2. Definitions: Universe – composed of the system and the surroundings System – part being studied; that which we are focusing on Surroundings – everything else in the universe outside the system Open System – where both energy and matter can be transferred to and from surroundings Closed System – only energy can be transferred to and from surroundings Isolated System – one where neither energy nor matter can be transferred to and from surroundings Exothermic – Energy (as heat) flows out of the system Endothermic – Energy (as heat) flows into the system

3. Thermodynamics – the study of energy and its interconversions Energy – the capacity to produce work or heat Kinetic Energy – the energy of motion; KE = ½ mass x (velocity) 2 Potential Energy – Energy that can be converted to useful work Heat – involves transfer of energy between two objects (from hot to cold) Work – Force x distance State Functions – depend only upon the initial and final state of a substance Examples: Δ H, Δ S, Δ G A property that is independent of pathway. That is, it does not matter how you get there, the difference in the value is the same. For example, you can drive from New York to Los Angeles via many different routes. No matter which one you take, you are still going from NY to LA. The actual distance between the two cities is the same.

4. Enthalpy and Calorimetry ΔH = H products – H reactants The change in enthalpy ( Δ H) of the system is equal to the energy flow as heat at constant pressure. Δ H = q p If Δ H>0, the reaction is endothermic (Heat is absorbed by the system) If Δ H<0, the reaction is exothermic (Heat is given off by the system) Endothermic **Enthalpy of products > enthalpy of reactants ** Δ H = positive (energy must be put into reaction to occur) Exothermic **Enthalpy of products< enthalpy of reactants ** Δ H = negative (energy released from reaction as it occurs)

5. Know the energy diagrams: Endothermic Exothermic Δ H = + Δ H = - ** Spontaneous reactions will tend toward conditions of lower enthalpy (more negative Δ H)

6. Calorimetry -the experimental technique used to determine the heat exchange (q) associated with a reaction. -the amount of heat exchanged in a reaction depends upon: 1. The net temperature change during a reaction 2. The amount of substance. The more you have, the more heat can be exchanged. 3. The heat capacity (C) of a substance. heat absorbed C = increase in temperature = J/°C -some substances can absorb more heat that others for a given temperature change. There are three ways of expressing heat capacity: 1. Heat capacity (as above) = J /°C 2. Specific heat capacity = heat capacity per gram of substance = J/g °C or J/g K 3. Molar heat capacity = heat capacity per mole of substance = J/ mol °C or J/ mol K The specific heat of water = 4.18 J/g°C

7. Specific Heat -the amount of heat energy required to raise one gram of a substance by one degree celsius or kelvin C p = specific heat (p = constant pressure) q = heat lost or gained m = mass in grams ΔT = change in temperature (°C or K)

8. -Standard Enthalpy of formation ( Δ H° f ) – the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states -the degree symbol indicates that the corresponding process has been carried out under standard conditions -For a Compound -standard state for gas is exactly 1 atm. -for a pure substance in a condensed state (liquid or solid), the standard state is the pure liquid or solid. -For a substance present in solution, the standard state is a concentration of exactly 1M. -For an Element -The standard state of an element is the form in which the element exists under conditions of 1 atmosphere and 25° C. (The standard state for oxygen is O 2 (g) at a pressure of 1 atm; the standard state for sodium is Na(s); the standard state for mercury is Hg(l); and so on.) Conditions of Enthalpy

9. For example: Δ H° f for C 2 H 5 OH (l) = -279 kJ/mol means 2 C (graphite) + 3 H 2(g) + ½ O 2(g)  C 2 H 5 OH (l) Δ H° f = -279 kJ/mol -Standard Enthalpy of Combustion ( Δ H° c ) – the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions. For example: Δ H° c for C 2 H 6(g) = -1565 kJ/mol means C 2 H 6(g) + 7/2 O 2(g)  2 CO 2(g) + 3 H 2 O (g) Δ H° c = -1565 kJ/mol

10. Δ H° reaction = Σ n p Δ H° f (products) - Σ n r Δ H° f (reactants) -When a reaction is reversed, the magnitude of Δ H remains the same, but its sign changes -When the balanced equation for a reaction is multiplied by an integer, the value of Δ H for that reaction must be multiplied by the same integer. -Elements in their standard states are not included in the Δ H reaction calculations. That is, H° f for an element in its standard state is zero. Using standard enthalpies of formation, calculate the standard enthalpy change for the following overall reaction: 4NH 3 (g) + 7 O 2 (g)  4NO 2 (g) + 6H 2 O (l) Δ H° f NH 3 (g) = -46 kJ/mol NO 2 (g) = 34 kJ/mol O 2 (g) = you figure it out H 2 O (l) = -286 kJ/mol

11. Bond Energy & Enthalpy Δ H = Σ D (bonds broken) - Σ D (bonds formed) Energy required Energy released the sum of the energies required to break old bonds (positive signs) plus the sum of the energies released in the formation of new bonds (negative signs) Using bond energies, calculate Δ H for the following reaction: CH 4 (g) + 2Cl 2 (g) + 2F 2 (g)  CF 2 Cl 2 (g) + 2HF (g) + 2HCl (g) Average bond energies (kJ/mol) C-H = 413 Cl-Cl = 239 F-F = 154 C-F = 485 C-Cl = 339 H-F = 565 H-Cl = 427 C=C = 614 C-C = 347 Calculate the energy required to break the bonds of 3-methyl-4-octene

12. First law of thermodynamics – The energy of the universe is constant Second law of thermodynamics – In any spontaneous process there is always an increase in the entropy of the universe (the entropy of the universe is increasing) Δ S univ = Δ S sys + Δ S surr The sign of Δ S surr depends on the direction of the heat flow What can you tell me about an exothermic reaction? What can you tell me about an endothermic reaction? The magnitude of Δ S surr depends on the temperature Δ S surr = - Δ H T

13. Δ G° = Δ H° - T Δ S° The standard free energy of formation and the standard free energy change the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states. Δ G° = Σ n p Δ G° f (products) - Σ n r Δ G° f (reactants) Consider the reaction 2SO 2 (g) + O 2 (g)  2SO 3 (g) carried out at 25°C and 1 atm. Calculate Δ H°, Δ S°, and Δ G° using the following data: Substance Δ H° f (kJ/mol) S°(J/K·mol) SO 2 (g) -297 248 SO 3 (g) -396 257 O 2 (g) 0 205

14. The Dependence of Free Energy on Pressure G = G° + RT ln P where R = 8.3145 J/K·mol G° is the free energy of the gas at a pressure of 1 atm G is the free energy of the gas at a pressure P atm T is the Kelvin temperature can also be written as Δ G = Δ G° + RT ln (Q) One method for synthesizing methanol involves reacting carbon monoxide and hydrogen gases: CO(g) + 2H 2 (g)  CH 3 OH(l) Calculate Δ G at 25°C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol. Standard free energies of formation: CH 3 OH(l) = -166 kJ H 2 (g) = 0 CO(g) = -137 kJ

15. Free Energy and Equilibrium the equilibrium position represents the lowest free energy value available to a particular reaction system Δ G° = - RT ln (K) The overall reaction for the corrosion (rusting) of iron by oxygen is 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s) Using the following data, calculate the equilibrium constant for this reaction at 25°C. Substance Δ H° f (kJ/mol) S° (J/K·mol) Fe 2 O 3 (s) -826 90 Fe(s) 0 27 O 2 (g) 0 205

16. Born – Haber Cycle 5 step process 1. Sublimation (enthalpy of sublimation) 2. Ionization to form ion (1 st ionization energy) 3. Dissociation of nonmetal (energy to break bond) 4. Formation of Anions (electron affinity) 5. Form solid compound (lattice energy)

17. Example Overall desired reaction: Li (s) + ½ F 2(g)  LiF (s) 1. Li (s)  Li (g) 161 kJ/mol 2. Li (g)  Li + (g) + e - 520 kJ/mol 3. ½ F 2(g)  F (g) 154/2 = 77 kJ/mol need to form 1 mole of F atoms 4. F (g) + e -  F - (g) -328 kJ/mol 5. Li + (g) + F - (g)  LiF (s) -1047 kJ/mol The sum yields the desired overall reaction: 161 + 520 + 77 + (-328) + (-1047) = -617 kJ/mol of LiF Li (s) + ½ F 2(g)  LiF (s)

18. Practice Problem Calculate the Lattice Energy for the following reaction: Na (s) + ½ F 2(g)  NaF (s) Write out the reaction for each step. Enthalpy of sublimation = 109 kJ/mol 1 st ionization energy = 494 kJ/mol Energy to break bond = 79 kJ/mol Electron affinity = -348 kJ/mol Lattice energy = ? kJ/mol Overall energy for the desired reaction = -569 kJ/mol

19. Born-Haber Cycle Diagram Link: http://upload.wikimedia.org/wikipedia/commons/a/af/BornHaberLiF.PNG

20. Lattice Energy • The principle reason for ionic stability is due to the attraction between oppositely charged ions. • The physical attraction between ions in an ionic compound releases energy as the ions are drawn together. • Once the ions reach the lowest energy possible, distance is minimum given electron repulsion, a crystal lattice is formed. The strength of the ion attraction is described by the lattice energy between the ions. • Lattice Energy (aka lattice enthalpy) = the energy required to separate one mole of a solid ionic compound into its gaseous ions. = k (Q 1 Q 2 ) r k = proportionality constant dependent on structure of solid and on electron configuration of the ions. (8.99 x 10 9 Jm/C 2 )

21. Q 1 and Q 2 = charges on the ions r = the shortest distance between the centers of the cation and anion • Lattice Energy is influenced much more by the product of Q 1 and Q 2 than by the value of r