Ch 15 - Chemical Equilibrium Compatibility Mode

8/2/2019 Ch 15 - Chemical Equilibrium Compatibility Mode

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Chemical Equilibrium

Brown, LeMay Ch 15

AP ChemistryMonta Vista High School

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15.1: Chemical Equilibrium

Occurs when opposing react ions areproceeding at t he same rat e Forward rate = reverse rate of reaction

Ex:

Vapor pressure:rate of vaporization = rate

of condensationSat urat ed solut ion:rate of dissociation =

rate of crystallization

Expressing concentrations: Gases:partial pressures, PX Solut es in li quids: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:

RTPkRate Af

RT

PkRate Br

nRTPV

R = 0.0821 LatmmolK

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http://www.kentchemistry.com/links/Kinetics/Equilibrium/equilibrium.swf

[B]k[A]k rf

r

f

k

k

[A]

[B]

RT

Pk

RT

Pk Br

Af or

eq

r

f

A

B Kconstantk

k

P

Por


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Kc= kf/kr, at equilibrium, if K> 1, thenmore products at equilib. And if k


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Equilibrium

The equilibrium position is the same at agiven temperature, no matter from whichdirection it is approached.

It is possible to force an equilibrium one wayor the other temporarily by altering thereaction conditions, but once this stress isremoved, the system will return to its

original equilibrium.

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PX

or[X

]

Time

[B] or PB/ RT Equilibrium is established

Figure 1: Reversible reactions

[A] or PA/ RT

[A]0or PA0/ RT


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Reversible Reactions and Rate

Reaction

Rate

TimeBackward rate

Forward rate

Equilibrium is established:

Forward rate = Backward rate

When equi l ibr ium 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 reactionin equilibrium:

a A + b B c C + d D

Equilibrium constant expression (Keq):

ba

dc

c[B][A]

[D][C]K

b

B

a

A

d

D

c

Cp

)(P)(P

)(P)(PK

Keq is strictly based on stoichiometry of the reaction (isindependent 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

B

a

A

d

D

c

Cp

)(P)(P

)(P)(PK

baba

dcdc

(RT)[B][A]

(RT)[D][C]

(RT)KKb)(a-d)(c

cp

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

(RT)Kn

c

RT

P

V

nM RTAPA ][

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Magnitude of Keq

Since Keq [products]/[reactants], the magnitude ofKeq 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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Relationship Between Q and K

Reaction Quotient (Q): The particular ratioof concentration terms that we write for aparticular reaction is called reactionquotient.

For a reaction, A B, Q= [B]/[A]

At equilibrium, Q= K

Reaction Direction: Comparing Q and K

QK, reaction proceeds to left, until Q=K

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Value of K

For thereference

rxn, A>B,

For thereverse rxn,B>A,

For thereaction,

2A> 2B

For the rxn,

A> C

C> B

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

K= 1/K(ref) K= K(ref)2 K (overall)=K1 X K2


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15.3: Types of Equilibria

Homogeneous:all components in samephase (usually gor aq)

N2 (g) + H2 (g) NH3 (g)

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H

1

N

2NH

P)(P)(P

)(PK

22

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b

B

a

A

d

D

c

CP

)(P)(P

)(P)(PK

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Fritz Haber(1868 1934)

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Het erogeneous:different phases

CaCO3 (s) CaO (s) + CO2 (g)

Definit ion: What we use:

][CaCO

)(P[CaO]K

3

CO

eq2

2

COp PK

Even though the concentrations of the solids or liquids donot appear in the equilibrium expression, the substancesmust be present to achieve equilibrium.

Concentrations of pure solids and pure liquids are notincluded in Keq expression because their concentrations donot 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 equilibriumconcentrations of all species in equilibriumexpression

2. C = Determine the concentration change forthe species where initial and equilibrium areknown

Use stoichiometry to calculateconcentration changes for all otherspecies involved in equilibrium

3. E = Calculate the equilibrium concentrations

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Ex: Enough ammonia is dissolved in 5.00 L of waterat 25C to produce a solution that is 0.0124 Mammonia. The solution is then allowed to come toequilibrium. Analysis of the equilibrium mixtureshows that [OH1-] is 4.64 x 10-4 M. Calculate Keq at

25C for the reaction:NH3 (aq) + H2O (l) NH4

1+ (aq) + OH1- (aq)


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

Initial

Change

Equilibrium

][NH

]][OH[NHK

3

-11

4c

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

101.81x0.0119

)10(4.64

NH3 (aq)H2O

(l)NH4

1+ (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 ofH2 and 1.000 x 10

-2 mol of I2 at 448C. The value ofKeq is 1.33. What are the concentrations of eachsubstance 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 eqn: 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 Chteliers Principle

If a system at equilibrium isdisturbed by a change in: Concentration of one of the

components,

Pressure, or

Temperature

the system will shift its equilibriumposition to counteract the effect ofthe disturbance.

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

Henri Le Chtelier(1850 1936)


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4 Changes that do not affect Keq:

1. Concentrat ion

Upon addition of a reactant or product,equilibrium shifts to re-establish equilibriumby consuming part of the added substance.

Upon removal of reactant or product,equilibrium shifts to re-establish equilibriumby producing more of the removed substance.

Ex: Co(H2O)

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2+ (aq) + 4 Cl1-CoCl4

2- (aq) + 6 H2O (l)

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

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2. Volume, wi th 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 NH

3(g)

If V of container is decreased, equilibrium shifts right. XN2 and XH2

dec

XNH3 inc

3

HN

2

NH

P)(PP

)(PK

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THTN

2

TNH

)P)(XP(X

)P(X

22

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T

3

HN

2

T

2

NH

P)(XX

P)(X

22

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T

3

HN

2

NH

P)(XX

)(X

22

3

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HN

2

NH

P)(

)(K

22

3

Since PT also inc, KP remains constant.


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3. Pressure, but not Volume

Usually addit ion of a noble gas, p. 560 Avogadros 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, sothere is no Le Chtelier shift.

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4. Catalysts

Lower the activation energy of both forward andreverse rxns, therefore increases both forward andreverse 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 whichdecreases Keq

2. Endothermic equilibria: Reactants + heat

Products Inc T increases forward reaction rate increases Keq

Ex: Co(H2O)62+ (aq) + 4 Cl1-CoCl4

2- (aq) + 6 H2O (l); H=+?Inc T temporarily inc forward rateDec T temporarily inc reverse rate

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Vant Hoffs Equation

Vant Hoffs equation shows mathematicallyhow the equilibrium constant is affected bychanges in temp.

ln K2 = -d H0rxn (1 1)

K1 R T2 T1


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Effect of Various Changes on Equilibrium

Disturbance Net Direction ofRxn

Effect of Valueof K

Concentration

Increase (reactant) Towards formation ofproduct

None

Decrease(reactant) Towards formation ofreactant

None

Increase (product) Towards formation of

reactant

None

Decrease (product) Towards formation ofproduct

None

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Effect of Pressure on Equilb.

Pressure

Increase P

(decrease V)

Towardsformation offewer moles ofgas

None

Decrease P

(Increase V)

Towards

formation ofmore moles ofgas

None

Increase P

( Add inertgas, no changein V)

None,concentrationsunchanged

None


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Effect of Temperature on Equilb

Temperature

Increase T

Towardsabsorption ofheat

Increases ifendothermic

Decreases ifexothermic

Decrease T Towards releaseof heat

Increases ifexothermic

Decreases ifendothermic

Catalyst Added None, forwardand reverseequilibriumattained sooner

None

Ch 15 - Chemical Equilibrium Compatibility Mode

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