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Unformatted text preview: Chem 260/Lecture 33 F11 Today in Chemistry 260 Chemistry 260 •
•
•
•
• 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
ln ⎜
⎜ Kp
⎝ ⎞
⎟ → 0, ΔGr → 0
⎟
⎠ Gibbs Energy ΔGr = − RT ln K p + RT ln Q
⎛Q⎞
ΔGr = RT ( ln Q − ln K p ) = RT ln ⎜
⎜ Kp ⎟
⎟
⎝
⎠
Q
as Q → K p ,
→ 1,
Kp Not at equilibrium: ⎛Q⎞
ΔGr = RT ln ⎜
⎜ Kp ⎟
⎟
⎝
⎠ Not at equilibrium: ..is the differential/variance
of Gibbs free energy with
the change of the number
of moles ΔGr = ΔGr ° + RT ln Q
Gibbs Energy Q vs. K NH4Cl ( s ) É NH3 ( g ) + HCl ( g ) ΔGr The Effect of Concentration Changes
As the reaction approaches
equilibrium, Q K ΔGr = ? ΔGr = ? ΔGr = 0
Q<K Pure
reactant ΔGr = 0 Pure
reactant 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 ConcepTest * 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 ConcepTest At T=1200 C, the reaction: At T=1200 C, the reaction: P4(g) ⇆ 2 P2(g) P4(g) ⇆ 2 P2(g) has an equilibrium constant K = 0.612 has an equilibrium constant K = 0.612 If the initial partial pressure of P4 is 5 atm and that of P2 is 2 atm,
which way will the reaction proceed?
A) left
B) right
C) neither Q= 2*2/5= 0.8
If the initial partial pressure of P4 is 5 atm and that of P2 is 2 atm,
which way will the reaction proceed?
A) left
B) right
C) neither 2 Chem 260/Lecture 33 F11 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. 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
2
eq
⎜ PNH3 ⎟
PNH3
2 ⋅ 23
⎝2
⎠
= 2⋅
3
eq
eq
2
1 eq ⎞⎛ 1 eq ⎞
⎛
PN2 PH 2
⎜ PN2 ⎟⎜ PH 2 ⎟
⎝2
⎠⎝ 2
⎠ 3 3 ⎞
Po ⎟
⎠ 2 eq
eq
⎛ PN2
⎞⎛ PH 2 ⎞
⎜
P o ⎟⎜
Po ⎟
⎝
⎠⎝
⎠ 3 The Presence of a Catalyst = 4K p ⎛Q⎞
⎛ 4K p ⎞
ΔGr = RT ln ⎜
= RT ln ⎜
⎜ Kp ⎟
⎟
⎜ Kp ⎟
⎟
⎝
⎠
⎝
⎠
= RT ln ( 4 ) = 1.387 RT Gibbs Energy (P )
(P )(P )
2 Q= ⎜
⎝ Q will decrease until Q = K
\ NH3(g) will be converted to N2(g) + H2(g) Q<K Q>K
K=Q 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)
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 !G = " RT ln K
! N2 ( g ) + 3H2 ( g )
The Effect of Temperature 2NH3 ( g ) The presence of a catalyst does not change the
equilibrium constant of a reaction. 3 Chem 260/Lecture 33 F11 The Effect of Temperature The van t Hoff equation According to Le Châtelier s Principle, when temperature is increased ΔGr ° = ΔH r ° − T ΔSr ° The equilibrium composition of an exothermic
reaction will shift toward reactants.
The equilibrium composition of an endothermic
reaction will shift toward products. reactants + heat If exothermic, ΔH < 0, reactants products products + heat The origin of this effect is the dependence of ΔG (and therefore K) on T. ⎡ −ΔGro ⎤
K P = exp ⎢
⎥
⎣ RT ⎦ The van t Hoff equation ⎛ K ⎞ −ΔH r ° ⎛ 1 1 ⎞
ln ⎜ ⎟ =
⎜−⎟
R ⎝ T T′ ⎠
⎝ K′ ⎠ ⇒ ln K = − At another
temperature, T ln K ′ = − ⎛ −ΔH r ° ⎛ 1 1 ⎞ ⎞
K
= exp ⎜
⎜ − ⎟⎟
K′
⎝ R ⎝ T T′ ⎠⎠ ⎛ −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′ ⎠ 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′ ⎠
When ΔH > 0 (endothermic)
K is larger more product
favored ln K = − ⎛ −ΔH r ° ⎛ 1 1 ⎞ ⎞
K
= exp ⎜
⎜ − ⎟⎟
K′
⎝ R ⎝ T T′ ⎠⎠
 + K′ > K ln K = ΔGr °
RT K is smaller more reactant
favored + + intercept =
ln(K) ⎛ −ΔH r ° ⎛ 1 1 ⎞ ⎞
K
= exp ⎜
⎜ − ⎟⎟
K′
⎝ R ⎝ T T′ ⎠⎠ K′ < K ln K ; K = PX ⎛ K ⎞ −ΔH ⎛ 1 1 ⎞
ln ⎜ ⎟ =
⎜−⎟
R ⎝ T T′ ⎠
⎝ K′ ⎠
o
r ⎛ P ⎞ −ΔH vap ⎛ 1 1 ⎞
ln ⎜ X ⎟ =
⎜−⎟
′
R ⎝ T T′ ⎠
⎝ PX ⎠
o ClausiusClapeyron Equation Using Normal Boiling Point at 1atm
o
⎛ P ⎞ −ΔH vap ⎛ 1 1 ⎞
ln ⎜ X ⎟ =
⎜−⎟
R ⎝ T Tb ⎠
⎝ 1atm ⎠ ln ( PX ) = ConcepTest o
−ΔH vap ⎛ 1 1 ⎞
⎜−⎟
R ⎝ T Tb ⎠ Use to determine vapor
pressure at some
temperature Use to determine
boiling point at some
external pressure B A C ln K () −ΔH ro
R 1/T Vapor Pressure: Special case is the Clausius Clapeyron equation () Jacobus Henricus
van 't Hoff
(1852  1911)
First Nobel Prize in
Chemistry ΔSro
R
slope = >1 X l !X g ΔGr ° = ΔH r ° − T ΔSr ° −ΔH r ° ΔSr ° −ΔH r ° ⎛ 1 ⎞ ΔSr °
+
=
⎜ ⎟+
RT
R
R ⎝T ⎠
R <1 When ΔH < 0
(exothermic) ΔGr ° −ΔH r ° ΔSr °
=
+
RT ′
RT ′
R The difference between the two is the van t Hoff equation: Why do reactions at equilibrium respond this way? ΔGr = ΔGr ° + RT ln Q ΔGr °
ΔH r ° − T ΔSr ° −ΔH r ° ΔSr °
=−
=
+
RT
RT
RT
R At a certain
temperature, T ln K If endothermic, ΔH > 0, Now let s consider the effect of temperature on K itself. 1/T 1 /T 1/T Which statement(s) is(are) true:
I.
A is an endothermic reaction
II.
DSr=0 for reaction B
III.
Reaction C is product favored at high temperature
A. I only
B. III only
C. I and III only
D. I, II, and III
E. None of these are true 4 Chem 260/Lecture 33 B March, 1952 Pressure Cooking ln K C ln K A ln K F11 () () H 2O l ! H 2O g
1/T 1 /T 1/T Which statement(s) is(are) true:
I.
A is an endothermic reaction
II.
DSr=0 for reaction B
III.
Reaction C is product favored at high temperature
(endothermic)
A. I only B. III only C. I and III only D. I, II, and III E. None of these are true ln(K) intercept = ΔS
R o
⎛ P ⎞ −ΔH vap ⎛ 1 1 ⎞
ln ⎜ X ⎟ =
⎜−⎟
R ⎝ T Tb ⎠
⎝ Po ⎠ ( slope = Double the Pressure ) 3
−1
1⎞
⎛ P ⎞ − 40.66 × 10 J mol ⎛ 1
ln ⎜
⎜−
⎟
⎟=
8.314 J K −1 mol −1 ⎝ Th 373.15 ⎠
⎝ 1atm ⎠ o
r −ΔH
R o
r ⎛1
1⎞
⎛ 2atm ⎞
ln ⎜
⎟
⎟ = −4890 K ⎜ −
⎝ 1atm ⎠
⎝ Th 373.15 ⎠ ⎛1
1⎞
⎛P ⎞
ln ⎜ X ⎟ = −4890 K ⎜ −
⎟
⎝ 1atm ⎠
⎝ Th 373.15 ⎠ 0.693 = Th = 394.1K = 121o C 1/T Mountain Top Cooking ⎛P
ln ⎜ X
⎝ Po ⎞ −ΔH
⎟=
R
⎠ o
vap ⎛1 1 ⎞
⎜−⎟
⎝ T Tb ⎠ Barometric Pressure Ph = Po e − mgh () ⎛
Pe
ln ⎜ o
⎜
Po
⎜
⎝ −mgh kbTa kbTa ⎞
o
⎟ = −ΔH vap
⎟
R
⎟
⎠ = = −ΔH
R (1.381 × 10 kbT −23 ) m2 kg s −2 K −1 Ta = ( ) − 40.657 × 103 J mol −1 ⎛ 1
1⎞
⎜−
⎟
8.314472 J K −1 mol −1 ⎝ Th 373.15 ⎠ ⎛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 ⎠ kbTa () H 2O l ! H 2O g o
vap ⎛1 1⎞
⎜−⎟
⎝ Th Tb ⎠
⎛
1.661 × 10 −27 kg ⎞
− ⎜ 29amu *
⎟ 9.8m / s * h
1amu
⎝
⎠ −mgh Mt. McKinley
Denali
h = 6194 m o
⎛ P ⎞ −ΔH vap ⎛ 1 1 ⎞
ln ⎜ h ⎟ =
⎜−⎟
Po ⎠
R ⎝ Th Tb ⎠
⎝
− mgh () Mountain Top Cooking () H 2O l ! H 2O g −4890 K
+ 13.1
Th On top of Mt. McKinley at external temp of 0o C 6194m 1.43 ×105 K
=
− 383.2
273.15K
Th 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 5 ...
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 Equilibrium, Catalyst

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