Equi Petrcuci

8/10/2019 Equi Petrcuci

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Chemistry 140 Fall 2002

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Yrd. Do. Dr. Burak ESAT

Fatih University

General ChemistryPrinciples and Modern Applications

Petrucci Harwood Herring9th Edition

Chapter 15: Principles of Chemical Equilibrium

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Contents

15-1 Dynamic Equilibrium15-2 The Equilibrium Constant Expression15-3 Relationships Involving Equilibrium Constants

15-4 The Significance of the Magnitude of anEquilibrium Constant15-5 The Reaction Quotient, Q: Predicting the

Direction of a Net Change


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Contents

15-6 Altering Equilibrium Conditions:Le Chtelliers Principle

15-7 Equilibrium Calculations:Some Illustrative ExamplesFocus On The Nitrogen Cycle and theSynthesis of Nitrogen Compounds

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15-1 Dynamic Equilibrium

Equilibrium twoopposing processestaking place at equalrates.

H2O(l) H 2O(g)

NaCl(s) NaCl(aq)H2O

I2(H2O) I 2(CCl 4)

CO(g) + 2 H 2(g) CH 3OH(g)


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15-2 The Equilibrium Constant Expression

Forward: CO(g) + 2 H 2(g) CH 3OH(g)

Reverse: CH 3OH(g) CO(g) + 2 H 2(g)

At Equilibrium:

Rfwrd = k 1[CO][H 2]2

Rrvrs = k -1[CH 3OH]

Rfwrd = Rrvrs

k 1[CO][H 2]2

= k -1[CH 3OH]

[CH 3OH]

[CO][H 2]2=

k 1k -1

= K c

CO(g) + 2 H 2(g) CH 3OH(g)k 1

k -1

k 1

k -1

You may refer to Chapter 14for discussion of rates of reaction

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Three Approaches to Equilibrium


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Three Approaches to the Equilibrium

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Three Approaches to Equilibrium

[CH 3OH]

[CO][H 2]2

K c(1) = 14.2 M -2

K c(2) = 14.2 M -2

K c(3) = 14.2 M -2

[CH 3OH]

[CO][H 2]K c =

[CH 3OH]

[CO]( 2[H2])

0.596 M -1

1.09 M -1

1.28 M -1

1.19 M -1

2.17 M -1

2.55 M -1

CO(g) + 2 H 2(g) CH 3OH(g)k 1

k -1


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General Expressions

a A + b B . g G + h H .

Equilibrium constant = K c= [A] a[B] b .[G] g[H] h .

Thermodynamic

Equilibrium constant = K eq=(aG)g(aH)h .

(aA)a(aB)b .

aB =[B]cB0

cB0 is a standard reference state

= 1 mol L -1 (ideal conditions)

= B[B]

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15-3 Relationships Involving theEquilibrium Constant

Reversing an equation causes inversion of K . Multiplying by coefficients by a common factor

raises the equilibrium constant to thecorresponding power.

Dividing the coefficients by a common factorcauses the equilibrium constant to be taken tothat root.


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Combining Equilibrium Constant

Expressions

N2O(g) + O 2 2 NO(g) K c= ?

N2(g) + O 2 N2O(g) K c(2)= 2.7x10 +18

N2(g) + O 2 2 NO(g) K c(3)= 4.7x10 -31

K c= [N 2O][O 2][NO] 2

=[N2][O 2]

[N 2O][N2][O 2]

[NO] 2

K c(2)

1K c(3)= = 1.7x10 -13

[N 2][O 2]

[NO] 2=

[N2][O 2][N2O]=

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Gases: The Equilibrium Constant, K P

Mixtures of gases are solutions just as liquids are.

Use K P, based upon partial pressures of gases.

2 SO 2(g) + O 2(g) 2 SO 3(g) K c = [SO 2]2[O 2][SO 3]

2

[SO 3]=V

nSO3 = RT

P SO3 [SO 2]=V

nSO2 = RT

P SO2

[O2] = V

nO2 = RT

P O2


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The Equilibrium Constant, K P

2 SO 2(g) + O 2(g) 2 SO 3(g)

K c =[SO 2]2[O 2]

[SO 3] RT

P SO3

2

RT

P SO2 RT

P O2

=

2

= RT P SO3

2P SO2 P O2

2

K c = K P( RT ) K P = K c( RT )-1

K P = K c( RT ) n

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Pure Liquids and Solids

Equilibrium constant expressions do not containconcentration terms for solid or liquid phases of a singlecomponent (that is, pure solids or liquids).

C(s) + H 2O(g) CO(g) + H 2(g)

K c =[H2O]

[CO][H 2] =P H2O

P COP H2 ( RT )-1


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15-5 The Reaction Quotient, Q: Predicting

the Direction of Net Change.

Equilibrium can be approached various ways. Qualitative determination of change of initial

conditions as equilibrium is approached isneeded.

CO(g) + 2 H 2(g) CH 3OH(g)k 1

k -1

At equilibrium Qc = K cQc = [A] tm[B] tn[G] tg[H] th

mA + nB gG + hH

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Reaction Quotient


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15-6 Altering Equilibrium Conditions:Le Chtelliers Principle

When an equilibrium system is subjected to achange in temperature, pressure, or concentrationof a reacting species, the system responds byattaining a new equilibrium that partially offsetsthe impact of the change.

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Le Chtelliers Principle

Q = = K c[SO 2]2[O 2]

[SO 3]2 Q > K c

2 SO 2(g) + O 2(g) 2 SO 3(g)k 1

k -1

K c = 2.8x10 2 at 1000K


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Effect of Temperature on Equilibrium

Raising the temperature of an equilibriummixture shifts the equilibrium condition in thedirection of the endothermic reaction .

Lowering the temperature causes a shift in thedirection of the exothermic reaction.

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Effect of a Catalyst on Equilibrium

A catalyist changes the mechanism of areaction to one with a lower activation energy.

A catalyst has no effect on the condition ofequilibrium. But does affect the rate at which equilibrium is

attained.


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15-7 Equilibrium Calculations:Some Illustrative Examples.

Five numerical examples are given in the textthat illustrate ideas that have been presented inthis chapter.

Refer to the comments which describe themethodology. These will help in subsequentchapters.

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Predicting the Effect of a Change in Concentrationon the Position of the Equilibrium

Problem: Carbon will react with water to yield carbon monoxide andand hydrogen, in a reaction called the water gas reaction that was usedto convert coal into a fuel that can be used by industry.

C (s) + H 2O (g) CO (g) + H 2 (g)What happens to:(a) [CO] if C is added? (c) [H

2O] if H

2is added?

(b) [CO] if H 2O is added? (d) [H 2O] if CO is removed?Plan: We either write the reaction quotient to see how equilibrium willbe effected, or look at the equation, and predict the change in directionof the reaction, and the effect of the material desired.Solution: (a) No change, as carbon is a solid, and not involved in the

equilibrium, as long as some carbon is present to allow the reaction.(b) The reaction moves to the product side, and [CO] increases.(c) The reaction moves to the reactant side, and [H 2O] increases.(d) The reaction moves to the product side, and [H 2O] decreases.


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Synthesis of Ammonia

The optimum conditions areonly for the equilibriumposition and do not take intoaccount the rate at whichequilibrium is attained.

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Chapter 15 Questions

Develop problem solving skills and base your strategy noton solutions to specific problems but on understanding .

Choose a variety of problems from the text as examples.

Practice good techniques and get coaching from people whohave been here before.

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