Chemical equilibrium occurs when opposing reactions are occuring at equal rates.  Consider the following reaction
                                                   N₂O₄(g)     ==  2NO₂(g)
                                                     Colorless           Brown
When a mixture of these two gases are confined to a container, colorless N₂O₄ will break down to form brown NO₂.  However a time will occur when the concentration of N₂O₄ and NO₂ will remain constant over an indefinite period of time - a dynamic equilibrium will be established.  The rate at which is NO₂ formed will equal the rate at which N₂O₄ breaks down.  If the reaction vessel appears to be a dark brown then there are mostly NO₂ molecules present and the equilibrium is said to lie to the right.  If, however, the vessel appears to be a very light brown then there are mostly N₂O₄ molecules and the equilibrium lies to the left.

Now consider a reaction between A (red molecules) and B (yellow molecules) to form C (green molecules) and D (blue molecules)
                                               A  +  B  =  C  +  D

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 The rate constants determine in which direction the equilibrium lies.
              If  k_f  > k_r  then the equilibrium lies to the right
              If  k_f  < k_r  then the equilibrium lies to the left

 The equilibrium constant is defined as K = k_f/ k_r  = [C][D]/[A][B]
              If concentrations of products > reactants - to the right
              If concentrations of products < reactants - to the left

 Another reaction called the Haber process:

                 N₂  +  3H₂  =  2NH₃     K =  [NH₃]²/[ N₂][ H₂]³

Consider the following aspects of chemical equilibrium
          1. Concept of dynamic equilibrium

          2. Disturbing an equilibrium

          3. Equilibrium constant

1. Concept of dynamic equilibrium
          Systems involving simultaneous reversible reactions will come to a
          state of dynamic equilibrium in which:

                    (a) the concentration of reactants and products remain constant

                    (b) there is continual interchange between reactants and
                    products so that:

                       rate of forward reaction = rate of reverse reaction

2. Disturbing an equilibrium

    If an equilibrium is upset, it will automatically re-establish itself - Le Chatelier's principle

    Concentration, volume, pressure & temperature changes can affect the position of an equilibrium

 Specifically:
    a.  Addition of a reactant, that is increasing the concentration of a reactant
    b.  Removal of a product, that is lowering the concentration of a product
    c.  Addition of a product
    d.  Changing the pressure of a gas phase reaction
    e.  Changing the temperature.

Le Chatelier's principle: If a system at equilibrium is disturbed, then the system adjusts itself so as to minimize the disturbance

    a. If a reactant or product is added to a solution or gas phase reaction mixture at equilibrium, then the equilibrium moves in the direction which consumes some of the added substance.
    b. If the pressure on a reaction system is increased, the equilibrium moves in the direction which tends to reduce pressure; this is the direction which corresponds to a decrease in the number of moles of gas
    c. If the temperature of a reaction mixture at equilibrium is increased, the equilibrium moves in the direction which absorbs heat. Exothermic reactions move to the left as temperature increases; endothermic to the right.

Forcing a reversible reaction to completion:
     Removal of a product as it forms, and use of excess of the common of cheap reactant, are the most usual methods of forcing reversible reactions to completion.

Catalysts alter the time taken to reach equilibrium, but do not alter the equilibrium position.

3. Equilibrium constant is expressed as a function of the molar concentrations of the reactants & products at equilibrium

 Define the equilibrium constant, K:

                                      aA + bB  =  cC + dD

     Any chemical reaction where A and B are reactants, C and D are products (molecules, ions or atoms); a, b, c, and d are the number of each species involved. When this reaction is at equilibrium, the expression
                                            [C]^c[D]^d / [A]^a[B]^b
has a constant value, regardless of the starting concentrations of the substances involved. The constant is called the equilibrium constant, K.

     Reaction quotient, Q: A reaction is at equilibrium if its reaction quotient, Q, is equal to the equilibrium constant, K.

                              If Q < K, reaction goes from left to right until Q = K.

                              If Q > K, reaction goes from right to left until Q = K.

                              If Q = K, reaction is at equilibrium.

Rules & units for equilibrium constants

  1.  The reaction quotient (and hence the equilibrium expression) is always written with products (right-hand side) in the numerator (top line) and reactants (left-hand side) in the denominator (bottom line)
  2.  The equilibrium expression always uses the coefficients of the reaction as written
  3.  The equilibrium constant remains the same (at constant temperature) even if the equilibrium concentrations of reactants or products are varied. Equilibrium constants may increase or decrease with a change in temperature

Equilibrium constants and Le Chatelier's principle

     At constant temperature, pressure is proportional to concentration
     Increasing the pressure is equivalent to increasing the concentration of all substances present.
     Decreasing the pressure corresponds to decreasing all the concentrations.
     If DH is negative, K decreases as temperature increases.
     If DH is positive, K increases as temperature increases.

Expressing K in terms of pressure for N₂  +  3H₂  =  2NH₃
                                                                                            K_p =  P_NH3²/P_N2 P_H2³

Heterogeneous Equilibrium
        Usually involves compounds in the gas phase and in the solid phase in equilibrium.  e.g.  2C(s) + O₂(g)  =  2CO(g)
The expression for the equilibrium constant does not contain the concentration or partial pressure of those compounds in
the solid phase.  Therefore,
  K_p  =  P_CO²/P_O2    or   K_c  =  [CO]²/[O₂]

Another reaction is the decomposition of limestone, CaCO₃.

CaCO₃(s)   =  CaO(s)  +  O₂(g)
                                                          K_p  =  P_O2   or   K  = [O₂]

Calculating equilibrium constants
        If one knows the equilibrium concentrations of reactants and products, simple substitution into the equilibrium
        expression yields K_c.
         If one knows the equilibrium partial pressures of reactants and products, simple substitution into the equilibrium
        expression yields K_p.

        For example, if the partial pressure of O₂ over a mixture of CaCO₃ and CaO is 350 torr in a closed container then
        K_p = 350 torr = 350/760 atm = 0.46 atm.

        Consider the reaction N₂  +  3H₂  =  2NH₃ , where the partial pressures of  N₂, H₂,  and NH₃ are 20 torr,
        10 torr and 250 torr respectively.   K_p =  P_NH3²/P_N2 P_H2³  =  (250)²/{(20)(10)³}=  3.125 torr^-2.

        Consider another reaction, H₂(g)  +  I₂(g)  =  2HI(g).  Given that P_H2, P_H2 andP_H2 are 0.2 atm, 0.3 atm and
        0.6 atm respectively, then K_p =  (0.6)²/(0.2)(0.3)  =  6.0.  Notice that the equilibrium constant in this case is unitless.

    The problems above were fairly easy.  Obtaining an equilibrium constant from equilibrium concentrations or partial pressures is relatively simple.  However the determination of equilibrium concentrations or partial pressures knowing the equilibrium constant and the initial concentrations (partial pressures) of reactants and products is a bit more challenging.

Calculating equilibrium concentrations

        Consider another reaction, H₂(g)  +  I₂(g)  =  2HI(g) and K_p = 6.0.  Calculate the equilibrium partial pressures of the reactants and products given that the initial partial pressure of H₂(g)  and  I₂(g) are 1.2 atm and 0.80 atm respectively and that of HI is zero.  Let me offer a stepwise approach to these kinds of problems.

    1.  DESCRIBE (write down all the equilibria needed)   H₂(g)  +  I₂(g)  =  2HI(g)

    2.  DEFINE (write down all the definitions for the K's)  K_p =  P_HI²/P_H2 P_I2

   3.  ASSIGN  (this is the most challenging step and depends upon how the problem is worded)  In the above problem
                        you are told the initial partial pressure of H₂(g)  and  I₂(g) are 1.2 atm and 0.80 atm respectively. These
                        are values before equilibration took place. When equilibration takes place some of the H₂(g)  and  I₂(g)
                        are "lost" to form HI.  Since you are not told P_HI, you assign it an unknown value "x" or P.
                        This unknown value P can be solved for using the definition above. Now you assign equilibrium
                        values         P_HI= 2P          P_H2 = 1.2 - P          P_I2 =  0.80 - P

    4.  PLUG IN (substitute values and unknowns into the definition)
                        K_p =  P_HI²/P_H2 P_I2  =  (2P)²/{(1.2 -P)(0.80 - P)}= 6.0

    5.  SOLVE (solve for the unknowns)  In this case the unknown is P.  It appears that you need to solve a quadratic
                        equation.  4P² = (1.2 -P)(0.80 - P) 6.0
                                        4P² = (1.2)(0.80)(6.0) -1.2(6.0)P- 0.80(6.0)P + 6.0P²
                            2P² -12.0P + 5.76 = 0        P = [12.0 - {144 - 4(2)(5.76)}^1/2]/(2)(2) = 0.526

    6.  RELATE (now use this value of P to determine the partial pressures of all the species)
                         P_HI= 2P = 1.052 atm                                        P_HI=  1.052 atm
                         P_H2 = 1.2 - P = 1.20 - 0.526 = 0.674 atm        P_H2 = 0.674 atm
                         P_I2 =  0.80 - 0.526 = 0.274 atm                       P_I2 =  0.274 atm

                Upon substituting these values into the equilibrium expression:  K_p = (1.052)²/(0.674)(0.274) = 5.99
                The value of 6.0 was not obtained because of the rounding off of the value of P.

   This approach will be used later when studying acid base equilibrium reactions.