You will learn how to calculate the equilibrium concentrations or partial pressures of reactants and products when you know the equilibrium constant for the reaction.
Instead of having to calculate the K, we might be given K and have to calculate either the equilibrium concentration or the equilibrium pressure. This process is a very straight-forward modification of the above process, all that changes is what is known and what is unknown. The algebra can get a little tricky for some of the problems. Let's look at the reaction H2 + F2 --> 2HF for which we are given that Kc is 1.15 x 102 at the temperature of the reaction. Suppose that we put 4.000 mol of hydrogen and 8.000 mol of fluorine in a 4.000-liter flask and we are asked to determine the equilibrium concentration of all components. First we set up a table as we did before:
|Initial conc. (mol/liter)||1.000||2.000||0|
|Change as reacts||-x||-x||2x|
|Equil. Conc.||1.000 -x||2.000 -x||2x|
The equilibrium constant is then:
Kc = 1.15 x 102 = [HF]2/[H2][F2] = (2x)2/[(1.000 - x)(2.000 - x)] or
4x2 = 1.15 x 102 [2 - 3.000x + x2] = 2.30 x 102 - 345 x + 115 x2 or
111x2 - 345x + 230. = 0
The above is a quadratic equation (ax2 + bx + c = 0) which has the solutions
x = [-b +/- sqrt[(b2 -4ac)1/2]/2a
Putting our a, b, and c values into this equation gives the two values for x: 2.14 and 0.968. How do we decide which answer to choose for x? For such problems we always choose the root that gives an answer that corresponds to a physical reality. In this case we want the equilibrium concentrations of all components. Note that if we put x = 2.14 into the equilibrium expression for H2 we would get an equilibrium concentration of - 1.14 and such a value is physically impossible; so, we throw out the 2.14 root. So the root to choose is x = 0.968. Therefore, the equilibrium concentrations will be 0.032 mol/liter for H2, 1.032 mol/liter for F2 and 1.936 mol/liter for HF.
Web Author: Dr. Leon L. Combs
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