Introduction to
Equilibrium
AP Chemistry
Equilibrium
Defined as the state when the
concentrations of all reactants and
products remain constant over time.
 This occurs when the rate of the
forward = rate of the reverse
reaction.
 Equilibrium happens because reactants
and products never stop colliding!

Equilibrium Graph
•After 60 seconds
the reaction has
reached equilibrium
•The concentrations
of the reactants and
products are
constant, but not
necessarily equal!!!
•Rate of forward
reaction = rate of
reverse reaction




Not all reactions will go to completion. The equilibrium
expression will tell us how much of each species is
present at equilibrium
The Keq is constant at a given temperature and depends
on the equilibrium concentrations of products and
reactants.
In a general reaction:
aA
+
bB
⇄
cC
+
dD

The lower case letters represent the stoichiometric coefficients and
the upper case letters represent reactants and products.

The law of Mass Action states that the

Equlibrium constant is:
Keq
=
[C]c[D]d
A]a[B]b


Example 1
What is the equilibrium expression (Keq)
for the following dissociation?
 2N2O5(g) ⇄ 4NO2(g) + O2(g)

Keq =
[NO2]4[O2]
[N2O5]2
Equilibrium Constants

When Keq > 1 then products are favored
in the reaction
– Which means that they are more products
than reactants at equlibrium

When Keq < 1 then the reactants are
favored in the reaction
– Which means there are more reactants then
products at equlibrium.
Equilibrium Constants
When a reaction is multiplied by a factor
of “n”, then the Keq is raised to the ‘nth’
power.
 2H2O  2H2 + O2
Keq = [H2]2[O2]

[H2O]2


Multiply by 2
(Keq)2
4H2O  4H2 + 2O2
= [H2]4[O2]2
[H2O]4
Equilibrium Constants

The equilibrium constant for the reverse
reaction is
Keq(rev) = 1/Keq(fwd)
Heterogenous Equilibria
Pure solids and pure liquids, and pure
water are not included in equilibrium
expressions because there concentrations
remain essentially constant over time.
 Write the equilibrium expression for the
following
 PCl5 (s) ↔ PCl3 (l) + Cl2(g)

Keq = [Cl2]
Kp versus Kc






In homogeneous gas equilibria, partial pressure may be
used in place of the concentrations—the format remains
the same.
The term Kp is often used to describe the equilibrium
constant in terms of partial pressure.
Kc may be used when the concentrations of the
reactants and products are expressed in terms of
molarities.
Numerical values can be calculated for any of these.
By convention, units are usually omitted for any
equilibrium constant.
Water is excluded in aqueous media because the
concentration of water does not change and/or water as
a product of a reaction is insignificant in aqueous media.
Example 2
In the homogeneous gas equilibrium,
dinitrogen tetraoxide decomposes as
follows:
 N2O4 (g)
⇄ 2NO2(g)


Kc =
[NO2]2
[N2O4]
Kp =
(PNO2)2
PN2O4
The Relationship Between Kp and Kc
For the general reaction:
 aA(g)
⇄ bB


Using the Ideal Gas Law:

pA V = nRT
and
pA = nA RT

V

Similarly: pB = nB RT


V
Substituting these terms into the
expression for Kp:
Kp =
nB RT
V
b
nART
V
a
Since n/V is concentration (Molarity):
Since n/V is concentration (Molarity):
Kp =
[B]b
(RT)
a
[A]
Δn
Kp = Kc(RT) Δn
Δn = Σ (moles products) – Σ(moles reactants)
Δn = b - a
Sample Problem
For the reaction:

N2(g) +
3H2(g)
⇄
2NH3(g)
 The following equilibrium pressures at 472oC
were obtained: H2 at 7.38atm, N2 at 2.46 atm
and NH3 at 0.166 atm. From these data,
calculate the Kp and the Kc at this temperature.

Determining Keq :Quantitative Equilibium
What is the equilibrium constant for the following
reaction in terms of Kc?

2NO(g)
+O2(g) ⇄ 2NO2(g)
 The reaction was run at 230o C and the
following concentrations were obtained:
 [NO]= 4.065x 10-2 M
 [O2] =9.525 x 10-2M
 [NO2] = 11.625 M
Sample Problem 2

The Kp for phosgene gas (COCl2) at 800K is 3.3. At
equilibrium, the pressure of CO was found to be 0.455
atm and that of Cl2 was found to be 0.356 atm. What is
the partial pressure of the phosgene at equilibrium? The
overall reaction is:

COCl2(g)
⇄
CO(g) +
Cl2(g)