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Chemical Equilibrium

Brown, LeMay Ch 15AP Chemistry

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15.1: Chemical Equilibrium

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

Ex: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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If equilibrium: A ↔ Bforward rate = reverse rate

  [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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PX or [X]

Time →

[B] or PB / RT

Equilibrium is established

Figure 1: Reversible reactions

[A] or PA / RT

[A]0 or PA0 / RT

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Reversible Reactions and Rate

Reaction Rate

Time

Backward rate

Forward rate

Equilibrium is established:

Forward rate = Backward rate

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

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15.2: Law of Mass Action

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

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Relating Kc and Kp

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

(RT) K xc

RT

P

V

nM RTAPA ][

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Magnitude of Keq

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”

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15.3: Types of Equilibria

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

N2 (g) + H2 (g) ↔ NH3 (g)

3H

1N

2NH

P )(P)(P

)(PK

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3

bB

aA

dD

cC

P )(P)(P

)(P)(PK

3 21

Fritz Haber(1868 – 1934)

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Heterogeneous: different phasesCaCO3 (s) ↔ CaO (s) + CO2 (g)

Definition: What we use:

][CaCO

)(P [CaO] K

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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).

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15.4: Calculating Equilibrium Constants

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

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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) ↔ NH41+ (aq) + OH1-

(aq)

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

(aq)

 

Initial

Change

Equilibrium

][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

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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)

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H2 (g) + I2 (g) ↔ 2 HI (g)

 

Initial

Change

Equilibrium

]][I[H

[HI] K

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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)

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15.6: Le Châtelier’s Principle

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.

Henri Le Châtelier(1850 – 1936)

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4 Changes that do not affect Keq:

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

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

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 inc3HN

2NH

P )(PP

)(PK

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THTN

2TNH

)P)(XP(X

)P(X

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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.

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3. Pressure, but not Volume

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.

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4. Catalysts

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

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1 Change that does affect Keq: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