Chemical Equilibrium
Part 1
Ch. 15 in Textbook
Fanpop.com
I. Static vs. Dynamic
Generally speaking,
an equilibrium is a
state of balance.
 A static equilibrium is
one in which there is
no motion.

Tutorvista.com
Wirednewyork.com
A dynamic equilibrium
is one in which there is
motion despite there
being no net change.
 A chemical equilibrium
is an example of a
dynamic equilibrium.

Ene.on.gov.ca
II. Chemical Equilibria




Suppose we have the
gaseous reactants I2 and
H2. They undergo a
synthesis reaction to
form HI:
H2 + I2 → 2HI
As HI accumulates, some
molecules have enough
energy to decompose to
H2 and I2:
2HI → H2 + I2
Coolchaser.com

As this process
continues,
eventually the
rate of the
forward reaction
equals the rate of
the reverse
reaction.
Link
Tutorvista.com

The final
equilibrium mixture
will contain both
reactants and
products.
Tutorvista.com

Although the amounts are
NOT necessarily equal, the
amounts must eventually
remain constant since the
forward and reverse
reactions are occurring
simultaneously and at the
same rate.
Mmsphyschem.com

Although there is no
noticeable net change
in reactants or
products, the reaction
is still proceeding in
the forward and
reverse directions,
making chemical
equilibrium a dynamic
equilibrium.
Tutorial
Freefoto.com
HW: 15.2
III. Kinetics Flashback!!!!
Given: A  B
 Assume that the
forward and reverse
reactions are both
elementary steps.
 Rate (forward) = kf [A]
 Rate (reverse) = kr [B]
 At equilibrium: kf [A] =
kr [B]

Motivatedphotos.com



Rearranging:
[B]/[A] = kf/kr = constant
What is the meaning of
this?!
At a given temp. the
products and reactants will
ALWAYS be in the same
ratio at equilibrium, no
matter the starting point…
Blog.al.com
HW: 15.3
IV. The Law of Mass Action
This proportion can be
represented by the
equilibrium expression.
 The Law of Mass Action
states that the
equilibrium expression
depends on the
equilibrium
concentrations of
reactants and products.

Marcvallee.wordpress.com
For the reaction:
aA + bB  pP + qQ
the equilibrium expression
is written as:
K = [P]p [Q]q
[A]a [B]b
where K is the equilibrium
constant.

Oudaily.com

Ex) What is the
equilibrium expression
for the synthesis of
hydrogen iodide?
Threadless.com
HW: 15.6,
15.20
In KINETICS, the rate law
did NOT depend on the
stoichiometry of the net
equation, only on the ratedetermining step.
Yes,
typo
 In EQUILIBRIUM, the
equilibrium expression
DOES depend on the
stoichiometry of the net
equation.

Efuse.com
V. The Equilibrium Constant, K,…
A) Dependence





…does not depend on the
reaction mechanism.
…does not depend on the
initial concentrations of
reactants and products.
…depends on the equilibrium
concentrations of reactants and
products.
..also depends on temperature!
(more on that later…)
We generally omit the final
units.
Tampabayrun.com
B) Magnitude

Since products are divided by
reactants in the expression:


A larger K value (>>1) means
that products dominate the
final reaction mixture and we
say “the equilibrium lies to the
right.”
A smaller K value (<<1) means
that reactants dominate the
final reaction mixture and we
say “the equilibrium lies to the
left.”
Balthaus.org
HW: 15.10
C) Kc
This is the constant when
concentrations are
expressed in molarity.
 It is the most common
version of K.

Dummidumwit.wordpress.com
D) Kp
This is the constant when
concentrations (of gases)
are expressed in terms of
partial pressures (atm).
 K = (PP) p (PQ)q
(PA)a (PB)b

Lankapeacewatch.com
E) Relationship between Kc
and Kp
Note: My notes are WAY
better…this PowerPoint slide
doesn’t even have a stupid picture in
it…or a picture of another chemistry
PowerPoint slide…of another
PowerPoint Slide…of another
PowerPoint slide…ad infinitum.
Files.chem.vt.edu
E) Relationship between Kc
and Kp (fo’ real dis time)
PV = nRT
 P = (n/V) RT
 For substance A:
PA = [A] RT
 Do this for all partial pressures in an
equilibrium mixture and we get:
Kp = Kc (RT)Δn
where Δn is the moles of gaseous products the moles of gaseous reactants.

Link
Ex) Given: 2SO3(g)  2SO2(g) + O2(g)
 Write the Kp expression.

W3.org

Calculate Kp if Kc = 4.08 x 10-3 at 1000 K.
HW: 15.8 (a)
& (b), 15.12
F) Direction
Equilibrium can be
reached from any
direction (all reactants
or all products or any
mixture of both).
 Kc (forward) = 1/Kc
(reverse)

Fotolia.com
HW: 15.16
Breakdancing Baby Break
VI. Heterogeneous Equilibria



These involve one or more
reactants or products present
in a different phase.
Many equilibria do not involve
dissolved species or gases.
The concentration of a pure
solid or liquid is constant and
therefore does not appear in
the equilibrium expression.
Erroraccessdenied.com
Ex) 3Fe(s) + 4H2O(g)  Fe3O4(s) + 4H2(g)
 Write the Kp expression.

Pig Iron from
Bbc.co.uk
Thus, only the partial
pressures of water and
hydrogen affect the
equilibrium expression.
 Don’t forget though, the
solids must be present in
order for the equilibrium
to be established.

Open.salon.com
HW: 15.18
VII. Calculating K




Use an ICE chart: Initial,
Change, and Equilibrium.
First, determine all initial
concentrations.
Second, determine any
available equilibrium
concentrations.
Third, use the reaction
stoichiometry to determine
the change in concentration
from the initial to equilibrium.
Burnblog.burningman.
com
Fourth, determine all
equilibrium
concentrations.
 Fifth, plug and chug
into equilibrium
expression and find K.

Nitrocotton.com
Ex) 2SO3 (g)  2SO2(g) + O2(g)
 A vessel at 1000 K contains 6.09
x 10-3 M SO3. At equilibrium,
the SO3 concentration is 2.44 x
10-3 M. What is the value of Kc?

Portlandsentinel.com
HW: 15.24, 15.26
VIII. The Reaction Quotient
(Q)


Given the initial
concentrations of reactants
and products, we can
determine the direction the
reaction will proceed in.
Substitute concentrations into
the reaction quotient
expression, which is the same
as K except NOT at
equilibrium.
Mommylounge.com
K
uses equilibrium
concentrations!!!!!!
Q uses initial
concentrations!!!!!!
Erie.gov
K-Dubs, 6th grade



If Q>K, then there is an
excess of products which will
react to form reactants to
establish equilibrium (shift
left).
If Q<K, then there is an
excess of reactants which will
react to form products to
establish equilibrium (shift
right).
If Q=K, equilibrium is
established.
Ajobi.net
HW: 15.28
IX. Calculating Equilibrium
Concentrations

Ex 1) At 500 K the value of
Kp is 0.497 for the following
reaction:
PCl5(g)  PCl3(g) + Cl2(g)
If the partial pressure of PCl5
is 0.860 atm and the partial
pressure of PCl3 is 0.350 atm
at equilibrium, what is the
partial pressure of Cl2 in the
mixture?
Blogs.nature.com

Ex 2) Using the equilibrium
from Ex. 1, what are the
equilibrium partial pressures
of all species if PCl5 has an
initial pressure of 1.66 atm?
Desperatelyseeking
suddenlysusan.
wordpress.com
HW: 15.34, 15.36, 15.38, 15.40
TO BE CONTINUED…
Intlxpatr.wordpress.com
Equilibrium Part 2
Ch. 15 in Textbook
Bethedream.com
X) Le Chatelier’s Principle
A) Stress



Once an equilibrium is
established, the forward and
reverse reactions will proceed
at equal rates and the
amounts of reactants and
products will not change.
Equilibrium will continue
indefinitely unless conditions
are somehow changed to
disrupt it.
A disturbance to the
equilibrium is called a stress.
Wolfescape.com
B) The Principle


When you deal with stress in your
own life (e.g. taking this class) you
do something to rid yourself of the
stress and re-establish your personal
equilibrium.
Similarly, French industrial chemist
Henri-Louis Le Chatelier stated that
when a stress is applied to a
chemical equilibrium, the equilibrium
will shift its position to eliminate
the disturbance.
Biocrawler.com
C) Changes in Concentration


Adding a reactant or a
product to an equilibrium
mixture causes the
equilibrium to speed up in
the direction that will
remove the excess.
This shift is only temporary
and a new equilibrium is
established with new
equilibrium concentrations
(although no change in K).
Chem.ufl.edu

Similarly, removing a
reactant or product from an
equilibrium mixture causes
the equilibrium to speed up
in the direction that will
make up for the deficit.
Stevenwolffinearts.com

How can we maximize our
ammonia output in the Haber
Process?
N2(g) + 3H2(g) ↔ 2NH3(g)
Zazzle.com
Link
D) Changes In Volume/Pressure


The pressure may be
increased by reducing
the container volume or
the addition of a nonreacting gas.
The equilibrium will
speed up in the
direction that will
reduce the pressure
through the production
of fewer moles of gas.
En.wikibooks.org



Similarly, the pressure may be
decreased by increasing the
container volume or the
removal of a non-reacting gas.
The equilibrium will speed up
in the direction that will
increase the pressure through
the production of greater
moles of gas.
This shift is only temporary
and a new equilibrium is
established with new
equilibrium concentrations
(although no change in K).
Iforgottocitethis.com

How can we maximize our
ammonia output in the Haber
Process?
N2(g) + 3H2(g) ↔ 2NH3(g)
Cafepress.com
Link
E) Changes in Temperature

Increasing the
temperature results in
an excess of heat,
causing the equilibrium
to speed up in the
direction that removes
the heat
(endothermic).
C-p-p.co.uk

Similarly, lowering
the temperature
results in a deficit of
heat, causing the
equilibrium to speed
up in the direction
that produces heat
(exothermic).
Tutorvista.com




Unlike all other stresses,
however, the temperature
changes the value of K.
In other words, a completely
different equilibrium mixture
is created at the different
temperature.
If a reaction shifts to the
right, then K increases.
Justhavingfunaroundthe
If a reaction shifts to the left,
house.com
then K decreases.

How can we maximize
our ammonia output in
the Haber Process?
N2(g) + 3H2(g) ↔
2NH3(g) + heat
Wow.com
Link
F) Adding Catalysts



A catalyst lowers the
activation energy for both
the forward and the reverse
reactions.
A catalyst therefore speeds
up the forward and reverse
reactions equally.
A catalyst speeds up the
rate at which equilibrium is
achieved, but the
equilibrium mixture and K
remain exactly the same.
En.wikivisual.com

Haber found that
iron mixed with
metal oxides allowed
the Haber Process to
run at sufficiently
low temperatures.
Dmotiv8.blogspot.com
HW: 15.44, 15.46
Mmonla.wordpress.com
Drumming Chimp Outro

