chapter 12 chemistry foundation 2014

8/12/2019 Chapter 12 chemistry foundation 2014

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CHAPTER 12 15)Chemical 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 Constants12.6 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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Consider Colorless frozen N2O4.

At room temperature, it decomposes tobrown NO2.

N2O4(g) 2NO2(g) At some time, the color stop changing and

we have a mixture of N2O4and NO2.

the concentration of all reactants andproducts 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 whichthe concentrations of all species are

constant. (concentrations of reactants andproducts cease to change with time)

Chemical equilibrium occurs

opposing reactions are proceeding at equalrate.

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 = kf[A] kf= rate constant (forward reaction)

Reverse reaction: B A

Rate = kr[B] kr= rate constant (reverse reaction)


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For gaseous substances we can use the ideal

gas equation to convert between concentration

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 closed

container.

As A reacts to form B, the partial pressure of A

will decrease and the partial pressure of B willincrease.

Expect forward reaction rate to slow and reverse

reaction rate to increase.

Eventually, we get to equilibrium where forwardand reverse rates are equal.


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Haber Process

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

The reaction is carried

out under conditions ofhigh pressure and high

temperature. Equilibrium

can be established either

by starting with N2andH2or by starting only

with NH3.


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At Equilibrium:

Rearrange:

a constant

This mixture is called an equi l ib r ium m ixture. This is an example of dynamic equi l ibr ium.

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 of the reactants and

products at equilibrium.

This expression is based on the law of massaction.

For a general reaction:

aA + bBcC + dD Equilibrium expression:

badc

BA

DCK


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When the reactions and products are all in

gaseous form, equilibrium constant is expressed

in terms of partial pressures of the gases.

The value of Keqdoes not depend on the initial

concentrations of reactants and products.

bB

a

A

d

D

c

C

PP

PPK


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When the reactants and products are all

in aqueous form, Kcis used for equilibrium

constant.

cindicates the molar concentration (M)

When the reactants and products are all

in gaseous form, Kpis used for equilibrium

constant.

p indicates the partial pressures of the

gases.


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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 1, then reactants dominate at equilibrium

and equilibrium lies to the left.


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The range of equilibrium

constants, K A: reaches equilibrium, little product.

K= 1/49

B:reaches equilibrium, nearly all product

K= 49/1

C: reaches equilibrium with significant

concentrations of reactant & 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 (at100C) 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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Equilibria in which all reactants and products arepresent in the same phase are calledhomogeneous equ i l ibr ia.

Equilibria in which one or more reactants orproducts are present in a different phase arecalled heterogeneous equ i l ibr ia.

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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The concentration of a pure solid or pure liquid

equals its density divided by its molar mass.

Neither density nor molar mass is a variable

thus the concentrations of solids and pureliquids are constant.

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CaCO

PCaOK

CO


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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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If a pure solid or pure liquid is involved in the

heterogeneous equilibrium, its concentration 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

CaCO3present.

However, they do participate in the reaction and

must be present for an equilibrium to be

established.


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f


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12.4 Calculating Equilibrium Constants, 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 a species, 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 of the equilibrium constant.


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Example 1

A mixture of hydrogen and nitrogen in a

reaction vessel is allowed to attain

equilibrium at 472C. The equilibrium

mixture of gases was analysed and found tocontain 7.38 atm H2, 2.46 atm N2, and 0.166

atm NH3. From these data, determine Kpfor

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 then

allowed to come to equilibrium. Analysis ofthe equilibrium mixture shows that theconcentration of OH- is 4.64 10-4 M.Calculate Kcat 25C for the reaction:

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


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Answer 2


Initial

Change

Eqm

4.64 10-4M

0.0124 M 0 M 0 M


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Answer 2


Initial

Change

Eqm

4.64 10-4M

0.0124 M 0 M 0 M

+4.6410-4M +4.6410-4M-4.6410-4M

0.0119 M 4.64 10-4M


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Answer 2


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). An

equilibrium mixture in a 2.00 L vessel is found to

contain 0.0406 mol CH3OH, 0.170 mol CO, and0.302 mol H2 at 500 K. Calculate Kp at this

temperature.

(Answer = 6.22

10

-3

)


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12.5 Application of Equilibrium Constants, K

The equilibrium constant can be used:

(i) to predict the direction in which a reaction

mixture will proceed to achieve

equilibrium.

(ii) to calculate the concentrations of

reactants and products once equilibriumhas been reached.


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12.5.1 To Predict the Direction of Reaction: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 system is 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 N2and2.00 mol of NH3 in a 1.00L container is at

472C. Kpfor the reaction at this temperature

is 2.75 10-5.

Will N2and H2react to form more NH3?


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Answer

N2(g) + 3H2(g)

2NH3(g)

First, write the equilibrium-constant expression, Kp(orQ):

To calculate the partial pressure of each gases, use

PV=nRT (PH2= 122 atm, PN2= 61.2 atm, PNH3= 122atm)

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 Kpvalue, Kp=2.75 10-5.

Therefore the quotient will need to decrease for thesystem to achieve equilibrium.

Can be achieved if decrease the pressure of NH3 orincrease the pressures of H2and N2.

Thus the reaction proceeds toward equilibrium byproducing H2 and N2 from NH3, i.e. the reactionproceeds from right to left.


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12.5.1 To Predict the Direction of Reaction

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 the equilibrium

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 I2at 448C. Kp for

the reaction at 448C is 50.5.

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

Question:

What are the partial pressures of H2, I2and 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 partialpressures.

Initial 59.17 atm 118.4 atm 0 atm

ChangeEquilibrium

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

nswer


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From the equation, 1 mol of H2

+ 1 mol of I2

willproduce 2 moles of HI.

The partial pressures of H2 and I2will decrease and thepartial pressure of HI will increase.

Initial (atm) 59.17 118.4 0

Change (atm) - x -x +2xEquilibrium (atm) 59.17x 118.4x 2x

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

nswer


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

nswer

5.504.11817.59

2 22

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 for equilibriumpartial pressures.

Using x = 137.6 will lead to negative partial pressuresof H2and I2which are not chemically meaningful. So x= 55.3.

PH2= 59.17x = 3.87 atm, PI2= 118.4x = 63.1 atm,PHI= 2x = 110.6 atm.

Check:

nswer

1.501.6387.3

6.110 22

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 reduce the effectof the disturbance.

Changes in concentration or pressure cause shifts in

equilibrium but K remains constant.

Change in temperature increases or decreases the K value.


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12.6.1 Change in Reactant or Product

Concentrations

A system at equilibrium:

- if we add a substance (a reactant or aproduct) the reaction will shift

reestablish equilibrium by consuming

part of the added substance.

- removal of a substance reestablish

by forming more of the substance.


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

i) Addition of H2 shift to reduce H2conc.

More NH3will form N2will be reduced

ii) Addition of NH3 shift to reduce NH3conc.

added NH3will decompose

to form N2 and H2

iii) Removal of NH3 shift from left to right -

formation of more NH3.

Conce

ntration

Time

Initial

Equilibrium

H2

NH3

N2

H2added at this

time

Equilibrium

reestablished

Example:


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12.6.2 Change in Pressure

Pressure will change if volume of thecontainer changes.

Volume decreases pressure increases

equilibrium will shift towards the lessnumber of gas molecules

Volume increases pressure decreases

equilibrium shift towards the more numberof gas 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 producing PCl3(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 changing the value ofKeq.

Applying Le Chteliersprinciple:

Consider heat as a chemical reagent.

Endothermic reaction (heat as a reactant)

A + B + HeatC

Exothermic 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 that absorbsheat.

In an endothermic reaction, H>0:

- increasing temperature or adding heat, the

equilibrium shifts to the right (product) and Kincreases.

In an exothermic reaction,

H


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Endothermic

A + B + HeatC

Exothermic

A + 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) NO2isremoved

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) NO2isremoved

Concentration of NO2 decreases, reaction

shifts 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. Itshifts to the right, 1 mol N2O4 moleculesgenerates 2 mol NO2molecules

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 the timetaken to reach equilibrium.

The catalyst has no effect on the equilibriumposition.

A catalyst does not effect the composition ofthe 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 (CH3OH) 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 or REMAIN THE SAME when anequilibrium mixture of reactants and product is subjected tothe 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 is exothermic.Reaction goes to the left (endothermic) toreduce back the T, Kp decreases, soconcentration decreases.

Increase. Because CO is a reactant, increasingCO, shift equilibrium to right.

Increase. When the volume decreases, pressureincreases. The reaction shifts to the sides withfewer 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 chemistry foundation 2014

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