Chemical Equilibrium

Brown, LeMay Ch 15AP Chemistry

Monta Vista High School1

15.1: Chemical Equilibrium

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Occurs when opposing reactions are proceeding at the same rate Forward rate = reverse rate of

reactionEx:Vapor pressure: rate of vaporization

= rate of condensationSaturated solution: rate of

dissociation = rate of crystallization

Expressing concentrations: Gases: partial pressures, PX Solutes in liquids: molarity, [X]

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Forward reaction: A → B Rate = kforward [A] Reverse reaction: B → A Rate = kreverse [B]

or

RT

P

V

nM

RT

PA A][

RT

PB B][

Forward reaction:

Reverse reaction:

RT

PkRate A

f

RT

PkRate B

r

nRTPV

R = 0.0821 L•atm

mol•K

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http://www.kentchemistry.com/links/Kinetics/Equilibrium/equilibrium.swf ( Good link on Equilibrium)

[B]k [A]k rf

r

f

k

k

[A]

[B]

RT

Pk

RT

Pk B

rA

f or

eqr

f

A

B Kconstantk

k

P

Por

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Kc= kf/kr, at equilibrium, if K> 1, then more products at equilib. And if k<1, then reactants favored at equilb. K=1 (conc. Of reactants and products nearly same at equilibrium)

The magnitude of Kc gives us an indication of how far the reaction has proceeded toward the formation of products, when the equilibrium is achieved.

The larger the value of K, the further the reaction will have proceeded towards completion when equilibrium is reached (more products present at equilibrium).

Equilibrium

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Equilibrium is dynamic. The forward and reverse rxns occur at the same rate.

There is a spontaneous tendency towards equilibrium. This does not mean that equilibrium will occur quickly, it simply means that there is always a drive TOWARD the equilibrium state. The amount of drive is measured as Free Energy (D G)

The driving force towards equilibrium diminishes as equilibrium is approached. Thus the appearance of products actually decreases the forward impetus of the reaction, making the free energy change less negative.

http://www.youtube.com/watch?v=CMs2WhGY3NE

Equilibrium

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The equilibrium position is the same at a given temperature, no matter from which direction it is approached.

It is possible to force an equilibrium one way or the other temporarily by altering the reaction conditions, but once this “stress” is removed, the system will return to its original equilibrium.

Figure 1: Reversible reactions

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PX o

r [X

]

Time →

[B] or PB / RT Equilibrium is established

[A] or PA / RT

[A]0 or PA0 / RT

Reversible Reactions and Rate

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Reacti

on

R

ate

Time

Backward rate

Forward rate

Equilibrium is established:

Forward rate = Backward rate

When equilibrium is achieved:[A] ≠ [B] and kf/kr = Keq

15.2: Law of Mass Action

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Derived from rate laws by Guldberg andWaage (1864) For a balanced chemical reaction

in equilibrium:a A + b B ↔ c C + d D

Equilibrium constant expression (Keq):

ba

dc

c [B] [A]

[D] [C] K b

Ba

A

dD

cC

p )(P)(P

)(P)(PK

Keq is strictly based on stoichiometry of the reaction (is independent of the mechanism).

Units: Keq is considered dimensionless (no units)

Cato Guldberg Peter Waage

(1836-1902) (1833-1900)

or

Relating Kc and Kp

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Convert [A] into PA:

ba

dc

([B]RT)([A]RT)

([D]RT)([C]RT)b

Ba

A

dD

cC

p )(P)(P

)(P )(PK

baba

dcdc

(RT)[B][A]

(RT)[D][C]

(RT) K K b)(a - d)(ccp

where n == change in coefficents of products – reactants (gases only!)= (c+d) - (a+b)

(RT) K nc

RT

P

V

nM RTAPA ][

Magnitude of Keq

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Since Keq [products]/[reactants], the magnitude of Keq predicts which reaction direction is favored:

If Keq > 1 then [products] > [reactants]and equilibrium “lies to the

right”

If Keq < 1 then [products] < [reactants]and equilibrium “lies to the

left”

Relationship Between Q and K

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Reaction Quotient (Q): The particular ratio of concentration terms that we write for a particular reaction is called reaction quotient.

For a reaction, A B, Q= [B]/[A] At equilibrium, Q= K Reaction Direction: Comparing Q and K Q<K, reaction proceeds to right, until

equilibrium is achieved (or Q=K). Why? Q>K, reaction proceeds to left, until Q=K.

Why?

Value of K

For therxn, A>B,

For the reverse rxn, B >A,

For the reaction,2A > 2B

For the rxn,A > CC > B

K(rxn)= [B]/[A]

K= 1/K(rxn)

K= K(rxn)2 K (overall)= K1 X K2

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15.3: Types of Equilibria

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Homogeneous: all components in same phase (usually g or aq)

N2 (g) + H2 (g) ↔ NH3 (g)

3H

1N

2NH

P )(P)(P

)(PK

22

3

bB

aA

dD

cC

P )(P)(P

)(P)(PK

3 21

Fritz Haber(1868 – 1934)

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Heterogeneous: reactants and products in different phases

CaCO3 (s) ↔ CaO (s) + CO2 (g)

Definition: What we use:

][CaCO

)(P [CaO] K

3

COeq

22COp P K

Even though the concentrations of the solids or liquids do not appear in the equilibrium expression, the substances must be present to achieve equilibrium.

Concentrations of pure solids and pure liquids are not included in Keq expression because their concentrations do not vary, and are “already included” in Keq (see p. 548).

15.4: Calculating Equilibrium Constants

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ICE tables are used to calculate equilibriumconcentrations of the reactants and productsfrom the initial concentrations.Steps to use “ICE” table:

1. “I” = Tabulate known initial and equilibrium concentrations of all species in equilibrium expression

2. “C” = Determine the concentration change for the species where initial and equilibrium are known

• Use stoichiometry to calculate concentration changes for all other species involved in equilibrium

3. “E” = Calculate the equilibrium concentrations

Sample Problem 1 on Equilibrium:

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Ex: Enough ammonia is dissolved in 5.00 L of water at 25ºC to produce a solution that is 0.0124 M ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that [OH1-] is 4.64 x 10-4 M. Calculate Keq at 25ºC for the reaction:NH3 (aq) + H2O (l) ↔ NH4

1+ (aq) + OH1- (aq)

NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)

 

Initial

Change

Equilibrium

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][NH

]][OH[NH K

3

-114

c

0.0124 M

- x

0.0119 M

0 M 0 M

+ x + x

4.64 x 10-4 M 4.64 x 10-4 M

5-2-4

101.81x 0.0119

)10(4.64

NH3 (aq)H2O (l)

NH41+ (aq) OH1- (aq)

XXX

x = 4.64 x 10-4 M

Sample Problem 2 on Equilibrium

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Ex: A 5.000-L flask is filled with 5.000 x 10-3

mol of H2 and 1.000 x 10-2 mol of I2 at 448ºC. The value of Keq is 1.33. What are the concentrations of each substance at equilibrium?

H2 (g) + I2 (g) ↔ 2 HI (g)

H2 (g) + I2 (g) ↔ 2 HI (g)

 

Initial

Change

Equilibrium

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]][I[H

[HI] K

22

2

c

1.000x10-3 M

- x M

(1.000x10-3 – x) M

2.000x10-3 M 0 M

- x M + 2x M

(2.000x10-3 – x) M 2x M

33.1x)-10x)(2.000-10(1.000

(2x)3-3-

2

4x2 = 1.33[x2 + (-3.000x10-3)x + 2.000x10-6]

0 = -2.67x2 – 3.99x10-3x + 2.66x10-6

Using quadratic eq’n: x = 5.00x10-4 or –1.99x10-3; x = 5.00x10-4

Then [H2]=5.00x10-4 M; [I2]=1.50x10-3 M; [HI]=1.00x10-3 M

H2 (g) I2 (g) HI (g)

15.6: Le Châtelier’s Principle

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If a system at equilibrium is disturbed by a change in: Concentration of one of the components, Pressure, or Temperature

…the system will shift its equilibrium position to counteract the effect of the disturbance.

http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/lechv17.swf

Henri Le Châtelier(1850 – 1936)

4 Changes that do not affect Keq:

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1. Concentration Upon addition of a reactant or product,

equilibrium shifts to re-establish equilibrium by consuming part of the added substance.

Upon removal of reactant or product, equilibrium shifts to re-establish equilibrium by producing more of the removed substance.Ex: Co(H2O)6

2+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l)

•Add HCl, temporarily inc forward rate•Add H2O, temporarily inc reverse rate

2. Volume, with a gas present (T is constant)

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Upon a decrease in V (thereby increasing P),equilibrium shifts to reduce the number of moles of gas.

Upon an increase in V (thereby decreasing P),equilibrium shifts to produce more moles of gas.

Ex: N2 (g) + 3 H2 (g) ↔ 2 NH3 (g) If V of container is decreased, equilibrium shifts

right. XN2 and XH2

dec XNH3 inc3

HN

2NH

P )(PP

)(PK

22

33

THTN

2TNH

)P)(XP(X

)P(X

22

3 4T

3HN

2T

2NH

P)(XX

P)(X

22

3 2T

3HN

2NH

P)(XX

)(X

22

3

23HN

2NH

P )(

)(K

22

3

Since PT also inc, KP remains constant.

3. Pressure, but not Volume

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Usually addition of a noble gas, p. 560 Avogadro’s law: adding more non-reacting

particles “fills in” the empty space between particles.

In the mixture of red and blue gas particles, below, adding green particles does not stress the system, so there is no Le Châtelier shift.

4. Catalysts

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Lower the activation energy of both forward and reverse rxns, therefore increases both forward and reverse rxn rates.

Increase the rate at which equilibrium is achieved, but does not change the ratio of components of the equilibrium mixture (does not change the Keq)Energy

Rxn coordinate

Ea, uncatalyzed

Ea, catalyzed

1 Change that does affect Keq:

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Temperature: consider “heat” as a part of the reaction Upon an increase in T, endothermic reaction is

favored (equilibrium shifts to “consume the extra heat”)

Upon a decrease in T, equilibrium shifts to produce more heat.Effect on Keq

1. Exothermic equilibria: Reactants ↔ Products + heat

• Inc T increases reverse reaction rate which decreases Keq

2. Endothermic equilibria: Reactants + heat ↔ Products

• Inc T increases forward reaction rate increases Keq

Ex: Co(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l);

H=+?•Inc T temporarily inc forward rate•Dec T temporarily inc reverse rate

Vant Hoff’s Equation

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Vant Hoff’s equation shows mathematically how the equilibrium constant is affected by changes in temp.

ln K2 = -d H0 rxn (1 – 1)

K1 R T2 T1

Effect of Various Changes on Equilibrium

Disturbance Net Direction of Rxn

Effect of Value of K

Concentration Increase (reactant)

Towards formation of product

None

Decrease(reactant)

Towards formation of reactant

None

Increase (product)

Towards formation of reactant

None

Decrease (product)

Towards formation of product

None

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Effect of Pressure on Equilb.

Pressure Increase P (decrease V)

Towards formation of fewer moles of gas

None

Decrease P (Increase V)

Towards formation of more moles of gas

None

Increase P ( Add inert gas, no change in V)

None, concentrations unchanged

None

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Effect of Temperature on Equilb

Temperature Increase T

Towards absorption of heat

Increases if endothermicDecreases if exothermic

Decrease T Towards release of heat

Increases if exothermicDecreases if endothermic

Catalyst Added None, forward and reverse equilibrium attained sooner

None

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