Chemical Equilibrium

1. Chemical Equilibrium Chapter 13 Introduction

2. Intro: • Many chemical reactions go to completion (where all reactants are consumed to form product) • This unit deals with reactions that do not completely convert all reactants to products.

3. Chemical Equilibrium • A state where the rate of the forward chemical reaction is equal to the rate of the reverse chemical reaction • Double displacement arrow show that a system is at equilibrium

5. Characteristics of a system at equilibrium • Reversible reaction must be possible • There is a dynamic state of balance between both the forward and backward reaction • There is no change in concentration of reactants or products once chemical equilibrium is reached • There is no bulk change in properties of the system (ex: no colour or pressure change)

6. It is a closed system (no heat or matter in or out) • The temperature of the system remains constant • It can be reached from either direction • Any change to the system at equilibrium can be reversed if restored back to original equilibrium conditions.

7. Examples at Equilibrium • Water evaporating and condensing in a jar with lid • A supersaturated solution with a few crystals added • A bottle of unopened pepsi, carbon dioxide is in equilibrium in solution and air above pop.

9. Changing the temperature affects the rates of forward and reverse reactions differently because they have different activation energies • Kc = expressed in terms of molar concentration (mol/L)

13. Calculating Kc N2 + 3Cl2<--> 2NCl3 Temp remains constant in 5 L flask. Equilbrium was reached and the following was found: 0.007 mol N2 , 0.0022 mol Cl2 and 0.95 mol ofNCl3. Calculate the Kc for this reaction. • Find molar concentration of everything (mol/L) • Use equilibrium expression based on chemical equation

14. Answer • [N2] = 0.007 mol / 5 L = 0.0014 M [Cl2] = 0.0022 mol / 5 L = .00044 M [NCl3] = 0.95 mol / 5 L = 0.19 M • Kc = [NCl3]2= [0.19]2 [N2][Cl2]3 [0.0014][0.00044]3 = 3.0 x 10 11

15. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) 1. Write an expression for Keq 2. Calculate K at a given temperature if [CH4] = 0.020 M, [O2] = 0.042 M, [CO2] = 0.012 M, and [H2O] = 0.030 M at equilibrium. (include units)

16. Practice: • Pg 493 #1-5 • Pg 497 #1-3 • Pg 499 # 6-8

17. The value of K • The value or magnitude of K tells us the extent to which reactants have been converted into products. • Remember in the ratio for K, the concentration of products divided by the concentration of reactants.

18. A small value for K means that very little of the reactants were converted into products before equilibrium was reached. • This is stated as “reactants are favoured”. • A large value of K means that most of the reactants were converted into products before equilibrium was reached. • This is stated as “products are favoured”.

19. The Magnitude of Equilibrium Constants If K >1, then products dominate at equilibrium and equilibrium lies to the right. If K <1, then reactants dominate at equilibrium and the equilibrium lies to the left.

20. If K = 1 neither reactants nor products are favoured. The value of • K does not indicate how long it takes for equilibrium to be reached. The value of K varies with temperature and that’s why its usually mentioned with K

21. 2NO2 (g) <--> N2O4 (g) • Four experiments were performed. • The initial concentrations of the two chemicals were different in each experiment and the concentration of each gas was measured once the system reached equilibrium.

22. Calculate • Using the data on the previous page, calculate the K for each experiment. • What did you find?

23. Problems • Work on the reverse side of the handout previously passed out. • Tell me which direction the reaction favors for each question.