15.1 The Concept of Equilibrium

Chemical equilibrium exists when 2 opposing reactions
    occur simultaneously at the same rate.
(It is important to realize that very few reactions go to 100% completion)

Frozen N₂O₄ is colorless at room temperature; it
        decomposes to brown NO₂

                              N₂O₄(g) ---->  2NO₂(g)
After some time, the color stops changing and we have
     a mixture of N₂O₄(g) and NO₂(g).

Chemical equilibrium is the point at which the
    concentrations of all species are constant.
    (Review rate plot).

 The point at which the rate of decomposition:
                            N₂O₄ --->  2NO₂

 equals the rate of dimerization:
                            2NO₂ --->  N₂O₄

 is called dynamic equilibrium.
                             2NO₂  ---->  N₂O₄
                                  ^<----

The reaction has not stopped; the opposing rates are
        equal.
 

For the reaction:        A  ---->  B
                                        <----

Forward reaction:       A  ---->  B      Rate = k_f[A]
Reverse reaction:        B  ---->  A      Rate = k_r[B]
 At equilibrium,     k_f[A] = k_r[B]
 Therefore,

                  [B]/[A] = k_f/k_r = constant

 For an equilibrium, we write:
                                    A  --->  B
                                       <---

 As the reaction progresses, [A] decreases to a constant

    value, [B] increases from zero to a constant value.
 When [A] and [B] are constant, equilibrium is achieved.

Alternatively:
       k_f[A] decreases to a constant,
       k_r[B] increases from zero to a constant.
When k_f[A] = k_r[B], equilibrium is achieved.

    Rearranging:  k_r[B]/k_f[A]  = k
             [B]/[A] = k_f/k_r = K_c = equilibrium constant

 A and B are still reacting, but both processes occur at
    the same rate!!

     (Don't use <---->  this means resonance ONLY!!)

Ex.:  At 25^oC, k_f is 7.3 x 10³M^-1s^-1 and k_r is 0.55M^-1s^-1.
 Calculate the equilibrium constant K_c for this reaction:
        ClNO₂(g) + NO(g) ----> NO₂(g) + ClNO(g)

At equilibrium, the rates of the forward and reverse
    reactions are the same:
            therefore,  k_f[ClNO₂][NO] = k_r[NO₂][ClNO]

      K_c = k_f/k_r =[ClNO₂][NO]/[NO₂][ClNO]
        = 7.3 x 10³M^-1s^-1/0.55M^-1s^-1 = 1.3 x 10⁴
                                                                                            (NOTE:  no units!!!)
   

The Equilibrium Constant K

                        N₂(g) + 3H₂(g) ----> 2NH₃(g)
                                                  <----

If we start with a mixture of nitrogen and hydrogen
    (in any proportion), the reaction will reach equilibrium
    with a constant concentration of nitrogen, hydrogen,
    and ammonia.

**However, if we start with just ammonia and no nitrogen
    or hydrogen, the reaction will proceed to produce N₂
    and H₂ until equilibrium is achieved; i.e., the ammonia
    concentration will decrease and nitrogen and hydrogen
    concentrations will increase until the forward and
    reverse reactions are at equilibrium.

No matter what the starting composition of reactants and
    products is, the same ratio of concentrations is
    achieved at equilibrium (at the same temperature!!).

 For a general reaction:

                        aA + bB ^----> cC + dD
                                     <----

the equilibrium constant expression is:

             K_c = [C]^c[D]^d/[A]^a[B]^b

      where K_c is the equilibrium constant.

Called LAW OF MASS ACTION.

Terms for pure liquids and pure solids are NOT included
        in K_c

K_c is based on the molarities of reactants and products.
  We generally omit the units of the equilibrium constant.
  Note that the equilibrium constant expression always
  has [products] over [reactants].
 

             K_p =   (P_C)^c(P_D)^d/(P_A)^a(P_B)^b

    If K >> 1, then products dominate at equilibrium, and
            the equilibrium lies to the right.
    If K << 1, then reactants dominate at equilibrium, and
            the equilibrium lies to the left.

NOTE:  K for a reaction in one direction is the reciprocal
    of the equilibrium constant K' in the opposite direction.

Ex.:  2 O₃(g) --->  3 O₂(g)
                     <---
Given: at equilibrium, [O₃(g)] = 1.0 x 10^-2M and
            [O₂(g)] = 3.2 x 10^-1M.   Calculate K_c

 K_c = [O₂(g)]³/[O₃(g)]² = (3.2 x 10^-1M)³/(1.0 x 10^-2M)²
      = 3.3 x 10²
 

    Calculate K_c for 3 O₂(g) --->  2 O₃(g)
                                            <---

 K_c'= [O₃(g)]²/[O₂(g)]³ = (1.0 x 10^-2M)²/(3.2 x 10^-1M)³
      = 3.0 x 10^-3

 Therefore, K_c' = 1/K_c = 1/3.3 x 10²= 3.0 x 10^-3

 Or, K_p = P_O ³/P_O ², where these are the partial
                   ^{2           3}             pressures of the gases.

and    K_p' = P_O ²/P_O ³     for the reverse reaction.
                                    ^{3           2}

Heterogeneous Equilibria
              MgO(s) + SO₂(g) ---->  MgSO₃(s)
                                           <----
              K_c = 1/[SO₂(g)]       NO SOLIDS!!!!!!  (OR PURE LIQUIDS)
 

              NH₄Cl(s)  ----> NH₃(g) + HCl(g)
                              <----

              K_c = [NH₃(g)][HCl(g)]      NO SOLIDS!!!!! (OR PURE LIQUIDS)
****************************************************************************

Given:  N₂(g) + O₂(g) ---->  2 NO(g)
                                   <----
    Starting with 1.0 mole N₂(g) and 1.0 mole O₂(g) in
    a 1.0 L container, let the reaction go to equilibrium.
    At equilibrium, we analyze and find that there are
    0.40 moles of NO(g).  Calculate K_c  for this reaction.

         K_c = [NO(g)]²/[N₂(g)][O₂(g)]

                   N₂(g)         O₂(g)        NO(g)
Init []        1.0 M          1.0 M        0.0 M
D[]            -0.20 M      -0.20 M      0.40 M
Equil []      0.80 M        0.80 M      0.40 M

Therefore, K_c = [NO(g)]²/[N₂(g)][O₂(g)]
                       = (0.40M)²/(0.80M)(0.80M) = 0.25

  If we need the equilibirum constant for the reverse
         reaction:    K_c' = 1/K_c = 1/0.25 = 4.0
 

Also, for the reaction:
                1/2N₂(g) + 1/2 O₂(g) ----> NO(g)
                                                 <----

    K_c= [NO(g)]/[N₂(g)]^1/2[O₂(g)]^1/2
         = (0.40M)/(0.80M)^1/2(0.80M)^1/2= 0.50

Therefore, K_c (for THIS LAST reaction) = K_c^1/2
                                         K_c =(0.25)^1/2 =0.50

BACK HOME