14.6
Catalysis
Homogeneous Catalysis
Heterogeneous Catalysis
Enzymes
Chapter 15
Chemical Equilibrium
15.1
The Concept of Equilibrium
15.2
The Equilibrium Constant
Equilibrium Constants in Terms of Pressure
The Magnitude of Equilibrium Constants
The Direction of the Chemical Equation and K
15.3
Heterogeneous Equilibria
15.4
Calculating Equilibrium Constants
15.5
Applications of Equilibrium Constants
Predicting the Direction of Reaction
Calculation of Equilibrium Concentrations
15.6
Le Châtelier’s Principle
Change in Reactant or Product Concentrations
Effects of Volume and Pressure Changes
Effect of Temperature Changes
The Effect of Catalysts
C2H4 (g) + H2 (g) à C2H6 (g)
catalyzed reaction on a Ni, Pd, or Pt surface
Fig. 14.29 -- Decomposition of HCOOH (g) alone (red line)
and with ZnO (s) added (blue line).
Be sure to work through the associated integrative problem on p 546.
There are two reaction mechanisms that deserve special mention:
(1) Consecutive first order reactions
A à B
B à C
k1
k2
This introduces the idea of a ‘steady state for B’.
(2) A common mechanism for enzyme catalysisà Michaelis-Menten eqn
E + S à ES
ES à E + S
k1
k -1
ES à E + P
k2
where E = enzyme (the catalyst), S = substrate,
ES = complex
and P = product
1
Chapter 15
Chemical Equilibrium
15.1
The Concept of Equilibrium
15.2
The Equilibrium Constant
Equilibrium Constants in Terms of Pressure
The Magnitude of Equilibrium Constants
The Direction of the Chemical Equation and K
15.3
Heterogeneous Equilibria
15.4
15.5
15.6
Calculating Equilibrium Constants
Applications of Equilibrium Constants
Predicting the Direction of Reaction
Calculation of Equilibrium Concentrations
Le Châtelier’s Principle
Change in Reactant or Product Concentrations
Effects of Volume and Pressure Changes
Effect of Temperature Changes
The Effect of Catalysts
From last chapter—note that if we followed the reaction long enough,
the amount of A would go to zero.
IF the reaction mechanism is simple:
and
N2 O4 à 2 NO2
2 NO2 à N2 O4
forward, kf
reverse, kr
rate forward = rate reverse
kf[N2 O4 ] = kr [NO2 ]2
Note the limiting values for Concentrations and for Rates
for an equilibrium reaction.
and
k f [ NO2 ]2
=
= Kc
k r [ N 2O4 ]
The last part is always true—
the first part is NOT always true
Consider the following reaction of gases in a 10.0 l flask at 12 00 K
CO
+
3 H2
à
CH 4
+
H2O
ini
1.000
1.000
0
0
d
-x
- 3x
x
x
Eq
1.000-x
1.000-3x
0+x
0+x
Obs: CH4 = 0.387
Eq
M
0.613
1.839
0.387
0.387
0.0613
0,1839
0.0387
0.0387
2
Similar reactions, starting with different concentrations
Initial
[CO] eq
[H2 ]eq
0.100 M CO
0.300 M H2
0.0613 0.1839
0.0387 0.0387
0.2000 M CO
0.3000M H2
0.1522 0.1566
0.0478 0.0478
0.1000 M CH4
0.1000 M H2 O
0.0613 0.1839
0.0387 0.0387
Each of these yield the following interesting result:
Kc =
The generalization is simple
[CH4 ]eq [H2 O] eq
and
aA + bB = cC + dD
Kc =
important
(overall reaction)
[C ]c [D ]d
[ A]a [B ]b
Examples
[ CH 4 ][ H 2O ]
= 3 .92 ± 0. 01
[ CO][ H 2 ] 3
The Law of Mass Action -- or the Chemical Equilibrium Constant
Heterogeneous Equilibria are independent of the amount of solid or liquid !!!
[So long as you have some present, the concentration is a constant.]
thus,
CaCO3 (s) = CaO (s) + CO2 (g)
Kc = [CO2 ]
Announcements
1.
Dr. Zellmer’s tutorial should be operating fully by now.
Use the link to Chem 125 or the
independent link under Additional Notes.
2.
Quiz next week will cover 14.6 through all of
Chapter 15, but the emphasis will be on Chapter 15.
14.6
Catalysis
Homogeneous Catalysis
Heterogeneous Catalysis
Enzymes
Chapter 15
Chemical Equilibrium
15.1
The Concept of Equilibrium
15.2
The Equilibrium Constant
Equilibrium Constants in Terms of Pressure
The Magnitude of Equilibrium Constants
The Direction of the Chemical Equation and K
15.3
Heterogeneous Equilibria
15.4
Calculating Equilibrium Constants
15.5
Applications of Equilibrium Constants
Predicting the Direction of Reaction
Calculation of Equilibrium Concentrations
15.6
Le Châtelier’s Principle
Change in Reactant or Product Concentrations
Effects of Volume and Pressure Changes
Effect of Temperature Changes
The Effect of Catalysts
3
2 H 2S(g) º 2 H 2(g) + S2(g)
Consider
After putting 0.100 mole H 2S in a 10.0 L vessel
and heating to 1132 oC, the equilibrium mixture
contained 0.0285 mol H2. What is K c ?
Change to concentrations.
Set up calculations.
K=
[CO][ H 2 ]
x2
=
[ H2 O]
(0.100 − x)
Solve for eq. concentrations
Solve for Kc
If you know x, you can solve for K,
or if you know K you can solve for x.
(a) Calculate one concentration given others and
a known Kc (= 3.92 in the following example).
CO(g) + 3 H 2(g) º CH4(g) + H 2O(g)
0.30
0.10
?
(b) Consider CO + H2O = CO2 + H2 which has an eq
constant of 0.58 at 1000 oC. If we start with 0.0200 M
CO and 0.0200 M H2O, what are the final eq concs?
0.020
set up summary of conditions under equation
Kc =
[ CH4 ][ H2O]
= 3.92
3
[ CO][ H2 ]
solve for x
Kc =
[CH4 ][ H2O] [CH4 ][ 0.020]
=
= 3.92
3
3
[ CO][ H2 ]
[ 0.30][ 010
. ]
substitute for each component to find eq. conc.
[CH ] =
4
( 3.92)( 0.30)( 010
. )
= 0.059 M
( 0.020)
3
(c) and finally the whole bit
0 = 0.920 x2 - 3.00 x + 2.00
o
Consider @ 458 C
H2
i
1.00
*
-x
+
I2
2 HI
+ bx + c
Kc = 49.7
2.00
0.00
Use the quadratic expression to solve for x
-x
2x
x = 1.63 ± 0.70
eq (1.00-x) (2.00-x)
Kc =
º
0 = a x2
2x
[HI ]2
[2 x ]2
=
= 49.7
[ H2 ][ I2 ] [100
. − x ][2.00 − x ]
x = 2.33 (impossible)
or
0.93
solve for each of the concentrations
4
Predicting the Direction of Reaction
• We define Q, the reaction quotient, for a
general reaction
as
aA + bB(g)
Q=
pP + qQ
[P ] p [Q]q
[A]a [B]b
where [A], [B], [P], and [Q] are molarities at
any time.
• Q = K only at equilibrium.
Predicting the Direction of Reaction
• If Q > K then the reverse reaction must occur
to reach equilibrium (i.e., products are
consumed, reactants are formed, the
numerator in the equilibrium constant
expression decreases and Q decreases until
it equals K).
• If Q < K then the forward reaction must occur
to reach equilibrium.
Use of Q in comparison to Kc
A 50.0 L vessel contains 1.00 mol N 2, 3.00 mol
H2, and 0.500 mol NH3 at 500 oC where the
equilibrium constant Kc = 0.500 for the reaction
N2(g) + 3 H 2(g) º 2 NH3(g). Will the amount of
NH3 remain the same, decrease, or increase?
Write the expression for Q and solve for its
value.
Compare it to Kc.
recall that if Q > K c reaction goes to left
Q < K c reaction goes to right
Q = K c reaction is at equilibrium
Le Châtelier’s Principle
• Consider the production of ammonia
N2(g) + 3H2(g)
2NH3(g)
• As the pressure increases, the amount of
ammonia present at equilibrium increases.
• As the temperature decreases, the amount of
ammonia at equilibrium increases.
• Can this be predicted?
• Le Châtelier’s Principle: if a system at
equilibrium is disturbed, the system will move
in such a way as to counteract the
disturbance.
5
Change in Reactant or Product Concentrations
• Consider the Haber process
N2(g) + 3H2(g)
2NH3(g)
• If H2 is added while the system is at equilibrium, the
system must respond to counteract the added H2 (by
Le Châtelier).
• That is, the system must consume the H2 and produce
products until a new equilibrium is established.
• Therefore, [H2] and [N2] will decrease and [NH3]
increases.
• Notice ‘swamping’ effect
Change in Reactant or Product
Concentrations
• N2 and H2 are pumped into a chamber.
• The pre-heated gases are passed through a
heating coil to the catalyst bed.
• The catalyst bed is kept at 460 - 550 °C
under high pressure.
• The product gas stream (containing N2, H2
and NH3) is passed over a cooler to a
refrigeration unit.
• In the refrigeration unit, ammonia liquefies but
not N2 or H2.
Effects of Volume and Pressure on Gaseous Rxns
Change in Reactant or Product
Concentrations
• The unreacted nitrogen and hydrogen are recycled
with the new N2 and H2 feed gas.
• The equilibrium amount of ammonia is optimized
because the product (NH3) is continually removed
and the reactants (N2 and H2) are continually being
added.
• As volume is decreased pressure increases.
• Le Châtelier’s Principle: if pressure is increased the
system will shift to counteract the increase.
• That is, the system shifts to remove gases and
decrease pressure.
• An increase in pressure favors the direction that has
fewer moles of gas.
• In a reaction with the same number of product and
reactant moles of gas, pressure has no effect.
• Consider
N2 O4 (g)
2NO 2 (g)
6
Effects of Volume and Pressure
Effect of Temperature Changes
• An increase in pressure (by decreasing the volume)
favors the formation of colorless N2O4.
• The equilibrium constant is temperature
dependent.
• For an endothermic reaction, ∆H > 0 and heat can
be considered as a reactant.
• For an exothermic reaction, ∆H < 0 and heat can
be considered as a product.
• Adding heat (i.e. heating the vessel) favors away
from the increase:
– if ∆H > 0, adding heat favors the forward
reaction,
– if ∆H < 0, adding heat favors the reverse
reaction.
• The instant the pressure increases, the system is
not at equilibrium and the concentration of both
gases has increased.
• The system moves to reduce the number moles of
gas (i.e. the forward reaction is favored).
• A new equilibrium is established in which the
mixture is lighter because colorless N2O4 is favored.
Effect of Temperature Changes
• Removing heat (i.e. cooling the vessel),
favors towards the decrease:
– if ∆H > 0, cooling favors the reverse reaction,
– if ∆H < 0, cooling favors the forward reaction.
• Consider
Cr(H2O)6(aq) + 4Cl-(aq)
CoCl42-(aq) + 6H2O(l)
for which ∆H > 0.
– Co(H2O) 62+ is pale pink and CoCl 42- is blue.
Effect of Temperature Changes
Cr(H2O)6 (aq) + 4Cl-(aq)
CoCl42-(aq) + 6H2O(l)
– If a light purple room temperature equilibrium mixture is
placed in a beaker of warm water, the mixture turns deep
blue.
– Since ∆H > 0 (endothermic, heat is a reactant ), adding
heat favors the forward reaction, i.e. the formation of blue
CoCl42-.
– If the room temperature equilibrium mixture is placed in a
beaker of ice water, the mixture turns bright pink.
– Since ∆H > 0, removing heat favors the reverse reaction
which is the formation of pink Co(H2O)62+ .
The Effect of Catalysts
• A catalyst lowers the activation energy barrier for
the reaction.
• Therefore, a catalyst will decrease the time taken to
reach equilibrium.
• A catalyst does not effect the composition of the
equilibrium mixture.
7
ln K =
ln
K T2
K T1
=
− ∆H
+ Cnst
RT
− ∆H  1 1 
 − 
R  T2 T1 
The van’t Hoff Equation -- (not his i-factor)
8