AP Notes Chapter 16 Equilibrium Dynamic chemical system in which two reactions, equal and opposite, occur simultaneously

Properties 1. Appear from outside to be inert or not functioning 2. Can be initiated in both directions

Pink to blue Co(H 2 O) 6 Cl 2 Co(H 2 O) 4 Cl H 2 O Blue to pink Co(H 2 O) 4 Cl H 2 O Co(H 2 O) 6 Cl 2

Equilibrium achieved Product conc. increases and then becomes constant at equilibrium Reactant conc. declines and then becomes constant at equilibrium

At any point in the reaction H 2 + I 2 2 HI

Equilibrium achieved In the equilibrium region

Kinetics Definition R f = R r

At equilibrium, the rates of the forward and reverse reactions are equal.

aA + bB cC R f (eq) = k f [A] a [B] b R r (eq) = k r [C] c R f = R r k f [A] a [B] b = k r [C] c

By convention

K is a concentration quotient for a system at equilibrium. K = Q

For a system NOT at equilibrium Q K

if Q > K The reverse reaction will occur until equilibrium is achieved.

if Q < K The forward reaction will occur until equilibrium is achieved.

Achieving equilibrium is a driving force in chemical systems and will occur when possible. It cannot be stopped (spontaneous)

1.For the equilibrium system, 2NO 2 (g) N 2 O 4 (g), the equilibrium constant, K C, is 8.8 at 25 0 C. If analysis shows that 2.0 x mole of NO 2 and 1.5 x mole of N 2 O 4 are present in a 10.0 L flask, is the reaction at equilibrium?

Types of Reactions 1. one way (goes to completion) NaOH (s) Na + (aq) + OH - (aq)

Types of Reactions 2. Equilibrium (two opposite reactions at same time) a. dimerization 2NO 2 (g) N 2 O 4 (g)

b. dissociation of a weak electrolyte CH 3 COOH + H 2 O CH 3 COO - + H 3 O +

c. saturated aqueous solutions AgCl (s) Ag + (aq) + Cl - (aq) C 6 H 12 O 6 (s) C 6 H 12 O 6 (aq)

By convention EQUILIBRIUM CONSTANT K eq

if K eq > 1 [products] coeff > [reactants] coeff the forward reaction proceeded to a greater extent than the reverse reaction to achieve equilibrium (i.e. the products predominate at equilibrium)

if K eq < 1 [products] coeff < [reactants] coeff the forward reaction proceeded to a lesser extent than the reverse reaction to achieve equilibrium (i.e. the reactants predominate at equilibrium)

How are k f and k r related to temperature? k f and k r are temperature dependent thus, K eq is temperature dependent

N 2 O 4 + heat 2 NO 2 (colorless) (brown) H o = kJ K c (273 K) = K c (298 K) =

N 2 O 4 (g) 2NO 2 (g) Examples of Equilibrium Expressions

CH 3 COOH + H 2 O CH 3 COO - +H 3 O +

AgCl (s) Ag + (aq) + Cl - (aq)

Concentrations of pure liquids and solids are NOT included in equilibrium expressions, as their concentrations are themselves constants.

The value of K eq may appear to change based on way equation is balanced.

A value that is mathematically related to another (eg. temp) is NOT considered a new value

Multiple Equilibria H 3 PO H 2 O PO H 3 O + H 3 PO 4 + H 2 O H 2 PO H 3 O + H 2 PO H 2 O HPO H 3 O + HPO H 2 O PO H 3 O +

K eq = K 1. K 2. K 3 for the complete dissociation of phosphoric acid

So far, K eq has been studied as a function of concentration, or expressed with appropriate notation, K c

But, what about equilibrium systems where all components are gases? Partial pressures mole distribution

where V = a container parameter (constant for all gases) T = constant for given values of K R = constant

aA(g) + bB(g) cC(g)

Substituting for a gas

Let c - (a + b) = n where n is the change in # of moles of gas (product - reactant) for the forward reaction.

If we express the equilibrium constant as a function of partial pressures

Thus K C = K P (RT) - n or K P = K c (RT) n

2.When 2.0 moles of HI(g) are placed in a 1.0 L container and allowed to come to equilibrium with its elements, it is found that 20% of the HI decomposes. What is K C and K P ?

Applications of the Equilibrium Constant & LeChateliers Principle

mol of n-butane is placed in a 0.50 L container and allowed to come to equilibrium with its isomer isobutane. K C at 25 0 C is 2.5. What are the equilibrium concentrations of the two isomers?

Set up an ICE table Initial [ ] of components Change in [ ] Equilibrium [ ]

n-butane isobutane I C -x +x E x x

solve

4.2.0 mols Br 2 are placed in a 2.0 L flask at 1756 K, which is of sufficient energy to split apart some of the molecules. If K C = 4.0 x at 1756 K, what are the equilibrium concentrations of the bromine molecules and atoms?

Br 2 (g) 2 Br(g) I C -x +2x E x 2x

solve

if K <<< [A] 0, then can assume amount that dissociated to reach equilibrium is VERY small, thus

solve

5.Calculate [OH - ] at equilibrium of a solution that is initially M nicotine.

2H 2 O + Nic NicH OH - I C -x +x +2x E x x 2x

K C = K 1. K 2 K C = (7.0 x )(1.1 x ) K C = 7.7 x

LeChatliers Principle When a stress is placed on a system at equilibrium, the system will adjust so as to relieve that stress.

Stress Factors 1. Change in concentration of reactants or products 2. Change in volume or pressure (for gases) 3. Change in temperature

Responses to Stress 3 H 2 (g) + N 2 (g) 2 NH 3 (g) 1. Change concentration a. add either H 2 or N 2 b. remove NH 3

Responses to Stress 3H 2 (g) + N 2 (g) 2 NH 3 (g) 2. Change in volume or pressure a. increase volume b. Increase pressure c. add He

Responses to Stress 3H 2 (g) + N 2 (g) 2 NH 3 (g) H = -92 kJ 3. Change in temperature a. increase temperature

AgCl(s) Ag + (aq) + Cl - (aq) a. add AgCl b. add H 2 O c. add NaCl d. add NH 3 (aq)

K eq is a temperature dependent constant, similar to the rate constant, k f or k r slope of line is different

1/T ln K eq m = - H R /R

Change T – change in K – therefore change in P or concentrations at equilibrium Use a catalyst: reaction comes more quickly to equilibrium. K not changed. Add or take away reactant or product: –K does not change –Reaction adjusts to new equilibrium position Le Chateliers Principle