Download ICE Box Diagrams - AP Chemistry Period 5

Applications of ICE Box Diagrams
A+B⇌C
Now, we can complicate this problem just a bit by having the concentrations of A and B be initial, rather
than equilibrium, quantities. The problem is different because we put some of A and B into the reaction
vessel, and some of each reacts (i.e., goes away) to form some C. The initial concentration of C is zero.
The technique in this case is to set up an ICE-Box diagram
Component
A
B
C
Initial
[A]o
[B]o
[C]o
Change
-x
-x
+x
Equilibrium [A]o - x [B]o - x [C]o + x
The finish line is found by adding the start and the change entries in each column. We are assuming the
reaction goes forward when we used a "-x" for A and B; if we are wrong (the reaction actually goes
backwards) the only thing that will happen is that x will be negative (which is no big deal). The entries on
the finish line are the equilibrium concentrations of each component:
[A] = [A]o – x
[B] = [B]o – x
[C] = [C]o + x
Plugging these into the equilibrium equation yields
If Kc , [A]o , [B]o , and [C]o are given, then this equation becomes simply a quadratic equation in x and is
solved via the familiar quadratic formula
Example Basic Start-Change-Finish problem
Suppose we start with equal initial concentrations of A and B, [A] o= [B]o = 0.14 M, and do the same
reaction at the same temperature so that that Kc = 96.2 M-1. What is the new equilibrium concentration of
the product C?
Solution It’s easier to do algebra with symbols than numbers, so let’s call  = [A]o = [B]o = 0.14 M, and
[C]o . Since there is no C initially, [C]o = 0 and the equation becomes
According to our algebra formulae, ( - x)2 = 2- 2x + x2 and so
This last expression is in quadratic form with
a = Kc = 96.2 M-1
b = - (1 + 2Kc) = - (1 + 2(0.14 M)( 96.2 M-1)) = -27.94
c =  Kc = (0.14 M)2(96.2 M-1) = 1.89 M
Note that the units come out so that every quantity is in molarity – this is correct since x must be in units of
molarity. Putting these numbers into the quadratic formula gives us
The quadratic formula gives two solutions – which is the correct one? Notice that the first solution is larger
than the initial amount of reactant – so even if 100% reacted, the most product we could get is 0.14M, so
we reject that solution and take
x = 0.107M
[A] = [A]o – x = 0.14 – 0.107 = 0.033M
[B] = [B]o – x = 0.14 – 0.107 = 0.033M
[C] = [C]o + x = 0 + 0.107 = 0.107M
The last number is our desired answer; notice we solved for all the other numbers in the problem, too – this
is a real good habit, since you maximize learning for each problem solved. Another good habit is to run a
check on the final numbers by calculating Kc again:
The answer is a little different from the original number, because we rounded the answers in the
denominator. This is normal, and shows how rounding can introduce error. For our purposes here, this is
close enough. When precision matters more, we’ll be more careful.
There are several things that should be pointed out now; the most important is the algebra got complicated.
This will get worse in other problems we consider below, so the reader is advised to go slow and be
extremely careful. Mistakes in algebra are easy to make and hard to find. The other important observation
is that organization and neatness drastically improve your chances of getting the answer right the first time,
and make it much simpler to find mistakes when you make them. You will make mistakes, so do yourself a
favor and make these problems look good on paper. You’ll be very glad you took the extra effort.
Now, let’s take a look at what happens when we increase the precision of the calculation: in the above
calculation, we could have used more digits:
With this increased precision, we get for the equilibrium constant
The answer is worse when we used more digits! The lesson here is that increased precision doesn’t
necessarily mean more accurate answers. Use the rules of significant digits carefully and don’t worry too
much if the answers are a little different (a few percent error, say less than 5%, is OK). There are ways to
get better answers by careful choice of calculation method. Suffice to say that small errors in small
denominators can cause relatively large errors – this is a good place to be cautious.