Editing revision 32 of ChemicalEquilibrium

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Here are the basic procedures for solving equilibrium problems: (from Zumdahl textbook)

<li>Write the balanced equation for the reaction.
<li>Write the equilibrium expression using the law of mass action.
<li>List the initial concentrations.
<li>Calculate "Q," and determine the direction of the shift to equilibrium.
<li>Define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations. 
<li>Substitute the equilibrium concentrations ionto the equilibrium expression, and solve for the unknown. 
<li>Check your calculated equilibrium concentrations by making sure they give the correct value of "K."

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)

As you can see, this is going ot be a bit messy. Because the right side of this equation is, unfortunately, not a perfect square, you will have to use the quadratic formula, b+/- the square root of b2 - 4ac all divided by 2a.

However, first, cross multiply. (1.0 - x)(2.0 - x)(2.3 x 102) = (2x)2

or (2.3 x 102)x2 - 1.0(2.3 x 102x) + 2.0(2.3 x 102)

Next, collect your terms so that your equation can take on the quadratic form, then solve. The number you obtain will equal "x." Plug in "x" to equilibrium concentration expressions containing variables to find the equilibrium concentrations.

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But...

sometimes, the math doesn't have to be as complicated. If your K value is 10-3 or less, you can employ what is known as the "5% rule." In this case, your equilibrium concentrations that look like "1.0 - x" or "1.0 + x" can be simplified to, in these cases, 1.0 M. However, if you use the 5% rule, you must be sure to check your answer. When you obtain a value for "x," you must plug x into all of the original equilibrium concentrations, divide by x, and multiply by 100. If the result is less than 5%, your answer is valid. Otherwise, you must redo the entire problem using the quadratic formula and keeping the concentrations in the form "1.0 - x."

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