Calculate Equilibrium Concentrations from Initial Concentrations.

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

By the way, I'm going to stop using K_{eq} and will start using K_{c}, where 'c' stands for 'concentration.' In the future, you will also study K_{p}, where 'p' stands for 'pressure.'

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

Hwhen the equilibrium concentrations at 25.0 °C were found to be:_{2}(g) + I_{2}(g) ⇌ 2HI(g)

[H_{2}] = 0.0505 M

[I_{2}] = 0.0498 M

[HI] = 0.389 M

**Solution:**

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

[HI] ^{2}K _{c}=–––––––– [H _{2}] [I_{2}]

2) Now, all you have to do is substitute numbers into the equilibrium expression:

(0.389) ^{2}K _{c}=–––––––––––––– (0.0505) (0.0498)

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

What are the units of the K_{c}? And, for that matter, K_{p}. For reasons beyond the scope of this lesson, the answer is none. The equilibrium constant does not have any units.

Some advice: your teacher may insist on putting units on the equilibrium constant. Please do not march up to him/her and announce they are wrong because some guy on the Internet says so. That will NOT make your teacher happy!

**Example #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:**

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

(0.0127) ^{2}K _{c}=–––––––––––––––––– (0.00560) (0.0000590)

2) 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 the first edition of 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, please note that the above problem is routine for me. Yet, I made a mistake and solely on the basis of experience I rejected my first answer as being wrong (it "felt" wrong). You guys don't have much chemistry experience, so your chemistry feel has some room to grow.

So, BE CAREFUL.

Here endth the lecture.

**Example #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_{c}.

**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: 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_{c} 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_{c} that get published are actually an average of many careful experiments.

**Example #4:** The following reaction:

2SOwas allowed to come to equilibrium and the following concentrations were measured:_{2}(g) + O_{2}(g) ⇌ 2SO_{3}(g)

[SO_{2}] = 0.600 M

[O_{2}] = 0.820 M

[SO_{3}] = 1.86 M

Determine the value of the equilibrium constant, K_{c}

**Solution:**

1) Write the equilibrium expression

[SO _{3}]^{2}K _{c}=––––––––– [SO _{2}]^{2}[O_{2}]

2) Insert values and solve:

(1.86) ^{2}K _{c}=–––––––––––––– (0.600) ^{2}(0.820)K

_{c}= 11.7

**Example #5:** Using the concentrations in the previous problem, determine the K_{c} for this reaction at equilibrium:

SO_{2}(g) +^{1}⁄_{2}O_{2}(g) ⇌ SO_{3}(g)

**Solution:**

1) Write the equilibrium expression

[SO _{3}]K _{c}=––––––––– [SO _{2}] [O_{2}]^{½}

2) Insert values and solve:

(1.86) K _{c}=–––––––––––––– (0.600) (0.820) ^{½}K

_{c}= 3.42

I want you to compare the two expressions just above. I want you to see that the second one (the one with the [O_{2}] having a one-half power) is the square root of the first expression. Then, look at the square root of 11.7 and discover it to be 3.42.

When you divide the coefficients of the equation by two, the new K_{c} is the square root of the old value.

**Example #6:** Using the same set of concentrations given just above, determine the equilibrium constant for this reaction:

SO_{3}(g) ⇌ SO_{2}(g) +^{1}⁄_{2}O_{2}(g)

**Solution:**

1) Write the equilibrium expression

[SO _{2}] [O_{2}]^{½}K _{c}=––––––––– [SO _{3}]

2) Insert values and solve:

(0.600) (0.820) ^{½}K _{c}=–––––––––––––– (1.86) K

_{c}= 0.292

I want you to compare the two expressions (the two with the one-half power) just above. I want you to see that one is simply the inverse of the other. One has a K

Calculate Equilibrium Concentrations from Initial Concentrations.