Transcript Physical Properties - Winthrop University

Chapter 9: Chemical Equilibrium
The forward and reverse reaction are both taking place
at the same rate
Production and Decomposition of
Ammonia
Forward Reaction: N2 (g) + 3H2 (g)  2NH3 (g)
Reverse Reaction: 2NH3 (g)  N2 (g) + 3H2 (g)
Equilibrium Reaction: N2 (g) + 3H2 (g)  2NH3 (g)
Note the double headed arrow!
The ammonia is decomposing as fast as it is
being made at equilibrium
Equilibrium and the Law of Mass Action
2SO2 (g) + O2 (g)  2SO3 (g)
5 mixtures of different initial compositions of gases were
made and allowed to reach equilibrium at 1000K
At first, you don’t see a trend in the data…
Equilibrium and the Law of Mass Action
No trends, but if you calculate:
PSO3 
2
  
PSO3 

 P 
K
or K=
2
PSO2 2PO2 
PSO2  PO2 

     
 P   P 
2
You get the same
value, regardless of
initial concentration
Note: K is unitless!
The Equilibrium Constant
Product Concentration

n
Reactant Concentration

n
K =

K is the equilibrium constant for
the reaction
The Equilibrium Constant
At equilibrium, the composition of the
reaction mixture can be expressed in
terms of an equilibrium constant where:
Product Concentration

n
Reactant Concentration

n
K =

For ideal gases, the concentrations are the
partial pressures of the individual gases
For solutions, the concentrations are the molar
values of the individual atoms/ions/molecules
Examples of K setup
aA (g) + bB (g)  cC (g) + dD (g)
PC  PD 

K
a
b
Pa  PB 
c
d

Units and Equilibrium Constants
When working with equilibrium Constants,
we’ll use the following unit
conventions:
1. Gases: Units are bar
2. Aqueous Solutions: Unit is Molarity
3. Solids: The number 1
Solids have a single value (1) because the
concentration of a solid doesn’t change.
Thermodynamic Origin of Equilibrium
Constants
The Free Energy changes
as the composition of
the reaction mixture
changes
• All reactions will
proceed towards
equilibrium (by either
forward or reverse
reaction)
 Gº is the free energy
difference b/w the pure
products and pure
reactants
Thermodynamic Origins of Equilibrium
Constants
• We can calculate the Free Energy change at
any point along the reaction coordinate with
the equation
aA (g) + bB (g)  cC (g) + dD (g)
c
d 

[C]
[D]
G r  Gr  RT ln a b 
[A] [B] 
or
G r  Gr  RT lnQ
Gr° is the textbook Free Energy of reaction
Gr is the Free
 Energy of value when the reactants and products
are at particular concentrations

Example:
The standard free energy of reaction for:
2SO2 (g) + O2 (g)  2SO3 (g)
Is Gr°= -141.74 kJ/mole at 25°C.
What is the Gibbs Free Energy of
reaction when the partial pressure of
each gas is 100.0 bar?
Example:
The Standard Gibbs Free Energy of Reaction
for
N2O4 (g) --> 2NO2 (g)
Is Gr° = +4.73 kJ/mole at 298K. What is the
value of Gr when the partial pressures are
PN2O4 = 0.8 bar and PNO2 = 2.10 bar?
Free Energy of a Reaction at
Equilibrium
• Q=K at equilibrium
• At equilibrium, G=___
• Therefore,
G = Grº + RTlnK
Grº = -RTlnK (only at equilibrium)
• We can use this to compute equilibrium
constants from Grº values
K and the Extent of Reactions
• When K is very
large, the reaction
favors the products
• When K is very
small, the reaction
favors the reactants
• When K=1, the
reaction is neither
reactant nor
product favored
(Equilibrium)
The Direction of Reaction
How can we tell if a reaction will
continue towards the products or
back towards the reactants at a
given point along the reaction
coordinate?
Q = Reaction quotient used at
any point in the coordinate
K = Equilibrium constant
When Q<K, G is negative (product favored)
When Q=K, G = 0
When Q>K, G is positive (reactant favored)
Equilibrium Calculations
Toolbox 9.1: Know it. Love it. Use it.

Example:
Under certain conditions, nitrogen and oxygen
react to form dinitrogen oxide, N2O. Suppose
that 0.482 moles of N2 and 0.933 moles of O2
are transferred to a reaction vessel of volume
10.0L and allowed to form N2O @ 800K. At
this temperature, K=3.2x10-28 for the reaction:
2N2 (g) + O2 (g)  2N2O (g)
What are the partial pressures of the gases at
equilibrium?
Example:
Chlorine and fluorine react at 2500K to produce
ClF and reach the equilibrium:
Cl2 + F2  2ClF
With an equilibrium constant value of 20. If a
gaseous mixture of 0.2 bar Cl2, 0.1 bar F2 and
0.1 bar ClF is allowed to reach equilibrium,
what is the partial pressure of ClF in the
mixture?
LeChatelier’s Principle
When the equilibrium
composition is perturbed
by adding or removing a
reactant of product, the
reaction tends to proceed
in the direction that brings
Q closer to that of K.

Consider the Equilibrium Reaction:
4NH3 (g) + 3O2 (g)  2N2 (g) + 6H2O (g)
What would result from the:
a) Addition of N2
b) Removal of NH3
c) Removal of H2O
Effects of the Environment on Equilibria
• Compressing a Gas Phase Reaction
– The reaction shifts so as to decrease the pressure
• Decrease the number of gas molecules
• Changing the Temperature of a Reaction
– For exothermic reactions, lowering the
temperature causes a shift towards the products
– For endothermic reactions, increasing the
temperature causes a shift towards the products