Chapter 12 May11

8/6/2019 Chapter 12 May11

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CHAPTER 12(15)Chemical EquilibriumChemical Equilibrium


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CONTENTS

12.1 The Concept of Equilibrium

12.2 The Equilibrium Constant

12.3 Heterogeneous Equilibria

12.4 Calculating Equilibrium Constants

12.5 Application of Equilibrium Constants

12.6 Le Chteliers Principle


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Learning Outcomes Student should be able to use concept of

equilibrium to calculate equilibrium constant and

equilibrium concentrations. Able to apply Le Chaterliers Principle to predict

the direction of a reaction hence apply it incommercial sense in increasing yield, reduce cost

etc. Differentiate homogenous and heterogeneous

equilibria, reaction quotient and Equilibriumconstant


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12.1 The Concept ofEquilibrium

Consider Colorless frozen N2O4.

At room temperature, it decomposes to brownNO2.

N2O4(g) 2NO2(g)

At some time, the color stop changing and wehave a mixture of N2O4 and NO2.

the concentration of all reactants and products

no longer change with time.N2O4 (g) 2 NO2 (g)


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N2O4 (g) 2 NO2 (g) Chemical equilibrium is the point at which

the concentrations of all species are constant.(concentrations of reactants and products cease

to change with time)

Chemical equilibrium occurs opposing reactions are proceeding at equal rate.

Rate forward = Rate reverse


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N2O4 (g) - colorless

NO2 (g) - reddish brown

Concentration (color) remains constant at

equilibrium


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Consider a simple reaction: A(g) B(g) We can write rate expressions for each reaction:

Forward reaction: A B Rate = k

f[A] k

f= rate constant (forward

reaction)

Reverse reaction: B A

Rate =k

r[B]k

r= rate constant (reversereaction)


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For gaseous substances we can use theideal gas equation to convert betweenconcentration and pressure:

PV=nRT so M(Molarity) = (n/V) =(P/RT) For substances A and B:

[A] = (PA

/RT) and [B] = (PB

/RT)

Ratefwd = kf(PA/RT) and Raterev = kr(PB/RT)


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Place some pure compound A into a closedcontainer.

As A reacts to form B, the partial pressure of Awill decrease and the partial pressure of B willincrease.

Expect forward reaction rate to slow and reversereaction rate to increase.

Eventually, we get to equilibrium where forward

and reverse rates are equal.


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Haber Process

N2(g) + 3H2(g) 2NH3(g)

The reaction is carried out

under conditions of highpressure and high

temperature. Equilibrium

can be established either by

starting with N2

and H2

or

by starting only with NH3.


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At Equilibrium:

Rearrange:

a constant

This mixture is called an equilibrium mixture.

This is an example ofdynamic equilibrium.

RT

Pk

RT

Pk Br

Af =

( )

( )===

r

f

A

B

A

B

k

k

P

P

RTP

RTP

/

/


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12.2 The Equilibrium Constant,

K We can write an expression for the

relationship between the concentration ofthe reactants and products at equilibrium.

This expression is based on the law ofmass action.

For a general reaction:

aA + bB cC + dD Equilibrium expression:[ ] [ ]

[ ] [ ]ba

dc

BA

DCK=


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K When the reactions and products are all in

gaseous form, equilibrium constant isexpressed in terms of partial pressures of

the gases.

The value of Keq does not depend on the

initial concentrations of reactants andproducts.

( ) ( )

( ) ( )bBa

A

d

D

c

C

PP

PPK=


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K When the reactants and products are all

in aqueous form, Kc is used for

equilibrium constant.

c indicates the molar concentration(M)

When the reactants and products are all

in gaseous form, Kp is used forequilibrium constant.

p indicates the partial pressures of thegases.


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12.2.1 The Magnitude of

Equilibrium Constants, K K is the ratio of products to reactants.

The larger K the more products are present

at equilibrium. The smaller the K the more reactants are

present at equilibrium.

If K >> 1, then products dominate at

equilibrium and equilibrium lies to the right. K


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The range of equilibrium

constants, K A: reaches equilibrium, little

product.

K= 1/49

B:reaches equilibrium, nearly allproduct

K= 49/1

C: reaches equilibrium with

significant concentrations ofreactant & product.

K= 25/25 =1


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12.2.2 The Direction of the Chemical

Equation and K An equilibrium can be approached from

any direction:

Eg: N2O4 (g) 2 NO2 (g) The equilibrium constant for this reaction

(at 100C) is:

49.6)(

)(

42

2

2

==

ON

NOp

P

PK


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12.2.2 The Direction of the Chemical

Equation and Keq

For the reverse reaction:

2 NO2 (g) N2O4 (g) The equilibrium constant for this reaction (at

100C) is:

The equilibrium constant for a reaction in onedirection is the reciprocal of the equilibriumconstant of the reaction in the opposite direction.

154.0)(

)(2

2

42==

NO

ON

pP

PK


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12.3 Heterogeneous

Equilibria Equilibria in which all reactants and

products are present in the same phase arecalled homogeneous equilibria.

Equilibria in which one or more reactants orproducts are present in a different phase arecalled heterogeneous equilibria.

Consider:

CaCO3(s)

CaO(s) + CO2(g) Experimentally, the amount of CO2 does not

depend on the amounts of CaO and CaCO3.WHY?


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12.3 Heterogeneous

Equilibria The concentration of a pure solid or pure

liquid equals its density divided by itsmolar mass. Neither density nor molar

mass is a variable thus the concentrationsof solids and pure liquids are constant.

[ ]

[ ]3

2

CaCO

PCaOK

CO

=


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12.3 Heterogeneous

Equilibria CaO and CaCO3 are pure solids and have

constant concentrations.

Rearrange:

( )( )2constant

P1constantK 2CO

=

( )( )

( )2

21constant

2constantKK'

COp

CO

PKTherefore

P

=

==


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12.3 Heterogeneous

Equilibria If a pure solid or pure liquid is involved in

the heterogeneous equilibrium, itsconcentration is not included in the

equilibrium constant expression. The amount of CO2 formed (pressure of

CO2) will not depends on the amounts of

CaO and CaCO3 present.

However, they do participate in thereaction and must be present for anequilibrium to be established.


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12.3 Heterogeneous

Equilibria f


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12.4 Calculating EquilibriumConstants, K

Steps: Tabulate initial and equilibrium concentrations (or partial

pressures) for all species in the equilibrium.

If an initial and an equilibrium concentration is given for aspecies, calculate the change in concentration.

Use the coefficients in the balanced chemical equation to

calculate the changes in concentration of all species. Deduce the equilibrium concentrations of all species.

Use these equilibrium concentrations to calculate the value ofthe equilibrium constant.


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Example 1

A mixture of hydrogen and nitrogen in areaction vessel is allowed to attainequilibrium at 472C. The equilibrium

mixture of gases was analysed and foundto contain 7.38 atm H2, 2.46 atm N2, and

0.166 atm NH3. From these data, determine

Kp for

N2(g) + 3H2(g) 2NH3(g)


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Answer 1

N2(g) + 3H2(g) 2NH3(g)( )

( )( )( )

( ) ( )5

3

2

2

1079.2

38.746.2

166.0

322

3

=

=

=

HN

NH

p PP

P

K


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Example 2

Enough ammonia is dissolved in 5.00 L ofwater at 25C to produce a solution that is0.0124 M in ammonia. The solution is thenallowed to come to equilibrium. Analysis of

the equilibrium mixture shows that theconcentration of OH- is 4.64 10-4 M.Calculate Kc at 25C for the reaction:

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)


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Answer 2

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)Initial

Change

Eqm

4.64 10-4M

0.0124 M 0 M 0 M


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Answer 2

NH3(aq) + H2O(l)

NH4+

(aq) + OH-

(aq)

Initial

Change

Eqm

4.64 10-4M

0.0124 M 0 M 0 M

+4.64 10-4M +4.64 10-4M-4.64 10-4M

0.0119 M 4.64 10-4M


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Answer 2

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

[ ][ ][ ]

( )( )

5

24

3

4

1081.10119.0

1064.4

+

=

=

=

c

c

K

NH

OHNH

K


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Exercise 1

Methanol is produced commercially by thecatalysed reaction of carbon monoxide and

hydrogen: CO(g) + 2H2(g) CH3OH(g). Anequilibrium mixture in a 2.00 L vessel is found to

contain 0.0406 mol CH3OH, 0.170 mol CO, and

0.302 mol H2 at 500 K. Calculate Kp at this

temperature.

(Answer = 6.22 10-3)


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12.5 Application of EquilibriumConstants, K

The equilibrium constant can be used:

(i) to predict the direction in which areaction mixture will proceed toachieve equilibrium.

(ii) to calculate the concentrations of

reactants and products onceequilibrium has been reached.


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12.5.1 To Predict the Direction ofReaction: Writing Reaction

Quotient Consider: aA + bB pP + qQ

Q is the called reaction quotient.

a,b,p and q are the coefficients in the

balanced chemical equation. The reaction quotient will equal the

equilibrium constant, K, only if the systemis at equilibrium: Q = K at equilibrium.

[ ] [ ]

[ ] [ ]ba

qp

BA

QPQ =


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12.5.1 To Predict the Direction ofReaction: Example

N2(g) + 3H2(g) 2NH3(g) A mixture of 2.00 mol H2; 1.00 mol of N2

and 2.00 mol of NH3 in a 1.00L container

is at 472C. Kp for the reaction at this

temperature is 2.75

10

-5

.

Will N2 and H2 react to form more NH3?


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Answer

N2(g) + 3H2(g) 2NH3(g) First, write the equilibrium-constant expression, Kp (or Q):

To calculate the partial pressure of each gases, use PV=nRT(PH2 = 122 atm, PN2 = 61.2 atm, PNH3 = 122 atm)

( )( )( )32

22

3

HN

NH

PPPQ =


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Answer

Inserting the values into the reaction quotient, Q= 1.34 10-4.

Compare with Kp value, Kp =2.75 10-5.

Therefore the quotient will need to decrease for

the system to achieve equilibrium. Can be achieved if decrease the pressure of NH3 or

increase the pressures of H2 and N2.

Thus the reaction proceeds toward equilibrium by

producing H2 and N2 from NH3, i.e. the reactionproceeds from right to left.


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12.5.1 To Predict the Direction ofReaction

If Q < K,.. moreproduct forms

If Q > K,. more

reactant forms

If Q = K, no net

change

]tan[Re

][Pr

]tan[Re

][Pr

tsac

oductsQ

tsac

oductsKeq

=

=


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12.5.2 Calculating EquilibriumConcentrations

The same step used to calculate theequilibrium constants.

Generally, we do not have a number for the

change in the concentration. Therefore we need to assume that x mol/L of

the species is formed (or used).

The equilibrium concentrations are given asalgebraic expressions.


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Example

A 1.000L flask is filled with 1.00 mol of H2and 2.00 mol of I2 at 448C. Kp for the

reaction at 448C is 50.5.

H2(g) + I2(g) 2HI(g)Question:

What are the partial pressures of H2, I2

and HI in the flask at equilibrium?


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First, calculate the initial partial pressures of H2

and I2 using PV=nRT. (PH2 = 59.17 atm, PI2 =118.4 atm).

Second, construct a table that consists of initial

partial pressures.

Initial 59.17 atm118.4 atm 0 atm

Change

Equilibrium

H2(g) + I2(g) 2HI(g)

Answer


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From the equation, 1 mol of H2

+ 1 mol of I2

willproduce 2 moles of HI.

The partial pressures of H2and I2 will decreaseand the partial pressure of HI will increase.

Initial (atm) 59.17 118.4 0

Change (atm) - x -x +2xEquilibrium(atm)

59.17 x 118.4 x 2x

H2(g) + I2(g) 2HI(g)

Answer


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Substitute the values into equilibrium-constantexpression, Kp (value is 50.5).

Answer

( )

( )( )

( )

( ) ( )5.50

4.11817.59

222

22

=

==

xx

x

PP

PK

IH

HIp

( )

3.556.137

2

4

01054.31097.85.46

1001.76.1775.504

2

532

322

orx

a

acbbxusethen

xx

xxx

=

=

=+

+=


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Substitute x values into the expressions forequilibrium partial pressures.

Using x = 137.6 will lead to negative partialpressures of H2 and I2 which are not chemically

meaningful. So x = 55.3. PH2 = 59.17 x = 3.87 atm, PI2 = 118.4 x = 63.1

atm, PHI = 2x = 110.6 atm.

Check:

Answer

( )

( )( )( )

( ) ( )1.50

1.6387.3

6.11022

22

===

IH

HIp

PP

PK


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12.6 Le Chteliers Principle

If a system at equilibrium is disturbed by a change in

temperature,

Pressure (by changing the volume),

the concentration of one of the components (adding orremoving)

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

Changes in concentration or pressure cause shifts inequilibrium but K remains constant.

Change in temperature increases or decreases the Kvalue.


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12.6.1 Change in Reactant or

Product ConcentrationsA system at equilibrium:

- if we add a substance (a reactant or a product)

the reaction will shift reestablish equilibriumby consuming part of the added substance.

- removal of a substance reestablish byforming more of the substance.


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N2(g) + 3H2(g)

2NH3(g)

i) Addition of H2 shift to reduce H2 conc.

More NH3 will form N2 will be reduced

ii) Addition of NH3 shift to reduce NH3 conc.

added NH3 will decomposeto form N2 and H2

iii) Removal of NH3 shift from left to right -

formation of more NH3.

Concen

tra

tio

n

Time

InitialEquilibrium

H2

NH3

N2

H2 added at

this time

Equilibriumreestablished

ExampleExample

::


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12.6.2 Change in Pressure

Pressure will change if volume of the containerchanges.

Volume decreases pressure increases

equilibrium will shift towards the less number ofgas molecules

Volume increases pressure decreases equilibrium shift towards the more number ofgas molecules


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12.6.2 Change in Pressure

E.g: PCl5(g) PCl3(g) + Cl2(g) No. of gas molecules is more on RHS

P , equilibrium shift to LHS producing morePCl5 (g)

P , equilibrium shift to RHS producingPCl3(g) + Cl2(g)

No change will occur if we increase the totalpressure by the addition of a gas that is notinvolved in the reaction.


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Changes in concentration or total pressure

cause shifts in equilibrium without changingthe value of Keq .

Applying Le Chteliers principle:

Consider heat as a chemical reagent.

Endothermic reaction (heat as a reactant)

A + B + Heat CExothermic reaction (heat as a product)

A + B C + Heat

12.6.3 Change in Temperature


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When the temperature is increased, the

equilibrium shifts in the direction thatabsorbs heat.

In an endothermic reaction, H>0:

- increasing temperature or adding heat,the equilibrium shifts to the right (product)and K increases.

In an exothermic reaction,

H


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Endothermic

A + B + HeatC

ExothermicA + B C +

Heat

IncreaseIncrease

Decrease

Reaction KTemperature

Decrease

Decrease

Increase

Increase

Decrease

12.6.3 Change in Temperature


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Example

Consider the following equilibrium.

N2O4(g) 2NO2(g) H = +58.0 kJ.Which direction will the equilibrium shift if:

a) N2O4 is added

b) NO2 isremoved

c) N2 is added

d) the volume is increased.

e) the temperature is decreased


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a) N2O4 is added

Concentration of N2O4 increases, reaction

shifts to the right to decrease back N2O4.

b) NO2 isremoved

Concentration of NO2decreases, reactionshifts to the right to increase back NO2.

c) N2 is added

No effect, no shift in equilibrium position

N2O4(g) 2NO2(g) H = +58.0 kJ( )

( )422

ON

2

p P

P

K

NO=

Answer


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d) The volume is increased.

V P , the equilibrium shifts in the direction toincrease P by producing more gas molecules. It

shifts to the right, 1 mol N2O4 moleculesgenerates 2 mol NO2 molecules

e) the temperature is decreased.

The reaction is endothermic, heat is a reactant.Equilibrium shifts to the left, forming more N2O4,decreasing Kp.

N2

O4

(g) 2NO2

(g) H = +58.0

kJ

Answer


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12.6.4 The Effect of Catalyst

A catalyst lowers the activation energybarrier for the reaction.

Therefore, a catalyst will decrease thetime taken to reach equilibrium.

The catalyst has no effect on theequilibrium position.

A catalyst does not effect thecomposition of the equilibrium mixture.


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Haber synthesis of ammoniaAmmonia is used for production of fertilizer,

explosives and polymers.


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N2 (g) + 3H2 (g) 2NH3 (g) H = -91.8 kJ/mol


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Haber Process


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ExampleMethanol (CH

3OH) is manufactured by reaction of carbon

monoxide with hydrogen in the presence of a ZnO/Cr2O3catalyst

CO (g) + 2H2

(g) CH3OH (g) Ho = - 91 kJ

Does the amount of methanol (CH3OH) INCREASE,

DECREASE orREMAIN THE SAME when an equilibriummixture of reactants and product is subjected to the following

changes? The temperature is increased.

CO is added.

The volume of the vessel is decreased.

The catalyst is removed


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Answer

Decrease. Because the reaction isexothermic. Reaction goes to the left(endothermic) to reduce back the T, Kpdecreases, so concentration decreases.

Increase. Because CO is a reactant,

increasing CO, shift equilibrium to right. Increase. When the volume increases,

pressure decreases. The reaction shifts to thesides with fewer molecules i.e. shifts to the

right. Remains the same. Additional or removal of acatalyst does not affect the equilibriumcomposition.


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END of CHAPTER 12

Chapter 12 May11

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