2009년도 제2학기 화 학 2 담당교수 신국조 textbook: p. atkins / l....

2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9

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CHAPTER 9 CHEMICAL EQUILIBRIA

□ Nitrates : Fertilizers, Explosives

▷Chilean saltpeter (초석,硝石), NaNO3

▷Nitrogen fixation from the air : Haber-Bosch process

Haber and Le Rossignol, Zeitschrift für Elektrochemie (1913),

"Über die technische Darstellung von Ammoniak aus Elementen"

(On the technical production of ammonia from the elements)

2 2 3N ( ) 3 H ( ) 2 NH ( )g g+ g at 400oC, 200 atm, Catalysts : Os, U (later Fe)

Fritz Haber (獨,1868-1934) Carl Bosch (獨,1874-1940)

Nobel Prize in Chemistry, 1918 Nobel Prize in Chemistry, 1931 "for the synthesis of ammonia from its elements" “High pressure chemistry” Poison gases during WWI

REACTIONS AT EQUILIBRIUM

Fig. 9.1 (a) (b) (c)

CH4(g) + 2O2 CO2 + 2H2O C6H12O6(s) + 6O2 6CO2 +6H2O 2NO2 N2O4

Open system Too slow In equiluibrium ▷ Dynamic Equilibrium :

Rate of forward reaction = Rate of reverse reaction

Responds to small changes in T, P and addition of small amount of reagents


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9.1 The Reversibility of Reactions

Fig. 9.2 Forward reaction (N2 + 3H2 2NH3) Reverse reaction (2NH3 N2 + 3H2)

9.2 Equilibrium and Law of Mass Action (질량작용 법칙)

► Cato Guldberg and Peter Waage in 1864

Discovered mathematical relation for the composition of equilibrium mixture

Law of Mass Action

2 2 32 SO ( ) O ( ) 2 SO ( )g g+ g (D)

( )( ) ( )

( )( )

3 3

2 2 2

2 2oSO SO

2 2o oSO O SO O

/

/ /

P P PK

P P P P P P= =

2

, where o 1 bar (standard pressure)P =

The value of K is nearly the same for every experiment (with different initial compositions)!


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◆ Law of mass action for the reaction :

At equilibrium, the composition of the reaction mixture can be expressed in terms of an

equilibrium constant ( ) ( )( ) ( )

C D

A B

c d

a

P PK

P P= b for the reaction A( ) B( ) C( ) D( )a g b g c g d g+ + .

Ex. 9.1 Writing the expression for an equilibrium constant

2 2 3N ( ) 3 H ( ) 2 NH ( )g g+ g

( )( )

3

2 2

2

NH3

N H

PK

P P=

Different concentration measures in K :

Ideal gas: Partial pressure

Ideal solution: Molarity [J] for a species J relative to the standard molarity o 11 mol Lc −= ⋅

Activity : general expression real gases Ja

Pure solid and liquid does not appear in K : J 1a =

3 2CaCO ( ) CaO(s) CO (g)s + 2

oCO /K P P=

▶ General expressions of K for reaction A B C Da b c d+ + :

( ) ( )( ) ( )

C D

A B

c d

a b

a aK

a a=

Homogeneous equilibrium: 2 2 3N ( ) 3 H ( ) 2 NH ( )g g+ g

aq

Heterogeneous equilibrium:

2+2Ca(OH) ( ) Ca ( ) 2OH ( )s aq −+

2

2

22 2Ca OH

Ca(OH)

1 for a pure solid

( )[Ca ][OH ]

a aK

a+ − + −= =


4

Reactants Equilibrium Products

Fig. 9.3 Equilibrium is reached regardless of the initial composition of reactants. Equilbrium composition is obtained according to the equilibrium constant that depends on temperature.


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RT P P= +

9.3 The Thermodynamic Origin of Equilibrium Constants

Gibbs free energy of reaction,

m m(products) (reactants)G nG nG∆ = −∑ ∑

changes during reaction

Standard Gibbs free energy of reaction,

o o of f(products) (reactants)G nG nG∆ = −∑ ∑

does not change during reaction

Molar Gibbs free energy of an ideal gas J

G G ( )o om m J(J) (J) ln /

Fig. 9.4 Variation of G∆ (slope) during reaction. At

equilibrium, 0G∆ = (Minimum). does not change.oG∆

In general, om m(J) (J) lnG G RT= + Ja

HOW DO WE DO THAT? Gibbs free energy of reaction

Ex. 9.2. Calculating the Gibbs free energy of reaction from the reaction quotient.

2 2 3 g r2 SO ( ) O ( ) 2 SO ( )g g+ → o o141.74 kJ/mol at 25.00 CG∆ = −

Initially, J 100 bar (for each species J)P =

r ?G∆ = , Spontaneous direction of reaction?

( )( )

3

2 2

22

SO 22 2

SO O

(100) 1.00 10(100) 100

PQ

P P−= = =

××


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ln

o or r ln 153.16 kJ/mol (< 0) at 25.00 CG G RT Q∆ = ∆ + = −

Forward reaction is spontaneous.

At equilibrium, . r 0G∆ = o or eq r0 lnG RT Q G RT K= ∆ + ≡ ∆ +

o lnG RT K∴ ∆ = −r Link between thermodynamic quantity and equilibrium composition

Ex. 9.3. Predicting the value of K from the standard Gibbs free energy of reaction.

1 12 22 2H ( ) I ( ) HI( )g g+ → g ,

o or 1.70 kJ/mol at 25.00 CG∆ = +

or

3

ln /1.70 10 J/mol 0.685

(8.3145 J/K/mol) (298.15 K)

K G RT= −∆

×= − = −

×

0.50K∴ =

▶ Contributions of and to K: orH∆ o

rS∆

Since , o or rG H T∆ = ∆ − ∆ o

rSo or rln G HK

orS

RT RT∆ ∆ ∆

= − = − +R

/ or

o o o or r r r/ / /H RT S R H RT S RK e e e−∆ +∆ −∆ ∆= =

K ↓ as while as orH∆ ↑ K ↑ o

rS∆ ↑

Endothermic reaction or 0H∆ > only if 1K > o

rS∆ is large and positive.

Exothermic reaction for or 0H∆ < 1K > o

r 0S∆ >

9.4 The Extent of Reaction

2 2H ( ) Cl ( ) 2 HCl( )g g+ g ×

g ×

, ( )2 2

2 18HCl H Cl(500 K) / 4.0 10K P P P= =

2 2N ( ) O ( ) 2 NO( )g g+ , ( )2 2

2 21NO N O(800 K) / 3.4 10K P P P −= =

Fig. 9.5 Value of K indicates whether the reactants or the products are favored.


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9.5 The Direction of Reaction

0Q K

G<

∆ <

0Q K

G=

∆ = 0Q K

G>

∆ > Fig. 9.7 Position of Gmin and the direction of reaction

Fig. 9.6 Relative sizes of Q and K determine the direction of reaction.

Ex. 9.5 Predicting the direction of reaction

2 2H ( ) I ( ) 2 HI( ), (783 K) 46g g g K+ =

2 2HI H I 55 kPa 0.55 barP P P= = = =

( )2 2

2 2HI

H I

(0.55) 1.00.55 0.55

PQ

P P= = =

×

Since , the reaction will tend to form more products. Q K<

EQUILIBRIUM CALCULATIONS

9.6 The Equilibrium Constant in Terms of Molar Concentrations of Gases

A B C Da b c d+ + 2 2 3N ( ) 3 H ( ) 2 NH ( )g g g+

[ ] [ ][ ] [ ]C DA B

c d

c a bK = [ ]

[ ][ ]

23

32 2

NHN H

cK =

HOW DO WE DO THAT? Relation between K and cK

Partial pressures in K Molar concentrations Generate cK

Write activities as and . oJ /P P o[J] / c


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( ) ( )( ) ( )

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )( ) ( )

o o o( )C D C D C Do

o o oA BA B A B

/ /

/ /

c d a b c d c da b c d

a b c d aa b

P P P P P P P P PK P

P PP P P P P P P

++ − +

+= = = b

Molar concentration: J[ ] /J n V=

Ideal gas law as J JPV n RT= J JJ [ ]n RT nP RT R

V V= = = T J

( ) ( )( ) ( )

( ) ( )( ) ( )

( ) ( )C D

A B

[C] [D] [C] [D][A] [B][A] [B]

c d c d c dc d a b

a b a b a b

P P RT RTRT

P P RT RT+ − += =

( ) ( )( ) ( )

( )( )

( )o o o

( )o

o o o

[C]/ [D]/ [C] [D] [C] [D][A] [B][A]/ [B]/ [A] [B]

c d a b c d c da b c d

c a b c d a ba b

c c cK c

c c c

++ − +

+= = =

( ) ( )o

[C] [D][A] [B]

c dc

a b c da b

K

c+ − +=

( ) ( )( ) ( )

( )( )

( ) ( )( ) oC D( )o

A B

c dc d a bc d a b c

ca b a b c d

P P KRT c RTP P c

+ − ++ − +

+ − += = K

( ) ( )( )o( ) ( )o o

o c d a b

a b c d c d a b

c cc RTK P c RT K K

P

+ − ++ − + + − + ⎛ ⎞

∴ = = ⎜ ⎟⎝ ⎠

o

o

n

cc RTK K

P

∆⎛ ⎞

= ⎜ ⎟⎝ ⎠

, where products reactantsn n n∆ = − (Change in the number of gas molecules)

Ex. 9.6 Converting between K and cK

2 2 32 SO ( ) O ( ) 2 SO ( )g g+ g

1

, o 4(400 ) 3.1 10K C = × o(400 ) ?cK C =

o 1 barP = , , o 11 mol Lc −= ⋅ 2 18.3145 10 L bar K molR − − −= × ⋅ ⋅ ⋅

( ) ( )o

o 2 1 1 1

1 bar 12.03 K8.3145 10 L bar K mol 1 mol L

PRc − − − −

= =× ⋅ ⋅ ⋅ × ⋅

12.03 K

n

cTK K

∆⎛ ⎞∴ = ⎜ ⎟⎝ ⎠

2 (2 1) 1n∆ = − + = − , (400 273.15) K 673 KT = + =

( )( 1)

4 6673 K 3.1 10 1.7 1012.03 K 12.03 K

n

cTK K

∆ − −⎛ ⎞ ⎛ ⎞∴ = = × × = ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

−


9

g

9.7 Alternative Forms of the Equilibrium Constant

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

12 2

[HI](700 K) 54[H ][I ]cK = =

Multiplication

2 22 H ( ) 2 I ( ) 4 HI( )g g g+ , 4

2 32 2 2

2 2

[HI](700 K) (54) 2.9 10[H ] [I ]cK = = = ×

g

Reverse reaction

2 22 HI( ) H ( ) I ( )g g + , 2 23 2

[H ][I ] 1(700 K) 0.019[HI] 54cK = = =

Addition

(1) , 2 32 P( ) 3 Cl ( ) 2 PCl ( )g g g+( )

( ) ( )3

2

2

PCl1 32

P Cl

PK

P P=

+

(2) , 3 2 5PCl ( ) Cl ( ) PCl ( )g g g 5

3 2

PCl2

PCl Cl

PK

P P=

+

(3) , 2 52 P( ) 5 Cl ( ) 2 PCl ( )g g g( )

( ) ( )5

2

2

PCl3 52

P Cl

PK

P P=

(1) 2 32 P( ) 3 Cl ( ) 2 PCl ( )g g+ g

5 g

g

2×(2) 3 22 PCl ( ) 2 Cl ( ) 2 PCl ( )g g+

(3) = (1) + 2×(2) 2 52 P( ) 5 Cl ( ) 2 PCl ( )g g+

( )( ) ( )

( )( ) ( )

( )( ) ( )

21 2

5 3 5

2 2 3 2

2 2 2

PCl PCl PCl 23 1 25 3 222 2

P Cl P Cl PCl Cl

K K

P P PK K

P P P P P P= = × = K

9.8 Using Equilibrium Constants

TOOLBOX 9.1 HOW TO SET UP AND USE AN EQUILIBRIUM TABLE

Conceptual basis

Composition changes until Q=K.

Change in one component is linked to changes in other components.


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Procedure

Step 1 Write down the balanced chemical equation and the expression for the equilibrium constant.

Construct an equilibrium table as

Step 2 Fill the second row, use x if necessary. Changes needed to reach equilibrium

Step 3 Fill the third row, use x for the unknown change. Equilibrium composition

Step 4 Determine x using the equilibrium constant.

Ex. 9.7 Calculating the equilibrium composition by approximation.

2 2 22 N ( ) O ( ) 2 N O( )g g+ g , ( )( )

2

2 2

2

N O 282

N O

(800 K) 3.2 10P

KP P

−= = ×

Initial mixture of 0.482 mol N2 and 0.933 mol of O2

Volume of reaction vessel: 10.0 L

Initially,

2

2 1 1

N(0.482 mol) (8.3145 10 L bar K mol ) (800 K) 3.21 bar

10.0 LP

− − −× × ⋅ ⋅ ⋅ ×= =

2

2 1 1

O(0.933 mol) (8.3145 10 L bar K mol ) (800 K) 6.21 bar

10.0 LP

− − −× × ⋅ ⋅ ⋅ ×= =

2N O 0P = .


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N2 O2 N2O

Step 1 Initial partial pressure 3.21 6.21 0

Step 2 Changes in partial pressure -2x -x +2x

Step 3 Equilibrium partial pressure 3.21-2x 6.21-x 2x

Step 4 Determine x from K 2 2

282 2

(2 ) (2 ) 3.2 10(3.21 2 ) (6.21 ) (3.21) (6.21)

x xKx x

−= ≈ = ×− × − ×

14 7.2 10x −∴ = ×

From , 2

2

2

2N 3.21 2 barP x= −

From , O 6.21 barP x= −

From , N O 2 barP x=

N 3.21 barP =

2O 6.21 barP =

2

13N O 1.4 10 barP −= ×

Ex. 9.8 Calculating the equilibrium composition by using a quadratic equation

5 3 2PCl ( ) PCl ( ) Cl ( ) g g + g , 3 2

5

PCl Clo

PCl

(250 C) 78.3P P

KP

= =

Initially, 3.12 g of PCl5 placed in 500 mL vessel. Equilibrium composition?

[Solution] 5

5

5

PCl 2PCl 1

PCl

3.12 g 1.50 10 mol208.24 g mol

mn

M−

−= = = ×⋅

5

5

PClPCl

2 2 1 1(1.50 10 mol) (8.3145 10 L bar K mol ) (523 K)0.500 L

1.30 bar

n RTP

V− − − −

=

× × × ⋅ ⋅ ⋅ ×=

=

PCl5 PCl3 Cl2

Step 1 Initial partial pressure 1.30 0 0

Step 2 Change in partial pressure –x +x +x

Step 3 Equilibrium partial pressure 1.30 – x x x


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From 3 2

5

2PCl Cl

PCl 1.30 1.30P P x x xK

P x×

= = =− − x

281.x =

2 78.3 1.30 0x x K+ − =

o79.6− r

From , 5

3

2

5PCl 1.30 barP x= −

From , PCl barP x=

From , Cl barP x=

PCl 1.30 1.28 bar 0.02 barP = − =

3PCl 1.28 barP =

2Cl 1.28 barP =

Fig. 9.8 Approach toward equilibrium for the reaction . 5 3PCl ( ) PCl ( ) Cl ( ) g g + 2 g

THE RESPONSE OF EQUILIBRIA TO CHANGES IN CONDITIONS

◆ Le Chatelier’s principle:

When a stress is applied to a system in dynamic equilibrium,

the equilibrium tends to adjust to minimize the effect of the stress.

Fig. 9.9. Henri Le Chatelier (1850-1936)


13

g

9.9 Adding and Removing Reagents

Fig. 9.10

2 2 3N ( ) 3 H ( ) 2 NH ( )g g+

(1) Additional H2 added. Formation of NH3

(2) Additional NH3 added.

Decomposition of NH3

Fig. 9.11 (a) Addition of reactants to an equilibrium mixture (Q=K). Q<K Q=K Formation of products

(b) Addition of products to an equilibrium mixture (Q=K). Q>K Q=K Formation of reactants

Ex. 9.10 Calculating the new equilibrium composition after addition of a reagent to an equilibrium mixture..

5 3 2PCl ( ) PCl ( ) Cl ( ) g g + g

5 3

Addition of 0.0100 mol of Cl2(g) to the equilibrium mixture of Ex. 9.8 in a 500 mL vessel.

Composition of new equilibrium mixture ?

Initial equilibrium composition: , PCl 0.02 barP = PCl 1.28 barP = , 2Cl 1.28 barP =

Additional partial pressure of Cl2 :


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2

2

ClCl

2 1 1(0.0100 mol) (8.3145 10 L bar K mol ) (523 K)0.500 L

0.870 bar

n RTP

V− − −

=

× × ⋅ ⋅ ⋅ ×=

=

PCl5 PCl3 Cl2

Step 1 Initial partial pressure 0.02 1.28 2.15

Step 2 Change in partial pressure +x –x –x

Step 3 Equilibrium partial pressure 0.02 + x 1.28 – x 2.15 – x

From 3 2

5

2PCl Cl

PCl

(1.28 ) (2.15 ) 2.75 3.430.02 0.02

P P x x x xx

KP x

− × − − += = =

+ +

.0144x =

2 81.7 1.18 0x x− + =

o81.7 r 0

From , 5PCl 0.02 barP x= +

From , 3PCl 1.28 barP x= −

From , 2Cl 2.15 barP x= −

5PCl 0.02 0.01 bar 0.03 barP = + =

3PCl 1.28 0.01 bar 1.27 barP = − =

2Cl 2.15 0.01 bar 2.14 barP = − =

Fig. 9.12 Approach toward a new equilibrium

for the reaction 5 3PCl ( ) PCl ( ) Cl ( ) g g + 2 gafter additional Cl2 is added.


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9.10 Compressing a Reaction Mixture

Le Chatelier’s principle on : 2I ( ) 2 I( )g g

Compression favors the reverse reaction minimize the pressure increase

Expansion favors the forward reaction minimize the pressure decrease

Fig. 9.13 Effect of compression and expansion on the dissociation equilibrium of a diatomic molecule.

Fig. 9.14 The high-pressure vessel (withstanding 250 atm) for the catalytic synthesis of ammonia.

HOW DO WE DO THAT? Discovering the effect of compression on the equilibrium

2 2 42 NO ( ) N O ( )g g , ( ) ( ) ( )

2 4 2 4 2 4

2 2 2

o oN O N O N O

2 2o oNO NO NO

/ /

/ /

P P n RT VP n P VK

P P n RT VP n R= = =

o

2T

Since is constant, o /P RT ( )2 4 2

2

N O NO/n n as V in order that K remains constant. ↓↑

Increase in and decrease in 2 4 2N On NOn


16

2 g2

9.11 Temperature and Equilibrium

Exothermic reaction: Favors the product formation by lowering T

Endothermic reaction: Favors the product formation by increasing T

3CaCO ( ) CaO( ) CO ( ) s s + , strongly endothermic, CO 0.22 atmP = at 800oC

Fig. 9.15 A radar image of the surface of Venus rich in carbonates. 2CO 87 atmP = at 500oC

Ex. 9.12 Predicting the effect of temperature on an equilibrium

2 5V O2 2 32 SO ( ) O ( ) 2 SO ( )g g+ g

Standard reaction enthalpy of the forward reaction:

o o of 3 f 2(2 mol) (SO , ) (2 mol) (SO , )

2( 395.72 kJ) 2( 296.83 kJ) 197.78 kJH H g H∆ = ×∆ − ×∆

= − − − = −g

Forward reaction is exothermic. Increasing temperature favors the reverse reaction.

HOW DO WE DO THAT? Relation between two equilibrium constants at two temperatures, T1 and T2

or,1 1 1lnG RT∆ = − K K,

or,2 2 2lnG RT∆ = −

or,1

11

lnG

KRT∆

= −

or,2

22

lnG

KRT∆

= −

o or,1 r,2

1 21 2

1ln lnG G

K KR T T⎧ ⎫∆ ∆⎪ ⎪− = − −⎨ ⎬⎪ ⎪⎩ ⎭

o o or,1 r,1 1 r,1G H T S∆ = ∆ − ∆ o o o

r,2 r,2 2 r,2G H T S∆ = ∆ − ∆,


17

o o or,1 1 r,1 r,2 2 r,2

1 21 2

o oo or,1 r,2

r,1 r,21 2

1ln ln

1

H T S H T SK K

R T T

H HS S

R T T

⎧ ⎫∆ − ∆ ∆ − ∆⎪ ⎪− = − −⎨ ⎬⎪ ⎪⎩ ⎭⎧ ⎫∆ ∆⎪ ⎪= − − −∆ + ∆⎨ ⎬⎪ ⎪⎩ ⎭

o

orH∆ and are nearly constant over the temperature range of interest.

orS∆

o2 r

1 1

1 1 ln K HK R T T

⎧ ⎫∆= −⎨

⎩ ⎭2⎬ van’t Hoff equation

Ex. 9.13 Predicting the value of an equilibrium constant at a different temperature

2 2 3N ( ) 3 H ( ) 2 NH ( )g g+ g

5298K 6.8 10K = ×

3 1

400K ?K =

o o 1r f 32 (NH , ) 2( 46.11 kJ mol ) 92.22 10 J molH H g − −∆ = ∆ = − ⋅ = − × ⋅

-192.22 kJ mol

o 4 -12 r

2 1 11 1 2

1 1 9.222 10 J mol 1 1 ln 9.498.3145 10 J K mol 298 K 400 K

K HK R T T

− ⋅

− − −

⎧ ⎫∆ − × ⋅ ⎧ ⎫= − = × − = −⎨ ⎬ ⎨ ⎬× ⋅ ⋅ ⎩ ⎭⎩ ⎭

Take antilogarithms ( ). xe

9.49 5 9.492 1 (6.8 10 ) 51K K e e− −= = × × =

cf. 400K (exp) 41K =

9.12 Catalysts and Haber’s Achievement

2 2 3N ( ) 3 H ( ) 2 NH ( )g g+ g

Haber’s catalyst: Os (and later U)

Bosch: Industrial scale

High-pressure synthesis

Fe catalyst

Fig. 9.16 Fritz Haber (1868-1934)

Nobel Prize in Chemistry, 1918


18

Fig. 9.17. The use of ammonia in industry.

9.13 The Impact on Biology

▶ Homeostasis (항상성,恒常性):

A mechanism similar to chemical equilibrium, allows living organisms to control biological processes

at a constant level.

▷ Transport of oxygen homeostatic biological process

2 2Hb( ) O ( ) HbO ( )aq g aq+

At lung tissues: forward reaction

At capillaries in muscles: reverse reaction lactic acid causes the release of oxygen from HbO2


19

Myoglobin (monomeric hemeprotein) Hemoglobin (tetrameric hemeprotein)

Muscle tissue, Color of meat Carrier of oxygen from lung to muscles

Fig. 9.18 Variation of the extent of saturation of Mb and Hb with . 2

2

2

2

OP

OP <105 Torr 98% saturation for Hb

OP = 40 Torr 75% saturation for Hb (in resting muscle tissue)

OP < 20 Torr 10% saturation for Hb, Mb starts to release its reserved oxygen

2009년도 제2학기 화 학 2 담당교수 신국조 textbook: p. atkins / l....

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