Examples:
1. Assume that the reaction for the formation of gaseous hydrogen iodide from hydrogen and iodine has an equilibrium constant of 2.3 x 102. In an experiment, 4.0 mol H2 and 2.0 mol of I2 were added to a 2.0 L flask. Find the equilibrium concentrations of all species involved in this reaction.
The balanced equation for the reaction is:
H2(g) + I2(g) <--> 2HI(g)
The equilibrium expression is:
K = 2.0 x 102 = [HI]2 / [H2][I2]
First, you must calculate the initial concentrations.
[HI]0 = [H2]0 = 4.0 mol / 2.0 L [I2]0 = 2.0 mol / 2.0 L = 1.0 M
Next, find the value of Q (initial conditions):
Q = [HI]20 / [H2]0[I2]0 = (2.0 M)2 / (2.0 M)(2.0 M) = 1.00
Since Q is less than K, the system will shift to the right (towards the products).
Now, determine what chang ein concentrations is necessary in order for this shift to take place. We do not yet know the amount we will need, so let that amount equal "x." In this problem, let "x" equal the number of moles per liter of H2 consumed to reach equilibrium. Take a look at the reaction.
H2(g) + I2(g) <--> 2HI(g)
The stoichiometry of this reaction indicates that x mol/L I2 will be consumed and 2x mol/L HI will be produced.
To help you see this problem more clearly, set up a chart that will allow you to express equilibrium concentrations in terms of "x."
Initial Concentration: [H2]0 = 2.0 M [I2]0 = 1.0 M [HI]0 = 0 M
Change: [H2] = -x [I2] = -x [HI] = +2x
Equilibrium Concentration: [H2] = 2.0 M - x [I2] = 1.0 M - x [HI] = 0 + 2x
Plug the equilibrium concnetrations into the equlibrium expression to give you:
K = 2.3 x 102 = [HI]2 / [H2][I2] = (2x)2 / (2.0 M - x)(1.0 M - x)