Levi Howard
Jordan Leach
In a Reaction
 When a reaction occurs, eventually the molarity
(concentration) of the reactants and products will
become constant, not necessarily equal. This
determines when the reaction is “over”, however the
reaction never truly stops. This is demonstrated
below.
In the Reaction
Which of the following on the graph shows when the
reaction has reached Equilibrium?
C
A
B
Equilibrium Cont.
The correct answer is C, because it is the only point at a
straight horizontal line. B shows where the two
concentrations are equal, but that has nothing to do
with equilibrium. A is just… wrong.
Equilibrium Constants (K)
 There are two very important constants that go with
equilibrium, and these two are Kc and Kp. Kc is for using
concentrations, which is associated with most problem,
but some will have the partial pressures for compounds,
where you would use the pressure constant, Kp. If these K
values are large, that means that the reaction is
spontaneous. If it is small, the reverse reaction would be
spontaneous. K also only depends on temperature, T.
Knowing which type of K to use helps to organize
information and use the correct equations. Since reactions
can be revered, the constant can be also. If a reaction is
reversed, you would use the reciprocal of the normal K
value. If it is multiplied by a number, the original K will
have an exponent of that number.
Equilibrium Expression (K)
 Writing an equilibrium expression is simple. The
important thing to remember is to write Products
over Reactants equals the constant, K. Phases are
important, because for equilibrium expressions, the
elements must be in either the gas phase or aqueous
(aq). Solids and liquids are removed from the
expression completely. If you are using Kc, then you
would put in the equilibrium concentrations for the
compounds in their respective spot. If you are using
Kp, then write products over reactants, and plug in the
partial pressures at equilibrium that are given.
Equilibrium Expressions Cont.
After taking note of the phases, and concentrations (or
partial pressures) of the reactants and products, be
sure to look at the coefficients (the big number before
the chemical formula). This coefficient decides the
exponent of the compound. For example, in the
reaction below,
N2 + 3H2 <-> 2NH3
You notice the 3 before hydrogen, and 2 before
ammonia. This just means that in an expression, H2 is
cubed, and NH3 is squared. Since N2 technically has a
1, it is to the first power which means it isn’t changed.
Writing Expressions
 When 0.40 mole of SO2(g) and 0.60 mole of O2(g) are placed in an evacuated
1.00-liter flask, the reaction represented below occurs. After the reactants and
the products reach equilibrium and the initial temperature is restored, the
flask is found to contain 0.30 mole of SO3. Based on these results, the
equilibrium constant, Kc, for the reaction is…
 2 SO2(g) + O2(g) <-> 2 SO3(g)
 A (o.30)^2 / [(o.45)(0.10)^2]
 B (0.30)^2 / [(o.60)(0.40)^2]
 C(2 x 0.30) / [(0.60)(0.40) ^2]
 D(0.30) / [(0.45)(0.10)]
Expressions Cont.
The answer is A. This is because in the reaction, 0.30
mol SO3 was found. SO3 also had a 2 as a subscript, so
this amount would be squared in an expression. This
means that the answer is going to have [0.30^2] on top.
This means C and D cant be the answer. Also, when a
reaction happens, reactants will be used up. In B, the
concentrations of the reactants are the exact same as
before the reaction happened, which is impossible.
This leaves A as your answer.
Expressions Cont.
Hydrofluoric acid, HF, dissociates in water as
represented by the equation below. Write the
equilibrium constant expression for the dissociation of
HF in water.
HF(aq) + H2O(l) <-> H3O+(aq) + F-(aq)
Expressions Cont.
The first thing to look for is the phases of all the
reactants and products. You will notice H2O as a
liquid, which means it will not be used in our
expression. This leaves H3O and F- on top, over HF on
the bottom.
[H3O+] [F-]
___________
[HF]
Is it done yet?
 You now remember that Equilibrium is when the concentrations are




remaining constant. But how exactly can you tell when it is when you
aren’t looking at a graph? Well, this is where Q comes in. Q is
calculated just like K, by putting products over your reactants.
However, with Q, you use whatever concentrations you have, which
aren’t necessarily at equilibrium. To know if it is at equilibrium, Q
would be equal to you K value, because the numbers you used for Q in
fact were the equilibrium concentrations. To put it simply…
Q = Products over Reactants for any concentration
Q<K means it is not at EQ, need to make more products.
Q=K means it is at Equilibrium
Q>K means it is not at EQ, need to make more reactants.
Organizing with ICE Tables
 ICE is an acronym. These tables are great for organizing information in
an equilibrium reaction by showing the change in concentration as the
reaction goes on to equilibrium. Like equilibrium expressions, liquids
and solids are omitted from ICE tables. It stands for…
 I – Initial concentration. This is before the reaction starts.
 C – The change in concentrations. This is usually where an unknown
variable (x) is put into. All changes in concentrations are proportional
to the coefficients.
 E – The concentration of the atom (or compound) at Equilibrium.
Ice Tables in Action
 Hydrofluoric acid dissociates in water according to the reaction below.
If the Ka of hydrofluoric acid is 7.2 x 10^-4, calculate the molar
concentration of H3O in a .40M HF solution.
 HF(aq) + H2O(l) <-> H3O+(aq) + F-(aq)
Ice Tables Cont.
The ICE table shown below is a great way to organize information. Hydrofluoric
acid, HF, was given to be 0.4M, so that will be put in the box for initial
concentrations. It shows that the products start out at 0, because they havent
been made yet. After the equation a certain amount, x, has been added to the
product’s size, while being taken away from the reactant side. To find the
molar concentration, we need to find the value for x. This can be done by
putting all of this information into an equilibrium expression.
[x] [x]
_______ = 7.2 x 10^-4
[0.4]
0.4M
0M
0M
-x
+x
+x
0.4-x
x
x
x^2 = 2.88 x 10^4
x = 0.0170M
Le Chatelier’s Principle
 The scientist Henry Louis Le Chatelier discovered how to change
equilibrium in a reaction. This is possible by doing a number of things,
such as changing the concentration, temperature, volume, or partial
pressure. Changing one of these will cause the reaction to shift left or
right, having a larger concentration of either products or reactants to
counteract the change.
Le Chatelier
 In the exothermic equation below,
HCO3(ap) + OH(aq) <-> H2O(l) + CO3(aq)
ΔH=-41.4kJ
Raising the temperature would shift the reaction how?
A. Left
B. Right
C. It would not shift
Why A?
 If a reaction is exothermic, it means that as it goes on,
heat is produced. In other words, heat is a product.
Raising the temperature would mean adding a
product, which would make the equation want to
make more products, and have the equation shift left.