Unformatted text preview: Chem 260/Lecture 33 F11 Today in Chemistry 260 Chemistry 260 •
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• Lecture 33 ΔG ° = − RT ln K p Evaluating the equilibrium composition
r
Predicting changes in equilibrium
Le Châtelier s Principle
Effects of concentration, pressure, and catalysts
Temperature dependence of K  the van t Hoff equation Later in Chemistry 260 November 23, 2011
Lake Michigan
Chicago, IL Mt. McKinley
Denali Reading: Mon, 15.115.4, 15.715.9 Wed, 11.3, 15.515.6, 16.116.4
Proton transfer equilibria
Equilibria of aqueous salts
Electrochemistry
c Evaluating the equilibrium composition
Predicting changes in equilibrium
Le Châtelier s Principle
Effects of concentration, pressure, and catalysts
Temperature dependence of K  the van t Hoff equation d eq
⎛ PCeq ⎞ ⎛ PD
⎞
⎜
Po ⎟ ⎜
Po ⎟
⎠ = exp ⎡ −ΔGr ° ⎤
⎝
⎠⎝
KP =
a
b
⎢ RT ⎥
eq
⎣
⎦
⎛ PA
⎞ ⎛ PBeq ⎞
o⎟ ⎜
o⎟
⎜
P⎠⎝
P⎠
⎝ ΔGr ° = − RT ln K p Le Châtelier s Principle Today in Chemistry 260 When a system at equilibrium is disturbed, the composition
adjusts so as to minimize the effect of the disturbance.
Example NH4Cl ( s ) É NH3 ( g ) + HCl ( g ) ; K = PNH3 × PHCl ΔGr ° = − RT ln K p
c d eq
⎛ PCeq ⎞ ⎛ PD
⎞
o⎟ ⎜
⎜
P⎠⎝
Po ⎟
⎠ = exp ⎡ −ΔGr ° ⎤
⎝
KP =
a
b
⎢ RT ⎥
eq
⎣
⎦
⎛ PA
⎞ ⎛ PBeq ⎞
⎜
Po ⎟ ⎜
Po ⎟
⎝
⎠⎝
⎠ effect of adding NH3(g)? effect of adding NH4Cl(s)? effect of compression?
Le Châtelier s Principle is only a rule of thumb can see why it works/
where it comes from by looking at the THERMODYNAMICS
Movement towards equilibrium is a spontaneous process
• Some changes in conditions affect the value of ΔG , K (temperature).
• Others changes that do not affect the numerical value of K (pressure). The Effect of Concentration Changes Q vs. K NH4Cl ( s ) É NH3 ( g ) + HCl ( g ) The Effect of Concentration Changes Q = ( PHCl )( PNH 3 ) ( K = ( PHCl )eq PNH3 ) eq Not at equilibrium: ΔGr = ΔGr ° + RT ln Q
Tells us if forward reaction is
spontaneous from the current
conditions of the system At equilibrium: Slope of line (change in G)
with change of n (moles) ΔGr = 0
ΔGr ° = − RT ln K p As the reaction approaches
equilibrium, Q K Gibbs Energy •
•
•
•
• •
•
•
•
• Pure
reactant ΔGr = 0 Pure
product 1 Chem 260/Lecture 33 F11 The Effect of Concentration Changes Q = ( PHCl )( PNH 3 ) ( K = ( PHCl )eq PNH3 ) eq ΔGr = ΔGr ° + RT ln Q ⎛Q⎞
ΔGr = RT ( ln Q − ln K p ) = RT ln ⎜
⎜ Kp ⎟
⎟
⎝
⎠
Q
as Q → K p ,
→ 1,
Kp
⎛Q
ln ⎜
⎜ Kp
⎝ ⎞
⎟ → 0, ΔGr → 0
⎟
⎠ Gibbs Energy ΔGr = − RT ln K p + RT ln Q
ΔGr = ? Pure
reactant ..is the differential/variance
of Gibbs free energy with
the change of the number
of moles Not at equilibrium: ⎛Q⎞
ΔGr = RT ln ⎜
⎜ Kp ⎟
⎟
⎝
⎠ Not at equilibrium: ΔGr ΔGr = ΔGr ° + RT ln Q
Gibbs Energy Q vs. K NH4Cl ( s ) É NH3 ( g ) + HCl ( g ) The Effect of Concentration Changes
As the reaction approaches
equilibrium, Q K ΔGr = ? ΔGr = 0
Q<K Pure
reactant ΔGr = 0 ΔGr < 0 Pure
product Effect of Concentration Changes H2 ( g ) + I2 ( g ) É 2HI ( g ) Pure
product ΔGr > 0 Effect of Concentration Changes H2 ( g ) + I2 ( g ) É 2HI ( g ) Q>K
K=Q Q= ( PHI )2
( PH2 )( PI2 ) Q= Pure
reactant ΔGr = ΔGr ° + RT ln Q
Gibbs Energy Gibbs Energy ΔGr = ΔGr ° + RT ln Q ( PHI )2
( PH2 )( PI2 ) Pure
product * If add reactants, drive the
reaction to the right.
* If remove products, drive
the reaction to the right. * If add products, drive the
reaction to the left.
* If remove reactants, drive
the reaction to the left.
As the reaction approaches
equilibrium, Q K Pure
reactant Pure
product The Effect of Pressure
According to Le Châtelier s Principle,
When a system at equilibrium is compressed, the composition of a gasphase
equilibrium adjusts so as to reduce the number of molecules in the gas phase. N2 ( g ) + 3H2 ( g ) É 2NH3 ( g )
The Effect of Pressure ΔG is the standard Gibbs energy of
reaction and is therefore defined for the
standard pressure Po = 1 atm !G ! = " RT ln K Hence, K is independent of the
pressure at which the reaction
is carried out. If the NH3 synthesis reaction is compressed isothermally
the equilibrium constant will remain unchanged. 2 Chem 260/Lecture 33 F11 The Effect of Pressure
eq
However, pressure CAN have a large effect on composition ⎛ PNH N2 ( g ) + 3H2 ( g ) É 2NH3 ( g ) Kp = 1
Imagine that the volume is doubled Px = Pxeq
2
2
1
⎛
⎞
NH 3 3 N2 H2 ( eq = ) eq
⎜ PNH3 ⎟
PNH3
2 ⋅ 23
⎝2
⎠
= 2⋅
3
eq
eq
2
⎛ 1 eq ⎞⎛ 1 eq ⎞
PN2 PH 2
⎜ PN2 ⎟⎜ PH 2 ⎟
⎝2
⎠⎝ 2
⎠ ⎞
Po ⎟
⎠ 2 eq
eq
⎛ PN2
⎞⎛ PH 2 ⎞
⎜
P o ⎟⎜
Po ⎟
⎝
⎠⎝
⎠ 3 2 The Presence of a Catalyst = 4K p ( )( ) ⎛Q⎞
⎛ 4K p
ΔGr = RT ln ⎜
= RT ln ⎜
⎜ Kp ⎟
⎟
⎜ Kp
⎝
⎠
⎝
= RT ln ( 4 ) = 1.387 RT 3 3 Gibbs Energy (P )
(P )(P )
2 Q= ⎜
⎝ ⎞
⎟
⎟
⎠ Q<K Q>K
K=Q Q will decrease until Q = K
\ NH3(g) will be converted to N2(g) + H2(g) ΔGr < 0 ΔGr > 0 The Presence of a Catalyst
A catalyst is a substance that accelerates a reaction
without itself appearing in the chemical reaction.
A catalyst speeds up a reaction,
by lowering the energy barrier.
(foreshadowing notice: kinetics) N2 ( g ) + 3H2 ( g )
The Effect of Temperature Catalysts alter reaction pathways.
However, ΔG of the reaction does not
change, since ΔG is a state function
Since ΔG is unaltered by the
catalyst, K is also unaltered 2NH3 ( g ) !G = " RT ln K
! The presence of a catalyst does not change the
equilibrium constant of a reaction. The Effect of Temperature The van t Hoff equation According to Le Châtelier s Principle, when temperature is increased The equilibrium composition of an endothermic
reaction will shift toward products. If endothermic, ΔH > 0, reactants + heat If exothermic, ΔH < 0, reactants ΔGr °
ΔH r ° − T ΔSr ° −ΔH r ° ΔSr °
=−
=
+
RT
RT
RT
R At a certain
temperature, T products products + heat Why do reactions at equilibrium respond this way?
The origin of this effect is the dependence of ΔG (and therefore K) on T. ΔGr = ΔGr ° + RT ln Q Now let s consider the effect of temperature on K itself. ΔGr ° = ΔH r ° − T ΔSr ° The equilibrium composition of an exothermic
reaction will shift toward reactants. ⎡ −ΔGro ⎤
K P = exp ⎢
⎥
⎣ RT ⎦ ln K = − At another
temperature, T ln K ′ = − ΔGr ° −ΔH r ° ΔSr °
=
+
RT ′
RT ′
R The difference between the two is the van t Hoff equation: ⎛ −ΔH r ° ΔSr ° ⎞ ⎛ −ΔH r ° ΔSr ° ⎞ −ΔH r ° ΔH r °
ln K − ln K ′ = ⎜
+
+
+
⎟−⎜
⎟=
R ⎠ ⎝ RT ′
R⎠
RT
RT ′
⎝ RT
⎛ K ⎞ −ΔH r ° ⎛ 1 1 ⎞
ln ⎜ ⎟ =
⎜−⎟
R ⎝ T T′ ⎠
⎝ K′ ⎠ 3 Chem 260/Lecture 33 F11 The van t Hoff equation ⎛ −ΔH r ° ⎛ 1 1 ⎞ ⎞
K
⇒
= exp ⎜
⎜ − ⎟⎟
K′
⎝ R ⎝ T T′ ⎠⎠ ⎛ K ⎞ −ΔH r ° ⎛ 1 1 ⎞
ln ⎜ ⎟ =
⎜−⎟
R ⎝ T T′ ⎠
⎝ K′ ⎠ The van t Hoff equation: (ln
(k) Temp. dependence) ⎛ K ⎞ −ΔH r ° ⎛ 1 1 ⎞
ln ⎜ ⎟ =
⎜−⎟
R ⎝ T T′ ⎠
⎝ K′ ⎠ ⎛1 1 ⎞
if T ′ > T , ⎜ − ⎟ > 0
⎝ T T′ ⎠ ln K = − ⎛ −ΔH r ° ⎛ 1 1 ⎞ ⎞
K
= exp ⎜
⎜ − ⎟⎟
K′
⎝ R ⎝ T T′ ⎠⎠ When ΔH > 0 (endothermic)
K is larger more product
favored  + K′ > K ΔGr °
RT ΔGr ° = ΔH r ° − T ΔSr ° −ΔH r ° ΔSr ° −ΔH r ° ⎛ 1 ⎞ ΔSr °
ln K =
+
=
⎜ ⎟+
RT
R
R ⎝T ⎠
R <1 ⎛ −ΔH r ° ⎛ 1 1 ⎞ ⎞
K
= exp ⎜
⎜ − ⎟⎟
K′
⎝ R ⎝ T T′ ⎠⎠ K is smaller more reactant
favored + + intercept =
ln(K) When ΔH < 0
(exothermic) K′ < K () () ; K = PX o () ⎛P
ln ⎜ X
⎝ Po ClausiusClapeyron Equation Use to determine vapor
pressure at some
temperature o
⎛ P ⎞ −ΔH vap ⎛ 1 1 ⎞
ln ⎜ X ⎟ =
⎜−⎟
R T Tb ⎠
⎝ 1atm ⎠ ln ( PX ) = −ΔH
R ⎛1 1 ⎞
⎜−⎟
⎝ T Tb ⎠ Mountain Top Cooking ⎛ P ⎞ −ΔH
ln ⎜ X ⎟ =
R
⎝ Po ⎠ o
vap ⎛1 1 ⎞
⎜−⎟
⎝ T Tb ⎠ Barometric Pressure Ph = Po e − mgh () −mgh kbTa ⎞
o
⎟ = −ΔH vap
⎟
R
⎟
⎠ = ) 3
−1
1⎞
⎛ P ⎞ − 40.66 × 10 J mol ⎛ 1
ln ⎜
⎜−
⎟
⎟=
8.314 J K −1 mol −1 ⎝ Th 373.15 ⎠
⎝ 1atm ⎠ ⎛1
1⎞
⎛P ⎞
ln ⎜ X ⎟ = −4890 K ⎜ −
⎟
⎝ 1atm ⎠
⎝ Th 373.15 ⎠ Double the Pressure ⎛1
1⎞
⎛ 2atm ⎞
ln ⎜
⎟
⎟ = −4890 K ⎜ −
⎝ 1atm ⎠
⎝ Th 373.15 ⎠ 0.693 = −4890 K
+ 13.1
Th Th = 394.1K = 121o C () Mountain Top Cooking () H 2O l ! H 2O g () H 2O l ! H 2O g o
−ΔH vap ⎛ 1 1 ⎞
−mgh
=
⎜−⎟
kbTa
R ⎝ Th Tb ⎠
⎛
1.661 × 10 −27 kg ⎞
− ⎜ 29amu *
⎟ 9.8m / s * h − ( 40.657 × 103 J mol −1 )
⎛1
1amu
1⎞
⎝
⎠
=
⎜−
⎟
−23
2
−2
8.314472 J K −1 mol −1 ⎝ Th 373.15 ⎠
(1.381 × 10 m kg s K −1 )Ta Mt. McKinley
Denali
h = 6194 m kbT ⎛1
h
1 ⎞ 1.43 ×105 K
= 1.43 ×105 K ⎜ −
− 383.2
⎟=
Ta
Th
⎝ Th 373.15 ⎠
Mt. McKinley
Denali
h = 6194 m ⎛1 1⎞
⎜−⎟
⎝ Th Tb ⎠ o
−ΔH vap ⎛ 1 1 ⎞
⎜−⎟
R ⎝ Th Tb ⎠ o
⎞ −ΔH vap ⎛ 1 1 ⎞
⎟=
⎜−⎟
R ⎝ T Tb ⎠
⎠ Use to determine
boiling point at some
external pressure o
⎛ P ⎞ −ΔH vap ⎛ 1 1 ⎞
ln ⎜ h ⎟ =
⎜−⎟
R ⎝ Th Tb ⎠
⎝ Po ⎠ ⎛ − mgh kbTa
Pe
ln ⎜ o
⎜
Po
⎜
⎝ () ( Using Normal Boiling Point at 1atm o
vap March, 1952 Pressure Cooking H 2O l ! H 2O g ⎛ K ⎞ −ΔH ⎛ 1 1 ⎞
ln ⎜ ⎟ =
⎜−⎟
R ⎝ T T′ ⎠
⎝ K′ ⎠
o
r ⎛ P ⎞ −ΔH vap ⎛ 1 1 ⎞
ln ⎜ X ⎟ =
⎜−⎟
′
R ⎝ T T′ ⎠
⎝ PX ⎠ −ΔH ro
R 1/T Vapor Pressure: Special case is the Clausius Clapeyron equation X l !X g ΔSro
R
slope = >1 Jacobus Henricus
van 't Hoff
(1852  1911)
First Nobel Prize in
Chemistry On top of Mt. McKinley at external temp of 0o C 6194m 1.43 ×10 K
=
− 383.2
273.15K
Th
5 On Lake Michigan at external temp of 10o C 177m
1.43 ×105 K
=
− 383.2
283.15K
Th Th = 352.3K
Lake Michigan
Chicago, IL
h = 177 m Th = 372.6 K = 79.2o C = 99.4o C Lake Michigan
Chicago, IL
h = 177 m 4 ...
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This note was uploaded on 01/19/2012 for the course CHEM 260 taught by Professor Staff during the Fall '08 term at University of Michigan.
 Fall '08
 STAFF
 Equilibrium, Catalyst

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