Equilibrium II

 

How writing the overall reaction differently changes the mass action expression.

All reactions are gas phase unless noted.

 

            N₂O₄ = 2 NO₂                         K_c = [NO₂]² / [N₂O₄]

 

            Say we wrote the reverse reaction

 

            2 NO₂  = N₂O₄                        K_c’ = [N₂O₄] / [NO₂]²

 

            So K_c’ = 1/K_c

 

            Say we multiple an overall reaction by a constant

 

            H₂ + ½ O₂ = H₂O                   K_c = [H₂O] / [H₂] [O₂]^1/2

 

                2 H₂ + O₂ = 2 H₂O                 K_c^’ = [H₂O]² / [H₂]² [O₂]

 

            So K_c’ = K_c²

 

Relating K_p  and K_c (14.2)

 

            Using the ideal gas law,  P = (n/V) RT = MRT

 

                N₂O₄ = 2 NO₂                         K_c = [NO₂]² / [N₂O₄]

 

                Expressed as partial pressures  K_p = P_NO2² / P_N2O4 (all pressure are in atmospheres)

 

            How do K_c  and K_p relate?               K_p = P_NO2² / P_N2O4 = ( [NO₂]RT )² / [N₂O₄]RT

 

                                                                        K_p = = [NO₂]² RT / [N₂O₄]

 

            In general,   K_p = K_c (RT)^Dng ,  where Dn_g = (#moles gas product - # moles gas reactant).  As usual, use Kelvin for T and R = 0.08205 Latm/K mol.  For the above reaction, Dn_g = 1.  For the production of gas phase water from its elements;

 

            2 H₂ + O₂ = 2 H₂O                 Dn_g = -1          so  K_p = K_c (RT)^-1

 

            K_c = 9.1 X 10⁸⁰ @ 25°C,  so K_p = 9.1 X 10⁸⁰ / (.0821 * 298) = 3.7 X 10⁷⁹

 

            Note this is a rather large equilibrium constant, and it tells us that this reaction greatly favors the products.

 

            Consider this reaction

 

            N₂ + O₂ = 2 NO                                  K_c = 4.5 X 10^-31 @ 25ºC

 

            Calculate the value of K_p.  Would you say that this reaction at equilibrium favors reactants or products?

 

            If K_c >> 1, (read the equilibrium constant is much greater than 1) the reaction favors products.  If K_c << 1, then the reaction favors reactants.

 

Heterogeneous Equilibria

 

            In heterogeneous equilibria, one or more reactants or products are solids or pure liquids.  Consider

 

(a)       CaCO_{3 (s)} = CaO_(s) + CO_{2 (g)}

 

            Consider calcium carbonate.  It has a density of 2.710 g/mL at 18°C.  Using this we could calculate the concentration of 1 mL of calcium carbonate.

 

            M = 2.710 g CaCO₃ X 1 mol/ 100.0869 g = 27.08 M

                        .00100 L

So solid calcium carbonate has a concentration of 27.08 M.  This will be constant over a wide temperature and pressure range.    Using the density of calcium oxide, we could calculate its concentration as well.  The mass action or equilibrium-constant expression for reaction (a) above is

 

            K_c = [CaO] [CO₂] / [CaCO₃]

           

            Since the two solid concentrations are constant, instead of calculating them and including them in the mass action expression, the following is done.

           

(b)       K_c [CaCO₃] / [CaO] = K_c’ = [CO₂]

 

Here I multiplied both sides by [CaCO₃] / [CaO].  The left side of the equation is a constant, K_c multiplied by a constant [CaCO₃] and divided by a constant [CaO].  All these constants combine to give a new constant, K_c’. (So, if a pure solid or pure liquid or a solvent is involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium constant expression). 

 

For example, write the equilibrium constant expression for the following reaction

            2 Hg_(L) + Cl_{2 (g)} = Hg₂Cl_{2 (s)}

 

Calculating equilibrium constants

 

            The calculations involve substituting in the values of the equilibrium concentrations into the equilibrium constant expression.  This will often involve some stoichiometric calculations to go from initial concentrations to equilibrium concentrations.

 

            When solid PbCl₂ is added to pure water at 25°C, the salt dissolves until [Pb⁺²] = 1.6 X 10^-2 M.  What is the equilibrium constant?  (Ref. Olmstead and Williams)

 

            PbCl_{2 (s)} = Pb⁺²_(aq) + 2 Cl^- _(aq)             K_c = [Pb⁺²] [Cl^-]²

 

            Example 2 - At a certain T a mixture of hydrogen and iodine were prepared by placing 0.200 mol of H₂ and 0.200 mol of I₂ in a 2.00 L bottle.  After a period of time the concentrations of the reactants and products appeared constant, suggesting we have obtained equilibrium

 

            H₂ + I₂ = 2 HI (all gas phase)

 

Absorption of light due to the purple color of iodine was used to monitor its concentration.  At equilibrium, the [I₂] dropped to .020 M.  What is K_c?

 

A concentration table (or ICE table) is helpful here and in many problems.

                                                H₂        +                      I₂                      =          2 HI

Initial [ ]	0.200 mol / 2.00 L = 0.100 M	0.200 mol / 2.00 L = 0.100 M	0
Change
Equilibrium		.020 M

 

Using stoichiometry, we know that for every mole of I₂ that reacted, one mole of H₂ reacted, and 2 moles of hydrogen iodide were produced.  So we add to our table.

 

Initial [ ]	.100 M	.100 M	0
Change	-.080	-.080	+.160
Equilibrium	.020	.020 M	.160

 

            Now we can calculate K_c

 

            K_c = [H₂] [I₂] / [HI]² = (.020) (.020) / (.160)² = 64

 

Calculating equilibrium concentrations

 

A more common problem is using an equilibrium constant to determine a concentration at equilibrium.  For example.

 

            The Haber process

 

            N₂ + 3 H₂ = 2 NH₃                  K_p = 1.45 X 10^-5 at 500°C (typical T for this process).  At equilibrium we were able to measure the pressure of nitrogen (.432 atm) and hydrogen (.928 atm).  What is the equilibrium pressure of ammonia?

 

At the end of my Equilibrium III notes I have two more examples using the ICE table.