2. Chapter 19: Chemical Thermodynamics 2 KClO 3 2 KCl + 3 O 2 2 KClO 3  2 KCl + 3 O 2 The reaction proceeds spontaneously once the KClO 3 has been heated to a high temperature. If the reaction were endothermic, you would have to continuously provide heat or the reaction would stop. Thus the reverse reaction, 2 KCl + O 2  2 KClO 3, is not likely to occur spontaneously. Spontaneity of endothermic and exothermic reactions.

3. A spontaneous process is one that proceeds without any outside assistance.  A process that is spontaneous in one direction is nonspontaneous in the reverse direction. ice water Spontaneous for T < 0  C Spontaneous for T > 0  C Recall that spontaneous ≠ a fast reaction rate Spontaneous only implies that the reaction can occur without outside assistance.

4. First premise: Reactants must collide in order to react and form products. A 2 + B 2 2 AB Collision Theory

5. First premise: Reactants must collide in order to react and form products. A 2 + B 2 2 AB Second premise: Reactants must have the correct orientation to form the products upon collision

6. Collision Theory First premise: Reactants must collide in order to react and form products. A 2 + B 2 2 AB Second premise: Reactants must have the correct orientation to form the products upon collision Third premise: Reactants must have sufficient energy for the collision to result in formation of products E < E a

7. Collision Theory First premise: Reactants must collide in order to react and form products. A 2 + B 2 2 AB Second premise: Reactants must have the correct orientation to form the products upon collision Third premise: Reactants must have sufficient energy for the collision to result in formation of products E > E a

8. Reaction progress Energy EaEa Activated complex: an unstable transition state between reactants and products. It can either fall back down on the reactant side or go on to the product side.  H rxn

9. Reaction progress Energy EaEa  H rxn EaEa A catalyst speeds up the rate of a reaction by lowering the activation barrier

10. Reaction progress Energy  H rxn EaEa A catalyst speeds up the rate of a reaction by lowering the activation barrier

11. Reaction progress Energy  H rxn EaEa The E a for an endothermic reaction is:  H rxn + energy barrier

12.  U is the same for both blocks (recall definition of a state function) However, q and w for both blocks will depend on how quickly the blocks are moved. If the blocks are moved using infinitesimally small steps, the amount of heat will be zero and the amount of work done will equal  U. A process is reversible when it can be restored exactly to its initial state without a loss of energy to the surroundings.

13. Reversible processes are NOT spontaneous. All spontaneous processes are irreversible. Entropy, S, is a measure of the degree of randomness, or disorder of a system, and has a positive, nonzero value for all substances above absolute zero. H 2 (l) < H 2 (g) < 2 H (g)  S, a state function, is a measure of the change in the degree of order in a system and can thus be positive, negative or equal to zero. (at constant Temp)

14. Temperature (°C) 0 100 Heating curve of water solid warming solid + liquid present liquid warming liquid + gas present Gas warming Heat (kJ/s)

15. Temperature (°C) 0 100 Heating curve of water melting/freezing point boiling/condensation point Temperature is constant during phase transitions!! All heat energy goes to changing the state of matter. Heat (kJ/s)

16. Temperature (°C) 0 100 Heating curve of water  H fus = the amount of heat needed to convert a solid into its liquid phase  H fus  H vap (heat of fusion) (heat of vaporization)  H vap = the amount of heat needed to convert a liquid into its gaseous phase

17. Entropy and phase changes Temperature is constant during any phase change. The process is reversible. The amount of heat energy involved is equal to  H. EX: Use Appendix C to calculate the boiling point of CCl 4.  H = H f (product) – H f (reactant) = -106.7 kJ – (-139.3 kJ) = 32.6 kJ CCl 4 (l)  CCl 4 (g)  S = S(product) – S(reactant) = 309.4 J/K – 214.4 J/K = 95.0 J/K T = 3.26 x 10 4 J/95.0 J/K = 343 K(lit: 350 K)

18. The Second Law of Thermodynamics: The total entropy of the Universe increases in any spontaneous process. Reversible process:  S univ =  S system +  S surroundings = 0 Irreversible process:  S univ =  S system +  S surroundings > 0 The Third Law of Thermodynamics: The entropy of a perfect crystal at absolute zero is zero.

19. Three Laws of Thermodynamics (paraphrased): First Law: You can't get anything without working for it. First Law: You can't win. OR… Second Law: The most you can accomplish by work is to break even. Third Law: You can't break even. Second Law: You can't even break even. Third Law:You can't get out of the game. THE LAW OF ENTROPY: The perversity of the universe tends towards a maximum.

20. Entropy Activity Boltzman’s Equation S = k lnW k = Boltzman’s constant = 1.38 × 10 −23 J/K W = number of microstates = number of equivalent ways that the particles in the system can be arranged. When you’re talking about 6.022 x 10 23 particles…that’s a LOT of microstates!! In general, W and thus entropy increase with: Temperature Volume Number of independently moving particles

21. (d) 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) Is  S positive or negative? (a) CO 2 (s)  CO 2 (g) (c) HCl(g) + NH 3 (g)  NH 4 Cl(s) (b) CaO(s) + CO 2 (g)  CaCO 3 (s)  S > 0 because gas is formed  S < 0 because gas was consumed by reaction  S < 0 because there are fewer moles of gas on the product side than on the reactant side.

22. Appendix C lists standard molar entropies, S o (1 atm, 298 K). Some guidelines to help with predicting the sign of  S: Unlike enthalpies of formation, the standard molar entropies of elements at the reference temperature of 298 K are not zero The standard molar entropies of gases are much, much greater than those of liquids and solids. Standard molar entropies generally increase with increasing molar mass. [Compare Li(s), Na(s), and K(s).] Standard molar entropies generally increase with an increasing number of atoms in the formula of a substance due to the greater number of modes of motion.

23. Metals Nonmetals Metalloid 1 234567 8 # Valence electrons

24. Periodicity: Atomic Trends There are a number of atomic characteristics that either increase or decrease along the periodic table. Ionization Energy & Electron Affinity Electronegativity

25. Atomic Radius: As you go down a group, the atomic radius increases with increasing energy level. Atomic radius decreases as you go left to right along a period because the greater nuclear charge pulls the electrons in closer to the nucleus. s-blockp-block

26. Shielding effect: The inner electron shells insulate the valence electrons from some of the electrical attraction with the positive charge of the nucleus. + - nucleus valence e- core electrons Why atomic diameter increases as you go down a group…

27. -- Increasing Effective Nuclear charge: Electrons in the outermost energy levels do not effectively screen each other from an increasingly positive nucleus. Why ionization energy increases and atomic/ionic radius decreases as go across a period: - nucleus +++++++ --

30. Ionization Energy: the amount of energy that is required to remove an electron. If an atom will easily lose an electron (like the alkali metals), the ionization energy will be low. If it will not lose electrons easily (noble gases, halogens), the ionization energy will be high.

31. Electronegativity: Ability of an atom to attract electrons when in a molecule.

32. Electronegativity increases going left to right, and decreases going down.

33. H 2 O (l)H 2 O (g) Energy H 2 O (l) H 2 O (g)  H vap = +40.7 kJ/mol Water will spontaneously evaporate at room temperature even though this process is endothermic. What is providing the uphill driving force?

34. a measure of the disorder or randomness of the particles that make up a system Water will spontaneously evaporate at room temperature because it allows the disorder of the water molecules to increase. The entropy, S, of gases is >> than liquids or solids. If S products > S reactants,  S is > 0 Predict the sign of  S: ClF (g) + F 2 (g)ClF 3 (g)  S < 0 CH 3 OH (l)CH 3 OH (aq)  S > 0

35. Are all +  S reactions spontaneous? 2 H 2 O (l) 2 H 2 (g) + O 2 (g)  S is large and positive… …but  H is large and positive as well. Gibb’s Free Energy,  G, allows us to predict the spontaneity of a reaction using  H AND  S. If –  G  spontaneous reaction Free Energy is energy that is available to do work.

36. 2 H 2 O (l) 2 H 2 (g) + O 2 (g) What is  G for this reaction at 25  C?  H rxn =  H f (products) –  H f (reactants)  H rxn = [2(0) + 0] - 2(-285.83 kJ/mol ) = 571.66 kJ/mol  S rxn = [2(130.58 J/molK ) + 205.0 J/molK ] - 2(69.91 J/molK )  S rxn =  S f (products) –  S f (reactants)  S rxn = 326.34 J/molK = 0.32634 kJ/molK  G rxn =  H rxn – T  S rxn = 571.66 kJ/mol - 298K(0.32634 kJ/molK )  G rxn = +474.41 kJ/mol+  G = nonspontaneous

37. 2 H 2 O (l) 2 H 2 (g) + O 2 (g)  G rxn =  H rxn – T  S rxn = 571.66 kJ/mol - T(0.32634 kJ/molK ) What is the minimum temperature needed to make this reaction spontaneous? Set  G rxn = 0 to find minimum temperature 0 = 571.66 kJ/mol - T(0.32634 kJ/molK ) T = (571.66 kJ/mol)/(0.32634 kJ/molK) = 1751 K T > 1478  C

38. ++ Spontaneous at high T HH SS GG + Always spontaneous Spontaneous at low T + Never spontaneous  G = -w max Remember, Free Energy is the energy that’s available to do work on the surroundings. Hence, Q: What does it mean if  G = 0? Rate of forward reaction = rate of reverse reaction The system is at equilibrium!

39. Chemical Equilibria: The basics Irreversible reactions: The reactants combine to form products and the reaction stops when the limiting reagent has been completely used up. “Only” the forward reaction occurs. Reversible reactions: Both the forward and the reverse reaction occur to a significant degree. Au 2 O 3 (s) + 2 Fe (s) 2 Au (s) + Fe 2 O 3 (s)Ex: 3 H 2 (g) + N 2 (g)  2 NH 3 (g) 3 H 2 (g) + N 2 (g)  2 NH 3 (g) Forward: Reverse: Equilibrium: A reaction is at equilibrium when Rate forward rxn = Rate reverse rxn 3 H 2 (g) + N 2 (g) ⇋ 2 NH 3 (g)

40. General form of equilibrium constant, K eq : aA + bB ⇋ cC + dD If Keq > 1, then more products than reactants present at equilibrium If Keq < 1, then more reactants than products present at equilibrium

41. Time Concentration N 2 (g) + 3 H 2 (g) 2 NH 3 (g) ⇋ H2H2 NH 3 N2N2 K eq = [NH 3 ] [N 2 ] [H 2 ] 2 3

42. Heterogeneous equilibria If the reactants and products are in more than one state, the equilibrium constant is for a heterogeneous equilibria. Because the concentration of a solid does not change significantly compared to gaseous or liquid reactants or products, solids NEVER appear in equilibrium constant calculations. Ex: BaCl 2 (s) ⇋ Ba 2+ (aq) + 2 Cl - (aq) Since this particular type of equilibrium involves the solubility of the product, it is given a special designation: K sp = solubility product constant

43. Heterogeneous equilibria (cont.) Since the concentration of a liquid solvent that the reaction takes place in also doesn’t change significantly, it also doesn’t appear in the equilibrium calculation. Ex: H 2 O (l) ⇋ H 2 O (g) What are the equilibrium constants for the following: 1.C 10 H 8 (s) ⇋ C 10 H 8 (g) 2.CaCO 3 (s) ⇋ CaO (s) + CO 2 (g) 3.C (s) + H 2 O (g) ⇋ CO (g) + H 2 (g) 4.FeO (s) + CO (g) ⇋ Fe (s) + CO 2 (g) K eq = [C 10 H 8 ] K eq = [CO 2 ] K eq = [CO][H 2 ] [H 2 O] K eq = [CO 2 ] [CO]

44. Le Chatelier’s Principle: When a stress is applied to an equilibrium, the equilibrium will shift to alleviate the stress. Fe +3 (aq) + SCN -1 (aq) ⇋ FeSCN +2 (aq) Colorless ⇋ Dark red Initial color Left shift = lighter color Right shift = darker color

45. Other ways to cause a Le Châtelier Shift: 3 H 2 (g) + N 2 (g) + heat ⇋ 2 NH 3 (g) What kind of shift would you see and how would  G (temporarily) be affected if: Pressure increased? Volume increased? Heating temperature increased? Right shift   G more negative  Left shift  G more positive Right shift   G more negative

46. CH 4 (g) + 2 Cl 2 (g) ⇋ CCl 4 (g) + 2 H 2 (g) + heat What kind of shift would you see if: Pressure increased? Heating temperature increased? No Change  G remains zero  Left shift  G more positive

47. “Of shoes and ships and sealing wax and cabbages and Kings…” …or “What if conditions aren’t quite standard?” 3 H 2 + N 2  2 NH 3  G = 0 at equilibrium Q: What is  G when you first mix the reactants together?  Recall that the equilibrium constant for a reaction, K, is: When the reaction first starts, it is NOT at equilibrium. We use a reaction quotient, Q, to evaluate how far from equilibrium it is.

48. EX: Is the reaction for the synthesis of ammonia at equilibrium when there are 2 moles of nitrogen and hydrogen, and 1 mole of ammonia in a 2 L flask? Step 1: Determine the concentrations of the reactants and products  Recall that the square brackets indicate that concentrations are in terms of moles/liter (M). [H 2 ] = 2 moles/2 L = 1 M [N 2 ] = 2 moles/2 L = 1 M [NH 3 ] = 1 moles/2 L = 0.5 M Step 2: Calculate the reaction quotient

49. K = 7.0 x 10 5 Step 3: Calculate the equilibrium constant from  G o 3 H 2 + N 2  2 NH 3 IF the reaction is being done under standard conditions, you can use the tabulated  G f values in Appendix C to calculate  G o for the reaction

50. Step 4: Compare Q and K Q = 0.25K = 7.0 x 10 5 Is this reaction anywhere near equilibrium?? Nope. EX: Calculate  G for the synthesis of ammonia given the previous reaction conditions.  G =  G  + RTlnQ  G = -3.7 x 10 4 J = -370 kJ  The fact that the reaction is far from equilibrium has increased the spontaneity of the reaction (increased its driving force).