1. Obj 15.1, 15.2

2. 15.1 The concept of Equilibrium A.) Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

3. Take the reaction: N 2 O 4 2NO 2 This in an elementary reaction at fast equilibrium, so how would you write the forward and reverse rate laws of this reaction? Rearranging this equation, you can solve for k 1 /k -1, and this is called your equilibrium constant k1k1 k -1

4. B.) As a system approaches equilibrium, both the forward and reverse reactions are occurring. C.) At equilibrium, the forward and reverse reactions are proceeding at the same rate.

5. D.) Once equilibrium is achieved, the amount of each reactant and product remains constant.

6. E.) Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow. F.) Extra information: 1.) for equilibrium to occur, neither products nor reactants can escape from the system. 2.) the equilibrium constant (K) may change if there is a change in temperature N 2 O 4 (g) 2 NO 2 (g)

7. 15.2 The Equilibrium Constant A.) The ratio of the rate constants is a constant at that temperature, and the expression becomes K eq = kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] =

8. © 2009, Prentice- Hall, Inc. B.) Consider the generalized reaction The equilibrium expression for this reaction would be K c = [C] c [D] d [A] a [B] b aA + bBcC + dD

9. Sample Exercise 15.1 Writing Equilibrium-Constant Expressions Write the equilibrium expression for K c for the following reactions: Practice Exercise

10. C.) Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written K p = (P C c ) (P D d ) (P A a ) (P B b )

11. © 2009, Prentice- Hall, Inc. D.) Relationship Between K c and K p 1.) From the Ideal Gas Law we know that 2.) Rearranging it, we get PV = nRT P = RT nVnV

12. 3.) Plugging this into the expression for K p for each substance, the relationship between K c and K p becomes where K p = K c (RT)  n  n = (moles of gaseous product) - (moles of gaseous reactant)

13. Sample Exercise 15.2 Converting between K c and K p In the synthesis of ammonia from nitrogen and hydrogen, K c = 9.60 at 300 °C. Calculate K p for this reaction at this temperature.

14. 4.) the ratio of [NO 2 ] 2 to [N 2 O 4 ] remains constant at this temperature no matter what the initial concentrations of NO 2 and N 2 O 4 are.

16. © 2009, Prentice-Hall, Inc. 5.) It doesn’t matter whether we start with N 2 and H 2 or whether we start with NH 3 : we will have the same proportions of all three substances at equilibrium (don’t have to be the same concentrations, just the same proportions).

17. © 2009, Prentice-Hall, Inc. 1.) If K>>1, the reaction is product-favored; product predominates at equilibrium. 2.) If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium. E.) What Does the Value of K Mean?

18. Sample Exercise 15.3 Interpreting the Magnitude of an Equilibrium Constant The following diagrams represent three different systems at equilibrium, all in the same size containers. (a) Without doing any calculations, rank the three systems in order of increasing equilibrium constant, K c. (b) If the volume of the containers is 1.0 L and each sphere represents 0.10 mol, calculate K c for each system.