2. The Equilibrium Constant For a general reaction the equilibrium constant expression for everything in solution is where K eq is the equilibrium constant.

3. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) 1. Write an expression for K eq 2. Calculate K at a given temperature if [CH 4 ] = 0.020 M, [O 2 ] = 0.042 M, [CO 2 ] = 0.012 M, and [H 2 O] = 0.030 M at equilibrium. (include units)

4. If K << 1, then reactants dominate at equilibrium and the equilibrium lies to the left.

5. Types of Equilibrium Constant K eq = K c = K p = Equilibrium concentrations pressures

6. When all reactants and products are in one phase, the equilibrium is homogeneous. If one or more reactants or products are in a different phase, the equilibrium is heterogeneous. Consider: –experimentally, the amount of CO 2 does not seem to depend on the amounts of CaO and CaCO 3. Why? Heterogeneous Equilibria

7. The concentration of a solid or pure liquid is its density divided by molar mass. Neither density nor molar mass is a variable, the concentrations of solids and pure liquids are constant. For the decomposition of CaCO 3 : We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions. The amount of CO 2 formed will not depend greatly on the amounts of CaO and CaCO 3 present.

8. Example – Write the K eq expression for the following equations: NH 4 Cl (s)NH 3 (g) + HCl (g) NH 3 (g) + HCl (g) NH 4 Cl (s)

9. Example – Write the K eq expression for the following equations: NH 4 Cl (s) ↔ NH 3 (g) + HCl (g) K eq = [NH 3 ][HCl] NH 3 (g) + HCl (g) ↔ NH 4 Cl (s) K eq = ([NH 3 ][HCl]) -1

10. Proceed as follows: –Tabulate initial and equilibrium concentrations (or partial pressures) given. –If an initial and equilibrium concentration is given for a species, calculate the change in concentration. –Use stoichiometry on the change in concentration line only to calculate the changes in concentration of all species. –Deduce the equilibrium concentrations of all species. Use “ ICE ” Charts Calculating Equilibrium Constants

11. Example – N 2 (g) + 3 H 2 (g)2 NH 3 (g) The initial concentration of N 2 is 0.25 M and of H 2 is 0.60 M. The equilibrium concentration of H 2 is 0.45 M. What are the equilibrium concentrations of N 2 and NH 3 ? What is the value of K eq ?

12. Example – N 2 (g) + 3 H 2 (g)↔ 2 NH 3 (g) The initial concentration of N 2 is 0.25 M and of H 2 is 0.60 M. The equilibrium concentration of H 2 is 0.45 M. What are the equilibrium concentrations of N 2 and NH 3 ? What is the value of K eq ? 1. [N 2 ] =.2 M [NH 3 ] =.1 M 2. K eq =.549 M -2

13. Predicting the Direction of Reaction We define Q, the reaction quotient, for a general reaction as Q = K only at equilibrium. Applications of Equilibrium Constants

14. If Q > K then the reverse reaction must occur to reach equilibrium (i.e., products are consumed, reactants are formed, the numerator in the equilibrium constant expression decreases and Q decreases until it equals K). If Q < K then the forward reaction must occur to reach equilibrium.

15. Consider the production of ammonia As the pressure increases, the amount of ammonia present at equilibrium increases. As the temperature decreases, the amount of ammonia at equilibrium increases. Can this be predicted? Le Châtelier ’ s Principle

16. Effects of Volume and Pressure Changes As volume is decreased pressure increases. Le Châtelier’s Principle: if pressure is increased the system will shift to counteract the increase. That is, the system shifts to remove gases and decrease pressure. An increase in pressure favors the direction that has fewer moles of gas. In a reaction with the same number of product and reactant moles of gas, pressure has no effect.

17. Increasing total pressure by adding an inert gas has no effect on the partial pressures of reactants and products, therefore it has no effect on the equilibrium. Effect of Temperature Changes The equilibrium constant is temperature dependent. For an endothermic reaction,  H > 0 and heat can be considered as a reactant. For an exothermic reaction,  H < 0 and heat can be considered as a product.

18. Adding heat (i.e. heating the vessel) favors away from the increase: –if  H > 0, adding heat favors the forward reaction, –if  H < 0, adding heat favors the reverse reaction. Removing heat (i.e. cooling the vessel), favors towards the decrease: –if  H > 0, cooling favors the reverse reaction, –if  H < 0, cooling favors the forward reaction.

19. The Effect of Catalysis A catalyst lowers the activation energy barrier for the reaction. Therefore, a catalyst will decrease the time taken to reach equilibrium. A catalyst does not effect the composition of the equilibrium mixture. Le Châtelier’s Principle

20. SO 3 (g) SO 2 (g) + 1 / 2 O 2 (g) ΔH = +98.9 kJ Determine the effect of each of the following on the equilibrium (direction of shift) – What happens to the concentration of SO 3 after each of the changes? A) Addition of pure oxygen gas. B) Compression at Constant Temperature C) Addition of Argon gas D) Decrease temperature E)Remove sulfur dioxide gas F)Addition of a catalyst

21. Calculating Equilibrium Concentrations The same steps used to calculate equilibrium constants are used. K is given. Generally, we do not have a number for the change in concentration line. Therefore, we need to assume that x mol/L of a species is produced (or used). The equilibrium concentrations are given as algebraic expressions. We solve for x, and plug it into the equilibrium concentration expressions.

22. Example – H 2 (g) + FeO (s)H 2 O (g) + Fe (s)K c = 5.20 If the initial concentration of H 2 is 0.50 M and the inintial concentration of H 2 O is 6.50 M, what will the equilibrium concentrations be? If the initial concentration of H 2 is 1.00 M (no H 2 O present), what will the equilibrium concentrations be?

23. The oxidation numbers of all atoms in a molecule or ion add up to the total charge (0 for a molecule, the ion’s charge for an ion) Elements by themselves are the charge shown O is almost always -2 (exception H 2 O 2 ) F is almost always -1 Group 1 elements are +1 Group 2 elements are +2 Determining Oxidation Numbers

24. Examples: Determine the oxidation numbers of each atom in the following: 1.NaC 2 H 3 O 2 2.Na 2 SO 4 3.C 2 O 4 2- 4.SO 3 2- Determining Oxidation Numbers

25. Half Reactions The half-reactions for Sn 2+ (aq) + 2Fe 3+ (aq)  Sn 4+ (aq) + 2Fe 2+ (aq) are Sn 2+ (aq)  Sn 4+ (aq) +2e - 2Fe 3+ (aq) + 2e -  2Fe 2+ (aq) Oxidation: electrons are products. Reduction: electrons are reactants. Loss of Gain of Electrons isElectrons is OxidationReduction

26. Balancing Equations by the Method of Half Reactions 1. Write down the two half reactions. 2. Balance each half reaction: a. First with elements other than H and O. b. Then balance O by adding water. c. Then balance H by adding H +. d. Finish by balancing charge by adding electrons. 3. Multiply each half reaction to make the number of electrons equal. 4. Add the reactions and simplify. 5.If it is in basic solution, remove H + by adding OH - 6.Check!

27. Example: MnO 4 - + C 2 O 4 2- → Mn 2+ + CO 2 The two incomplete half reactions are MnO 4 -  Mn 2+ C 2 O 4 2-  CO 2

29. Examples – Balance the following oxidation- reduction reactions: 1.Cr (s) + NO 3 - (aq)  Cr 3+ (aq) + NO (g) (acidic) 2.Al (s) + MnO 4 - (aq)  Al 3+ (aq) + Mn 2+ (aq) (acidic) 3.PO 3 3- (aq) + MnO 4 - (aq)  PO 4 3- (aq) + MnO 2 (s) (basic) 4.H 2 CO (aq) + Ag(NH 3 ) 2 + (aq)  HCO 3 - (aq) + Ag (s) + NH 3 (aq) (basic)

31. Voltaic cells consist of –Anode: Zn(s)  Zn 2+ (aq) + 2e - –Cathode: Cu 2+ (aq) + 2e -  Cu(s) –Salt bridge (used to complete the electrical circuit): cations move from anode to cathode, anions move from cathode to anode. Necessary to balance charge. The two solid metals are the electrodes (cathode and anode). Anode = oxidationCathode = reduction

32. As oxidation occurs, Zn is converted to Zn 2+ and 2e -. The electrons flow towards the cathode (through the wire) where they are used in the reduction reaction. We expect the Zn electrode to lose mass and the Cu electrode to gain mass. “Rules” of voltaic cells: 1. At the anode electrons are products. (Oxidation) 2. At the cathode electrons are reactants. (Reduction) 3. Electrons cannot swim.

33. Electrons flow from the anode to the cathode. Therefore, the anode is negative and the cathode is positive. Electrons cannot flow through the solution; they have to be transported through an external wire. (Rule 3.)

34. Standard Reduction (Half-Cell) Potentials Convenient tabulation of electrochemical data. Standard reduction potentials, E  red are measured relative to the standard hydrogen electrode (SHE). Oxidation Potentials are obtained by reversing a reduction half-reaction (sign of E° is changed)

36. The SHE is the cathode. It consists of a Pt electrode in a tube placed in 1 M H + solution. H 2 is bubbled through the tube. For the SHE, we assign 2H + (aq, 1M) + 2e -  H 2 (g, 1 atm) E  red of zero. The emf of a cell can be calculated from standard reduction potentials:

37. For any electrochemical process A positive E  indicates a spontaneous process (galvanic cell). A negative E  indicates a nonspontaneous process. Spontaneity of Redox Reactions

38. Electrolysis of Aqueous Solutions Nonspontaneous reactions require an external current in order to force the reaction to proceed. Electrolysis reactions are nonspontaneous. In voltaic and electrolytic cells: –reduction occurs at the cathode, and –oxidation occurs at the anode. –In electrolytic cells, electrons are forced to flow from the anode to cathode. Electrolysis

40. Example, decomposition of molten NaCl. Cathode: 2Na + (l) + 2e -  2Na(l) Anode: 2Cl - (l)  Cl 2 (g) + 2e -. Industrially, electrolysis is used to produce metals like Al.

41. Electroplating Active electrodes: electrodes that take part in electrolysis. Example: electrolytic plating.

43. Consider an active Ni electrode and another metallic electrode placed in an aqueous solution of NiSO 4 : Anode: Ni(s)  Ni 2+ (aq) + 2e - Cathode: Ni 2+ (aq) + 2e -  Ni(s). Ni plates on the inert electrode. Electroplating is important in protecting objects from corrosion.

44. Quantitative Aspects of Electrolysis We want to know how much material we obtain with electrolysis. 1 Ampere is 1 Coulomb per second (A= C/s) 1 mole of electrons = 96,485 C = 1 Faraday Use balanced half-equation to equate moles of substance to moles of electrons Molar mass (in g) = 1 mole of substance

45. Examples 1.A car bumper is to be electroplated with Cr from a solution of Cr 3+. What mass of Cr will be applied to the bumper if a current of 0.50 amperes is allowed to run through the solution for 4.20 hours? 2.What volume of H 2 gas (at STP) will be produced from the SHE after 2.56 minutes at a current of 0.98 amperes? 3.What volume of F 2 gas, at 25°C and 1.00 atm, is produced when molten KF is electrolyzed by a current of 10.0 A for 2.00 hours? What mass of potassium metal is produced? At which electrode does each reaction occur?