Imagine a jar of water open to the atmosphere. We know, that after some time has passes, that
the water will all evaporate and disappear. But what if we put a lid on the water container?
The water does not disappear, in fact, it could stay in the jar for years. We do, however, see
little beads of water form along the upper walls and on the top of the jar as the water
evaporates, and then condenses inside the container. The gas molecules are not free to “fly
away” and thus the volume change of water is minimal (depends on how tight we have that
lid!!), or perhaps even constant!
On the molecular level (which is where we should be thinking now) chemical and physical
changes invariably are composed of two processes, with each process undoing the work or the
progress of the other. When one process occurs more rapidly than the other, we see something
happen - a gas is evolved, a precipitate forms etc. . . When both processes occur at the same rate
we think that nothing is happening – and we oftentimes make the mistake of saying that
“nothing is happening”. In actuality, the system is at equilibrium such that the forward and the
reverse reactions are occurring at the same rate. We will now investigate this state when
equilibrium exists and we will see that equilibrium systems are actually very active and very
much “alive”.
Up until now we have only been able to use calculations that were described by the
stoichiometry of the chemical reaction. We were examining reactions that “go to completion”.
However, not all reactions go to completion. Reversible reactions reach the equilibrium state
when the forward and reverse reactions proceed at the same rate and to the same extent. Rate
is a tricky concept. Chemical equilibrium is actually unconcerned with the amount of time that
it takes for a reaction to reach the equilibrium state. We will be examining the concentrations of
components once they reach the equilibrium state. Later on, we will be examining how long it
takes for reactions to proceed, but now, we are unconcerned with how long. At equilibrium, we
are studying the EXTENT that the reaction has proceeded – how much of each species is
present. If it took 10 years to get to equilibrium or 10 seconds, that does not matter!
Dynamic equilibrium is a state in which no net change takes place between two opposing
processes occurring at the same rate. We have already examined equilibrium in terms of vapor
pressures (liquid gas equilibrium). There are two requirements for equilibrium to take place:
the process must be reversible (one that goes backwards as well as forwards)
the system must be closed (a system from which substances cannot escape)
True equilibria are rarely found in nature because few real systems are entirely closed. In an
open container, the products tend to disperse and the reverse reaction cannot take place.
1
In the state of equilibrium, the concentration of the reactants and the concentration of the
products no longer change with time. It appears to us – as the observer on the macroscopic
scale – that there is nothing going on.
Since no changes occur in the concentrations of reactants or products in a reaction system at
equilibrium, it may appear that everything has stopped. However, this is not the case. On the
molecular level, there is frantic activity. Equilibrium is not static but is a highly dynamic
situation. The concept of chemical equilibrium is analogous to the flow of cars across a bridge
connecting two island cities. Suppose the traffic flow on the bridge is the same in both
directions. It is obvious that there is motion, since one can see the cars traveling back and forth
across the bridge, but the number of cars in each city is not changing because equal numbers of
cars are entering and leaving. The result is no net change in the car population.
To see how this concept applies to chemical reactions, consider the reaction between steam and
carbon monoxide in a closed vessel at a high temperature where the reaction takes place
rapidly. Notice that the chemical reaction is written with a two sided arrow – not the typical
arrow that we have used in the past when writing a chemical reaction. For equilibrium, we use
the two sided arrow  or even two arrows
H2O(g) + CO(g)  H2(g) + CO2(g)
The changes in concentrations with time for the reaction H2O(g) + CO(g)  H2(g) + CO2(g)
when equimolar quantities of H2O(g) and CO(g) are mixed.
Assume that the same number of moles of gaseous CO and gaseous H2O are placed in a closed
vessel and allowed to react. The plots of the concentrations of reactants and products versus
time are shown in the figure above. Note that since CO and H2O were originally present in
equal molar quantities, and since they react in a 1:1 ratio, the concentrations of the two gases are
always equal. Also, since H2 and CO2 are formed in equal amounts, they are always present in
the same concentrations.
The figure above is a profile of the progress of the reaction. When CO and H2O are mixed, they
immediately begin to react to form H2 and CO2. This leads to a decrease in the concentrations of
the reactants, but the concentrations of the products, which were initially at zero, are increasing.
Beyond a certain time, indicated by the dashed line in the figure, the concentrations of reactants
and products no longer change-equilibrium has been reached. Unless the system is somehow
2
disturbed, no further changes in concentrations will occur. Note that although the equilibrium
position lies far to the right, the concentrations of reactants never go to zero; the reactants will
always be present in small but constant concentrations. A molecular view of the chemical
reaction is shown below:
Time
KEY:
H2O
CO
CO2
H2
(a) H2O and CO are mixed in equal numbers and begin to react (b) to form CO2 and H2. After
time has passed, equilibrium is reached (c) and the numbers of reactant and product molecules
then remain constant over time (d).
At equilibrium, the reactant and product concentrations – written [reactant] and [product] (the
brackets mean concentration) stop changing because the forward and reverse rates have become
equal. The system is at dynamic equilibrium on the molecular level, but on the macroscopic
level we “see” no further net change because the change in one direction is balanced by a
change in the reverse direction.
At a specific temperature, the system will reach equilibrium, and the product and reactant
concentrations will remain constant.
At equilibrium rateforward = ratereverse
Why does equilibrium occur? We have seen that molecules react by colliding with one another,
and the more collisions, the faster is the reaction. This is why reaction rates depend on
concentrations. In this case the concentrations of H2O and CO are lowered by the forward
reaction:
H2O + CO

3
H2 + CO2
As the concentrations of the reactants decrease, the forward reaction slows down. As in the
bridge traffic analogy, there is also a reverse direction:
H2O + CO

H2 + CO2
Initially in this experiment no H2 and CO2 were present, and this reverse reaction could not
occur. However, as the forward reaction proceeds, the concentrations of H 2 and CO2 build up,
and the rate of the reverse reaction increases as the forward reaction slows down. Eventually,
the concentrations reach levels where the rate of the forward reaction equals the rate of the
reverse reaction. The system has reached equilibrium.
The changes with time in the rates of forward and reverse reactions for
H2O(g) + CO(g)  H2(g) + CO2(g)
when equimolar quantities of H2O(g) and CO(g) are mixed. The rates do not change in the
same way with time because the forward reaction has a much larger rate constant than the
reverse reaction.
The equilibrium position of a reaction seen after some undetermined amount of time is
determined by many factors: the initial concentrations, the relative energies of the reactants and
products, and the relative degree of "organization" of the reactants and products. Energy and
organization come into play because nature tries to achieve minimum energy and maximum
disorder, as we will show later. For now, we will simply view the equilibrium phenomenon in
terms of the rates of opposing reactions.
When equilibrium of a system is reached, we may want to study some aspects of the chemical
reaction. We can do that by using the equilibrium constant (K) for that particular reaction at a
defined temperature. While the rates for the particular reactions are equal, that does not mean
that the concentrations are. The magnitude of K will tell us how far the reaction proceeds
towards the products at a given temperature. The K value is significant as its value is constant
no matter what the starting concentration of species used. It can be determined experimentally.
4
To write the Kc expression for species whose concentrations are known:
1.)
2.)
The product concentrations are multiplied together and divided by the reactant
concentrations. The products of the reversible reaction are still defined as the chemical
species on the right hand side of the equation while the reactants are defined as the
chemical species on the left hand side of the equation.
Each concentration is raised to a power equal to its coefficient in the balanced chemical
equation.
Kc = [products]coefficients
[reactants]coefficients
Concept Test:
Write the Kc expression for the following:
H2O(g) + CO(g)  H2(g) + CO2(g)
Kc =
4HCl(g) + O2 (g)  2H2O(g) + 2Cl2 (g)
Kc =
Concept Test:
Find the Kc for the following reaction:
N2O4 (g)  2NO2 (g)
The equilibrium concentrations for the experiment are
N2O4 = 0.0202 mole/L and NO2 = 0.00966 mole/L. Determine K
Kc =
Kc =
5
When the equilibrium constant is calculated a variety of answers can result. The magnitude of
K is an indication of how far the reaction proceeds towards the products (as written) at a given
temperature. Different chemical species have different K values, even if at the same
temperature. Therefore, the values of K are limitless.
aA + bB … eE + dD …
[E]e [D]d
K
[A]a [B]b
Some generalities can be made concerning the value of K:
1.) if K is small: then the concentrations of the reactants (A and B) must be significantly
larger than the concentrations of E and D. In fact, if the value of K is very very small, we
might even say that no appreciable amount (concentration) of product is made and
therefore we might label the reaction: No reaction. Double digit negative powers of ten
generally meet the requirement for “small K values”.
2.) if K is large: then the reaction reaches a state of equilibrium where the concentrations of
the products are much much larger than the concentration of the reactants. In this
particular chemical reaction, we might say that the reaction has gone to completion.
Double digit positive powers of ten generally meet the requirement for “large K values”.
In effect, these reactions may even be defined as not reversible.
3.) Intermediate K: significant amounts of both reactants and products are present and we
can say that the system is/can be at equilibrium state.
Let’s examine the significance of Kc using the following reaction:
2HI(g)  H2 (g)
Experiment Number
1
2
3
Initial
Concentrations, M
[HI] = 1.000
[H2] = 0.000
[I2] = 0.000
[HI] = 0.000
[H2] = 1.000
[I2] = 1.000
[HI] = 1.000
[H2] = 1.000
[I2] = 1.000
+ I2 (g)
at 698 K
Equilibrium
Concentrations, M
[HI] = 0.786
[H2] = 0.107
[I2] = 0.107
[HI] = 1.573
[H2] = 0.213
[I2] = 0.213
[HI] = 2.360
[H2] = 0.320
[I2] = 0.320
6
[H2][I2]
[HI]2
0.0185
0.0183
0.0184
We see that our Kc expression gives a constant value when we take the concentrations of the
products raised to their coefficients divided by the reactants raised to their coefficients. This
shows us that regardless of what starting concentration we begin with, eventually the system
will reach an equilibrium state where the ratio of the products to the reactants is a constant
number. Knowing this, when missing one variable and given the others (the concentrations of
the other reactants and products and/or the K value) you should be able to determine the
missing value!
Concept Test:
Consider the case in which the decomposition of HI(g) produces
equilibrium concentrations of H2 and I2 of 0.0250 M. Find the
equilibrium concentration of HI
Kc =
Solve for HI
HI =
At a given temperature, a chemical system reaches a state in which a particular ratio of reactant
and product concentrations has a constant value. In examining reactions (like the one given
above), scientists discovered that for a particular system and temperature, the same equilibrium
state is attained regardless of how the reaction is run. The above example shows us that
regardless of what the starting [reactant] and [product], the ratio of the chemical species attains
the same value (within some experimental error).
We can monitor the numerical change in the ratios of concentrations as the reaction proceeds
along, and in doing so, we are calculating the reaction quotient – or Q for a particular reaction.
The set-up for the Q expression is exactly the same as the K expression.
Q = [products]coefficients
[reactants]coefficients
As we monitor the chemical reactions as it proceeds through time, the numerical value of Q will
also change. At the beginning of the reaction, the concentrations of reactants are large, and the
Q value will be small. A moment later, the concentration of the reactants is smaller and the
concentration of the products has also increased. Thus, as the reaction proceeds, we will have a
7
gradual increase in the Q value for a reaction. The Q value will continue to change until the
reaction reaches equilibrium. Once equilibrium has been reached, the concentrations of the
reactants and products will be constant values. Thus, at equilibrium:
At equilibrium: Q = K
Since K only applies to the system at equilibrium, monitoring Q will allow us to examine a
system as it changes up to the point when it is at equilibrium. K is a special value of Q that
occurs when the reactant and product terms are at equilibrium.
The reaction quotient – Q – is a collection of terms based on the balanced equations exactly as
written for a given reaction. The value of Q will vary during the chemical reaction, while the
value of K, which is the constant value of Q at equilibrium, will also depend on how the
equation is balanced. Q and K, given that they are ratios, will be unitless numbers.
Write the equilibrium expression for the following reaction:
2NO(g) + O2 (g)  2NO2 (g)
Kc = [NO2]2 = 4.67 x 1013
[NO]2[O2]
And if the equation is reversed, what is the equilibrium expression then??
2NO2 (g)  2NO(g) + O2 (g)
Kc = [NO]2[O2]
[NO2]2
How are these two equilibrium expressions related to one another? We are examining the same
chemical reaction, it is just that the reaction has been reversed. When the reaction is reversed –
what happens to the value of K (or Q)?
When the chemical reaction with the equilibrium constant = Kc is reversed,
the reverse reaction will have the equilibrium constant = 1/Kc
8
Concept Test:
Calculate the numerical value for the equation written in reverse =
What if we write the reaction in terms of the formation of 1 mole of NO2?
NO(g) + ½ O2 (g)  NO2 (g)
What is the K of this reaction?
Kc = [NO2]
[NO][O2}1/2
How is THIS expression related to the original equation??
ORIGINAL EQUATION:
Kc =
[NO2]2
[NO]2[O2]
What is the Kc for the formation of 1 mole of NO2?
Kc 1 mole NO2 =
When the coefficients of an equation are multiplied by a common factor, n,
to produce a new equation, we raise the original Kc value to the power n
to obtain the new equilibrium constant.
9
Concept Test:
Calculate the Kc expression for the following chemical reaction:
NO2 (g)  NO(g) + ½ O2 (g)
Kc =
Mathematically you can determine the overall K expression and value for a target equation
when given known Kc values for the equations that will be summed.
Target Equation: N2O(g) +
3/2O2 (g)
 2NO2 (g)
Given: (1) N2O (g) + ½ O2 (g)  2NO(g)
(2) 2NO(g) + O2 (g)  2NO2 (g)
N2O(g) + 3/2O2 (g)  2NO2 (g)
K1 = 1.7 x 10-13
K2 = 4.67 x 1013
K3 = ??
Let’s examine what the individual K expressions look like:
K1 = 1.7 x 10-13 = [NO]2
[N2O][O2]1/2
K3 =
K2 = 4.67 x 1013 = [NO2]2
[NO]2[O2]
[NO2]2
[N2O][O2]3/2
Do you see a way to mathematically manipulate the given Kc
expressions to yield the missing Kc expression?
Are all the components – reactants and products present?
What if we add them?
What if we subtract them?
What if we divide them?
What if we multiply them?
10
[NO]2
[N2O][O2]1/2
X
[NO2]2
[NO]2[O2]
= [NO2]2
[N2O][O2]3/2
Which is the target Kc expression 
When we add equations for individual reactions to obtain a net equation,
we multiply their equilibrium constants to obtain the target
equilibrium constant which represents the net reaction.
The general feature of equilibrium constant expressions is that they conform to the following
idea:
The equilibrium constant expression does not include the terms for
pure solids or pure liquids because their concentrations
do not change in a reaction.
Remember that for solids, the volume component is not going to change. Thus its concentration
is going to change very little as well. Also, the same argument applies for the liquid phase –
liquids are also not very compressible, thus their volume component will not change by very
much either and can be said to be constant. Aqueous solutions are components (solutes)
dissolved into a liquid (the solvent, typically water) and thus their concentrations will change –
and of course, gases are susceptible to volume changes. For this reason, we will omit any
substances that are solids or liquids from the Kc and Q expressions.
Concept Test:
The reaction of steam and solid carbon produces a mixture of carbon
monoxide and hydrogen. Write the equilibrium constant expression
Kc for this reaction:
C(s) + H2O (g)  CO(g) + H2 (g)
Kc =
11
Molarity, defined as moles/L is not always the most convenient concentration to use. For gas
phase reactions, it is oftentimes easier to express concentrations in terms of the partial pressures
of each gas. Such that the partial pressure of gas A =
PA = nART = [A]RT
V
Remember, gas concentrations are spoken of in terms of their molar volume – or moles/L.
Thus, at a given temperature, the partial pressure of a gas is proportional to its molarity. If
partial pressures are used, then the reaction quotient and equilibrium constant are represented
by Qp and Kp (as opposed to Qc and Kc which represent the reactants and products in terms of
molarity). Unless otherwise specified, the pressures are in terms of atmospheres. Calculations
of Kp and Qc are analogous to Kc and Qc calculations.
12
Concept Test:
Write the Kp expression for the following reaction:
N2 (g) + 3H2 (g)  2NH3 (g)
Kp =
From the above equation PNH3 = [NH3]RT and PN2 = [N2]RT and PH2 = [H2]RT
Concept Test:
Substitute the above relationships into the Kp expression:
Kp
=
Kp = Kc(RT)n
where n is numerically equal to the change in the number of moles
of gas in the forward equation.
In the above equation, we start with 4
moles of gas and only form 2 – therefore that is a loss of two moles –
such that n would equal -2.
You can use Kp or Kc to express reactions involving gases, but know that the two values for K
are not numerically equivalent. They can be mathematically interconverted though!
13
Q and K values are related to one another. In fact, we talked about how K is just a special type
of Q value – when the system is at equilibrium. For other situations Q values are determined
which show us the reaction as is proceeds and as the concentrations of reactants and products
change in a chemical reaction. At any particular time during a reaction at some temperature we
could monitor the concentrations of species – perhaps by doing a titration, or by
spectrophotometry. And given the value of Q we could determine if the ratio of concentrations
of products is greater than the concentration of reactants, or if the ratio of concentrations of
reactants is greater than the concentration of products, or if the reaction has reached a state of
equilibrium.
We know that Q can be less than K, Q can be greater than K, or Q can be equal to K. But what
does that mean?
Q < K: Remember that the formula to determine Q is the same as the formula to determine K.
Therefore, if Q is smaller than K, then the species in the denominator – the reactant
concentrations – must be larger than the concentrations of the products. If we are comparing
this situation to equilibrium, then we have too much reactant and not enough product. The
reaction would have to shift towards the formation of more product in order to attain
equilibrium.
Q > K: If Q is larger than K that means that the concentration of products must be greater than
the concentration of reactants needed to reach the equilibrium value. In order to reach
equilibrium, the concentration of the products must decrease and the concentration of the
reactants must increases. This indicates a shift towards the left (or towards the reactants) in
order to attain equilibrium.
Q = K: The special situation which the system is at equilibrium. This situations exists only
when the concentrations or reactants and products have attained their equilibrium values. It
does NOT mean that the values are EQUAL! K = 1 = Q is not what we are talking about! Thus,
despite the system being dynamic with changes occurring on the nanoscopic scale, no net
change is occurring on the macroscopic level.
There are two general types of equilibrium problems to be aware of: the first is when you are
given the equilibrium quantities and asked to solve for K, and the second is when you are given
K and some of the equilibrium quantities and asked to solve for a missing equilibrium quantity.
The general problems can be modified to be more complex: Such as the equilibrium quantities
could be given to you directly and you “plug and chug” the values to determine K or to
determine the missing quantity. But really, what fun is that? Let’s look at an example of this
type of problem and we’ll see just how boring it is 
14
Concept Test:
Given the following equation, determine Kc
H2 (g) + I2 (g)  2HI(g)
Given 1.20 M of H2 and I2 and 0.347 M HI, calculate Kc
Kc =
Kc =
Concept Test:
If the Kc value for the above reaction = 8.36 x 10-2 and the concentrations
For H2 and I2 = 0.133 M, what is the concentration of HI?
Kc =
[HI] =
But!!! Those are boring! Plug and chug -  In not so simple situations it might be necessary to
set up a table to examine the changes that the system is undergoing in order to determine
equilibrium concentrations.
Steps to Creating an ICE table
1.) Write down the BALANCED chemical equation
2.) Write down the skeleton of the ICE table
3.) Use the given data to find the initial concentration of each species
4.) Use the balanced chemical equation to examine the change that each
species will undergo as the system comes to equilibrium
5.) Find the equilibrium “values” of each species by adding steps 3 and 4
15
Again, pure solids and pure liquids can be IGNORED!
Concept Test:
When phosgene (COCl2) , a poisonous gas that has been used as a chemical
warfare agent, is heated, it decomposes into carbon monoxide and chlorine gas
Write and balance the chemical equation!
____________________________________________________________________
When 2.00 moles of phosgene are put into a 1.00 L empty flask at 395oC
and allowed to come to equilibrium, the final mixture contains
0.0398 moles of chlorine. Find Kc
Remember - we need concentrations! We know the # moles and we
know the volume – calculate M
moles/
moles/
L=
M phosgene
L=
M Cl2
Make your ICE table: fill in what you KNOW
COCl2 (g)

CO(g)
+
Cl2 (g)
I
C
E
I = initial
C = change
E = equilibrium
Kc =
Filling in the table: We have a starting (known) concentration of phosgene. Initially, that means
that we have 0 moles of both carbon monoxide and chlorine gas.
The equilibrium value of chlorine gas was given. That value can also be filled in.
Then, performing some simple math – we see that we start with no chlorine gas, and end up
with 0.0398 M at equilibrium. The changes that chlorine undergoes will be the same – or related
16
by the stoichiometric coefficients to the changes that the other chemical species will undergo.
Since the compounds are in a 1:1 ratio with one another, then the changes that 1 undergoes will
be directly related to the changes that the other compounds undergo.
Amounts will be added to the side of the equation that starts with 0 amounts. Amounts will be
subtracted from the side of the equation that does NOT have a zero value.
Concept Test:
You are given 0.400 moles of HI gas is placed in a 1.00 L container and allowed
to dissociate until it reaches equilibrium.
At equilibrium, its concentration is 0.0156 M
Calculate Kc
Step 1: write balanced equation
Step 2: create ICE table
Step 3: fill in values
Step 4: calculate Kc
__________

__________
+
__________
I
C
E
[Initial] – [Change] = [Equilibrium]
0.400 M =
M
0.400 –
=
Write down all EQUILIBRIUM CONCENTRATIONS of each species, then solve for K
Kc =
17
On the flip side, given the Kc value, equilibrium concentrations of the chemical species can be
calculated. This variation, however, can be rather simple (plug and chug) or rather complicated
– meaning we will make assumptions about the concentrations or we will have to use the
quadratic formula to solve for the concentrations.
Concept Test:
When methane gas is mixed with water vapor in a 0.32 L flask,
carbon monoxide and hydrogen gas are formed at 1200 K.
At equilibrium, the flask contains 0.33 moles of CO, 0.088 moles
of H2 and 0.025 moles of CH4. What is the [H2O] at equilibrium if the
Kc = 0.33
Step 1: Write and balance the chemical reaction
Step 2: Calculate equilibrium concentrations!!
Step 3: Set-up equilibrium expression
Step 4: solve for the missing value
In more involved problems, you might be given initial concentrations and asked to find
equilibrium concentrations. In this case, the equilibrium concentrations are unknown values (or
x values). We will use ICE tables again to solve these types of equations. There will be two
types of problems, one in which x is small, and another when x is a significant number. If x is
small (also known as the 5% rule), then changes involving x will not be significant.
18
COCl2 (g)
Concept Test:
 CO (g) + Cl2 (g) Kc = 8.3 x 10-4 at 360oC
Calculate the equilibrium concentrations of each species at equilibrium
in a 10.0 L flask if:
a.) 5.00 moles of COCl2 decompose
Step 1: Make sure equation is balanced
Step 2: Make an ICE table
Step 3: Calculate concentrations
Step 4: Fill in the ICE table with values
Step 5: Set-up equilibrium expression
COCl2 (g)

CO (g)
+
Cl2 (g)
I
C
E
Kc =
Kc = 8.3 x 10-4 =
let’s assume that x is small. Obviously, the creation of product in
any amount is significant, but if x is small then we will not lose
a lot of starting COCl2. Kc is on the smaller side, so we will make this assumption
and see where it takes us.
How accurate is our assumption: 5% rule
Change in concentration x 100 < 5% then assumption is valid
Initial concentration
x 100 =
%
0.500
Therefore, our assumption that x is small is
valid
[COCl2] = 0.500 –
[CO] =
[Cl2] =
19
Concept Test:
COCl2 (g)  CO (g) + Cl2 (g) Kc = 8.3 x 10-4 at 360oC
Calculate the equilibrium concentrations of each species at equilibrium
in a 10.0 L flask if:
b.) 0.100 moles of COCl2 decompose
Step 1: Make sure equation is balanced
Step 2: Make an ICE table
Step 3: Calculate concentrations
Step 4: Fill in the ICE table with values
Step 5: Set-up equilibrium expression
COCl2 (g)

CO (g)
+
Cl2 (g)
I
C
E
Kc =
Kc = 8.3 x 10-4 =
if we assume x is small then x =
x 100 =
%
0.0100
Therefore we cannot make this assumption x is significant: use quadratic formula!
a quadratic expression:
x=
- b  b 2 - 4ac
2a
x=
x=
20
x cannot be negative!! Remember x is the change – you cannot
have a negative change! Product must be formed!
Therefore, the only value for x that makes any sense is
x=
Therefore, the equilibrium concentrations are:
[CO] =
M
[Cl2] =
M
[COCl2] =
M
The important note about reactions at equilibrium is that the reaction is and is able to proceed in
both directions at the same time. In order to get to equilibrium, the reactants reacted to form
products – whose concentrations increased until come amount resulted in the products reacting
(or dissociating) to re-form the reactants. Reactions are able to stay at equilibrium and “reequilibrize” whenever something is done to the reaction. The drive to re-attain equilibrium
after some action is take on the chemical reaction is known as LeChatelier’s Principle.
LeChatelier’s Principle: When a chemical reaction at equilibrium is disturbed or altered in some
way, the chemical system will re-attain equilibrium by undergoing a net reaction that
counteracts, or reduces the effect of the disturbance.
What does disturb really mean? Think about the expression that describes a reaction. When the
reaction is at equilibrium it is called the K expression. When reactions are not at equilibrium, it
is called the Q expression. Either expression is a ratio of the concentrations of products to
reactants. Thus, changing the concentrations of reactants or products will disturb the system at
equilibrium. Other disturbances are changes in pressure or volume, or a change in temperature.
When we are affecting reactions by disturbing them, we will use the terms “shift to the right” or
“shift to the left” to describe the result of the disturbance. As the reactants are on the left, if the
reaction is shifting to the left to counteract the disturbance then the reaction will be forming
more reactants. Conversely, as the products are on the right, if we say the reaction is shifting to
the right, then the reaction is forming more products as a result of the disturbance.
The shifting will occur such that the concentrations or pressures will change in a way to reduce
the disturbance and the system will attain a new equilibrium.
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LeChatelier’s Principle – while we will speak of it in chemical terms, is basically a fancy way
that describes the circle of life. What happens when the number of predator’s decrease? The
prey’s numbers increase dramatically. Eventually, their numbers will reach a point that they
will start dying by starvation, disease, and also by the predator. Thus, a new equilibrium has
been reached.
When a system at equilibrium is disturbed by changing the concentration of a component, the
system will react in the direction that will reduce the change:
If the concentration of a component increases, the system will react in the direction that
will reduce the added component
If the concentration of a component decreases, the system will react in the direction that
will create/produce more of that component.
Remember, no matter what we do to the reaction, we will be maintaining the same ratio of
products to reactants. We will be reestablishing a new equilibrium point in time and with new
concentrations, but their ratio will be the same!
Examine the following reaction:
PCl3 + Cl2  PCl5
If more Cl2 is added, then more Cl2 can react. If more Cl2 can react then it will react with more
PCl3. Thus, more products will be formed. This is an example of a shift to the right
You can think about this mathematically as well: Remember that we can write a Q expression
for a reaction (it will be Q since we are changing the concentrations of components the reaction
is not at K at this point).
Q = [PCl5]
[PCl3][Cl2]
If [Cl2] increases then Q decreases. In order for Q to equal K, the numerator
must increase – which means that if the concentration of
either reactant increases, the [PCl5] must increase in order for the
Q value to return back to equal K!
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Concept Test:
PCl3 + Cl2  PCl5
Which way will the reaction shift if PCl5 is siphoned out of the reaction vessel?
(PCl5 removed)
Which way will the reaction shift if more PCl5 is added to the reaction vessel?
Which way will the reaction shift if Cl2 is removed from the reaction vessel?
If PCl3 is removed from the reaction vessel?
Putting it All Together:
The reaction will shift to the RIGHT if reactant is added or product is removed. Use what we
add, or replace what we take away.
The reaction will shift to the LEFT if product is added or reactant removed. Again, must use
what we add, or replace what we take away.
In general: the system will react to consume any added species (the reaction will go away from
the added species) or replace/produce any species that is removed (the reaction will go towards
the removed species)
We could monitor this change in the system using an ICE table. Suppose the equilibrium
concentrations are [PCl3] = 0.200 M, [Cl2] = 0.125 M, and PCl5 = 0.600 M Additional Cl2 is added
such that the added amount correspond to a change of 0.075 M. Thus, a new starting
concentration for Cl2 is achieved which equals 0.125 M + 0.075 M = 0.200 M. All other
concentrations remain the same as the initial concentrations. Think about WHICH way the
reaction will proceed when you ADD additional substance – it will shift AWAY from the added
substance – therefore we will decrease (subtract) from the side of the equation that has the
added component:
PCl3
+
Cl2

PCl5
I
0.200
0.200
0.600
C
-x
-x
+x
E
0.200 – x
0.200 - x
0.600 + x
Known
0.637
Experimentally it was determined that [PCl5]eq = 0.637 M
Therefore: 0.600 + x = 0.637
x = 0.037
Thus: [Cl2]eq = [PCl3] = 0.200 – 0.037 = 0.163 M
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CHECK: verify that new Q = K!
Koriginal = [PCl5]
=
[PCl3][Cl2]
[0.600]
[0.200][0.125]
Koriginal = 24
Knew equilibrium = [PCl5]
[PCl3][Cl2]
Knew equilibrium = [0.637]
[0.163][0.163]
Knew equilibrium = 23.98 = 24
The system HAS reached equilibrium
Notice what has happened to the individual equilibrium concentrations now that equilibrium
has been re-established. The [Cl2] increased, [PCl3] decreased, and [PCl5] increased after the
addition of Cl2.
Changes in pressure will only affect the components if they are in the gaseous state – remember
– solids are incompressible and liquids compressible but the amount is negligible. Changes in
pressure can be achieved by changing the concentration of a gaseous component, adding an
inert gas that will not participate in a chemical reaction, changing the volume of the reaction
container.
We just examined the effect of changing the concentration – and the same rationale holds true
with gases. When calculating Kp values or Qp values we are looking at the partial pressures of
the gases. Adding an inert gas – one that does not participate in the chemical reaction – has no
effect on the partial pressures of the gases that do participate in the chemical reaction (the
volume of the container did not change and the number of moles of gas did not change – thus
the partial pressures did not change). Thus, adding an inert gas will have no effect on QP or Kp.
Changing the pressure by changing the volume of the container will have an effect on the
reaction. If the volume of the container increases, then the concentration of the components
decreases. If the volume of the container decreases, then the concentration of the components
increases. The number of moles of components stays constant, but the denominator value
(Liters) changes.
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Concept Test:
Given 5.00 moles of gas, which is put in a 1.00 L container
results in a concentration of 5.00 M
if the same moles of gas is placed in a 2.00 L container,
the concentration is?
M
if the same number of moles of gas is placed in a 0.500 L container
the concentration is?
M
In the case of chemical reactions with different numbers of moles of products and reactants,
increasing the pressure/ decreasing the volume will cause the reaction to shift in the direction
of the fewer number of molecules/moles. Increasing the pressure/decreasing the volume is a
stress on the system. Crowding more molecules into a smaller space is more difficult than
producing the fewer moles/molecules of gas. Decreasing the pressure/increasing the volume
has the opposite effect. In this case, more mole/molecules can be produced as there is more
room to accommodate them.
Again, we can use the Q expression to explain the shift in the reaction direction:
PCl3
+
Cl2  PCl5
Q = [PCl5]
[Cl2][PCl3]
Decreasing the volume by ½, doubles the concentrations. It will
double the concentration of the numerator but by doubling
the concentrations of both reactants, the denominator ends up being
quadrupled numerically in value. The larger denominator must be counteracted
By forming more PCl5 in order to maintain equilibrium – thus a shift to the right
Doubling the volume means that the concentrations are halved.
now we are halving the numerator and having each component
in the denominator – which makes the denominator smaller.
In order to achieve equilibrium, the reaction must shift to the left.
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Putting it All Together:
The reaction will shift towards the side that has fewer moles/molecules if the pressure is
increased (volume decreased)
The reaction will shift towards the side that has more moles/molecules if the pressure is
decreased (volume increased)
NOTE: if the number of moles/molecules is the same on each side of the equation, then
changes in pressure/volume will have no effect on the equilibrium position.
Only changes in temperature alter K values. Other changes DO NOT affect K. Changes in
temperature DO AFFECT the K value. The expression is the same – but the value of K changes.
If you consider heat as part of the reaction – or a component of the system (remember thinking
of exothermic/endothermic reactions as adding heat to some side of the equation??), then you
can predict the shift in the chemical reaction.
Adding heat will favor the endothermic reaction
Removing heat favors the exothermic reaction
Remember – the endothermic and exothermic reaction are tied together – it is just a matter of
HOW we write the equation!! So be careful!!
PCl3 (g) + Cl2 (g)  PCl5 (g)
PCl3 (g) + Cl2 (g) → PCl5 (g) + heat exothermic reaction
Heat + PCl5 (g) → PCl3 (g) + Cl2 (g) endothermic reaction
Adding heat will cause PCl5 to decompose into PCl3 and Cl2. In the original equation this
means that the concentrations of PCl3 and Cl2 will increase. Think about what that means for
Q. In the original equations, PCl3 and Cl2 would be in the denominator. That means that the Q
value will be smaller than the original K. If you maintain this new temperature the
concentrations will stabilize to new values. The ratio of concentrations will now NOT be equal
to the original K value – they will reach a new equilibrium at this new temperature!! Therefore,
causing a change in temperature will cause a reaction to reach a new equilibrium which will
have a new K value!
Notice the two reactions above. The reactions will shift in the direction that is AWAY from the
heat added. Keep in mind that if given individual reactions (like the ones above) the shift in
direction will be obvious – AWAY from the heat added. If given only the equilibrium
expression THEN you will have to examine whether the reaction is exothermic or endothermic
and if you are adding heat or taking heat away.
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Concept Test:
Given the following reactions predict which direction
they will shift if heat is ADDED to each of them:
CaO(s) + H2O(l)  Ca(OH)2 H = -82 kJ
CaCO3 (s)  CO2 (g) + CaO(s) H = 178 kJ
Another way of looking at this: If you are good at recognizing which side of the reaction the
heat should be added to when speaking of endothermic or exothermic reactions then you can
add the heat in yourself.
For example, in the reaction above: Since the first reaction is exothermic, heat is a product
CaO(s) + H2O(l)  Ca(OH)2 + heat
The heat is being added to the “product” side. The addition of heat will be on the same side as
the heat which indicates whether the reaction is exo or endothermic. The reaction will always
shift AWAY from the heat added. Thus, the reaction will shift to the left.
In the second example, the reaction as written is endothermic, that means that we are adding
heat as a reactant:
Heat + CaCO3 (s)  CO2 (g) + CaO(s)
Therefore, the addition of heat will be to the same side as the heat already present. The reaction
will shift AWAY from the heat added and therefore shifts to the right.
A catalyst is defined as a chemical species that increases the rate of the reaction but does NOT
participate in the chemical reaction – or is not used up in the chemical reaction. The catalyst
shortens the time it takes for the reaction to REACH equilibrium but has no effect on the
equilibrium concentrations. Thus, adding a catalyst has NO effect on K! The reaction will move
quicker. The time is takes to reach equilibrium will be shorter – but the concentrations and thus
K will be unaffected.
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