# Equilibrium Constant

Synopsis

We would like to be able to state the extent of a reaction involved in dynamic equilibrium and to determine what factors affect the extent of the reaction. One way to do this is with the equilibrium constant, K. This constant can be defined in terms of the equilibrium concentrations of the reactants and products as seen below for the reaction aA + bB --> cC + dD:

Kc = {[C]c[D]d / [A]a[B]b}eq

We used a "c" subscript here because the equilibrium constant can also be written in terms of partial pressures, Kp, or in terms of mole fractions, Kx. This constant is a function of temperature so we should write K(T). How does this constant impart to us information about the extent of a reaction? If Kc were equal to 6.6 x 1040, what would that communicate to us? Because the value is very large, that means that the numerator is much larger than the denominator and that tells us that the reaction is strongly shifted toward the products. If Kc were equal to 4.8 x 10-25, then we would know that the denominator was much larger than the numerator and the reaction was shifted strongly toward the reactants.

Note some of the special means of dealing with equilibrium constants:

• If a solid is involved as a product or a reactant, it does not appear in the equilibrium constant. So for the reaction CaCO3(s) --> CaO(s) + CO2(g), the equilibrium constant would be Kc =[CO2(g)].
• For reactions involving pure liquid water in which the molar concentration of water is virtually unchanged during the course of the reaction, the water concentration does not appear in the equilibrium constant.
• If the equilibrium reaction is written in a reverse manner, the new equilibrium constant is the inverse of the original K. For the reaction aA + bB --> cD + dD, the equilibrium constant is Kc = [C]c[D]d / [A]a[B]b

If we reverse the reaction: cD + dD --> aA + bB, then the equilibrium constant is inverted

K'c = 1/Kc = [A]a[B]b/[C]c[D]d

• If we multiply both sides of the reaction by any factor (2, 3, ½, 1/3 , etc) then the equilibrium constant is raised to that power. For example, if we multiply the reaction aA --> bB through by 2, the new equilibrium constant will be the square of the original. If we multiply the same equation by ½, the new K is the square root of the original

Review Questions

1. Write the equilibrium constant for the evaporation of water (a physical process).
2. Write the equilibrium constant for the reaction 2 SO2+ O2 --> 2 SO3 and then write the equilibrium constant for the reaction 2 SO3 --> 2SO2 + O2.
3. Write the equilibrium constant for the reaction SO2 + ½ O2 --> SO3.
4. The Kc for the reaction 2 SO2+ O2 --> 2 SO3 is 2.8 x 102 at 1000 K. What is the Kc for the reaction

SO2 + ½ O2 --> SO3 at 1000 K?

Web Author: Dr. Leon L. Combs