equilibrium.ppt

							Chemical Equilibrium
      Chemistry.
      Ms. Siddall
                  Standard 9b: equilibrium conditions

    Reversible Reactions
 Most reactions are ‘reversible’
 Forward reaction: reactants make products
                 e.x. 3O2(g)  2O3(g)
 Reverse reaction: products make reactants
                 e.x. 2O3(g)  3O2(g)
 In a reversible reaction the forward and reverse
  reactions occur at the same time
                 e.x. 3O2(g)  2 O3(g)
               Summary 1
 What   is a reversible reaction?
   Reversible reactions reach
    equilibrium: a balance between reactants
    and products

Conditions of Equilibrium:
1. rate of forward reaction = rate of reverse
   reaction
2. Concentration of reactants and products is
   constant (does not change)
NOTE:
Rate = speed
Concentration = number of particles or moles
example: [HCl] = concentration of HCl
            6M HCl = 6mole/L HCl
            = 6 moles of HCl per liter of solution
               Summary 2
 What   is equilibrium?
                 Reversible reaction: X  Y
                [X]                         XY
concentration




                                    rate
                 [Y]                        YX


                       time                        time
          Concentrations               Reaction rates are equal
          are constant
                              equilibrium
              Summary 3
 Describe  the part of each graph that
 illustrates equilibrium conditions.
                           9a: Le Chatelier’s Principle

  Le Chatelier’s Principle
A system in equilibrium will react to relieve
 stress (change) and re-establish equilibrium

 Stress:
     Adding reactants or products
     Removing reactants or products
     Changing temperature
     Changing pressure (for gases only)
              Summary 4
 Accordingto Le Chatelier’s Principle, what
 will happen to a system at equilibrium if
 more reactants or products are added?
                Example:
      N2(g) + 3H2(g)  2NH3(g) + heat

 Stress:  Add N2
 Stress relief:
   Forward reaction (→) to get rid of N2

   H2 is used up (↓)

   NH3 and Heat are produced (↑)
             Summary 5:
    N2(g) + 3H2(g)  2NH3(g) + heat

 Stress: remove N2
 Stress relief:
   Which way does equilibrium shift?

   What happens to [H2]?

   What happens to [NH3]?

   What happens to heat?
 Stress relief.
 Adding    products or reactants
     Equilibrium shifts to remove addition
 Removing     products or reactants
     Equilibrium shifts to replace what has been
      removed
 Gasses
     Equilibrium shifts to produce:
       • more gas at low pressure
       • Less gas at high pressure
            Summary 6
 Why would a gas equilibrium system
 produce more gas at low pressure and
 less gas at high pressure?
Haber Process: N2(g) + 3H2(g)  2NH3(g) + heat

Change    Equilib.   [N2]      [H2]    [NH3]

 ↑ [N2]
 ↓ [N2]
 ↑ [H2]
 ↓ [H2]
↑ [NH3]
↓ [NH3]
Haber Process: N2(g) + 3H2(g)  2NH3(g) + heat

Change      Equilib.   [N2]   [H2]     [NH3]

 ↑ heat

 ↓ heat

↑pressure

↓pressure
                 Summary 7
 According to Le Chatelier’s Principle:
 Increasing reactant concentration will
  cause:
     other reactants to __________?
     products to __________?
 Decreasing    reactant concentration will
 cause:
     other reactants to __________?
     products to __________?
A(g) + B(g)  AB(g) + heat
Summary 8                      A(g) + B(g)  AB(g) + heat

    Complete the table of
     equilibrium changes

Change      Equilib.     [A]           [B]        [AB]

 ↑ heat
 ↓ [A]
↑pressure
 ↓[AB]
         A(g) + B(g)  AB(g) + heat

Change   Equilib.   [A]       [B]     [AB]

↑ [A]
↓ [A]
↑ [B]
↓ [B]
↑ [AB]
↓ [AB]
            A(g) + B(g)  AB(g) + heat

Change      Equilib.   [A]       [B]     [AB]

 ↑ heat

 ↓ heat

↑pressure

↓pressure
HONORS Standard 9c: equilibrium constant



       Equilibrium Constant: Keq
 Atequilibrium concentrations are constant
 Keq represents concentrations of reactants
  and products at equilibrium
 Example: aA + bB  cC + dD
 Keq = [C]c[D]d
        [A]a[B]b
            Summary 9
      Keq expression for the Haber-Bosch
 write
 Process: N2(g) + 3H2(g)  2NH3(g)
 Concentrations   calculated in mol/L (M)
 Only solutions(aq) & gases(g) are considered
   No solids (s)

   No liquids (l)




Example: 2H2O(l)  2H2(g) + O2(g)

 Keq = [H2]2[O2]
                 Summary 10
   Fe(OH)2(aq) + 2HSO3(aq)  Fe(SO3)2(aq) + 2H2O(l)
 Find   Keq
What Keq tells us

 If Keq ≤ 1 There are more reactants than
  products at equilibrium
 If Keq ≤ 1/100 There are mostly reactants at
  equilibrium
 If Keq ≥ 1 There are more products than
  reactants at equilibrium
 If Keq ≥ 100 There are mostly products at
  equilibrium
                Summary 11
CO(g) + 2H2(g)  CH3OH(g) Keq=290 at 430°C

 Write   the expression for Keq

 Reaction   is…
  (mostly products or reactants?)
                 Solubility
 Ksp is the equilibrium constant for solubility
 Example: AgCl(s)  Ag+(aq) + Cl-(aq)
 Ksp AgCl = 1.77 x 10-10
   Does not really dissolve, mostly solid

 Example: AgNO3(s)  Ag+(aq) + NO3-(aq)
 Ksp AgNO3 ~ 1 x 1010
   Very soluble
             Summary 13
 Write the balanced equation for the
  dissolving of sodium sulfate.
 Write a Ksp expression for the reaction.