1. 1 Equilibrium Constant even though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between them the relationship between the chemical equation and the concentrations of reactants and products is called the Law of Mass Action for the general equation aA + bB  cC + dD, the Law of Mass Action gives the relationship below the lowercase letters represent the coefficients of the balanced chemical equation always products over reactants K is called the equilibrium constant unitless

2. 2 Writing Equilibrium Constant Expressions for the reaction aA (aq) + bB (aq)  cC (aq) + dD (aq) the equilibrium constant expression is so for the reaction 2 N 2 O 5  4 NO 2 + O 2 the equilibrium constant expression is:

3. 3 What Does the Value of K eq Imply? when the value of K eq >> 1, we know that when the reaction reaches equilibrium there will be many more product molecules present than reactant molecules the position of equilibrium favors products when the value of K eq << 1, we know that when the reaction reaches equilibrium there will be many more reactant molecules present than product molecules the position of equilibrium favors reactants

4. 4 A Large Equilibrium Constant

5. 5 A Small Equilibrium Constant

6. 6 Relationships between K and Chemical Equations when the reaction is written backwards, the equilibrium constant is inverted for the reaction aA + bB  cC + dD the equilibrium constant expression is: for the reaction cC + dD  aA + bB the equilibrium constant expression is:

7. 7 Relationships between K and Chemical Equations when the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor for the reaction aA + bB  cC the equilibrium constant expression is: for the reaction 2aA + 2bB  2cC the equilibrium constant expression is:

8. 8 Relationships between K and Chemical Equations when you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations for the reactions (1) aA  bB and (2) bB  cC the equilibrium constant expressions are: for the reaction aA  cC the equilibrium constant expression is:

9. K backward = 1/K forward, K new = K old n Ex 14.2 – Compute the equilibrium constant at 25°C for the reaction NH 3 (g)  0.5 N 2 (g) + 1.5 H 2 (g) for N 2 (g) + 3 H 2 (g)  2 NH 3 (g), K = 3.7 x 10 8 at 25°C K for NH 3 (g)  0.5N 2 (g) + 1.5H 2 (g), at 25°C Solution: Concept Plan: Relationships: Given: Find: N 2 (g) + 3 H 2 (g)  2 NH 3 (g)K 1 = 3.7 x 10 8 K’K 2 NH 3 (g)  N 2 (g) + 3 H 2 (g) NH 3 (g)  0.5 N 2 (g) + 1.5 H 2 (g)

10. 10 Equilibrium Constants for Reactions Involving Gases the concentration of a gas in a mixture is proportional to its partial pressure therefore, the equilibrium constant can be expressed as the ratio of the partial pressures of the gases for aA(g) + bB(g)  cC(g) + dD(g) the equilibrium constant expressions are or

11. 11 K c and K p in calculating K p, the partial pressures are always in atm the values of K p and K c are not necessarily the same because of the difference in units K p = K c when  n = 0 the relationship between them is:  n is the difference between the number of moles of reactants and moles of products

12. 12 Deriving the Relationship between K p and K c

13. 13 Deriving the Relationship Between K p and K c for aA(g) + bB(g)  cC(g) + dD(g) substituting

14. Ex 14.3 – Find K c for the reaction 2 NO(g) + O 2 (g)  2 NO 2 (g), given K p = 2.2 x 10 12 @ 25°C K is a unitless number since there are more moles of reactant than product, K c should be larger than K p, and it is K p = 2.2 x 10 12 KcKc Check: Solution: Concept Plan: Relationships: Given: Find: KpKp KcKc 2 NO(g) + O 2 (g)  2 NO 2 (g)  n = 2  3 = -1