Using an Equilibrium Constant

Now that we know how to calculate an Equilibrium Constant or Reaction Quotient, we can use either or both for more complex calculations. Equilibrium calculations use quantities in a system that may or may not be at equilibrium.

When a set of conditions is given, usually one or more quantities are to be evaluated. From a given set of conditions, we need to reinterpret the problem to prepare equations. Understanding what the problem is asking you is a major step in solving the problem. When the conditions you are given for a problem are incomplete, assumptions must be made to supplement the missing information. You have learned the various expressions of the equilibrium constant in the previous lectures. The expression gives the quantitative relationship of various compounds in a system. Any unknown quantity can be represented by a symbol (x for example). The relationship is then a mathematical equation. The unknown quantities can then be calculated. I will give you several examples, but you will need to practice in order to get a feel for those "assumptions" I mentioned above.

Let's start simple:

For the gas phase reaction H2(g) + I2(g) = 2 HI(g)
Kc = 50.3 at 731 K. Equal amounts (0.100 M each) is introduced to a container, and then the temperature is raised to 731 K.

Is the system at equilibrium? If not, what direction is the reaction?

Solution
When equal amounts are present, 

       [HI]2        (0.1)2    
Qc = ---------- = -------- = 1 < Kc (50.3)    
     [H2] [I2]      (0.1)2
There is a tendency to increase [HI] to reach equilibrium at 731 K.

 

Now let's make it more difficult:

For the gas phase reaction H2(g) + I2(g) = 2 HI(g) Kc = 50.3 at 731 K. Equal amounts (0.100 M each) is introduced to a container, and then the temperature is raised to 731 K.

Calculate the concentration of each when the system is at equilibrium.

Solution

For convenience, we write down the reaction equation and put the concentration of the species below the formula. Since the [HI] concentration will increase, we assume x M of H2 or I2 to react to give 2x M of HI. 
     H2(g) + I2(g) = 2 HI(g)      
    0.100    0.100   0.100   initial concentration     
    -x       -x       2x     amounts change      
   0.1-x    0.1-x   0.1+2x   equilibrium concentration
To simplify notation, we ignore the number of significant figure in the formulations. By definition of equilibrium constant, we have 
       [HI]2       (0.1+2x)^2
Kc = ---------- = -------------- = 50.3       
     [H2] [I2]    (0.1-x)(0.1-x)
Expanding the above equation gives, 46.3 x2 - 10.43x + 0.493 = 0

Please expand the equation to see if you get the same result. Many students have difficulty in deriving or simplifying this equation.

Solution of the quadratic equation gives x = 0.067 and x' = 0.158. When x' = 0.158, 0.1 - 0.158 = -0.058. Since concentration should not be a negative value, this solution does not agree with reality. The only acceptable solution is x = 0.067, and thus we have:

[H2] = [I2] = 0.100 - 0.067 = 0.033 M
[HI] = 2x = 0.100 + 0.134 = 0.234 M

It is always a good idea to check the validity of the answer. We use the result to calculate the equilibrium constant.

0.2342 / 0.0332 = 50.3 = Kc.

 

And another example:

At 1100 K, Kc = 4.20e-6 for the gas phase reaction,

2 H2S(g) = 2 H2(g) + S2(g) What concentration of S2 can be expected when 0.200 mole of H2S comes to equilibrium at 1100 K in an otherwise empty 1.00-L vessel?

Solution

We write the equation and place quantities at the initial condition and at various stages below the formula. We assume x mol of S2 is formed, and this leads to the formation of 2x moles of H2. Since -2x moles of H2S is required, the equilibrium concentration of H2S is 0.200 - 2x. Thus, we have 
  2 H2S(g) = 2 H2(g) + S2(g)    
  0.200       0         0   initial condition   
  -2x         2x        x   assume x mole of S2 is formed   
  0.200-2x    2x        x   equilibrium concentration

The definition of Kc leads to the equation, 

      [H2]2 [S2]       (2x)^2 (x)   
Kc = ----------- = ------------ = 4.20e-6          
       [H2S]^2       (0.200-2x)^2

This is a cubic equation, and there is no general method to solve this type of equations. Fortunately, since the equilibrium constant is very small, we expect x to be a small value. Thus, 0.200 - 2x is almost 0.200. With this approximation, we have a simpler equation to solve:

4 x3 = (0.200)2 * 4.20e-6
    = 1.68e-7
x3 = 1.68e-7 / 4 = 4.2e-8
x = (4.2e-8)1/3
    = 3.5e-3 = [S2]
[H2] = 2 x = 7.0e-3

Discussion:

Is the approximation justified? We can calculate the equilibrium constant from the equilibrium concentrations.

4 (0.0035)3 / (0.200-0.007)2 = 4.6e-6

The error in Kc = (4.6-4.2)e-6/4.2e-6 = 4.8%. This is within 5 % criterion, but we are approaching the limit for using approximation. In the approximation, we used 0.200 M instead of 0.193 M as [H2S].

In this example, you have learned the approximation technique.

For lon-capa, if you run up against a cubic equation, use the cubic solver at Cubic Calculator