1. Entry Task: Jan 7 th Monday Provide the Kc expression for this equation. 3 Ca 2+ (aq) + 2 PO 4 3- (aq) ↔ Ca 3 (PO 4 ) 2 (s) You have 5 minutes

2. Agenda: Discuss Kc Equilibrium constant ws HW: – Ch. 15 sec. 4- Calculating Equilibrium constants PLEASE MUDDLE THROUGH IT!!

3. 1. What are the 3 conditions necessary for equilibrium? Must have a closed system Must have a constant temperature Ea must be low enough to allow a reaction.

4. 2. What is a forward reaction versus a reverse reaction? In a forward reaction, the reactants collide to produce products and it goes from left to right. In a reverse reaction, the products collide to produce reactants and it goes form right to left.

5. 3. Why does the forward reaction rate decrease as equilibrium is approached? As the reaction goes to the right, the reaction concentration decreases and therefore, there are less reactant collisions causing the forward rate to decrease.

6. 4. What are the characteristics of equilibrium or define equilibrium? Forward rate is equal to the reverse rate. The concentration of reactants and products are constant.(not equal) Macroscopic properties are constant (color, mass, density, pressure, concentrations).

7. 5. As a reaction is approaching equilibrium describe how the following change. Explain what causes each change. a. Reactant concentration As the reaction goes to the right, the reaction concentration decreases. b. Products concentration As the reaction goes from left to right, the concentration of the products increases.

8. 5. As a reaction is approaching equilibrium describe how the following change. Explain what causes each change. c. Forward reaction rate. The reaction concentration decreases and therefore, there are less reactant collisions causing the forward rate to decreases. d. Reverse reaction rate The concentrations of the products increases, therefore there are more product collisions causing the reverse reaction rate to increase.

9. 6. What is equal at equilibrium? The forward and reverse rates are equal.

10. 7. What is constant at equilibrium? The reactant and product concentrations and the macroscopic properties are constant.

11. For the following three reactions, *Write the Kc expression AND calculate the value of K c. *State whether the equilibrium is product-favored, reactant- favored, or fairly even ([products]  [reactants]). At equilibrium:[N 2 ] = 1.50 M [H­ 2 ] = 2.00 M [NH 3 ]= 0.01 M 8.N 2 (g) + 3 H 2 (g)  2 NH 3 (g) Kc = 8.33 x 10 -6 Small Kc favors reactant

12. For the following three reactions, *Write the Kc expression AND calculate the value of K c. *State whether the equilibrium is product-favored, reactant- favored, or fairly even ([products]  [reactants]). 9. HF(aq)  H + (aq) + F - (aq) At equilibrium: [HF] = 0.55 M [H + ] = 0.001 M [F - ]= 0.001 M Kc = 1.8 x 10 -6 Small Kc favors reactant

13. For the following three reactions, *Write the Kc expression AND calculate the value of K c. *State whether the equilibrium is product-favored, reactant- favored, or fairly even ([products]  [reactants]). 10. Fe 3+ (aq) + SCN - (aq)  FeSCN 2+ (aq) Kc = 1.8 Kc is very close to 1 so even At equilibrium:[Fe 3+ ] = 0.55 M [SCN - ] = 0.001 M [FeSCN 2+ ]= 0.001 M

14. Summarize: Fill in the blanks with product-favored, reactant-favored, and approximately equal KcKc state of equilibrium K c >> 1 K c << 1 K c  1 Products favored Reactant favored ~Equal~

15. 11.Write equilibrium expressions for each of the following reactions: a)CaCO 3 (s)  CaO(s) + CO 2 (g) b)Ni(s) + 4CO(g)  Ni(CO) 4 (g) c)5CO(g) + I 2 O 5 (s)  I 2 (g) + 5CO 2 (g)

16. 11.Write equilibrium expressions for each of the following reactions: d)Ca(HCO 3 ) 2 (aq)  CaCO 3 (s) + H 2 O(l) + CO 2 (g) e)AgCl(s)  Ag + (aq) + Cl - (aq)

17. 12. Write the equilibrium expression in terms of partial pressures (K p ) for each of the following reactions. Rate the reactions in order of their increasing tendency to proceed toward completion (products): ___ ___ ___ ___ (a)4NH 3 (g) + 3O 2 (g)  2N 2 (g) + 6H 2 O(g) Kp = 1 x 10 228

18. 12. Write the equilibrium expression in terms of partial pressures (K p ) for each of the following reactions. Rate the reactions in order of their increasing tendency to proceed toward completion (products): ___ ___ ___ ___ (b)N 2 (g) + O 2 (g)  2NO(g) Kp = 5 x 10 -31

19. 12. Write the equilibrium expression in terms of partial pressures (K p ) for each of the following reactions. Rate the reactions in order of their increasing tendency to proceed toward completion (products): ___ ___ ___ ___ (c)2HF(g)  H 2 (g) + F 2 (g) Kp = 1 x 10 -13

20. 12. Write the equilibrium expression in terms of partial pressures (K p ) for each of the following reactions. Rate the reactions in order of their increasing tendency to proceed toward completion (products): ___ ___ ___ ___ (d)2NOCl(g)  2NO(g) + Cl 2 (g) Kp = 4.7 x 10 -4

21. 12. Write the equilibrium expression in terms of partial pressures (K p ) for each of the following reactions. Rate the reactions in order of their increasing tendency to proceed toward completion (products): ___ ___ ___ ___ (d)2NOCl(g)  2NO(g) + Cl 2 (g) Kp = 4.7 x 10 -4 (c)2HF(g)  H 2 (g) + F 2 (g) Kp = 1 x 10 -13 (b)N 2 (g) + O 2 (g)  2NO(g)Kp = 5 x 10 -31 (a)4NH 3 (g) + 3O 2 (g)  2N 2 (g) + 6H 2 O(g) Kp = 1 x 10 228 Lowest value is reactant favored- least likely to complete reaction BCD Highest value is product favored- most likely to complete reaction A

22. 13.(a) Write the K c expression for 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) Calculate the value of K c : At equilibrium:[SO 2 ] = 1.50 M [O 2 ] = 1.25 M [SO 3 ]= 3.50 M Kc = 4.36

23. 13.(b)If we reverse the equation, it is: 2 SO 3 (g)  2 SO 2 (g) + O 2 (g) Write the K c expression for this equation and calculate the new value of K c : At equilibrium: [SO 2 ] = 1.50 M [O 2 ] = 1.25 M [SO 3 ]= 3.50 M Kc = 0.229

24. 13.(b)If we reverse the equation, it is: 2 SO 3 (g)  2 SO 2 (g) + O 2 (g) Write the K c expression for this equation and calculate the new value of K c : How does the expression and the value of K c in 7(b) compare with those in 7(a)? Kc = 1/Kc

25. (c)If we now multiply all of the coefficients by ½: SO 3 (g)  SO 2 (g) + ½ O 2 (g) Write the K c expression for this equation and calculate the new value of K c : At equilibrium: [SO 2 ] = 1.50 M [O 2 ] = 1.25 M [SO 3 ]= 3.50 M Kc  = 0.479

26. (c)If we now multiply all of the coefficients by ½: SO 3 (g)  SO 2 (g) + ½ O 2 (g) Write the K c expression for this equation and calculate the new value of K c : How do they compare with 7(b)? Kc = 0.229Kc  = 0.479 Kc  = square root of Kc

27. (d) What would happen to the K c expression and its value if we doubled the coefficients? Kc  = (Kc) 2

28. Summarize: EquationK c expression Value doubled reversed halved squared Inverse or 1/Kc Square root