Chemical Equilibrium

2

17 Chemical Equilibrium

3

Chapter Goals

1. Basic Concepts

2. The Equilibrium Constant

3. Variation of Kc with the Form of the Balanced Equation

4. The Reaction Quotient

5. Uses of the Equilibrium Constant, Kc

6. Disturbing a System at Equilibrium: Predictions

4

Chapter Goals

7. The Haber Process: A Commercial Application of Equilibrium

8. Disturbing a System at Equilibrium: Calculations

9. Partial Pressures and the Equilibrium Constant

10.Relationship between Kp and Kc

11.Heterogeneous Equilibria

5

Chapter Goals

12.Relationship between Gorxn and the

Equilibrium Constant

13.Evaluation of Equilibrium Constants at Different Temperatures

6

Basic Concepts

• Reversible reactions do not go to completion.– They can occur in either direction– Symbolically, this is represented as:

gggg D d + C cB b + A a

7

Basic Concepts

• Chemical equilibrium exists when two opposing reactions occur simultaneously at the same rate.– A chemical equilibrium is a reversible reaction

that the forward reaction rate is equal to the reverse reaction rate.

• Chemical equilibria are dynamic equilibria.– Molecules are continually reacting, even

though the overall composition of the reaction mixture does not change.

8

Basic Concepts

• One example of a dynamic equilibrium can be shown using radioactive 131I as a tracer in a saturated PbI2 solution.

solution. into go williodine eradioactiv theof Some

solution. filter the then minutes, few afor Stir 2

I 2 Pb PbI

solution. PbI saturated ain PbI solid Place 1-(aq)

2(aq)

OH

2(s)

2*22

9

Basic Concepts

• This movie depicts a dynamic equilibrium.

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Basic Concepts• Graphically, this is a representation of the

rates for the forward and reverse reactions for this general reaction.

gggg D d + C cB b + A a

11

Basic Concepts

• One of the fundamental ideas of chemical equilibrium is that equilibrium can be established from either the forward or reverse direction.

12

Basic Concepts

13

Basic Concepts

14

The Equilibrium Constant

• For a simple one-step mechanism reversible reaction such as:

• The rates of the forward and reverse reactions can be represented as:

rate. reverse therepresents which DCkRate

rate. forward therepresents which BAkRate

rr

ff

(g)(g)(g)(g) D C B A

15


• When system is at equilibrium:

Ratef = Rater

BA

DC

k

k

torearrangeswhich

DCkBAk

:give toiprelationsh rate for the Substitute

r

f

rf

16


• Because the ratio of two constants is a constant we can define a new constant as follows :

kk

K and

KC DA B

f

rc

c

17


• Similarly, for the general reaction:

we can define a constant

reactions. allfor validis expression This

BA

DCK

products

reactants ba

dc

c

D d C c B b A a (g)(g)(g)(g)

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The Equilibrium Constant• Kc is the equilibrium constant .• Kc is defined for a reversible reaction at a given

temperature as the product of the equilibrium concentrations (in M) of the products, each raised to a power equal to its stoichiometric coefficient in the balanced equation, divided by the product of the equilibrium concentrations (in M) of the reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation.

19


Example 17-1: Write equilibrium constant expressions for the following reactions at 500oC. All reactants and products are gases at 500oC.

5 3 2PCl PCl Cl

20


You do it!

HI 2 I + H 22

21


You do it!

OH 6+NO 4O 5 + NH 4 223

22


• Equilibrium constants are dimensionless because they actually involve a thermodynamic quantity called activity.– Activities are directly related to molarity

23


Example 17-2: One liter of equilibrium mixture from the following system at a high temperature was found to contain 0.172 mole of phosphorus trichloride, 0.086 mole of chlorine, and 0.028 mole of phosphorus pentachloride. Calculate Kc for the reaction.

Equil []’s 0.028 M 0.172 M 0.086 M

You do it!

235 ClPClPCl

24


Example 17-3: The decomposition of PCl5 was studied at another temperature. One mole of PCl5 was introduced into an evacuated 1.00 liter container. The system was allowed to reach equilibrium at the new temperature. At equilibrium 0.60 mole of PCl3 was present in the container. Calculate the equilibrium constant at this temperature.

25


Example 17-4: At a given temperature 0.80 mole of N2 and 0.90 mole of H2 were placed in an evacuated 1.00-liter container. At equilibrium 0.20 mole of NH3 was present. Calculate Kc for the reaction.

You do it!

26

Variation of Kc with the Form of the Balanced Equation• The value of Kc depends upon how the balanced

equation is written.• From example 17-2 we have this reaction:

• This reaction has a Kc=[PCl3][Cl2]/[PCl5]=0.53

235 ClPClPCl

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Variation of Kc with the Form of the Balanced EquationExample 17-5: Calculate the equilibrium constant for the reverse reaction by two methods, i.e, the equilibrium constant for this reaction.

Equil. []’s 0.172 M 0.086 M 0.028 M

The concentrations are from Example 17-2.

523 PClCl PCl

KPCl

PCl Cl

KK

or K K

c' 5

3 2

cc' c

'

c

0 0280172 0 086

19

1 1 10 53 19

.. .

.

. .

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The Reaction Quotient

29


30


• The mass action expression or reaction quotient has the symbol Q. – Q has the same form as Kc

• The major difference between Q and Kc is that the concentrations used in Q are not necessarily equilibrium values.

ba

dc

BA

DCQ

dD+cC bB+aA

:reaction general For this

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• Why do we need another “equilibrium constant” that does not use equilibrium concentrations?

• Q will help us predict how the equilibrium will respond to an applied stress.

• To make this prediction we compare Q with Kc.

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c

c

c

c

When:

Q=K The system is at equilibrium.

Q K The reaction occurs to the left to a greater extent.

Q K The reaction occurs to the right to a greater extent.

To help understand this think of Q and K

as fractions.

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Example 17-6: The equilibrium constant for the following reaction is 49 at 450oC. If 0.22 mole of I2, 0.22 mole of H2, and 0.66 mole of HI were put into an evacuated 1.00-liter container, would the system be at equilibrium? If not, what must occur to establish equilibrium?

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Uses of the Equilibrium Constant, Kc

Example 17-7: The equilibrium constant, Kc, is 3.00 for the following reaction at a given temperature. If 1.00 mole of SO2 and 1.00 mole of NO2 are put into an evacuated 2.00 L container and allowed to reach equilibrium, what will be the concentration of each compound at equilibrium?

(g)3(g)2(g)2(g) NO SO NO SO

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Uses of the Equilibrium Constant, Kc

Example 17-8: The equilibrium constant is 49 for the following reaction at 450oC. If 1.00 mole of HI is put into an evacuated 1.00-liter container and allowed to reach equilibrium, what will be the equilibrium concentration of each substance?

You do it!

(g)2(g)2(g) HI 2 I + H

36

Disturbing a System at Equilibrium : Predictions• LeChatelier’s Principle - If a change of conditions

(stress) is applied to a system in equilibrium, the system responds in the way that best tends to reduce the stress in reaching a new state of equilibrium.

– We first encountered LeChatelier’s Principle in Chapter 14.

• Some possible stresses to a system at equilibrium are:

1. Changes in concentration of reactants or products.

2. Changes in pressure or volume (for gaseous reactions)

3. Changes in temperature.

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Disturbing a System at Equilibrium : Predictions1 Changes in Concentration of Reactants and/or Products

• Also true for changes in pressure for reactions involving gases.– Look at the following system at equilibrium at 450oC.

2 (g) 2 (g) g

2

c2 2

H I 2 HI

HIK 49

H I

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2 (g) 2 (g) g

2

c2 2

2 c

H I 2 HI

HIK 49

H I

If some H is added, Q<K .

This favors the forward reaction.

Equilbrium will shift to the right or product side.

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2 (g) 2 (g) g

2

c2 2

2 c

H I 2 HI

HIK 49

H I

If we remove some H ,Q>K

This favors the reverse reaction.

Equilbrium will shift to the left, or reactant side.



40

Disturbing a System at Equilibrium : Predictions2 Changes in Volume

• (and pressure for reactions involving gases)– Predict what will happen if the volume of this system at equilibrium is

changed by changing the pressure at constant temperature:

22

42c

g42g2

NO

ON=K

ONNO 2

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Disturbing a System at Equilibrium : Predictions

gas. of molesfewer producesreaction forward The

reaction. forward or theformation product favors This

.K<Qpressure, theincreases which decreased, is volume theIf

NO

ON=K

ONNO 2

c

22

42c

g42g2

42

produced. are gas of moles More

reaction. reverse or the reactants thefavors This

.K>Qpressure, thedecreases which increased, is volume theIf

NO

ON=K

ONNO 2

c

22

42c

g42g2


43


44

Disturbing a System at Equilibrium : Predictions3 Changing the Temperature

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reaction.reactant or reverse thefavors This

products. thestresses emperaturereaction t theIncreasing

reaction thisofproduct a isHeat

kJ 198+SO 2O+SO 2 g3g2g2

reaction. forwardor reactants thefavors This

reactants. thestresses emperaturereaction t theDecreasing

reaction. thisofproduct a isHeat

kJ 198+SO 2O+SO 2 g3g2g2

Disturbing a System at Equilibrium : Predictions3 Changing the Reaction Temperature

– Consider the following reaction at equilibrium:

reaction? in thisproduct or reactant aheat Is

kJ/mol 198H SO 2 O SO 2 orxn3(g)2(g)2(g)

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• Introduction of a Catalyst– Catalysts decrease the activation energy of both the forward and

reverse reaction equally.

• Catalysts do not affect the position of equilibrium.– The concentrations of the products and reactants will be the

same whether a catalyst is introduced or not.– Equilibrium will be established faster with a catalyst.

47

Disturbing a System at Equilibrium : PredictionsExample 17-9: Given the reaction below at equilibrium in a closed container at 500oC. How would the equilibrium be influenced by the following?

o2(g) 2(g) 3(g) rxnN 3 H 2 NH H 92 kJ/mol

Factor Effect on reaction procedure

a. Increasing the reaction temperature

b. Decreasing the reaction tempe

2

3

rature

c. Increasing the pressure by decreasing the volume

d. Increase the concentration of H

e. Decrease the concentration of NH

f. Introducing a platinum catalyst

48

Disturbing a System at Equilibrium : PredictionsExample 17-10: How will an increase in pressure (caused by decreasing the volume) affect the equilibrium in each of the following reactions?

2 g 2 g g

2(g) 23 g g g

3 g 2 g 5 g

2 g 2

Reaction Effect on Equilibrium

a. H + I 2 HI

b. 4 NH +5 O 4 NO +6 H O

c. PCl +Cl PCl

d. 2 H O

2g g 2 H O

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Disturbing a System at Equilibrium : PredictionsExample 17-11: How will an increase in temperature affect each of the following reactions?

o2(g) 2 4(g) rxn

2 g 2 g g

2 g 2 g g

Reaction Effect on Equilibrium

a. 2 NO N O H 0

b. H Cl 2 HCl + 92 kJ

c. H + I 2 HI H 25 kJ

50

The Haber Process: A Practical Application of Equilibrium• The Haber process is used for the

commercial production of ammonia.– This is an enormous industrial process in the

US and many other countries.– Ammonia is the starting material for fertilizer

production.

• Look at Example 17-9. What conditions did we predict would be most favorable for the production of ammonia?

51

The Haber Process: A Practical Application of Equilibrium

dilemma. thisosolution t sHaber'

res. temperatulowat slow very are kineticsreaction eHowever th

e.unfavorabl is which 0<S favorable. also 0<H favorable. is which 0<G

atm. 1000 to200= N of P and C450 = T aat run isreaction This

gas. coal from obtained H air. liquid from obtained is N

kJ 22.92H NH 2 H 3N

2o

g2g2

og3

oxides metal & Feg2g2

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The Haber Process: A Practical Application of Equilibrium

kinetics! with thehelps and yieldreaction theincreases This

removed. is NH because mequilibriu reachesnever systemreaction The

right. ly toperiodical NH Remove 4

right. toN excess Use3

right. topressurereaction Increase 2

.decreased is yieldbut rate, increase toT Increase 1

dilemma. thisosolution t sHaber'

3

3

2

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The Haber Process: A Practical Application of Equilibrium• This diagram illustrates the commercial

system devised for the Haber process.

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Disturbing a System at Equilibrium: Calculations• To help with the calculations, we must determine

the direction that the equilibrium will shift by comparing Q with Kc.

• Example 17-12: An equilibrium mixture from the following reaction was found to contain 0.20 mol/L of A, 0.30 mol/L of B, and 0.30 mol/L of C. What is the value of Kc for this reaction?

ggg C B A

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Disturbing a System at Equilibrium: Calculations

45.0

20.0

30.030.0

A

CBK

0.30 0.30 0.20 s[]' Equil.

C B A

c

ggg

MMM

56

Disturbing a System at Equilibrium: Calculations• If the volume of the reaction vessel were

suddenly doubled while the temperature remained constant, what would be the new equilibrium concentrations?

1 Calculate Q, after the volume has been doubled

22.0

10.0

15.015.0

A

CB=Q

0.15 0.15 0.10 s[]' Equil.

C B A ggg

MMM

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Disturbing a System at Equilibrium: Calculations• Since Q<Kc the reaction will shift to the right to

re-establish the equilibrium.2 Use algebra to represent the new

concentrations.

MMM 0.15 0.15 0.10 s[]' initial New

C + B A ggg

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concentrations.

MxMxMx

MMM

+ + - Change

0.15 0.15 0.10 s[]' initial New

C + B A ggg

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concentrations.

MxMxMx

MxMxMx

MMM

+0.15 +0.15 -0.10 s[]' Equil. New

+ + - Change

0.15 0.15 0.10 s[]' initial New

C + B A ggg

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concentrations.

x

xx

MxMxMx

MxMxMx

MMM

10.0

15.015.045.0

A

CB=K

+0.15 +0.15 -0.10 s[]' Equil. New

+ + - Change

0.15 0.15 0.10 s[]' initial New

C + B A

c

ggg

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00225.075.0

+0.30+0.0225=0.45-0.045

equation quadratic thisSolve

2

2

xx

xxx

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Mx

x

x

0.03 and 78.02

81.075.0

12

0225.01475.075.0

2a

ac4bb-

2

2

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disturbed.been has mequilibriu the

after ionsconcentrat new theare These

18.0 15.0CB

07.0 )10.0(A

M. 0.03 is valueposibleonly The

answer.an as 0.78- discardcan we0.10,<<0 Since

MMx

MMx

x

x

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Disturbing a System at Equilibrium: CalculationsExample 17-13: Refer to example 17-12. If the initial volume of the reaction vessel were halved, while the temperature remains constant, what will the new equilibrium concentrations be? Recall that the original concentrations were: [A]=0.20 M, [B]=0.30 M, and [C]=0.30 M.

You do it!

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Disturbing a System at Equilibrium: CalculationsExample 17-14: A 2.00 liter vessel in which the following system is in equilibrium contains 1.20 moles of COCl2, 0.60 moles of CO and 0.20 mole of Cl2. Calculate the equilibrium constant.

You do it!

g2g2g COCl Cl CO

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Disturbing a System at Equilibrium: CalculationsAn additional 0.80 mole of Cl2 is added to the vessel at the same temperature. Calculate the molar concentrations of CO, Cl2, and COCl2 when the new equilibrium is established.

You do it!

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Partial Pressures and the Equilibrium Constant• For gas phase reactions the equilibrium

constants can be expressed in partial pressures rather than concentrations.

• For gases, the pressure is proportional to the concentration.

• We can see this by looking at the ideal gas law.– PV = nRT– P = nRT/V – n/V = M– P= MRT and M = P/RT

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Partial Pressures and the Equilibrium Constant• For convenience we may express the amount of

a gas in terms of its partial pressure rather than its concentration.

• To derive this relationship, we must solve the ideal gas equation.

ion.concentrat its toalproportiondirectly is

gas a of pressure partial theT,constant at Thus

[]RT=P

mol/L, units thehas V

nBecause

RTV

nP

nRTPV

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Partial Pressures and the Equilibrium Constant• Consider this system at equilibrium at

5000C.

2OH

2Cl

O4

HClp2

22

2

24

c

g2gg2g2

22

2

PP

PPK and

OHCl

OHClK

O+HCl 4OH 2+Cl 2

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Partial Pressures and the Equilibrium Constant

K mol

atm L0.0821 R useMust

(RT)K=Kor (RT)K=K

reaction for this so KK

PP

PPK

1cp

1-pc

RT1

pc

4RT1

5RT1

2OH

2Cl

O4

HCl

2

RT

P2

RT

P

RT

P4

RTP

c

22

2

O2H2Cl

2OHCl

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Relationship Between Kp and Kc

• From the previous slide we can see that the relationship between Kp and Kc is:

reactants) gaseous of moles of (#-products) gaseous of moles of (#=n

RTKKor RTKK npc

ncp

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Example 17-15: Nitrosyl bromide, NOBr, is 34% dissociated by the following reaction at 25oC, in a vessel in which the total pressure is 0.25 atmosphere. What is the value of Kp?

g2gg Br + NO 2 NOBr 2

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• The numerical value of Kc for this reaction can be determined from the relationship of Kp and Kc.

K = K RT or K = K RT n = 1

K

p cn

c pn

c

9 3 10 0 0821 298 38 103 1 4. . .

74


Example 17-16: Kc is 49 for the following reaction at 450oC. If 1.0 mole of H2 and 1.0 mole of I2 are allowed to reach equilibrium in a 3.0-liter vessel,

(a) How many moles of I2 remain unreacted at equilibrium?

You do it!

gg2g2 HI 2I H

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(b) What are the equilibrium partial pressures of H2, I2 and HI?

You do it!

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(c) What is the total pressure in the reaction vessel?

You do it!

77

Heterogeneous Equlibria• Heterogeneous equilibria have more than one

phase present.– For example, a gas and a solid or a liquid and a gas.

• How does the equilibrium constant differ for heterogeneous equilibria?– Pure solids and liquids have activities of unity.– Solvents in very dilute solutions have activities that are essentially unity.– The Kc and Kp for the reaction shown above are:

2COp2c P=K ][CO=K

C500at CO CaO CaCO og2ss3

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Heterogeneous Equlibria

2SO

p

2

32

c P

1 K

SO

SOH=K

You do it!

?K and K of forms theareWhat

solvent. theis OH

C)25at (SOHOHSO

:reaction For this

pc

2

o

aq322g2

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• What are Kc and Kp for this reaction?

K = Ca F K is undefinedc2 2

p

C)25at (F 2CaCaF o-1aq

2aqs2

You do it!You do it!

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• What are Kc and Kp for this reaction?

K =H

H O K

P

Pc

2

2p

H

H O

2

2

4

4

4

4

C)500at (H 4OFe OH 4Fe 3 og2s43g2s

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Relationship Between Gorxn

and the Equilibrium Constant Go

rxn is the standard free energy change. Go

rxn is defined for the complete conversion of all reactants to all products.

G is the free energy change at nonstandard conditions• For example, concentrations other than 1 M or pressures other

than 1 atm.

G is related to Go by the following relationship.

quotientreaction =Q

re temperatuabsolute = T

constant gas universal =R

Q log RT 303.2G=G

or lnQ RTG=Go

o

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and the Equilibrium Constant

• At equilibrium, G=0 and Q=Kc. • Then we can derive this relationship:

K log RT 2.303 -=G

orK ln RT -=G

: torearrangeswhich

K log RT 303.2G0

orK ln RTG0

0

0

0

0

83


and the Equilibrium Constant• For the following generalized reaction, the

thermodynamic equilibrium constant is defined as follows:

c d

C Da b

A B

A B

C D

aA + bB cC + dD

K= where

is the activity of A is the activity of B

is the activity of C is the activity of D

a a

a a

a a

a a

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• The relationships among Gorxn, K, and the

spontaneity of a reaction are:

Gorxn K Spontaneity at unit concentration

< 0 > 1 Forward reaction spontaneous

= 0 = 1 System at equilibrium

> 0 < 1 Reverse reaction spontaneous

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Relationship Between Gorxn and

the Equilibrium Constant

86


and the Equilibrium ConstantExample 17-17: Calculate the equilibrium constant, Kp, for the following reaction at 25oC from thermodynamic data in Appendix K.

• Note: this is a gas phase reaction.

g2g42 NO 2ON

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and the Equilibrium Constant• Kp for the reverse reaction at 25oC can be

calculated easily, it is the reciprocal of the above reaction.

kJ/mol 78.4G

ONNO 2orxn

4(g)22(g)

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and the Equilibrium ConstantExample 17-18: At 25oC and 1.00 atmosphere, Kp = 4.3 x 10-13 for the decomposition of NO2. Calculate Go

rxn at 25oC.

You do it.

2(g)(g)2(g)

O2NONO 2

89



• The relationship for K at conditions other than thermodynamic standard state conditions is derived from this equation.

Q log RT 2.303 GG

or

lnQ RT GG

o

o

90

Evaluation of Equilibrium Constants at Different Temperatures• From the value of Ho and K at one

temperature, T1, we can use the van’t Hoff equation to estimate the value of K at another temperature, T2.

21

o

T

T

12

12o

T

T

T

1

T

1

R

H

K

Kln

or

TT R

)T(TH

K

Kln

1

2

1

2

91

Evaluation of Equilibrium Constants at Different Temperatures

Example 17-19: For the reaction in example 17-18, Ho = 114 kJ/mol and Kp = 4.3 x 10-13 at 25oC. Estimate Kp at 250oC.

2(g)(g)2(g)

O2NONO 2

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Synthesis Question

Mars is a reddish colored planet because it has numerous iron oxides in its soil. Mars also has a very thin atmosphere, although it is believed that quite some time ago its atmosphere was considerably thicker. The thin atmosphere does not retain heat well, thus at night on Mars the surface temperatures are 145 K and in the daytime the temperature rises to 300 K. Does Mars get redder in the daytime or at night?

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Synthesis Question

The formation of iron oxides from iron and oxygen is an exothermic process. Thus the equilibrium that is established on Mars shifts to the iron oxide (red) side when the planet is cooler - at night. Mars gets redder at night by a small amount.

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Group Question

If you are having trouble getting a fire started in the barbecue grill, a common response is to blow on the coals until the fire begins to burn better. However, this has the side effect of dizziness. This is because you have disturbed an equilibrium in your body. What equilibrium have you affected?

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17 Chemical Equilibrium

Chemical Equilibrium

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