AP Notes Chapter 16Equilibrium

Dynamic chemical system in which two reactions, equal and opposite, occur simultaneously

Properties1. Appear from outside to

be inert or not functioning

2. Can be initiated in both directions

Pink to blueCo(H2O)6Cl2 Co(H2O)4Cl2 + 2 H2O

Blue to pinkCo(H2O)4Cl2 + 2 H2O Co(H2O)6Cl2

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 reactionH2 + I2 2 HI

Q reaction quotient = [HI]2

[H2 ][I2 ]

Equilibrium achieved

In the equilibrium region

[HI]2

[H2 ][I2 ] = 55.3 = K

K = equilibrium constant

Kinetics Definition

Rf = Rr

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

aA + bB cCRf(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

if Q > KThe 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, 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?

Types of Reactions1. one way

(goes to completion)

NaOH(s) Na+(aq) + OH-(aq)

Types of Reactions2. 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)

By convention EQUILIBRIUM CONSTANT Keq

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)

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

]O[N

][NO K

42

22

c ]O[N

][NO K

42

22

c

Kc (273 K) = 0.00077

Kc (298 K) = 0.0059

N2O4(g) 2NO2(g)

]ON[

]NO[K

42

22

eq

Examples of EquilibriumExpressions

CH3COOH + H2O CH3COO- +H3O+

]COOHCH[

]OH][COOCH[K

3

33eq

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

]Cl][Ag[Keq

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.

21

]ON[

]NO[K

NOON

42

2'eq

24221

eq'eq KK

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

value

Multiple EquilibriaH3PO4 + 3 H2O

PO43- + 3 H3O+

H3PO4 + H2O H2PO4- + H3O+

H2PO4- + H2O HPO4

2- + H3O+

HPO42- + H2O PO4

3- + H3O+

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

RT

P]gas[

V

n

nRTPV

whereV = a container parameter (constant for all gases)T = constant for given values of KR = constant

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

ba

c

c ]B[]A[

]C[K

Substituting for a gas

b

B

a

A

c

C

c

RTP

RTP

RTP

K

)ba(c

bB

aA

cC

C RT

1

PP

PK

Letc - (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

bB

aA

cC

PPP

PK

Thus

KC = KP(RT)-n

or

KP = Kc(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 it’s elements, it is found that 20% of the HI decomposes. What is KC and KP?

Applications of theEquilibrium Constant

& LeChatelier’s

Principle

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

x034.0

x5.2

etanbun

]etanisobu[KC

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

x0.1

)x2(100.4

]Br[

]Br[K

24

2

2

C

solve

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

small, thus

0eq ]A[]A[

0.1

)x2(100.4

]Br[

]Br[K

24

2

2

C

solve

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

)x020.0(

)x2)(x(107.7

]Nic[

]OH][NicH[K

217

22

C

LeChatlier’s PrincipleWhen a stress is placed

on a system at equilibrium, the system

will adjust so as to relieve that stress.

Stress Factors1. Change in

concentration of reactants or products

2. Change in volume or pressure (for gases)

3. Change in temperature

Responses to Stress3 H2(g) + N2(g) 2 NH3(g)

1. Change concentration

a. add either H2 or N2

b. remove NH3

Responses to Stress3H2(g) + N2(g) 2 NH3(g)

2. Change in volume or pressure

a. increase volume

b. Increase pressure

c. add He

Responses to Stress3H2(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)

Keq is a temperature dependent constant,

similar to the rate constant, kf or kr

slope of line is different

1/T

ln K

eq m = -HR/R

bT

1

R

H- K ln R

eq

12

R

2

1

T

1

T

1

R

H

K

K ln

• 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 Chatelier’s Principle