Go back to the first set of examples

Given this equation:

H_{2}+ Br_{2}⇌ 2 HBr

Calculate all three equilibrium concentrations when 0.500 mole each of H_{2} and Br_{2} are mixed in a 2.00 L container and K_{c} = 36.0.

This problem has a slight trick in it. Notice that moles are given and volume of the container is given. However, the calculations must be done in molarity. So you must divide 0.500 by 2.0 to get 0.250 mol/L. That is the number to be used.

In my classroom, I point this out over and over, yet some people seem to never hear. I promise them I will test it and when I do, many people use 0.500 for their calculation, not 0.250.

I hope you don't get caught in the same mistake.

The solution technique involves the use of an ICEbox. Here is the initial row, filled in:

[H _{2}][Br _{2}][HBr] Initial 0.250 0.250 0 Change Equilibrium

Remember, the last value of zero come from the fact that the reaction has not yet started, so no HBr could have been produced yet.

Now for the change row:

[H _{2}][Br _{2}][HBr] Initial 0.250 0.250 0 Change - x - x + 2x Equilibrium

The minus sign tends to mess people up, even after it is explained over and over. It is associated with the substances being used up as the reaction goes to equilibrium. Some people never seem to figure that something (in this case, H_{2} and Br_{2}) are going away and some new stuff (the HBr) is comming in.

x signifies that we know some H_{2} and Br_{2} get used up, but we don't know how much. What we do know is that an EQUAL amount of each will be used up. We know this from the coefficients of the equation. For every one H_{2} used up, one Br_{2} is used up also.

This also messes up a lot of people. I think it is because they do not have a good idea in their brain about what is happening during the chemical reaction. They have a hard time with the concept that the H_{2} splits into two separate H and the Br_{2} splits into two Br. The each of the two H and two Br hook together to make two different HBr molecules.

Now, I can just see you sitting there saying, "Geez, what a wasted paragraph." No way man, there are people who DO NOT GET IT. Those people are in your class and you know who they are. Go give them a bit of help.

The two is important. HI is being made twice as fast as either H_{2} or I_{2} are being used up.

In fact, always use the coefficients of the balanced equation as coefficients on the "x" terms. Even if you don't understand why, memorize the idea that the coefficients attach on front of each x. Another way: the coefficient of each substance in the chemical equation becomes the coefficient of its 'x' in the change row of the ICEbox.

The equilibrium row is:

[H _{2}][Br _{2}][HBr] Initial 0.250 0.250 0 Change - x - x + 2x Equilibrium 0.250 - x 0.250 - x 2x

Now we are are ready to put values into the equilibrium expression. For convenience, here is the equation again:

H_{2}+ Br_{2}⇌ 2 HBr

The equilibrium expression is:

K_{c}= [HBr]^{2}/ ([H_{2}] [Br_{2}])

Plugging values into the expression gives:

36.0 = (2x)^{2}/ ((0.250 - x) (0.250 - x))

Two points need to be made before going on:

1) Where did the 36.0 value come from? It was given in the problem.

2) Make sure to write (2x)^{2}and not 2x^{2}. As you well know, they are different. This mistake happens a LOT!! This mistake happens on the test even after 10-15 mentions of it in class.

Both sides are perfect squares (done so on purpose), so we square root both sides to get:

6.00 = (2x) / (0.250 - x)

From there, the solution should be easy. You can check for correctness by plugging back into the equilibrium expression.

Now for a second example. This example will involve the use of the quadratic formula.

Given this equation:

H_{2}+ Cl_{2}⇌ 2 HCl

Calculate all three equilibrium concentrations when K_{c} = 20.0 and [H_{2}]_{o} = 1.00 M and [Cl_{2}]_{o} = 2.00 M.

Here is the completed ICEbox:

[H _{2}][Cl _{2}][HCl] Initial 1.00 2.00 0 Change - x - x + 2x Equilibrium 1.00 - x 2.00 - x 2x

The equilibrium expression is:

K_{c}= [HCl]^{2}/ ([H_{2}] [Cl_{2}])

Substituting gives:

20.0 = (2x)^{2}/ ((1.00 - x) (2.00 - x))

After suitable manipulation (which you can perform yourself), we arrive at this quadratic equation in standard form:

16x^{2}- 60x + 40 = 0

Using the quadratic formula, which is x = (- b ± square root[b^{2} - 4ac]) / 2a, we obtain:

x = (60 +± square root[60^{2}- (4) (16) (40)]) / 32

After suitable calculations, we find x = 0.867.

In this problem, note that b was positive in the quadratic formula. That means both roos will be positive. Which one should you check first?

Always check the minus root first. it will give the smaller answer and that's usually what you want. In this problem, the larger root gives answer of 2.88, which leads to ridiculous results.

Determine the equilibrium amounts and checking for correctness by inserting back into the equilibrium expression is left to the student.