# Lecture: Chemical Equilibrium

**One way to introduce the concept.**

This beaker is filled with a mixture of water plus a little alcohol in which I plan to dissolve some Iodine. I added the alcohol to make the iodine a little more soluble so we should get good color in the solution phase as the process moves along. Now, I think some of you if not most of you know that the color will reach some maximum and then stay at that intensity rather than continuing to get darker. This will occur even with solid iodine still present on the bottom of the beaker. In elementary school and maybe even in middle school you (your teacher along with early chemist) would refer to this situation as saturation. Saturated suggest that the water alcohol mix is full of the iodine and no more can dissolve so things come to a standstill. The idea of saturation was fine for a time, but eventually someone noticed that if you added fresh iodine crystals that had sharp edges in a few minutes those sharp edges became more rounded like the others that had been in the container during the getting darker phase. The rounding of the edges suggested that dissolving was still going on and yet the solution did not get darker. Well, it had to be a mistake. Maybe we can’t see the solution getting darker but it of course must. In time however, chemist used ionic salts where the amount of salt in solution could be accurately monitored by checking the conductivity. The new crystals still got roundy edges along with no increase in conductivity. There is really no good explanation except that their must be a reverse process, will call it precipitation, that somehow offsets the dissolving process. This sounds way more complicated than saturation because it is not a stopped thing, but a dynamic process. It’s often represented as follows. ** **

** I**_{2(s)}** ** **I**_{2(in solution)}

The top arrow represents the process forward which in this case is dissolving and the bottom is the process reverse that we call precipitation. When they come into balance we have reached equilibrium. It may sound complex, but without dynamic chemical equilibriums even the simplest forms of life as we know them could not exist.

Were going to forget all this for a while and do a little exercise. I need your help and you will have to play your part well if we are going to get at some of the more complicated ideas inherent in understanding chemical equilibrium. Please tear out from your note book one blank sheet of paper, one that hopefully has the margin line down the left hand side. Fold the paper down that margin line and place it on you desk so that all you can see of the front of the sheet is the small area to the left of the margin line. We are going to write random numbers on each line down the sheet next to the margin line using only zero through four. Putting down random numbers for a person with a mind is not possible. It will be easier for some, but not possible for anyone. To do the best we can at this mindless activity I want you to work fast, jump around on the page and ignore what I’m saying while your doing this mindless work. Go ahead, but please don’t put 1-4 in order and please don’t not put 1-4 in order! Some won’t write as many fours as they should because your in a hurry and it takes two strokes to make the four and some of you will have too many fours because of the power of suggestion by me talking about it. When your done flip your sheet over and do the same thing down the right side of the margin line using all the number 0 through 9. Go fast, make sure you can’t see your first list and try not to think. When we unfold the papers each of you should have a two digit random number on each line with a range from zero to forty-nine. Fold your paper in half and on each side of this new fold do the same thing using 0-9 to create a second set of two digit random numbers ranging from zero to ninety-nine.

You need your sheet in order to play the equilibrium game. When we start the game use the wall clock’s second hand and starting with the left hand column of numbers you are to remain in your seat that many seconds then rise and walk about for the number of seconds on the first line in the middle of your page. Take you seat at that moment and move down one number on the left and continue as before. I’m going to keep track of how many of you are in your seats as time passes. My first number will be all of you unless one or more of you happen to have zero, zero as a starting value. No, all right lets start.

I was able to get a large lap clock from the track coach and set it in front of the room to make it easier for students to play their part properly. The more students playing the better so here is one case where a large class size is an advantage. There is a need for a little teacher license (like poetic license) in doing the counting. The number of students seated should level of at about one third of the number in class. It will jump around due to limited numbers of players so be a bit creative in your counting as necessary. The students are busy doing their part and never notice the fact that you might not count for a moment letting things settle back to the proper numbers.

Alright, we have the data so lets sit down and construct a simple graph of the results. Students seated started at 30 and as time past went down to 10 while the number standing stared a zero and moved up to 20. I have also calculated a ratio I’m going to call K which is number standing divided by the number seated. This is all well and good but why was it more difficult (took longer) for you to retake your seat after standing? Does this part of our model equilibrium game have any foundation in chemistry? Please recall reaction kinetics where for an exothermic reaction the activation energy going back in the endothermic direction is higher by the amount of heat given off in the exothermic process. In our model the process of getting out of your seat is the exothermic reaction and coming back is endothermic with a larger activation energy to overcome. Fewer collision will be able to bridge this higher value and so the chances of it occurring is less, hence 1 through 99 versus 1 through 49. Now for some questions.

Lets start with what is equal in an equilibrium? Certainly not the number of students standing and the number of students seated.

*The rate forward and the rate reverse. *

How could this happen when it was harder for the reverse process?

*When we reached equilibrium there were more students standing and less seated so this made up for the difference in the difficulties.*

Would it have mattered if we had started at the bottom of the page of numbers and worked our way up.

*No.*

What if we all started in the middle and some of you worked your way up and other went down their lists.

*No.*

What if we multiplied both side by 10 so the left set would range from 0 to 490 and the other from 0 to 990.

*Still come out much the same, but it would take longer to get to the equilibrium. *

If we divided both sides in half.

*Same, but it would not take as long to reach equilibrium. *

I hope you remember this answer when we talk about the effect of catalysts on a chemical equilibriums.

What would we have to do in order to get a different result? Maybe the easy way to state this question is; how could we get a different value for our K.

*You would need to change the random numbers so they were not representing a system that it is approximately twice as hard for the process reverse to occur compared to the process forward.*

If we made the reverse process even harder say between 0-150 how would this effect the value of our K?

*It would get bigger because we would get an equilibrium with fewer students seated and more standing.*

Now, I hope you’re thinking that in order to do this we would need to be modeling a different reaction because it would need to have a different reaction kinetics diagram. However, we will learn that it could be the same reaction at a new temperature. For example when you lower the temperature you slow both reactions but the more difficult process will be slowed more then the easier process.

Now for the most difficult and most important question of all. If you can answer this question you will begin to see what Henri Louis Le Chatelier (1850–1936) and others discovered long ago when equilibriums were first recognized and studied. What if, after we had reached equilibrium and were going along with the same number of students standing up and sitting down per minute I suddenly pull nine standers out of the game. This will throw things out of balance. Will we ever get back to an equilibrium? If so,can we predict how many standers and how many sitters will be present at this new equilibrium?

*We will get a new equilibrium. The loss of standers will cause the process reverse to slow, but corrections will occur. The one thing that won’t change is the value of K. The K value will still be two and this means now that only 21 students are in the game 14 will be standing and 7 will be seated at an given moment after equilibrium is re-established.*

Lets look at that by extending our graph. I have also added in what would happen if instead of taking out nine students I added six new stander. By studying these we can see Le Chatelier’s principle in action.

*When a equilibrium is subjected to a change processes occur that partially correct the imposed change.*

Dynamic chemical equilibriums respond to and try to correct for changes and this is the first sign of chemicals taking on behaviors that are life like.

Some students may point out that having an activation energy that is twice as high for the reverse reaction would make that process more than twice as unlikely to occur due to the fact the the kinetic energy distribution is not a “bell curve”. The threshold energy line would not have to be twice as large in order to make the process half as likely. Bravo!!!

About now some clever student may also point out that the game (model) is flawed if it’s trying to representing the Iodine equilibrium presented to the class at the start of the hour. It’s flawed in that if we added students to the seated group a change would take place shifting the equilibrium to produce more standers. If the seated students represents the Iodine in solid state there should be no shift and no increase in the red color in solution. The truth shall set you free and here this bit of truth sets in motion a needed discussion about models good and bad and how to use them wisely.

## Follow the Force!