Chapter 14 Chemical Equilibrium I.Equilibrium Conditions A.Example Reaction: NO 2 + NO 2 N 2 O 4 (brown) (colorless) 1.This reaction proceeds to the right at the start 2.But, it never goes all the way to completion. It reaches a point where the forward and reverse reactions are going at the same rate. a)  [NO 2 ] = 0 b)  [N 2 O 4 ] = 0 c)Chemical Equilibrium = The state where the concentrations of all reactants and products remain constant with time.

3.Some reactions almost reach completion before equilibrium is achieved. We say the equilibrium lies to the right. 2H 2 + O 2 2H 2 O 4.Some reactions barely get started before reaching equilibrium. We say the equilibrium lies to the left. 2CaO 2Ca + O 2 B.Equilibria are Dynamic 1.Even though  [NO 2 ] = 0, the forward and reverse reactions are both occurring. 2.At equilibrium, there is no net change in concentration, even though individual molecules are constantly reacting. 3.Reaction: 1H 2 O + 1CO 1H 2 + CO 2

a)At equilibrium, concentrations don’t change, even though individual molecules do b)The equilibrium lies to the right for this reaction. There are more products than reactants at equilibrium. C.Why does Equilibrium Occur? 1.Molecules react by colliding (Kinetic theory) 2.The number of collisions depends on concentration 3.[H 2 O] and [CO] are decreased by the forward reaction, so the forward reaction slows down. 4.[H 2 ] and [CO 2 ] are increased by the forward reaction, so the reverse reaction speeds up. 5.Eventually, the forward and reverse reactions reach the same rate

5.Factors that determine where the equilibrium position is: a)Initial concentrations b)Energies of the reactants and products c)Entropy (disorder is favored) d)Reaction rates of the forward and reverse reactions

D.Sample Reaction: N 2 + 3H 2 2NH 3 1.When we mix any concentrations of these gases at room temperature, there are no changes in concentration. a)We might be at equilibrium b)The reaction rates might be very slow at these conditions. i.The N≡N bond is very strong (432 kJ/mol) and hard to break ii.Entropy favors no forward reaction 2.When a catalyst is added and the reaction is heated, the concentrations do change until a real equilibrium is reached. a)H 2 disappears three times as fast as N 2 b)NH 3 appears twice as fast as N 2 disappears.

II.Equilibrium Constants A.The Law of Mass Action 1.This is an empirical law discovered in Every reaction has a constant associated with it telling us where the equilibrium position is. aA + bB cC + dD 3.K = Equilibrium Constant = tells us where the equilibrium position is a)K > 1 tells us the equilibrium lies to the right b)K < 1 tells us the equilibrium lies to the left 4.If we know the concentrations, we can find K from its equation 5.K is written without units, even in cases where there are units left not cancelled. This is correct for nonideal behavior of molecules. 6.Example: Write K for: 4NH 3 + 7O 2 4NO 2 + 6H 2 O K If K = 1, halfway

6.Example: N 2 + 3H 2 2NH 3 a)Find K if [NH 3 ] = 3.1 x 10 -2, [N 2 ] = 0.85, [H 2 ] = 3.1 x b)Find K for the reverse reaction c)Find K for 0.5 N H 2 NH 3 7.Conclusions: a)If K is for a forward reaction, and K’ is for the reverse: K = 1/K’ b)If you multiply a reaction by n, K’’ = K n 8.Equilibrium Position = the set of equilibrium concentrations a)There is only one equilibrium constant for a given reaction b)The infinite equilibrium positions depend on initial concentrations

9.Example: 2SO 2 + O 2 2SO 3 Show that K is the same for 2 different initial conditions B.Equilibrium Involving Pressures 1.Ideal gas law: PV = nRT Initial ConcentrationsEquilibrium ConcentrationsK [SO 2 ]=2M, [O 2 ]=1.5M, [SO 3 ]=3M[SO 2 ]=1.5M, [O 2 ]=1.25M, [SO 3 ]=3.5M? [SO 2 ]=0.5M, [O 2 ]=0, [SO 3 ]=0.35M[SO 2 ]=0.59M, [O 2 ]=0.045M, [SO 3 ]=0.26M?

2.N 2 + 3H 2 2NH 3 a)We can rewrite K using concentration C b)Since C is proportional to P, the partial pressure, we can write a similar equilibrium constant K P 3.Example: Find K P for: 2NO + Cl 2 2NOCl P NOCl = 1.2 atm, P NO = 0.05 atm, P Cl2 = 0.3 atm 4.How are K P and K related? a)K P = K(RT)  n b)  n = (sum of product coefficients) – (sum of reactant coefficients) 5.Example: Find K from K P in the example directly above

III.Heterogeneous Equilibria A.Types of Equilibria 1.Homogeneous = all species in the same phase (so far, gas phase) 2.Heterogeneous = species in multiple phases a)2CO(g) C(s) + CO 2 (g) b)Experiments show that K is not effected by how much pure solid or liquid is present. c)Pure solids and liquids are treated as constant concentrations, and therefore do not appear in the equilibrium expression

3.Examples: 2H 2 O(l) 2H 2 (g) + O 2 (g) 2H 2 O(g) 2H 2 (g) + O 2 (g) IV.Applications of Equilibrium Constants A.Equilibrium Constants (K) tell us many things 1.Will the reaction go forward? (K > 1, it will move forward) 2.Are current conditions at equilibrium? 3.What conditions will be present when equilibrium is achieved? B.Example: A(g) + B (g) C (g) + D (g) 1.Start with 9A molecules and 12B molecules 2.What are the conditions at equilibrium? a) Guess that 5C, 5D, 4A, and 7B

b)Can we find out mathematically instead of guessing? At Equilibrium: Try x = 8 Yes, we are at equilibrium when x = 8 This is a general way to solve an equilibrium problem: ICE Table ABCD Initial91200 Change-x +x Equilibrium9 – x12 – xxx