Notes
Outline
Molecular Equilibria: Gas Systems
Dynamic Equilibrium
Dynamic Equilibria-Two opposing prosesses that occur simultaneous at equal rates
aA  +  bB = cC
Rf = k1 [A]a [B]b
Rr = k-1[C]c
At equilibrium:
Rf = Rr
k1[A]a [B]b = k-1[C]c
K1 /k-1 = [C]c / [A]a [B]b
Kc =K-1 /k1 = [C]c / [A]a [B]b
T-148 conc vs time and rate vs time graph of equilibrium
T-149 The Haber Process
T-152 Haber Reactor
Law of Chemical Equilibrium
aA(g)  +  bB(g)  =  pP(g)  +  qQ(g)
Based on Molar Concentration
Kc = [P]p [Q]q  /  [A]a [B]b
Based on Partial Pressures
Kp = Pp Pq  /  Pa Pb
Kc and Kp remains constant at a fixed temperature
T-150 Equilibrium Data for NO2/ N2O4 System
Writing Equilibrium Expressions
3H2(g)  +  N2(g) = 2NH3(g)
Kc = [NH3]2 / [N2] [H2]3
Kp = (PNH3)2 / (PH2)3 PN2
Magnitude of Equilibrium Constant
K increases then components on the right side predominate in the equilibrium mixture
K decreases then the components on the left side predominate in the equilibrium mixture
Example
Reversed Equilibria
aA(g) = bB(g)  K1 = [B]b / [A]a
bB(g) = aA(g)  K2 = [A]a / [B]b
Therefore: K1 = 1 / K2
Heterogeneous Equilibria
Heterogeneous Equilibrium –equilibria with two or more physical states represented.
Liquid and solid states do not affect equilibria because of incompressibility
Examples
CaCO3(s) = CaO(s)  +  CO2(g)
Kc = [CO2]
Kp = PCO2
FeO(s)  +  H2(g) = Fe(s)  +  H2O(g)
Kc = [H2O] / [H2]
Kp = PH2O / PH2
Determining the Equilibrium Constant
Given the equilibrium concentrations(or Partial Pressures) of all components
Given the initial concentration (or partial pressure)  of components
Relationship Between Kc and Kp
Kp = Kc (RT) Dn
where Dn = mols of gaseous component on right – mols of gaseous components on left
Example:
Predicting The Direction A Reaction Must Proceed To Reach Equilibrium
For H2(g)  +  I2(g) = 2HI(g)
Q = quotient = [HI]2 / [H2] [I2]
If Kc = Q Rx in Equilibrium
If Kc < Q then Rx must proceed Left (favor reverse reaction) to reach equilibrium (Kc = Q)
If Kc > Q then Rx must proceed to Right (favor forward reaction) to reach equilibrium
Example
Determining Equilibrium Concentrations
Given the equilibrium concentrations(or partial pressures) of all components but one and the Kc(or Kp)
Given the initial concentration (or partial pressures) of the reactants and the Kc(or Kp)
LeChatlier’s Principle
When an external stress  is applied to a system in equilibrium, the equilibrium is upset and will respond by shifting in such a way as to undo the stress applied.
External Stresses
Change in the concentration of the components (adding or removing)
Change in the volume of the container (alters pressures of reacting gases)
Change in the temperature
Changing The Concentration of A Component
Increasing the concentration of a component (adding component) of a component drives the reaction to the opposite side (uses up the component and relieves the stress)
Decreasing the Concentration of a component(removing component) drives the reaction to the same side (produces more component thereby relieving the stress by replacing what was removed)
Examples
T-151  Changing Concentrations In An Equilibrium
Changing The Volume of The Container
Increasing the volume of a container decreases the pressures of the gaseous components(Boyles Law) which drives the reaction toward the side with the most total mols of gas(brings pressure back up)
Decreasing the volume of the container increases the pressure of the gaseous components (Boyles Law) which drives the reaction toward the side with the least total mols of gas (brings pressure back down)
Examples
T-14.7 Pressure Effect On Equilibrium
Temperature Effect On Equilibrium
Increasing the temperature will always favor the endothermic process (uses up energy added by increased temperature)
Decreasing the temperature will always favor the exothermic process(replaces the energy removed by stress)
Examples
T-14.8 Effect on Equilibrium on Changing The Temperature
Catalytic Effect On Equilibrium
Catalyst alters the Energy of Activation of both processes
Therefore equilibrium is not changed