1. Equilibrium Chapter 15 Chemical Equilibrium John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

2. Equilibrium The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

3. Equilibrium The Concept of Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate.

4. Equilibrium A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant. No observable changes are occurring to the system –Partial pressures, temperature, concentrations of products and reactants are all constant

5. Equilibrium Depicting Equilibrium In a system at equilibrium, both the forward and reverse reactions are being carried out; as a result, we write its equation with a double arrow N 2 O 4 (g) 2 NO 2 (g)

6. Equilibrium Real life Equilibrium Many reversible, observable processes occur –Evaporating/condensing (dew on grass) –Absorption and evaporation of gases from water –Dissolving/precipitation of a salt –Binding O 2 to hemoglobin in blood –Scent molecules binding to scent receptors in nose –Transfer C between atmosphere and biosphere (photosynthesis and cellular respiration) –Transfer of dissolved substances between atmosphere and hydrosphere

7. Equilibrium Reactions we have already talked about… Dissolution of a solid Transfer of protons during acid-base reactions Transfer of electrons during redox reactions

8. Equilibrium The Equilibrium Constant Forward reaction: N 2 O 4 (g)  2 NO 2 (g) Rate law: Rate = k f [N 2 O 4 ]

9. Equilibrium The Equilibrium Constant Reverse reaction: 2 NO 2 (g)  N 2 O 4 (g) Rate law: Rate = k r [NO 2 ] 2

10. Equilibrium The Equilibrium Constant Therefore, at equilibrium Rate f = Rate r k f [N 2 O 4 ] = k r [NO 2 ] 2 Rewriting this, it becomes kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] =

11. Equilibrium The Equilibrium Constant 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 ] =

12. Equilibrium The Equilibrium Constant To generalize this expression, consider the reaction The equilibrium expression for this reaction would be K c = [C] c [D] d [A] a [B] b aA + bBcC + dD c=concentration

13. Equilibrium What Are the Equilibrium Expressions for These Equilibria?

14. Equilibrium The Equilibrium Constant Because 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 p = pressure

15. Equilibrium Relationship between K c and K p From the ideal gas law we know that Rearranging it, we get PV = nRT P = RT nVnV

16. Equilibrium Relationship between K c and K p 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)

17. Equilibrium Equilibrium Can Be Reached from Either Direction As you can see, 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.

18. Equilibrium Equilibrium Can Be Reached from Either Direction This is the data from the last two trials from the table on the previous slide.

19. Equilibrium Equilibrium Can Be Reached from Either Direction It does not 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.

20. Equilibrium What Does the Value of K Mean? Since K = [products]/[reactants] If K >> 1, the reaction is product-favored; product predominates at equilibrium. If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

21. Equilibrium Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. 1 0.212 = K c = = 0.212 at 100  C [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) K c = = 4.72 at 100  C [N 2 O 4 ] [NO 2 ] 2 N2O4 (g)N2O4 (g) 2 NO 2 (g)

22. Equilibrium Manipulating Equilibrium Constants The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number. K c = = 0.212 at 100  C [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) K c = = (0.212) 2 at 100  C [NO 2 ] 4 [N 2 O 4 ] 2 2 N 2 O 4 (g) 4 NO 2 (g)

23. Equilibrium Manipulating Equilibrium Constants The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.

24. Equilibrium Heterogenous Equilibrium When more than one state of matter is present during the reaction –The Concentrations of Solids and Liquids Are Essentially Constant Both can be obtained by dividing the density of the substance by its molar mass—and both of these are constants at constant temperature.

25. Equilibrium The Concentrations of Solids and Liquids Are Essentially Constant Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression –Only aq and gases show up in the expression K c = [Pb 2+ ] [Cl − ] 2 PbCl 2 (s) Pb 2+ (aq) + 2 Cl − (aq)

26. Equilibrium As long as some CaCO 3 or CaO remain in the system, the amount of CO 2 above the solid will remain the same. Therefore, K c = [CO 2 ] CaCO 3 (s) CO 2 (g) + CaO (s)