Introduction to Equilibrium and Keq... Have your packet out.

On Equilibrium Island lived 20 single girls, 10 single guys, and 10 couples

Sometimes a couple split up….

….and just as frequently, sometimes a new couple would form…

RV PALMA One day, a ship sank not too far from Equilibrium Island….

The wreck of the RV Palma added 105 guys to the island. How will this change alter the number of “single” guys, “single” girls, and couples over time?

Example: 5 single girls, 100 single guys, and 25 couples

The End

Point 1 Reactions are reversible!! girl + guy  couple The forward and reverse reactions are happening at the same time.

Point 2 Changing amount of something in a reaction changes amount of everything else. Le Chatelier’s Principle If a stress is applied to a reaction at equilibrium, the system is no longer at equilibrium. The system will respond (change or shift) to partially relieve or undo the stress.

single+single  couples girlsguys Silly Suzy and Bozo Joe break up over the weekend……… Lazy Larry and Ditzy Daisy hold hands at the movie over the weekend…

At equilibrium: 1.The reactions are still taking place 2. No change in [reactants] or [products] 3. Rate of the forward rxn = rate reverse rxn

Assume this system is at equilibrium: A(g) + B(aq) C(s) + D(aq) kJ (exothermic) What happens to the equilibrium position when more of compound C is added to the reaction chamber? To relieve stress, the equilibrium position shifts to the left (the reactants side), causing the rate of the reverse reaction to speed up. What happens to equilibrium, A, C, and D when the [ ] of B is increased? Equilibrium shifts right (toward the product side). The forward reaction speeds up. More C and D are made, and more of A is consumed.

**refer to notes written out in packet** What happens to the equilibrium position when the temperature of the reaction is increased ? A(g) + B(aq) C(s) + D(aq) kJ (exothermic) Consider energy to be a product in this case! The equilibrium will have to shift to the left to relieve the stress What if pressure was increased (or volume decreased)? Equilibrium shifts towards the side with the fewest total number of moles of gas (in THIS case, the products side).  Equal moles gas? P or V changes have no effect. Heat is added

Predict the reaction shift (left, right, or none) when the following equilibria are stressed: Reaction #1: CO 2 (g) + C (s) + heat ↔ 2CO (g) Reaction #2: H 2 (g) + Cl 2 (g) ↔ 2HCl (g) + heat Reaction #3: N 2 (g) + O 2 (g) + heat ↔ 2NO (g) Increase temperature Decrease [O 2 ] Increase pressure add a catalyst Increase [H 2 ] Increase volume of the system Increase pressure left right No change right No change left No change

The Equilibrium Constant K, K c, or K eq : general reactions K sp : solubility product constant (dissociation constant for salts that aren’t very soluble) K a or K b : dissociation constant for acids or bases

The value of K eq tells us what direction the reaction prefers to go in. It’s calculated from the [products and reactants] at equilibrium. If K eq >>1, the system makes lots of products. Forward reaction goes almost to completion. Ag + (aq) + Cl - (aq)  AgCl(s) K sp = 1.0 x If K eq << 1, very little forward reaction occurs. The system favors reactants. CH 3 COOH  H + + CH 3 COO - K a = 1.8 x 10 -5

Writing K eq Expressions K eq = [C] c [D] d aA + bB cC + d D [A] a [B] b For the reaction: N 2 (g) + 3Cl 2 (g) 2NCl 3 (g) a. Write the equilibrium expression K eq = [NCl 3 ] 2 [N 2 ] 1 [Cl 2 ] 3 Rule: Only include g and aq substances; amounts of solid and liquids do not effect the eq position

K eq = [NCl 3 ] 2 plug in [N 2 ] 1 [Cl 2 ] 3 concentrations! K eq = (1.41 x ) 2 (1.04 x ) 1 (2.01 x ) 3 b. Calculate K eq if [N 2 ] = 1.04 x M [Cl 2 ] = 2.01x M [NCl 3 ] = 1.41x M …first simplify all values in parenthesis… K eq = ( x ) (1.04 x ) ( x ) K eq = 2.35 x 10 13

c. The container is opened and some chlorine gas escapes. Once equilibrium is reestablished, the [Cl 2 ] is 5.01x M and the new [N 2 ] is measured to be 3.24 x M. i. Explain why. N 2 (g) + 3Cl 2 (g)  2NCl 3 (g) ii. What is the new [NCl 3 ] ? K eq = 2.35 x = [NCl 3 ] 2 [N 2 ] 1 [Cl 2 ] x = [NCl 3 ] 2. (3.24 x ) 1 (5.01x ) 3

2.35 x = [NCl 3 ] 2. (3.24 x ) 1 (5.01x ) x = [NCl 3 ] 2. ( x ) x = [NCl 3 ] 2 [NCl 3 ] = 9.79 x M If you plug in all the new concentrations once equilibrium is reestablished, you will come up with the SAME equilibrium constant