Chemical Equilibrium The study of reactions that occur in both directions.

So far with reactions…. Looked at reactions that go to completion Used stoichiometry for calculation of many quantities Looked at spontaneity and rates of reactions

Now… Reactions can be reversible –They reach a state of equilibrium

Examples Vapor pressure –Evaporation begins –Over time, the system undergoes evaporation and condensation at the same rate

Examples Dissolving and Crystallization –A system could have an equal amount of precipitate (changes states at an equal rate) System of iron (II) dichromate at equilibrium

Examples NO 2 (g) + NO 2 (g) N 2 O 4 (g) –NO 2 = dark brown, N 2 O 4 = colorless –Ultimately ends up somewhere in between Equlibrium Change in Action NO 2 N2O4N2O4 Equilibrium!!

Equilibrium Defined Concentration of products and reactants remain constant over time –The reaction is reversible (can go both directions) –The rate of forward reaction equals the rate of the reverse reaction –Dynamic!! (looks the same when taking a snapshot, but constantly moving back and forth)

Demo, then Graph of Equilibrium Time [R] [P]

Explaining the Graph Time [R] [P] What’s happening to the products? What’s happening to the reactants? Which one is favored? Different reactions have different equilibria…

On a molecular level… Where does the following reach equilibrium?

Equilibrium Expressions Idea by Guldberg and Waage (1864) –Called the Law of Mass Action –Given a reaction: aA + bB  cC + dD

Equilibrium Expressions aA + bB  cC + dD = [C] c [D] d [A] a [B] b Equilibrium constant MUST use concentrations of products over reactants coefficients of balanced equation become exponents

Equilibrium Expressions Works only for GASES and IONS –No pure solids or liquids included Example –C 3 H 8 (g) + O 2 (g)  CO 2 (g) + H 2 O (g) = [C] c [D] d [A] a [B] b = [CO 2 ] 3 [H 2 O] 4 [C 3 H 8 ] [O 2 ] 5

Equilibrium Expressions What if it has solids or liquids? –Called heterogeneous equilibrium –The concentration of solids and liquids is assumed to always remain constant, so they are not included… –Example: Ca (s) + O 2 (g)  CaO (s)

Equilibrium Expressions More examples –H 2 (g) + I 2 (s)  2 HI (g) –CuSO 4  5 H 2 O (s)  CuSO 4 (s) + 5 H 2 O (g) –N 2 (g) + 3 H 2 (g)  2 NH 3 (g)

Values of K Equilibrium constant, K, is found by: [products] [reactants] –If K = 1….. equal ratio of products and reactants –If K > 1 …. reaction favors products –If K < 1 …. reaction favors reactants

What changes K? 1. Change the temperature. –Equilibrium is temperature dependent.

What changes K? 2. Change the reaction –Look at the original reaction and see the change that was made to it. Whatever you do to the reaction, you do to the power of K –Example 2 NO (g) + O 2 (g)  2 NO 2 (g) K= 4.67 x )Change the coefficients NO (g) + 1/2 O 2 (g)  NO 2 (g) What is the change? K’ = K 1/2 = (4.67 x ) 1/2 = 6.83 x 10 6

What changes K? 2. Change the reaction –Look at the original reaction and see the change that was made to it. Whatever you do to the reaction, you do to the power of K –Example 2 NO (g) + O 2 (g)  2 NO 2 (g) K= 4.67 x )Reverse the reaction 2 NO 2 (g)  2 NO (g) + O 2 (g) What is the change? K’ = 1/K = 1/(4.67 x ) = 2.14 x

Equilibrium Constant Can also find from reaction mechanism –OVERALL: N 2 O (g) + 3/2 O 2 (g)  2 NO 2 (g) K c = ?? –Steps: N 2 O (g) + 1/2 O 2 (g)  2 NO (g) K c 1 = 1.7 x NO (g) + O 2 (g)  2 NO 2 (g) K c 2 = 4.67 x –To find K c ? K c = K c 1 x K c 2

Equilibrium including Gases Look at partial pressures instead of molarities –aA (g) + bB (g)  cC (g) + dD (g) = (P C ) c (P D ) d (P A ) a (P B ) b

Equilibrium including Gases If it says K p, you must use pressures –Given moles and L, must use temperature and PV = nRT to get pressures of each species K or K c is still molarity

Practice Problems Write equilibrium expressions for: –a) 2 O 3 (g)  3 O 2 (g) –b) 2 NO(g) + Cl 2 (g)  2 NOCl(g) –c) BaSO 4 (s)  Ba +2 (aq) + SO 4 -2 (aq)

Practice Problems Changing K by changing the equation –Original equation: 2 NO (g) + O 2 (g)  2 NO 2 (g) K eq = NO (g) + ½ O 2 (g)  NO 2 2 NO 2 (g)  2 NO (g) + O 2 (g)