2. Chemical Equilibrium The big one!! This is your first required FRQ on the AP test.

3. Generally it is considered that all chemical reactions go to completion of the products and then stop. This is only true under a certain set of conditions. Most of the time, chemical reactions can go in the opposite direction as well if the condition changes made to the system are favorable. H 2 + O 2 --> H 2 O (synthesis) andH 2 O --> H 2 + O 2 (decomposition) become H 2 + O 2  H 2 O (both)

4. The equilibrium constant is a mathematical value that is constant for a specific chemical reaction. It, like the “k” value for kinetics is temperature dependent only. Concentrations may vary as much as necessary. The equilibrium constant, “k”, is determined from the equilibrium expression that can be determined by the stoichiometric equation: jA + kB  lC + mD leads to: [C] l [D] m [A] j [B] k

5. That is: Products coefficeints Reactants coefficients Write the equilibrium expressions for the following equilibrium reactions: 4NH 3 (g) + 7O 2 (g)  4NO 2 (g) + 6H 2 O (g) N 2 (g) + O 2 (g)  2NO (g) N 2 O 4 (g)  2NO 2 (g) 2PBr 3 (g) + 3Cl 2 (g)  2PCl 3 (g) + 3Br 2 (g) [NO 2 ] 4 [H 2 O] 6 [NH 3 ] 4 [O 2 ] 7 [NO] 2 [N 2 ][O 2 ] [NO 2 ] 2 [N 2 O 4 ] [PCl 3 ] 2 [Br 2 ] 3 [PBr 3 ] 2 [Cl 2 ] 3

6. Once the equilibrium expression has been written, you can input equilibrium concentration values into the formula and find the value of the equilibrium constant “k”. The following equilibrium concentration were observed for the synthesis of ammonia from its elements: N 2 (g) + H 2 (g)  NH 3 (g) a. Write the equilibrium expression for the reaction after balancing the equation. b. plug in the values for the given concentrations: [NH 3 ] = 3.1 x 10 -2 M [N 2 ] = 8.5 x 10 -1 M [H 2 ] = 3.1 x 10 -3 M

7. Find the value of “k” for the reverse reaction 2NH 3  N 2 + 3H 2 Calculate the value of “k” for the reaction given below 1/2 N 2 + 3/2 H 2  NH 3 [NH 3 ] = 3.1 x 10 -2 M [N 2 ] = 8.5 x 10 -1 M [H 2 ] = 3.1 x 10 -3 M

8. Now compare the equilibrium expressions in example b and c to a: What do you find is true?

9. For the reaction H 2 + Br 2 --> 2HBr K = 3.5 x 10 4 at 1495 K. What is the value of K under the following circumstances? a. HBr  1/2 H 2 + 1/2 Br 2 b. 2HBr  H 2 + Br 2 c. 1/2H 2 + 1/2 Br 2  HBr invert and square root invert only square root only

10. You can determine if the reaction will shift to the product side or the reactant side in a similar process: Q = [products] coefficients [reactants] coefficients Instead of using equlibrium concentrations, you will use beginning of reaction concentrations.

11. If k = Q, No shift in equilibrium will occur If Q > k the system will shift to the left, consuming products and turning them into reactants. If Q < k the system will shift to the right, consuming reactants and turning them into products

12. Equilibrium expressions that involve pressure instead of concentration. Because PV = nRT can be written P = (n/v)RT or P=CRT we can replace concentration with pressure. R is always a constant and for “k” to remain constant T must also be a constant for equilibrium expressions. Now you can put partial pressures of gases into your equilibrium expressions to calculate for a “k” value. Since “K” or K c are representative of concentration equilibrium constants, we will use K p when we solve with partial pressures.

13. Solve for K p of the reaction: 2NO (g) + Cl 2 (g)  2NOCl (g) when the partial pressures of each gas found at equlibrium are equal to: P NOCl = 1.2 atm P NO = 5.0 x 10 -2 atm P Cl2 = 3.0 x 10 -1 atm K p = P NOCl 2 = (1.2) 2. (P NO ) 2 (P Cl2 ) (5.0 x 10 -2 ) 2 (3.0 x 10 -1 ) = 1.9 x 10 3

14. Now because there is a difference in measurements between partial pressure measurements and concentration measurements of the same system at the same time K c ≠K p, but there is a relationship to convert from one to another. K p = K c (RT) D n (The derivation for this relationship can be found on page 619) K is equilibrium constant for concentration R is the ideal gas constant T is Temperature in Kelvin D n is sum of product coefficients - sum reactant coefficients

15. Convert the K p calculated earlier to a K value using the balanced equation and the mathematical relationship between K p and K c : 2NO (g) + Cl 2 (g)  2NOCl (g) K p = K(RT) D n The reaction is taking place at 25° C.

16. The states of matter become very important to calclulating K p when there are more than gas particles present in a system. If this is the case, only the gas particles in the system should be included in the equilibrium expression equation. 2H 2 O (l)  2H 2 (g) + O 2 (g) 2H 2 O (g)  2H 2 (g) + O 2 (g)

17. Write K c and K p expressions for the following: a. The decomposition of solid phosphous pentachloride to liquid phosphorous trichloride and chlorine gas. b. copper (II) oxide and carbon dioxide gas are combined to form solid copper (II) carbonate

18. Now since we can calculate K p, if we are given K p and have a missing concentration or pressure we can rearrange the equation to solve for that unknown if given K. N 2 O 4 (g)  2NO 2 (g) If the K p of the above reaction is 0.133 when the equilibrium pressure of N 2 O 4 is found to be 2.71 atm, find the pressure of NO 2 at equilibrium.

19. Now sometimes we will be asked to convert from original concentrations to equilibrium concentrations while being given a known k value. H 2 (g) + F 2 (g)  2HF (g) K = 1.15 x 10 2 The system starts with 3.000 mols of each component in a 1.5 L flask. Our job is find the equilibrium concentrations of each component (3variables!!!) We can do this by writing the change in every system as a function of the same variable.

20. 2NO(g) + Br 2 (g) 2NOBr (g) If 1.0 M and 0.8 M of NO and Br 2 gases were mixed, one quarter of the Br 2 would be consumed. Find the equilibrium concentrations and solve for the value of the equilibrium constant. ICEICE

25. We call this an ICE table H 2 F 2 2HF InitialI ChangeC EqulibriumE Equilibrium values, with variables included can now be plugged into the equation and you will solve for “x”.