Transcript Document 7926098

Equilibrium Basic Concepts
• Reversible reactions do not go to completion.
– They can occur in either direction

a Ag  + b Bg   c Cg  + d Dg 
• Chemical equilibrium exists when two opposing reactions
occur simultaneously at the same rate.
– A chemical equilibrium is a reversible reaction that the
forward reaction rate is equal to the reverse reaction rate.
• Chemical equilibria are dynamic equilibria.
– Molecules are continually reacting, even though the overall
composition of the reaction mixture does not change.
The Equilibrium Constant
• Kc is the equilibrium constant .
• Kc is defined for a reversible reaction at a given temperature
as the product of the equilibrium concentrations (in M) of the
products, each raised to a power equal to its stoichiometric
coefficient in the balanced equation, divided by the product of
the equilibrium concentrations (in M) of the reactants, each
raised to a power equal to its stoichiometric coefficient in the
balanced equation.
• Equilibrium constants are dimensionless because they actually
involve a thermodynamic quantity called activity.
– Activities are directly related to molarity
The Equilibrium Constant
Basic Concepts
• One of the fundamental ideas of chemical equilibrium is that
equilibrium can be established from either the forward or reverse
direction.

A(g)  B(g)  C(g)  D(g)
• The rates of the forward and reverse reactions can be represented as:
Rate f  k f A B which represents the forward rate.
Rate r  k r CD which represents the reverse rate.
• When system is at equilibrium:
Ratef = Rater
The Equilibrium Constant
• Write equilibrium constant expressions for the following
reactions at 500oC. All reactants and products are gases at
500oC.

PCl5  PCl3  Cl2

PCl3 Cl2 
Kc 
PCl5 
The Equilibrium Constant

H 2 + I 2  2 HI

HI 
Kc 

H2 I 2 
4 NH + 5 O  4 NO + 6 H O
2
3
2

NO H 2 O
Kc =
4
5
NH3  O 2 
4
6
2
The Equilibrium Constant
• One liter of equilibrium mixture from the following system at a
high temperature was found to contain 0.172 mole of
phosphorus trichloride, 0.086 mole of chlorine, and 0.028 mole
of phosphorus pentachloride. Calculate Kc for the reaction.

PCl5  PCl3  Cl2
• Equil []’s
0.028 M
0.172 M
0.086 M
The Equilibrium Constant
• At a given temperature 0.80 mole of N2 and 0.90 mole of H2 were introduced
in an evacuated 1.00-liter container after equilibrium was established the
equilibrium concentrations were determined top be 0.20 mole of NH3, 0.70
moles N2, and 0.06 moles H2. Calculate Kc for the reaction.
The Equilibrium Constant
Product-favored
K>1
3
Products favored10 >
Reactant-favored
K<1
K>
-3
10 Reactants favored
Variation of Kc with the
Form of the Balanced Equation
• The value of Kc depends upon how the balanced equation is
written. From the prior examples we have:
PCl5
Equil []’s 0.028 M
PCl3 + Cl2
0.172 M
0.086 M
Kc = [PCl3][Cl2]/[PCl5] = 0.53
PCl3 + Cl2
Equil. []’s 0.172 M 0.086 M
PCl5
0.028 M
Kc’ = [PCl5]/[PCl3][Cl2] = 1.89
PCl3 + Cl2
Equil. []’s 0.172 M
0.086 M
PCl5
0.028 M
Kc “ = [PCl5]2/[PCl3]2[Cl2]2 = 3.56
2PCl5
Equil []’s 0.028 M
2PCl3 + 2Cl2
0.172 M
0.086 M
Kc ’’’ = [PCl3] 2[Cl2] 2/[PCl5] 2 = 0.28
Partial Pressures and the Equilibrium
Constant
• For gas phase reactions the equilibrium constants can be
expressed in partial pressures rather than concentrations.
• For gases, the pressure is proportional to the concentration.
• We can see this by looking at the ideal gas law.
– PV = nRT
– P = nRT/V
– n/V = M
– P= MRT and M = P/RT
Partial Pressures and the Equilibrium Constant
• Consider this system at equilibrium at 5000C.
2SO2(g) + O2(g)
2SO3(g)
Relationship Between Kp and Kc
• Kc is 49 for the following reaction at 450oC. If 1.0 mole of H2 and
1.0 mole of I2 are allowed to reach equilibrium in a 3.0-liter
vessel,

H 2g   I 2g   2 HI g 
(a) How many moles of I2 remain unreacted at equilibrium?
(b) What are the equilibrium partial pressures of H2, I2 and HI?
(c) What is the total pressure in the reaction vessel?

H 2g   I 2g   2 HI g 
(a) How many moles of I2 remain unreacted at equilibrium?

H 2g   I 2g   2 HI g 
(b) What are the equilibrium partial pressures of H2, I2 and HI?

H 2g   I 2g   2 HI g 
(c) What is the total pressure in the reaction vessel?
Relationship Between Kp and Kc
•
Nitrosyl bromide, NOBr, is 34% dissociated by the following reaction at 25oC, in a vessel
in which the total pressure is 0.25 atmosphere. What is the value of Kp?
2 NOBrg  
 2 NOg  + Br2g 
Heterogeneous Equlibria
• Heterogeneous equilibria have more than one phase present.
– For example, a gas and a solid or a liquid and a gas.
CaCO3s  
 CaOs   CO2g 
•
o
at 500 C
How does the equilibrium constant differ for heterogeneous equilibria?
– Pure solids and liquids have activities of unity.
– Solvents in very dilute solutions have activities that are essentially unity.
– The Kc and Kp for the reaction shown above are:
K c = [CO 2 ]
K p = PCO 2
Heterogeneous Equlibria
For this reaction :
 H SO
SO 2aq   H 2 O   
2
3aq 
(at 25 C)
H 2 O   is the solvent.
What are the forms of K c and K p?
You do it!
o
Heterogeneous Equlibria
• What are Kc and Kp for this reaction?
2
1
CaF2s   Ca aq   2 Faq 
o
(at 25 C)
Heterogeneous Equlibria
• What are Kc and Kp for this reaction?

3 Fes   4 H 2Og   Fe3O4s   4 H 2g 
o
(at 500 C)
An aside: Similar to Hess’ Law we can
add several reactions together to get to an
overall reaction not listed in a table.
To determine K for the sum of the
reactions simply multiply all the values
for each intermediate step to get the K for
the overall reaction.
Uses of the Equilibrium Constant, Kc
• The equilibrium constant, Kc, is 3.00 for the following reaction at
a given temperature. If 1.00 mole of SO2 and 1.00 mole of NO2
are put into an evacuated 2.00 L container and allowed to reach
equilibrium, what will be the concentration of each compound at
equilibrium? SO
 NO 
 SO
 NO
2(g)
2(g)
3(g)
(g)
Uses of the Equilibrium Constant, Kc
•
The equilibrium constant is 49 for the following reaction at 450oC. If 1.00 mole of HI is put
into an evacuated 1.00-liter container and allowed to reach equilibrium, what will be the
equilibrium concentration of each substance?
 2 HI
H 2 g   I 2 g  
g 
The Reaction Quotient
• The mass action expression or reaction quotient has the
symbol Q.
– Q has the same form as Kc
• The major difference between Q and Kc is that the
concentrations used in Q are not necessarily equilibrium
values.
• Why do we need another “equilibrium constant” that does not
use equilibrium concentrations?
• Q will help us predict how the equilibrium will respond to an
applied stress.
• To make this prediction we compare Q with Kc.
The Reaction Quotient
• The equilibrium constant for the following reaction is 49 at 450oC.
If 0.22 mole of I2, 0.22 mole of H2, and 0.66 mole of HI were put
into an evacuated 1.00-liter container, would the system be at
equilibrium? If not, what must occur to establish equilibrium?
Disturbing a System at
Equlibrium: Predictions
•
•
LeChatelier’s Principle - If a change of conditions (stress) is
applied to a system in equilibrium, the system responds in the
way that best tends to reduce the stress in reaching a new state
of equilibrium.
– We first encountered LeChatelier’s Principle in Chapter 14.
Some possible stresses to a system at equilibrium are:
1. Changes in concentration of reactants or products.
2. Changes in pressure or volume (for gaseous reactions)
3. Changes in temperature.
Disturbing a System at Equlibrium: Predictions
1
Changes in Concentration of Reactants and/or Products
• Also true for changes in pressure for reactions involving
gases.
Disturbing a System at Equlibrium: Predictions
2 Changes in Volume
• (and pressure for reactions involving gases)
– Predict what will happen if the volume of this system at
equilibrium is changed by changing the pressure at constant
temperature:
Disturbing a System at
Equlibrium: Predictions
3 Changing the Reaction Temperature
Disturbing a System at Equlibrium: Predictions
• Introduction of a Catalyst
– Catalysts decrease the activation energy of both the forward and reverse
reaction equally.
• Catalysts do not affect the position of equilibrium.
– The concentrations of the products and reactants will be the same
whether a catalyst is introduced or not.
– Equilibrium will be established faster with a catalyst.
Disturbing a System at Equlibrium: Predictions
• Given the reaction below at equilibrium in a closed container at
500oC. How would the equilibrium be influenced by the
 2 NH H o  92 kJ/mol
following? N  3 H 
2(g)
2(g)
3(g)
rxn
Factor
Effect on reaction
a. Increasing the reaction t emperature

b. Decreasing the reaction t emperature

c. Increasing the pressure by decreasing the volume 
d. Increase the concentrat ion of H 2

e. Decrease the concentrat ion of NH3

f. Introducin g a platinum catalyst
no
Disturbing a System at
Equlibrium: Predictions
• How will an increase in pressure (caused by decreasing the
volume) affect the equilibrium in each of the following reactions?
Reaction
Effect on Equilibriu m
 2 HI
a. H 2g  + I 2g  
g 
 4 NO + 6 H O
b. 4 NH3g  + 5 O 2(g) 
g 
2 g 
c. PCl + Cl 
 PCl
 left
2H O
d. 2 H 2g   O 2g  
2 g 
 right
3 g 
2 g 
5 g 
no effect
 right
Disturbing a System at
Equlibrium: Predictions
• How will an increase in temperature affect each of the
following reactions?
Reaction
Effect on Equilibriu
o

a. 2 NO2(g)  N 2O 4(g) H rxn  0
 2 HCl + 92 kJ
b. H  Cl 
2 g 
2 g 
g 
 left
 left
c. H 2g  + I 2g   2 HI g  H  25 kJ  right
m
Disturbing a System at Equilibrium:
Calculations
• A 2.00 liter vessel in which the following system is in
equilibrium contains 1.20 moles of COCl2, 0.60 moles of CO
and 0.20 mole of Cl2. Calculate the equilibrium constant.
 COCl
COg   Cl2g  
2 g 
Disturbing a System at Equilibrium:
Calculations
• An additional 0.80 mole of Cl2 is added to the vessel at the same
temperature. Calculate the molar concentrations of CO, Cl2, and
COCl2 when the new equilibrium is established.
The Haber Process: An Application of Equilibrium
•
The Haber process is used for the commercial production of ammonia.
– This is an enormous industrial process in the US and many other countries.
– Ammonia is the starting material for fertilizer production.
Fritz Haber
1868-1934
Nobel Prize, 1918
Carl Bosch
1874-1940
Nobel Prize, 1931
∆G, ∆G˚, and Keq
• ∆G is change in free energy at non-standard
conditions.
• ∆G is related to ∆G˚
• ∆G = ∆G˚ + RT ln Q
where Q = reaction quotient
• When Q < K or Q > K, reaction is spontaneous.
• When Q = K reaction is at equilibrium
• When ∆G = 0 reaction is at equilibrium
• Therefore, ∆G˚ = - RT ln K
Relationship Between Gorxn and the
Equilibrium Constant
• The relationships among Gorxn, K, and the spontaneity of a
reaction are:
Gorxn
<0
K
>1
Spontaneity at unit concentration
Forward reaction spontaneous
=0
=1
System at equilibrium
>0
<1
Reverse reaction spontaneous
∆G, ∆G˚, and Keq
Product Favored, ∆G˚ negative, K > 1
But systems can reach
equilibrium when
reactants have NOT
converted completely
to products.
In this case ∆Grxn is <
∆Gorxn , so state with
both reactants and
products present is
MORE STABLE
than complete
conversion.
∆G, ∆G˚, and Keq
•
•
•
•
Product-favored
2 NO2 ---> N2O4
∆Gorxn = – 4.8 kJ
State with both reactants and
products present is more stable
than complete conversion.
• K > 1, more products than
reactants.
∆G, ∆G˚, and Keq
• Reactant-favored
• N2O4 --->2 NO2
∆Gorxn = +4.8 kJ
• State with both reactants and
products present is more stable
than complete conversion.
• K < 1, more reactants than
products
Thermodynamics and Keq

Keq is related to reaction favorability.

When ∆Gorxn < 0, reaction moves energetically “downhill”

∆Gorxn is the change in free energy when reactants convert
COMPLETELY to products.
Relationship Between Gorxn and the
Equilibrium Constant
• The relationship for K at conditions other than thermodynamic
standard state conditions is derived from this equation.
G  G  RT lnQ
or
o
G  G o  2.303 RT log Q