1. THERMODYNAMICS

2. THERMODYNAMICS: Basic Concepts Thermodynamics: (from the Greek therme, meaning "heat" and, dynamis, meaning "power") is the study of energy conversion between heat and mechanical work, and subsequently the macroscopic variables such as temperature, volume and pressureGreekheatpower macroscopictemperature volumepressure Energy: capacity to do work Work: motion against an opposing force System: part of the world which we have interest or being investigated. Surrounding: is where we make our observations

3. Definition of Work: Work is motion against opposing force. Work is defined as a force acting through a displacement x, the displacement being in the direction of the force.

4. Consider a Gas Expansion Work: Initial State Final State P 1, V 1, TP 2, V 2, T

5. Consider a Gas Expansion Work:

7.  the external pressure is infinitetissimally smaller than the internal pressure at all stages of the expansion - reversible process. A reversible change in thermodynamics is a change that can be reversed by an infinitesimal modification of a variable.  in reversible processes, there is maximum amount of work done that could possible extracted from a process

8. Gas Expansion Work: Sample Problem A sample of 4.50 g of methane occupies 12.7 L at 310 K. (A)Calculate the work done when the gas expands isothermally against a constant external pressure of 200 Torr until its volume has increased by 3.3 L. (B) Calculate the work that would be done if the same expansion occurred reversibly.

9. Consider a Gas Expansion Work:

10. Definition of HEAT: Heat is the transfer of energy between two bodies that are at different temperatures. Heat appears at the boundary of the system. heat is transferred from the hotter object to the colder one. Heat is path dependent.

11. THE FIRST LAW OF THERMODYNAMICS: “Energy can neither be created nor destroyed but can be converted from one form to another.” Law of Conservation of Energy. ENERGY OF THE UNIVERSE: If a system undergoes an energy change, the surroundings must also undergo a change equal in magnitude but opposite in direction

12. Total ENERGY of a system; Internal Energy, U TOTAL ENERGY: Internal Energy, U:  energy associated with the chemical system  depends on the thermodynamic parameters such as T, P, V, composition, etc.  consists of translational, rotational, vibrational, electronic energies, as well as intermolecular interactions

13. Internal Energy, U: For a given system, we do not know the exact nature of U, we will be only interested in the changes in U for a particular process that the system undergoes. Mathematical expression of the first law of thermodynamics: or for an infinitesimal change: The change in the internal energy of a system in a given process is the sum of the heat exchange, q, between the system and its surroundings and the work done on or by the system.

14. HEAT and WORK sign conventions: ProcessSign Work done by the system on the surroundings - Work done on the system by the surroundings + Heat absorbed by the system from the surroundings + Heat absorbed by the surroundings from the system -

27. Constant volume adiabatic bomb calorimeter Adiabatic means no heat exchange with the surroundings Sample inside an oxygen filled container is ignited by an electrical discharge. Heat released is measured by the increase in the temperature of the water.

28. ENTHALPHY In processes carried out under constant P, Total enthalpy of a system can not be measured directly, so the change in enthalpy, ΔH, is a more useful value than H itself. For a constant P process;

29. Entalphy,  H vs. Internal Energy,  U Consider the following reaction: 2Na(s) + 2H 2 O(l) 2NaOH(aq) + H 2 (g) Heat evolved by the reaction is 367.5 kJ, reaction takes place at constant pressure, q p =  H = -367.5KJ. Internal Energy  U: The volume of the H 2 (1 mole) generated by the reaction occupies 24.5 L, therefore – P  V = -24.5 L at or -2.5 kJ Finally,  U = -367.5 kJ – 2.5 kJ  U = -370 kJ. - Difference is due to expansion work.

34. Constant Pressure Calorimeter: A constant-pressure calorimeter made of two plastic cups. The outer cup helps insulate the reacting mixture from the surroundings. Two solutions of known volume containing the reactants at the same temperature are carefully mixed in the calorimeter. The heat produced or absorbed by the reaction can be determined by the temperature change, the quantities and specific heats of the solutions used, and the heat capacity of the calorimeter.

35. HEAT CAPACITIES: Calorimetry: measurement of heat changes in chemical and physical processes. When heat is added to the system, the corresponding temperature rise will depend on a)The amount of heat delivered, b)The amount of the substance, c)The chemical nature and the physical state of the substance, d)The conditions at which the energy is added to the system.

36. HEAT CAPACITIES: Thus for a given amount of a substance, change in temperature is related to the heat added:

37. HEAT CAPACITY at constant volume

38. HEAT CAPACITY at constant pressure processes for gases, C p > C v because of the work to be done to the surroundings in constant pressure processes for condensed phases, C p and C v are identical for most purposes for ideal gases,

43. THERMOCHEMICAL EQUATIONS:

44. SAMPLE PROBLEM: How much heat is released when 75 g of octane is burned completely if the enthalpy of combustion is –5,500 kJ/mol C 8 H 18 ? C 8 H 18 + 25/2 O 2  8CO 2 + 9H 2 O

47. THERMOCHEMISTRY: Heat of reaction is the heat change in the transformation of reactants at a given temperature and pressure to products at the same conditions. For constant pressure processes, the heat of reaction q p is equal to enthalpy change of the reaction,  r H.

48. THERMOCHEMISTRY: 2H 2 (g) + O 2 (g) 2H 2 O (g) 241.8 kJ of heat is given off. The enthalpy change for this process is called standard enthalphy of reaction,  r H o In general, standard enthalpy change of a chemical reaction is the total enthalpy of the products minus the total enthalpy of the reactants

49. THERMOCHEMISTRY: - is the enthalpy change when 1 mole of a compound is formed from its constituent elements at 1 bar and 298 K.

50. THERMOCHEMISTRY: H 2 (g) + 1/2O 2 (g) H 2 O(g)

51. FORMATION REACTIONS: formation of 1 mole of a compound from its constituent elements at 1 bar and 298 K. Example: Formation reaction of NaOH: Na(s) + ½ O 2 (g) + ½ H 2 (g) NaOH(s)

55. MEASUREMENTS OF DIRECT METHOD: measure  f H o of direct synthesis from their elements: 2H 2 (g) + O 2 (g) H 2 O(g)

59. MEASUREMENTS OF INDIRECT METHOD: HESS’S LAW: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. for carbon monoxide: C(graphite) + 1/2O 2 (g) CO(g) C(graphite) + O 2 (g) CO 2 (g) CO + 1/2O 2 (g) CO 2 (g)

60. MEASUREMENTS OF for carbon monoxide: C(graphite) + O 2 (g) CO 2 (g) CO 2 (g) 1/2O 2 (g) + CO(g) C(graphite) + 1/2O 2 (g) CO(g)

61. MEASUREMENTS OF C(graphite) + O 2 (g) CO 2 (g)+ 393.5kJ C(graphite) + O 2 (g) CO 2 (g) Enthalpy changes are additive. the of the reverse reaction will have the opposite sign multiplying a reaction by a factor, you also multiply the by the same factor. Hess’s Law can be rephrased as: ‘ The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which the reaction might be divided.’

62. SAMPLE PROBLEM: Calculate the standard molar enthalpy of formation of acetylene (C 2 H 2 ) from its elements. 2C(graphite) + H 2 (g) C 2 H 2 (g) The equations for combustion and the corresponding enthalpy changes are: (1) C(graphite) + 1/2O 2 (g) CO 2 (g) (2) H 2 (g) + 1/2O 2 (g) H 2 O(l) (3) 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O(l)

63. SAMPLE PROBLEM: for carbon monoxide: C(graphite) + O 2 (g) CO 2 (g) CO 2 (g) 1/2O 2 (g) + CO(g) C(graphite) + 1/2O 2 (g) CO(g)

64. SAMPLE PROBLEM:

68. TEMPERATURE DEPENDENCE OF ENTHALPY CHANGE KIRCHHOFFS’s LAW: The difference in enthalpies of a reaction at two different temperatures, T 1 and T 2, is just the difference in the enthalpies of heating the products and reactants from T 1 to T 2. Where  C p is the difference in molar heat capacities between the products and reactants

69. Bond Dissociation Enthalpy: Bond Enthalpy: change in enthalpy when bonds form or break in diatomic molecule Bond Dissociation Enthalpy: average change in enthalpy in polyatomic molecules when individual bonds form or break. N 2 (g) 2N(g) H 2 O(g) H(g) + OH(g)  r H o = 502 kJ/mol OH(g) H(g) + O(g)  r H o = 427 kJ/mol

70. SAMPLE PROBLEM:

71. ENTHALPY CHANGE and BOND ENERGIES: The Enthalpy of a Reaction can be estimated by the enthalpies of the total number of bonds broken and formed in the reaction.

72. SAMPLE PROBLEM: Estimate the enthalpy of combustion of methane: CH 4 (g) + O 2 (g) CO 2 (g) + H 2 O(g) Using bond enthalpies in Table 4.4. Compare your result with that calculated from the enthalpies of formation of products and reactants. C-H410 kJ/mole O=O494 kJ/mole C=O563 kJ/mole H-O460 kJ/mole

73. The Second Law of Thermodynamics

74. LIMITATION OF THE FIRST LAW: Does not address whether a particular process is spontaneous or not. Deals only with changes in energy. Consider this examples: Drop a rock from waist-high height, rock will fall spontaneously Plunger of a spray is presses, gas comes out spontaneously Metallic sodium is placed in a jar with chlorine gas, reaction occurs

75. Spontaneous Processes: Why does the color spread when placing a drop of dye in a glass of clean water?

76. SPONTANEOUS PROCESSES: Spontaneous processes are those that can proceed without any outside intervention. The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously go back to B.

77. SPONTANEOUS PROCESSES: Processes that are spontaneous in one direction are not spontaneous in the reverse direction.

78. SPONTANEOUS PROCESSES Processes that are spontaneous at one temperature may be not spontaneous at other temperatures. Above 0  C it is spontaneous for ice to melt. Below 0  C the reverse process is spontaneous

79. SPONTANEOUS PROCESSES: Changes in the extent of disorder. When we spill a bowl of sugar, why do the grains go everywhere and cause such a mess? natures’s way to seek disorder. It is easy to create disorder; difficult to create order.

80. ENTROPY: Entropy can be thought of as a measure of the randomness or disorder of a system. It is related to the various modes of motion in molecules.

81. The Concept of Entropy (S) Entropy refers to the state of order. A change in order is a change in the number of ways of arranging the particles and dispersing their energy of motion, and it is a key factor in determining the direction of a spontaneous process. solid liquid gas more orderless order crystal + liquid ions in solution more orderless order more orderless order crystal + crystal gases + ions in solution

82. PROCESSES THAT RESULT IN PREDICTABLE ENTROPY CHANGES: 1.Phase Changes 1.Temperature Changes 1.Volume Changes 1.Mixing of substances 1.Increase in number of particles 1.Changes in the number of moles of gaseous components 7. Atomic size or molecular complexity

88. Spontaneity and the Sign of  S Why does a room with a fragrance bottle at the other end of the room is suddenly filled with the aroma? Limonene (l) Limonene (g)  S(process) = S (final state) - S(initial state) A spontaneous process is accompanied by  S of positive sign.

89. Reaction Spontaneity by Inspection Why do damp clothes become dry when hung outside? S(g) >> S(l) > S(s) By inspection alone, decide whether the sublimation of solid carbon dioxide is spontaneous or not. How about the condensation of water?

90. THERMODYNAMIC DEFINITION OF ENTROPY: Although for the expansion of gases, they are applicable to all types of processes at constant temperature

91. The Second Law of Thermodynamics: The entropy of an isolated system increases in an irreversible processes and remain unchanged in a reversible process.

92. ENTROPY CHANGES PHASE CHANGE:

93. SAMPLE PROBLEM The enthalpy of vaporization of methanol is 35.27 kJ/mol at its normal boiling point of 64.1 o C. Calculate (a) the entropy of vaporization of methanol at this temperature

94. SAMPLE PROBLEM: Calculate the change in entropy when 1.0 mole of ice at -10 o C is heated until it is a liquid at 25 o C.

95. THIRD LAW OF THERMODYNAMICS: Every substance has a finite positive entropy, but at absolute zero the entropy maybe come zero, and it does in case of pure perfect crystalline substance.

96. THIRD LAW OF THERMODYNAMICS: The entropy of a perfect crystal at 0 K is zero. It is impossible to reach a temperature of absolute zero It is impossible to have a (Carnot) efficiency equal to 100% (this would imply Tc = 0).

97. THIRD LAW OF THERMODYNAMICS: T = 0, S = 0 T > 0, S > 0

98. ENTROPY OF CHEMICAL REACTIONS:

102. SAMPLE PROBLEM: Calculate the value of the standard molar entropy changes for the following reactions at 298 K.

103. The GIBBS Energy:

104. THE GIBBS ENERGY: Factors affecting HH SS GG ++ Positive at low Temp; negative at high Temp; Reaction is spontaneous at forward at high T and spontaneous in reverse direction at low temperature. +- Positive at all temperatures, Reaction is spontaneous in the reverse reaction at ll temperatures -+ Negative at all temperatures. Reaction is spontaneous in the forward direction at all T. -- Negative at low temperatures; positive at high temperatures. Reaction is spontaneous at Low temperatures. Tends to reverse at high temperatures.

105. The HELMHOLTZ Energy:

106. STANDARD MOLAR GIBBS ENERGY OF FORMATION

107. GIBBS FREE ENERGY, G ∆G o = ∆H o - T∆S o ∆G o = ∆H o - T∆S o Two methods of calculating ∆G o A. Determine ∆H o rxn and ∆S o rxn and use GIbbs equation. B. Use tabulated values of free energies of formation, ∆G f o. ∆G o rxn =  ∆G f o (products) -  ∆G f o (reactants)

108. FREE ENERGIES OF FORMATION Note that ∆G˚ f for an element = 0

112. SAMPLE CALCULATION, ∆G o rxn For the combustion of acetylene C 2 H 2 (g) + 5/2 O 2 (g) --> 2 CO 2 (g) + H 2 O(g) a)by inspection is the reaction spontaneous or not? b)Calculate the ∆G o rxn using standard molar enthalpies and entropies. c) Is the reaction spontaneous or not? Is it entropy or enthalpy driven? ∆H (acetylene) 52.26 s = 219.56 g = 68.15 O g = 205.14 Co2 h=-393.51 s = 213.74 g = -394.36 H20 h = -241.82 s 188.83 g = -228.57

113. FREE ENERGY AND TEMPERATURE Iron metal can be produced by reducing its ore (Iron(III) oxide with graphite: 2 Fe 2 O 3 (s) + 3 C(s) ---> 4 Fe(s) + 3 CO 2 (g) ∆H o rxn = +467.9 kJ ∆S o rxn = +560.3 J/K ∆G o rxn = +300.8 kJ A) Is the reaction spontaneous or not? B) At what temperature will the reaction become spontaneous? At what T does ∆G o rxn just change from being (+) to being (-)? When ∆G o rxn = 0 = ∆H o rxn - T∆S o rxn

118.  FACT: ∆G o rxn is the change in free energy when pure reactants convert COMPLETELY to pure products.  FACT: Product-favored systems have K eq > 1.  Therefore, both ∆G˚ rxn and K eq are related to reaction favorability. Thermodynamics and K eq

119. K eq is related to reaction favorability and so to ∆G o rxn. The larger the value of K the more negative the value of ∆G o rxn ∆G o rxn = - RT lnK ∆G o rxn = - RT lnK where R = 8.31 J/Kmol THERMODYNAMICS AND K eq