1. Chemical Equilibrium

2. aA + bB ↔ cC + dD a, b, c and d are the stoichiometric coefficients for the reacting molecules. A, B, C and D are the reacting molecules. K is the equilibrium constant: for a reversible reaction at equilibrium and a constant temperature, a certain ratio of reactant and product concentrations has a constant value, K

3. Equilibrium Constant K is defined by a quotient, wherein the numerator is the products and the denominator is the reactants This constant is vital for solving stoichiometry problems involving equilibrium.

4. Homogeneous Equilibria Reactions where all reacting species (the molecules) are all in the same phase. There are two ways we can describe our equilibrium: –K c (concentration: expressed in molarity - mol/L) –used with liquids (the solvent is not expressed - l) –K p (pressure: expressed in partial pressure - atm) –used with gases

5. The relationship between K p and K c Kp = Kc (0.0821 T)^Δn

6. Example All the reacting species are liquids (use K c ) Intermediary step (that's why the apostrophe or "prime" symbol is there) - water is the solvent and there is so much of it as compared to the other species present in any solution (1 L = 1000g; 18.02 g/mol = 55.5 M). There is not much change that will occur to the water (meaning: there is not enough of the other substances for water to react with...) Therefore, water (the solvent) is treated as a constant. It has no effect. We remove it...

7. Example II These substances are gases; use K p

8. Putting it together The concentrations of the reacting species at equilibrium are found to be [NO] = 0.0542 M, [O 2 ] = 0.127 M, and [NO 2 ] = 15.5 M Use K c because Molarity concentrations are given d

9. Try one... The equilibrium constant Kp, for the decomposition of phosphorus pentachloride to phosphorus trichloride and molecular chlorine is found to be 1.05 at 250°C. If the equilibrium partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2 at 250°C?

10. solution 1.05 = [(0.463)( P Cl 2 )]/(0.875) P Cl 2 =[(1.05)(0.875)]/(0.463) P Cl 2 = 1.98 atm

11. Due now (practice) p. 633-644 14.16, 14.17, 14.19, 14.21 Due at the end of the day...

12. Heterogeneous Equilibria a reversible reaction involving reactants and products that are in different phases for example: What do we do here?

13. Heterogeneous Equilibria SOLIDS: –the concentration of a solid is expressed as density –density is an intensive property and will not change as long as some of each are present at equilibrium –Therefore, we can exclude CaCO3 and CaO from the equilibrium constant equation –This applies to reactions where the other species are non-solids (liquids, aqueous solutions and/or gases) We can rewrite the equilibrium constant as: Kc = [CO 2 ] (works for Pc as well) Kp = P CO 2

14. Heterogeneous Equilibria LIQUIDS –The same applies to liquids in situations where they are reacting with solids, aqueous solutions and gases –We can remove it from the Equilibrium constant –For example:

15. REWRITE...

16. HOMEWORK p. 633-634 # 14.8 and 14.22