Equilibrium Calculations
ICE Tables:
Sometimes you are given information about initial reaction concentrations and are asked about more
than just the Reaction Quotient Q. Depending on the information given, you can either calculate the Keq or
the equilibrium concentrations. To do so you must organize your information into what is called an ICE
table, Initial, Change, Equilibrium.
There are two different ways to use the ICE table:
1. Calculation of Equilibrium Constant
This is by far the easier calculation of the two and the one you are most likely to see on the test.
Here they will give you initial concentrations for everything, and the equilibrium concentration of
1 thing, and ask you for the Keq.
Ex: For the following reaction 0.250 M of each of the reactants and products was placed in a 1 L
flask and allowed to come to equilibrium. At equilibrium, the concentration of NO2 was measured
to be 0.261 M. Calculate the Kc for this reaction.
NO2 (g) + SO2 (g)  NO (g) + SO3 (g)
We set the table up this way with the information we have:
NO2 (g) +
Initial Conc: 0.250
SO2 (g) 
NO (g) +
SO3 (g)
0.250
0.250
0.250
Change:
Equil Conc:
0.261
To find the equilibrium concentrations of everything else in the table, we have to know how much
the concentration change is and then apply that to everything in the reaction. For this reaction,
the balanced equation has a coefficient of 1 for everything so the change is the same for
everything. We can see that NO2 got bigger, so we know a positive change occurred for NO2 and
SO2 and we can assume a negative change occurred for NO and SO3.
We add this information to the table:
NO2 (g) +
SO2 (g) 
NO (g) +
SO3 (g)
Initial Conc: 0.250
0.250
0.250
0.250
Change:
+x
+x
-x
-x
Equil Conc:
0.261
0.250 + x
0.250 – x
0.250 – x
We can easily solve for x:
0.250 + x = 0.261, x = 0.011
So plugging x back in the Equil Conc line gives the equilibrium concentrations for everything: SO2
– 0.261, NO = 0.239, SO3 = 0.239.
Then you simply plug these numbers into the Keq and get your answer.
The only thing to remember is that you must drop down any coefficient in the balanced equation
to join the x in the table.
2. Calculation of Equilibrium Concentrations
This is the more difficult form of problem simply because the algebra involved is a little bit more
advanced. But conceptually, you do the same thing. The table is used in the exact same manner.
In this type of problem, you will be given initial concentrations and the Keq, and be asked to solve
for the equilibrium concentrations.
Ex: The reaction Br2 (g) + Cl2 (g)  2BrCl (g) has an equilibrium constant of 6.90. If 0.100 mol of
BrCl is placed in a 500 mL flask, calculate the equilibrium concentrations of Br2, Cl2, and BrCl.
You must first calculate the concentration of BrCl: M = mol/L = 0.100/0.5 = 0.200 M
Set up the initial ICE table as follows:
Br2 (g) +
Initial Conc: 0
Cl2 (g) 
2BrCl (g)
0
0.200
Change:
Equil Conc:
Now, since only BrCl is initially present, it must decrease, and Br2 and Cl2 must increase. Place the
x’s in the table appropriately, making note of the coefficient in front of BrCl.
Br2 (g) +
Cl2 (g) 
2BrCl (g)
Initial Conc: 0
0
0.200
Change:
+x
+x
-2x
Equil Conc:
x
x
0.200 – 2x
We don’t have any information about the equilibrium concentrations so we cannot solve for x as
we did in the previous problem. But we do have the Kc for this problem and can plug all of the
equilibrium concentrations into the equilibrium expression to solve for x.
Kc =
[BrCl]2
[Br2] [Cl2]
6.90 = (0.200 – 2x)2
(x)(x)
Now the easiest way to solve this particular problem is to square root both sides to clear out the
squares, but every problem is going to be different.
2.63 = 0.200 – 2x
x
2.63x = 0.200 – 2x
4.63x = 0.200
x = 0.200/4.63 = 0.0432
This answer for x can then be plugged back in to the Equil Conc line of the table to solve for each of
the reactants and products. Br2 = 0.0432 M, Cl2 = 0.0432 M, BrCl = 0.200 – 2(0.0432) = 0.114 M.
The problem can be solved by carrying through the square, and plugging into the quadratic
equation, or by using the solver function on a graphing calculator too. But they will normally give
you something that has an easy out like taking the square root here.
Possible Simplification:
If you set up your problem and the math looks just awful, there is probably a simplification that
can be made.
This ONLY works if Keq is either really SMALL (like x10-12 or smaller) or really BIG (like x1012 or
larger).
For really SMALL Keq, it means that products are barely formed and reactants barely change, that’s
how you get a small Keq. We can therefore ASSUME that the concentration of the reactant DOES
NOT CHANGE hardly at all, and you drop the x off in the calculation.
So 2.4 x10-25 = (2x)2(x)
(2.00 – 2x)2
becomes
2.4 x10-25 = (2x)2(x)
(2.00)2
The equation on the left is solvable, but really ugly. The equation on the right just requires a cube
root to solve for x. If you are going to use this simplification, YOU MUST state that you are doing
so in your problem. You would say “Because K is so small and favors reactants, we can assume
that 2x is going to be much smaller than 2.00, and that 2.00 – 2x = 2.00.” You still have your ICE
table and everything else. Just write that line somewhere, get your point, and go on with the
problem.
For really LARGE Keq, it means reactants barely exist and there is almost exclusively products,
that’s how you get a large Keq. We can therefore ASSUME that the concentration of the products
DOES NOT CHANGE, and you drop the x off in the calculation in the same manner as above.
This type of simplification only comes up as needed every few years on the first free response
question. Anticipate that you will have to use the ICE table in the problem. The simplification you
might or might not need.
Different Types of Keq (other than the Kc that was used in all previous examples):
Kp – Used when gases are present. Instead of concentrations in molarity, partial pressures are used. This
would be combined with some PV=nRT type calculations most likely, and Dalton’s Law of Partial
Pressures.
Kp is written the same way except we use PA instead of [A].
Ex:
2S (g) + 3O2 (g)  2SO3 (g)
Kc =
[SO3]2
[S]2 [O2]3
Kp =
P2SO3
P2S P3O2
Ksp – Solubility Product
Ksp is used for solids that are only slightly soluble in water, meaning that the only partially dissociate.
We use the equilibrium expression to calculate the “molar solubility” which is how much of the solid will
dissolve in water. Because we have a solid and ions in solution, the Ksp will never have a denominator
because we don’t include solids. This simplifies the math A LOT.
Ex: Calculate the molar solubility of Fe(OH)3 if the Ksp = 1.6 x10-39.
You will typically be asked to write the balanced equation. All you are doing is splitting the solid
up into its ions, using the subscripts as coefficients to balance it.
Fe(OH)3 (s)  Fe3+ (aq) + 3OH- (aq) and Ksp = [Fe3+] [OH-]3
If you set up your ICE table with the known information, we get:
Fe(OH)3 (s)  Fe3+ (aq) + 3OH- (aq)
Initial
solid
0
0
Change
-x
+x
+3x
Equil
solid
x
3x
So Ksp = [x] [3x]3 = (x)(27x3) = 27x4 = 1.6 x10-39
x = 8.8 x10-11
From here you could answer questions about the concentration of either of the ions, or state that
the molar solubility of the solid is 8.8 x10-11.
Sometimes you might then be given information about having one of the ions ADDED at the equilibrium
point. You would then just add that to the Equil line in your table and recalculate.
Ksp is one of the times you will probably have such small values for K that you can make that simplifying
assumption from above.
Ka & Kb – Weak Acids and Bases:
By definition, acids and bases are considered weak if they DO NOT fully dissociate in water. This means
that they are actually in a state of equilibrium. So when we write the equilibrium expression for a
reaction that is a weak acid or base, it is a type of Kc that will use molarity concentrations, but will always
have water as a reactant as a liquid, which can be dropped out.
Acetic Acid (most common weak acid)
CH3COOH (aq) + H2O (l)  CH3COO- (aq) + H3O+ (aq)
Ammonia (most common weak base)
NH3 (aq) + H2O (l)  NH4+ (aq) + OH- (aq)
All weak acids and bases will have the same format as these two examples, and any problems with them
are solved the same way as any other Kc problem.
The only additional thing that makes acids and bases special is that they involve water which itself can be
slightly dissociated giving us Kw, the dissociation constant for water. This is 1.0 x10-14 by definition and
we can use that fact for some problems.
Kw = Ka x Kb for acid-base pairs. Sometimes you will be given a Ka and asked for the Kb. You use this
relationship to solve that problem.
Le Chatelier’s Principle
Once something is at a state of equilibrium, it can be poked and proded and manipulated in ways that will
change or shift that state of equilibrium. The balance of reactants and products will change. Le
Chatelier’s Principle can give us guidance on what kinds of changes will occur. The trick is to think of
everything as part of the reaction. Not just the products and reactants, but temperature (or heat),
volume, and pressure too.
The easiest adjustments to make are to the actual reactants and products them selves. The general rules
of thumb are this:
If something gets taken away, equilibrium will shift to replace it.
If something gets added, equilibrium will shift to use it up.
The next most common adjustment is to increase or decrease the pressure (normally by changing
volume). This only has an affect on reactants and products that are in a gas state. To determine the
effect, you must look at the TOTAL amount of moles of gas on the reactant and the product side.
Increasing pressure shifts reaction to side with less moles of gas.
Decreasing pressure shifts reaction to side with more moles of gas.
Adding an inert gas to the mix does absolutely NOTHING to the equilibrium.
If there is the same number of moles of gas on each side, then pressure has no effect.
The last way to adjust a reaction is to alter the temperature. You already know that reactions can either
be endothermic or exothermic, and that endothermic reactions have a + ΔH which exothermic reactions
have a – ΔH. For equilibrium, think of the ΔH as a part of the reaction that can either be a reactant (for
endothermic) or product (for exothermic).
Thinking of it this way allows you to then look at it just like you make adjustments to concentration
above.
An endothermic reaction has heat as part of the reactants, so raising the temperature would be
like adding more heat and equilibrium will shift to products to use it up.
An endothermic reaction has heat as part of the reactants, so dropping the temperature would be
like taking away heat and equilibrium will shift to reactants to replace it.
An exothermic reaction has heat as part of the products, so raising the temperature would be like
adding more heat and equilibrium will shift to reactants to use it up.
An exothermic reaction has heat as part of the products, so dropping the temperature would be
like taking away heat and equilibrium will shift to products to replace it.
The other thing to note is that changes in temperature are THE ONLY thing that will change the actual
value of Keq. The change in Keq will depend on whether the reaction is endo or exothermic, and whether
temperature is raised or lowered. But in general, shifting to reactants lowers Keq and shifting to products
raises Keq.