The easiest way to explain is by looking at a few problems.

**Problem #1:** Calculate the equilibrium constant (K_{eq}) for the following reaction:

Hwhen the equilibrium concentrations at 25 °C were found to be:_{2}(g) + I_{2}(g) ⇌ 2 HI(g)

[H_{2}] = 0.0505 M

[I_{2}] = 0.0498 M

[HI] = 0.389 M

**Solution:**

The first thing to do is write the equilibrium expression for the reaction as written in the problem. This is what to write:

$\mathrm{Keq}=\frac{\mathrm{[HI]2}}{\mathrm{[H2]}\mathrm{[I2]}}$

Now, all you have to do is substitute numbers into the problem. K_{eq} is what we want to find, so that's our "x."

Here is what we get:

$x=\frac{\mathrm{(0.389)2}}{\mathrm{(0.0505)}\mathrm{(0.498)}}$

Solving this and rounding to the correct number of sig figs (remember those??), we get 60.2

Something of a side issue is what are the units of the K_{eq}? For reasons beyond the scope of this lesson, the answer is none. The equilibrium constant does not have any units.

**Problem #2:** The same reaction as above was studied at a slightly different temperature and the following equilibrium concentrations were determined:

[H_{2}] = 0.00560 M

[I_{2}] = 0.000590 M

[HI] = 0.0127 M

From the data, calculate the equilibrium constant.

**Solution:**

Same technique as above, write the equilibrium expression and substitute into it. Then solve. So, we get this:

$x=\frac{\mathrm{(0.0127)2}}{\mathrm{(0.00560)}\mathrm{(0.000590)}}$

The answer is 48.8

Time for a small lecture:

Please be very careful in using your calculator to solve these problems. When I solved this problem while writing this tutorial (on December 28, 1998), I first got some really weird looking answer that didn't feel right, so I did it again. Sure enough, I have made an entry error somewhere in the problem.

Underscoring my plea for carefulness is the difference between you and me in problem solving. The above problem is routine for me and solely on the basis of experience did I reject my first answer as being wrong (it "felt" wrong). You guys don't have that experience, so you don't have the feel. Yet!!

So, BE CAREFUL.

Here endth the lecture.

**Problem #3:** Using the same equation as above and with the following equilibrium concentrations:

[H_{2}] = 0.00460 M

[I_{2}] = 0.000970 M

[HI] = 0.0147 M

calculate the K_{eq}.

**Solution:**

I'm not going to write the set-up, but I want you to write it down on your paper. Then solve it. The answer is 48.4.

An important point: did you remember to square the numerator. This is the number one rookie problem in solving these things - forgetting the exponent.

The number two error is wanting to change the concentrations. For example, when [HI] = 0.0147, the rookie will want to double it, saying "Well, there is a 2HI in the equation. No, No, No!! Use the concentrations as given.

One more discussion point: you may have noticed the K_{eq} answers for #2 and #3 are slightly different when they are supposed to be the same. The answer: experimental error. One can never be perfect, so the values for K_{eq} that get published are actually an average of many careful experiments.

This is the next tutorial: Calculate Equilibrium Concentrations from Initial Concentrations. Notice that K_{eq} (a generic term) has been replaced in the tutorial by K_{c}. The "c" simply means "concentration." So K_{c} is an equilibrium constant based on concentration values. If you go on in chemistry (but not right now), you'll learn about K_{p}, where the "p" means "pressure." So K_{p} is an equilibrium constant based on pressure values.