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(H2O)6Cl2  Co(H2O)4Cl2 + 2 H2O Blue to pink Co(H2O)4Cl2 + 2 H2O  Co(H2O)6Cl2

Product conc. increases and then becomes constant at equilibrium 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 H2 + I2 2 HI

In the equilibrium region Equilibrium achieved In the equilibrium region

Kinetics Definition Rf = Rr

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

aA + bB  cC Rf(eq) = kf [A]a [B]b Rr(eq) = kr [C]c Rf = Rr kf [A]a [B]b = kr [C]c

By convention

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

For a system NOT at equilibrium Q ≠ K

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

The forward 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, 2NO2 (g)  N2O4 (g) , the equilibrium constant, KC , is 8.8 at 250C. If analysis shows that 2.0 x 10-3 mole of NO2 and 1.5 x 10-3 mole of N2O4 are present in a 10.0 L flask, is the reaction at equilibrium?

2NO2 (g)  N2O4 (g) , the equilibrium constant, KC , is 8. 8 at 250C 2NO2 (g)  N2O4 (g) , the equilibrium constant, KC , is 8.8 at 250C. If analysis shows that 2.0 x 10-3 mole of NO2 and 1.5 x 10-3 mole of N2O4 are present in a 10.0 L flask, is the reaction at equilibrium? 2NO2 (g)  N2O4 (g) KC = 8.8 so… Q = [product]m = [reactant]n

2NO2 (g)  N2O4 (g) , the equilibrium constant, KC , is 8. 8 at 250C 2NO2 (g)  N2O4 (g) , the equilibrium constant, KC , is 8.8 at 250C. If analysis shows that 2.0 x 10-3 mole of NO2 and 1.5 x 10-3 mole of N2O4 are present in a 10.0 L flask, is the reaction at equilibrium? 2NO2 (g)  N2O4 (g) KC = 8.8 so… Q = [product]m = (2.0 x 10-3 mole NO2/ 10.0L) [reactant]n (1.5 x 10-3 mole of N2O4 /10.0L)2 =

2NO2 (g)  N2O4 (g) , the equilibrium constant, KC , is 8. 8 at 250C 2NO2 (g)  N2O4 (g) , the equilibrium constant, KC , is 8.8 at 250C. If analysis shows that 2.0 x 10-3 mole of NO2 and 1.5 x 10-3 mole of N2O4 are present in a 10.0 L flask, is the reaction at equilibrium? 2NO2 (g)  N2O4 (g) KC = 8.8 so… Q = [product]m = (2.0 x 10-3 mole NO2/ 10.0L) [reactant]n (1.5 x 10-3 mole of N2O4 /10.0L)2 = 8888 KC = 8.8 so… Q = 8888 Is K = Q? Concentrations and orders must be used!! NO Q > K so [product] is BIGGER than is should be to be at Equilibrium with given K value so products will convert to reactants to reach 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 2NO2(g)  N2O4(g)

b. dissociation of a weak electrolyte CH3COOH + H2O  CH3COO- + H3O+

c. saturated aqueous solutions AgCl(s)  Ag+(aq) + Cl-(aq) C6H12O6(s)  C6H12O6(aq)

EQUILIBRIUM CONSTANT Keq By convention

[products]coeff > [reactants]coeff if Keq > 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)

[products]coeff < [reactants]coeff if Keq < 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 kf and kr related to temperature? kf and kr are temperature dependent thus, Keq is temperature dependent

N2O4 + heat 2 NO2 (colorless) (brown) ∆Ho = + 57.2 kJ Kc (273 K) = 0.00077 Kc (298 K) = 0.0059

Examples of Equilibrium Expressions N2O4(g)  2NO2(g)

CH3COOH + H2O CH3COO- +H3O+

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 Keq may appear to change based on way equation is balanced.

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

Multiple Equilibria H3PO4 + 3 H2O  PO43- + 3 H3O+ H3PO4 + H2O  H2PO4- + H3O+ H2PO4- + H2O  HPO42- + H3O+ HPO42- + H2O  PO43- + H3O+

for the complete dissociation of Keq = K1. K2. K3 for the complete dissociation of phosphoric acid

So far, Keq has been studied as a function of concentration, or expressed with appropriate notation, Kc

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 KC = KP(RT)-n or KP = Kc(RT)Dn

2. When 2. 0 moles of HI(g) are placed in a 1 2. When 2.0 moles of HI(g) are placed in a 1.0 L container and allowed to come to equilibrium with it’s elements, it is found that 20% of the HI decomposes. What is KC and KP?

Applications of the Equilibrium Constant & LeChatelier’s Principle

3. 017 mol of n-butane is placed in a 0 3. 0.017 mol of n-butane is placed in a 0.50 L container and allowed to come to equilibrium with its isomer isobutane. KC at 250C 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 0.034 0 C -x +x E 0.034-x x

solve

4. 2.0 mols Br2 are placed in a 2.0 L flask at 1756 K, which is of sufficient energy to split apart some of the molecules. If KC = 4.0 x 10-4 at 1756 K, what are the equilibrium concentrations of the bromine molecules and atoms?

Br2(g)  2 Br(g) I 1.0 0 C -x +2x E 1.0 - 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 0 5. Calculate [OH-] at equilibrium of a solution that is initially 0.020 M nicotine.

2H2O + Nic  NicH22+ + 2 OH- I 0.020 0 0 C -x +x +2x E 0.020 - x x 2x

KC = K1 . K2 KC = (7.0 x 10-7)(1.1 x 10-10) KC = 7.7 x 10-17

LeChatlier’s 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 1. Change concentration a. add either H2 or N2 3 H2(g) + N2(g)  2 NH3(g) 1. Change concentration a. add either H2 or N2 b. remove NH3

Responses to Stress 2. Change in volume or pressure a. increase volume 3H2(g) + N2(g)  2 NH3(g) 2. Change in volume or pressure a. increase volume b. Increase pressure c. add He

Responses to Stress 3. Change in temperature a. increase temperature 3H2(g) + N2(g)  2 NH3(g) H = -92 kJ 3. Change in temperature a. increase temperature

AgCl(s)  Ag+(aq) + Cl-(aq) a. add AgCl b. add H2O c. add NaCl d. add NH3(aq)

slope of line is different Keq is a temperature dependent constant, similar to the rate constant, kf or kr slope of line is different

m = -HR/R ln Keq 1/T

Le Chatelier’s Principle 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”