO 2 (g) + 2 H 2 O(g) ∆H o rxn =  ∆H f o (prod) -  ∆H f o (react) ∆H o rxn =  ∆H f o (prod) -  ∆H f o (react)

53. Measuring Heats of Reaction CALORIMETRYCALORIMETRY Constant Volume “Bomb” Calorimeter Burn combustible sample. Measure heat evolved in a reaction. Derive ∆E for reaction. Constant Volume “Bomb” Calorimeter Burn combustible sample. Measure heat evolved in a reaction. Derive ∆E for reaction.

54. CalorimetryCalorimetry Some heat from reaction warms water q water = (sp. ht.)(water mass)(∆T) Some heat from reaction warms “bomb” q bomb = (heat capacity, J/K)(∆T) Total heat evolved = q total = q water + q bomb

55. Calculate heat of combustion of octane. C 8 H 18 + 25/2 O 2 --> 8 CO 2 + 9 H 2 O Burn 1.00 g of octaneBurn 1.00 g of octane Temp rises from 25.00 to 33.20 o CTemp rises from 25.00 to 33.20 o C Calorimeter contains 1200 g waterCalorimeter contains 1200 g water Heat capacity of bomb = 837 J/KHeat capacity of bomb = 837 J/K Calculate heat of combustion of octane. C 8 H 18 + 25/2 O 2 --> 8 CO 2 + 9 H 2 O Burn 1.00 g of octaneBurn 1.00 g of octane Temp rises from 25.00 to 33.20 o CTemp rises from 25.00 to 33.20 o C Calorimeter contains 1200 g waterCalorimeter contains 1200 g water Heat capacity of bomb = 837 J/KHeat capacity of bomb = 837 J/K Measuring Heats of Reaction CALORIMETRY

56. Step 1Calc. heat transferred from reaction to water. q = (4.184 J/gK)(1200 g)(8.20 K) = 41,170 J Step 2Calc. heat transferred from reaction to bomb. q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 J = (837 J/K)(8.20 K) = 6860 J Step 3Total heat evolved 41,170 J + 6860 J = 48,030 J 41,170 J + 6860 J = 48,030 J Heat of combustion of 1.00 g of octane = - 48.0 kJ Step 1Calc. heat transferred from reaction to water. q = (4.184 J/gK)(1200 g)(8.20 K) = 41,170 J Step 2Calc. heat transferred from reaction to bomb. q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 J = (837 J/K)(8.20 K) = 6860 J Step 3Total heat evolved 41,170 J + 6860 J = 48,030 J 41,170 J + 6860 J = 48,030 J Heat of combustion of 1.00 g of octane = - 48.0 kJ Measuring Heats of Reaction CALORIMETRY

57. Second Law of Thermodynamics: the disorder (or entropy) of a system tends to increase ENTROPY (S) Entropy is a measure of disorder Low entropy (S) = low disorder High entropy (S) = greater disorder hot metal block tends to cool gas spreads out as much as possible

58. Total entropy change entropy change of system entropy change of surroundings += Dissolving disorder of solution disorder of surroundings must be an overall increase in disorder for dissolving to occur

59. 1. If we freeze water, disorder of the water molecules decreases, entropy decreases ( -ve  S, -ve  H) 2. If we boil water, disorder of the water molecules increases, entropy increases (vapour is highly disordered state) ( +ve  S, +ve  H)

60. A spontaneous change is a change that has a tendency to occur without been driven by an external influence e.g. the cooling of a hot metal block to the temperature of its surroundings A non-spontaneous change is a change that occurs only when driven e.g. forcing electric current through a metal block to heat it

61. A chemical reaction is spontaneous if it is accompanied by an increase in the total entropy of the system and the surroundings Spontaneous exothermic reactions are common (e.g. hot metal block spontaneously cooling) because they release heat that increases the entropy of the surroundings. Endothermic reactions are spontaneous only when the entropy of the system increases enough to overcome the decrease in entropy of the surroundings

62. System in Dynamic Equilibrium A + B C + D Dynamic (coming and going), equilibrium (no net change) no overall change in disorder   S  0 (zero entropy change)

63. انجام كارهاي يكسان و انتظار نتايج متفاوت ، ديوانگي محض است. آلبرت انيشتن