2009년도 제2학기 화 학 2 담당교수 신국조 Textbook: P. Atkins / L....
Transcript of 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
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
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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)!
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
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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+ − + −= =
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
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.
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
5
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−= = =
××
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
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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.
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
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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
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
8
( ) ( )( ) ( )
( ) ( ) ( )( ) ( ) ( )
( ) ( ) ( )( ) ( )
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
∆ − −⎛ ⎞ ⎛ ⎞∴ = = × × = ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
−
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
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.
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
10
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 = .
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
11
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
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
12
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)
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
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 :
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
14
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.
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
15
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
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
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∆ = ∆ − ∆,
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
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
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
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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
2009년도 제2학기 화 학 2 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 9
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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