Solving Equilibrium Problems

There are, of course, many ways to solve equilibrium problems.  In this section we will focus on a few useful approaches.  The key to solving these problems is to recognize that Q = K at equilibrium; that the reactant concentration is changed by some amount, say "x", multiplied by the correct coefficient for that reactant from the balanced equation.  Product concentration changes in the opposite direction by the same factor, "x" multiplied by the coefficient for that product from the balanced equation. 

Many chemists find it convenient to construct a reaction table that shows the initial amounts of all reactants and products, the amount by which they change, and the equilibrium amounts. You may also find it useful to experiment with the equilibrium reaction:

H₂(g) + I₂(g) 2HI(g)

         COCl₂(g)   CO(g)  +   Cl₂(g)                             K=8.3 x 10 ^-4 at 360^oC

Calculate [CO], [Cl₂] and [COCl₂] when 5.00 mol COCL₂ decomposes and reaches equilibrium in a 10.0 L flask?

Solution: 1. Balanced equation provided 2. Q = [CO] [Cl2] / [COCl2] 3.  [COCl2] = 5.00 mol / 10.0 L = 0.500 M       ( M is molarity or mol / L) 4. Since the problem begins with only phosgene, all other concentrations = 0 and equilibrium direction must be away from reactants towards products. 5. Concentration (M) COCl2(g)       CO(g)   +    Cl2(g) Initial 0.500 0 0 Change -x +x +x Equilibrium 0.500 - x x x 6. COCl2 is reduced by x amount every time CO and Cl2 concentrations are increased by the same amount.  7.  Q = x * x / (0.500 - x) 8.   0.500 - x is approximately the same as 0.500 because x is assumed to be much smaller than  0.500 M, then       Q = x2 / 0.500 9. At equilibrium Q = K and K was given to be 8.3 x 10-4 .  Therefore Q = K = 8.3 x 10-4 = x2 / 0.500.  x = 2.0 x 10-2 10. [CO] =  [Cl2] = x = 2.0 x 10-2 M             [COCl2] =  0.500 - x = 0.500 - 2.0 x 10-2 = 0.480 M