Solving Big X Problems

As stated before, big X problems are rare and you may never have to solve one, but the ideas are important for understanding certain situations, especially the difference between strong and weak acids and how we deal with them.

Big X problems are those where the change between the current concentrations and those at equilibrium is significant. (Remember, that with respect to “x” big does NOT mean big, it means significant...that is, not tiny.)

Let’s look at an example:

The system A ⇄ C + D has a Keq = 4.88x1027. If a solution is made with [A] = 1.50 M, what will the concentration of C be when equilibrium is reached?

Our Equilibrium expression will look like this:

And our ICE table will look like this:

Substituting in our “E” values gives an equilibrium expression that looks like this:

Now, however, things are very different than what we’ve worked with before. We can see that by checking whether “x” will be big or small.

In this case, since K is large and x is on the numerator of the fraction, the only way that the equation can work is if x is big. (Again, it is important to remember that in this context, big only means not tiny. So a value of 0.1 is considered big.)

Solving a big x problem is “impossible.” I put that in quotes because, of course, if you had enough computing power, you can certainly solve an algebra equation with only one variable. However, the amount of computing power needed is surprising. Even your TI-8? graphing calculator (which has an algebraic solving function) does NOT have enough power to solve this problem.

So, what do we do? We simply flip the reaction on its head.

Let’s think about an equilibrium system as existing on a number-line. The left end of the number line is when the system is entirely products and no reactants. The right end of the number line is the opposite - all products and no reactants. Equilibrium ALWAYS lies somewhere between these two extremes.

If K is small, then equilibrium lies toward the reactant end. If K is large, equilibrium lies near the right end of this scale.

In all cases, “x” is the change from the starting conditions to those of equilibrium. So, little x problems are those where the starting concentrations are close to equilibrium - that is a little x problem either starts with mostly (or entirely) reactants and has a little K or it starts with mostly (or entirely) with products and has a big K.

Any other situation will result in a big X problem. So, for example if K is small, but you start with mostly (or entirely) products, or (as in our case) you have a large K and start with mostly (or entirely) reactants. Visually, those situations would look like this:

So, knowing this, how do we “flip a reaction”? We are going to assume that the reaction shifts entirely from one end of the scale to the other (in this case from only reactants and no products, not no reactants and all products). We’ll do that using our ICE table and a little bit of logic.

Here is the ICE table again. If we were ignoring equilibrium, we could imagine that the reaction happens completely, that is it reacts until we run out of the reactant. 

So, we’ll run the reaction until there is no A left. In order for that to happen, x must be 1.50. Plugging that value back into the ICE table gives us this:

This is obviously not at equilibrium since we cannot end with [A] = 0. Let’s create a new ICE table, using the bottom row of our current ICE table as the “I” row of the new one. It will look like this:

The reaction clearly needs to run to the left, and the ICE table will fill out like this. We’ll use “y” instead of “x” so that we don’t confuse ourselves at the end of the problem.

Our Equilibrium expression will now be:

Since K is large and we want the numerator to be as big as possible, which will only be true if y is small. That means that we have turned a big X problem into a little y problem. Those we can solve! In this case we end with y = 4.61-28.

Now (hopefully) you can see why we used "y". Our last step is always to plug our value back into the last row of the ICE table. Had we used "x" there would have been TWO tables with missing "x" values and we would have had to choose which to work with. Having used "y" it is quite clear where we need to substitute our solution.

So, we end the problem with:

The simple version:

When you have a big X problem, run the reaction to the other end

(until you end up with a 0 concentration) and then start again with a little x problem.