Equilibrium
Reactions are reversible
A+B
C + D ( forward)
C+D
A + B (reverse)
Initially there is only A and B so only the
forward reaction is possible
As C and D build up, the reverse reaction
speeds up while the forward reaction
slows down.
Eventually the rates are equal
Reaction Rate
Forward Reaction
Equilibrium
Reverse reaction
Time
What is equal at Equilibrium?
Rates are equal
Concentrations are not.
Rates are determined by concentrations
and activation energy.
The concentrations do not change at
equilibrium.
Or if the reaction is verrrry slooooow.
Law of Mass Action
For any reaction
jA + kB
lC + mD
K = [C]l[D]m
[A]j[B]k
PRODUCTSpower
REACTANTSpower
K is called the equilibrium constant.
is how we indicate a reversible reaction
Only gases and aqueous solutions are included
in the law of mass action
Playing with K
If we write the reaction in reverse.
lC + mD
jA + kB
Then the new equilibrium constant is
K’ = [A]j[B]k = 1/K
[C]l[D]m
Playing with K
If we multiply the equation by a constant
njA + nkB
nlC + nmD
Then the equilibrium constant is
n
nj[B]nk
j[B]k)n
[A]
([A]
K
K’ =
=
=
[C]nl[D]nm ([C]l[D]m)n
Manipulating K
From a practical point of view, here is how
K gets manipulated:
If a reaction gets doubled, the value of K is
squared. If it is tripled, K is raised to the
third power.
If the reaction gets cut in ½ then K is found by
taking the square root of K.
The units for K
This one is simple.
K has no units.
K is CONSTANT
Temperature affects the equilibrium
constant.
The equilibrium concentrations don’t
have to be the same only K.
Equilibrium position is a set of
concentrations at equilibrium.
There are an unlimited number of K
values so the temperature needs to be
stated.
Equilibrium Constant
One for each Temperature
Calculate K
N2 + 3H2
Initial
[N2]0 =1.000 M
[H2]0 =1.000 M
[NH3]0 = 0 M
3NH3
At Equilibrium
[N2] = 0.921M
[H2] = 0.763M
[NH3] = 0.157M
Calculate K
N2 + 3 H2
2 NH3
(all gases)
Initial
At Equilibrium
[N2]0 = 0 M
[N2] = 0.399 M
[H2]0 = 0 M
[H2] = 1.197 M
[NH3]0 = 1.000 M
[NH3] = 0.157M
K is the same no matter what the amount
of starting materials
Equilibrium and Pressure
Don’t forget your gas laws, they are often
needed to get some information for the
problem. Especially PV = nRT
Some reactions are gaseous
P = (n/V)RT
P = CRT
C is a concentration in moles/Liter
C = P/RT
Equilibrium and Pressure
2 SO2(g) + O2(g)
Kp =
(PSO3)2
(PSO2)2 (PO2)
K=
[SO3]2
[SO2]2 [O2]
2 SO3(g)
Equilibrium and Pressure:
For those that are a glutton for punishment and want to see the full treatment.
K=
K=
(PSO3/RT)2
(PSO2/RT)2(PO2/RT)
(PSO3)2
(1/RT)2
(PSO2)2(PO2) (1/RT)3
K = Kp
(1/RT)2
(1/RT)3
= Kp RT
General Equation
jA + kB
lC + mD
Kp= (PC)l (PD)m= (CC x RT)l (CD x RT)m
(PA)j (PB)k (CA x RT)j(CB x RT)k
Kp= (CC)l (CD)m x (RT)l+m
(CA)j(CB)k x (RT)j+k
Kp = K (RT)(l+m)-(j+k) = K (RT)Dn
Dn=(l+m)-(j+k) = Change in moles of gas
Homogeneous Equilibria
So far every example dealt with reactants
and products where all were in the
same phase.
We can use K in terms of either
concentration or pressure.
Units depend on reaction.
Heterogeneous Equilibria
If the reaction involves pure solids or pure
liquids the concentration of the solid or
the liquid doesn’t change.
As long as they are not used up we can
leave them out of the equilibrium
expression.
For an example, check out the next slide.
For Example
H2(g) + I2(s)
K = [HI]2
[H2][I2]
2 HI(g)
But the concentration of I2 does not change,
so it factors out of the equilibrium expression
and looks like this.
K=
[HI]2
[H2]
Solving Equilibrium
Problems
Type 1
The Reaction Quotient
Tells you the direction the reaction will go
to reach equilibrium
Calculated the same as the equilibrium
constant, but for a system not at
equilibrium
Q = [Products]coefficient
[Reactants] coefficient
Compare value to equilibrium constant
What Q tells us
If Q < K
Not enough products
Shift to right
If Q > K
Too many products
Shift to left
If Q = K system is at equilibrium. We like
these ones – there is nothing to do !!
Example
For the reaction:
2 NOCl(g)
2 NO(g) + Cl2(g)
K = 1.55 x 10-5 M at 35ºC
In an experiment 0.10 mol NOCl, 0.0010
mol NO(g) and 0.00010 mol Cl2 are
mixed in 2.0 L flask.
Which direction will the reaction proceed
to reach equilibrium?
Solving Equilibrium Problems
Given the starting concentrations and
one equilibrium concentration.
Use stoichiometry to figure out other
concentrations and K.
Learn to create a table of initial and final
conditions.
Check out the next slide for an actual
problem.
Consider the following reaction at 600ºC
2 SO2(g) + O2(g)
2 SO3(g)
In a certain experiment 2.00 mol of SO2,
1.50 mol of O2 and 3.00 mol of SO3 were
placed in a 1.00 L flask. At equilibrium 3.50
mol of SO3 were found to be present.
Calculate the equilibrium concentrations of
O2 and SO2, K and KP
The ICE BOX
The table we create is called the ICE Box.
2 SO2(g) + O2(g)
2 SO3(g)
Here is the information given in the problem. The blank spaces are
what we have to determine. Some are determined and others are
calculated.
Initial
2
1.5
3
Change
Equilibrium
3.5
The ICE BOX
The table we create is called the ICE Box.
2 SO2(g) + O2(g)
2 SO3(g)
Here is the information given in the problem. The blank spaces are
what we have to determine. Some are determined and others are
calculated.
Initial
Change
Equilibrium
2
1.5
- 2x
-x
3
+ 2x
3.5
Where are we at ?
The values in the CHANGE row are a
combination of stoichiometry and equilibrium.
Notice that the numbers are the same as the
coefficients. We can’t forget stoichiometry.
The [SO3] increases so it has to be a + 2x.
This tells us it shifts to the right and that the
value of the reactants has to decrease.
We combine the + or – sign along with the
stoichiometry to set up the equation.
The ICE BOX
The table we create is called the ICE Box.
2 SO2(g) + O2(g)
2 SO3(g)
Here is the information given in the problem. The blank spaces are
what we have to determine. Some are determined and others are
calculated.
Initial
2
1.5
Change
- 2x
-x
Equilibrium
2 – 2x
1.5 – x
3
+ 2x
3.5
Where are we at ?
Part 2
The values of the equilibrium are known
algebraically. Here is our Law of Mass
Action. We can solve that x = .25
K=
[ SO3] 2
=
[ O2] [ SO2]2
K = 4.36
(3.5) 2
(1.25) (1.5)2
A New Problem
Consider the same reaction at 600ºC
2 SO2(g) + O2(g)
2 SO3(g)
In a different experiment .500 mol SO2 and
.350 mol SO3 were placed in a 1.000 L
container. When the system reaches
equilibrium 0.045 mol of O2 are present.
Calculate the final concentrations of SO2
and SO3 and K.
Let’s set up the ICE Box
2 SO2(g) + O2(g)
Initial
Change
.5
+ 2x
0
+x
2 SO3(g)
.35
- 2x
Notice the signs. Since there is no oxygen present at the start, the reaction must shift to
the left and that means products consumed and reactants created.
Equilibrium .5 +2x
.045
.35 – 2x
This tells us that x = .045 and therefore the concentrations must be as follows.
[SO2] = .5 + 2x or .5 + .09 = .59
[SO3] = .35 – 2x or .35 - .09 = .26
K = (.26)2 / (.59)2 (.045) = 4.32
What if you’re not given
equilibrium concentration?
The size of K will determine what
approach to take.
First let’s look at the case of a LARGE
value of K ( >100).
We can make simplifying assumptions.
5 % rule
Now is as good a time as any to introduce
the 5 % rule.
Equilibrium concentrations are often a bit of
an approximation. Therefore it’s allowable to
simplify the math to make it more workable.
The 5 % rule states that if x is small in
relation to the stated amount, and the K,
then it can be ignored in the calculation.
Generally works if the Keq is less than 10-3
or greater than 103
Example
H2 (g) + I2 (g)  2 HI (g) K = 7.1 x 102 at 25ºC
Calculate the equilibrium concentrations if a
5.00 L container initially contains 15.9 g of H2
and 294 g I2 .
[ H2 ]0 = (15.7g/2.02)/5.00 L = 1.56 M
[ I2 ]0 = (294g/253.8)/5.00L = 0.232 M
[ HI ]0 = 0
Let’s set up the Ice Box !!
Q > K so more product will be formed.
Since K is large, rxn. will go to completion.
Another great clue is that when there is a
component with a concentration of zero, the
reaction must shift in that direction.
Stoichiometry tells us I2 is LR, it will be
smallest at equilibrium let it be x
Set up table of initial, final and change in
concentrations.
This equation needs manipulating
H2(g) + I2(g)
 2 HI(g)
initial
1.56 M 0.232 M
0M
change
-x
-x
+ 2x
equilibrium 1.56 - x .232 - x
2x
Choose x so it is small.
Since we know the I2 is going to lose most of
its concentration, x is actually going to be a
very high percentage of the amount of I2
There is a way around this dilemma.
Here we go.
In the work below, we are going to assume
that the reaction goes to completion.
H2(g) + I2(g)  2 HI(g)
initial
1.56 M
.232 M
0M
change
- .232
- .232
+ .464
equilibrium 1.328
0
.464
Let’s combine the following facts
and ideas and solve the problem.
Piecing it together
1. We didn’t change the conditions of the rxn. (ie.
Temp. ) so the Keq remains unchanged.
2. The Keq is large so we know there will be more
product than reactant at equilibrium.
3. Since we assumed the reaction went to completion
we now have no Iodine and the reaction must
move to the left.
4. X must be small because there needs to be more
product to satisfy Keq. X will be negligible
compared to the amounts given.
5. However, x can’t be ignored for the I2 because it is
at zero and x is 100% of the total amount of I2
H2(g) + I2(g)  2 HI(g)
initial
1.328
0
.464
change
+x
+x
-2x
equilibrium 1.328 + x
x
.464 – 2x
Here is our new Law of Mass Action after we
remove the ‘x’ according to the 5% rule.
Remember we can’t remove the x in the I2
because that is all that there is of it.
710 = (.464)2 / (1.328) ( x ) = 2.28 x 10-4
Checking the assumption
The rule of thumb is that if the value of X is
less than 5% of all the other
concentrations, our assumption was valid.
If not we would have had to use the
quadratic equation
More on this later.
Our assumption was valid.
Practice
For the reaction Cl2 + O2  2 ClO(g)
K = 156
In an experiment 0.100 mol ClO, 1.00 mol
O2 and 0.0100 mol Cl2 are mixed in a
4.00 L flask.
If the reaction is not at equilibrium, which
way will it shift? Q = ?
Calculate the equilibrium concentrations of
all components.
Problems with small K
K< .01
Process is the same
Set up table of initial, change, and final
concentrations. Your ICE Box
Choose X to be small.

Sounds familiar so far, I hope !!
For this case it will be a product.
For a small K the product concentration is
small.
Let’s start the set up.
For the reaction
2 NOCl  2 NO + Cl2
K= 1.6 x 10-5 Make note the small K favors more reactant
If 1.20 mol NOCl, 0.45 mol of NO and 0.87
mol Cl2 are mixed in a 1 L container
What are the equilibrium concentrations ?
Q = [NO]2[Cl2] = (0.45)2(0.87) = 0.15
[NOCl]2
(1.20)2
Q > K so it shifts to the left.
Here is the ICE Box set up
Once again we can manipulate the components to make the
math easier and we do this by assuming the reaction goes to
completion according to stoichiometry.
2 NOCl  2 NO + Cl2
Initial
1.20
0.45
0.87
Change
+.45
-.45
-.225
Equilibrium
1.65
0
.645
We can now use the Equilibrium concentrations to rework the problem. Remember that
the K does not change. K favors the reactants. Having [ NO ] = 0 means that the
reaction will shift to the right. Combining the two means that the change will be small so
x will be small and we can use the 5 % simplification rule. This is a Happy Day !
Where are we at ?
We can now use the Equilibrium
concentrations to rework the problem.
Remember that the K does not change.
K favors the reactants.
Having [ NO ] = 0 means that the reaction
will shift to the right. Combining the two
means that the change will be small so x
will be small and we can use the 5 %
simplification rule.
This is truly a Happy Day !
Here is the new ICE Box
2 NOCl  2 NO + Cl2
Initial
1.65
0
.645
Change
- 2x
+2x
+x
Equilibrium
1.65 – 2x 2x
.645 + x
Equilibrium (5%) 1.65
2x
.645
Solving: 1.6 x 10-5 = (.645) (2x)2
(1.65)2
The algebra is trickier but doable x = .0042
Always verify the 5 % rule

If x = .0042, we compare that to the
concentration. .0042 / .87 = .48 %

We are good !!
Practice Problem
For the reaction
2 ClO(g)  Cl2 (g) + O2 (g)
K = 6.4 x 10-3
In an experiment 0.100 mol ClO(g), 1.00
mol O2 and 1.00 x 10-2 mol Cl2 are mixed
in a 4.00 L container.
What are the equilibrium
concentrations.
Mid-range K’s
.01<K<10
No Simplification
Choose X to be small.
Can’t simplify so we will have to solve the
quadratic formula
H2 (g) + I2 (g)  2 HI (g) K = 38.6
What is the equilibrium concentrations if
1.800 mol H2, 1.600 mol I2 and 2.600 mol HI
are mixed in a 2.000 L container?
Here is the set - up
Q = 2.34 so Q < K & it shifts to the right.
Initial
Change
Equilibrium
H2 (g) + I2 (g) 
.9
.8
-x
-x
.9 – x .8 – x
2 HI (g)
1.3
+ 2x
1.3 + 2x
K = 38.6 = (1.3 + 2x)2 / (.9 – x) (.8 – x) = ??
1. Fortunately, we don’t see them this type often.
2. You are encouraged to discover the many uses
of your TI – 84 or equivalent.
Problems Involving Pressure
Solved exactly the same, with same rules for
choosing X but use pressures due to Kp
For the reaction N2O4 (g)  2NO2 (g)
Kp = .131 atm. What are the equilibrium pressures
if a flask initially contains 1.000 atm N2O4?
Initial
Change
Equilibrium
N2O4 (g)  2 NO2 (g)
1
0
-x
+ 2x
1-x
1+ 2x
Le Chatelier’s Principle
If a stress is applied to a system at
equilibrium, the position of the equilibrium
will shift to reduce the stress, and
establish a new equilibrium.
3 Types of stress



Temperature changes
Pressure changes
Concentration changes
The Basic Steps
Here is the best way to always get
LeChatelier’s problems correct.
Sample Problem A + B  C + D
How is the system going to change if more B
is added ?
1. Identify the stress: The addition of B
2. What is the opposite: The removal of B
3. Which way must the reaction shift to
achieve step 2: It must shift to the right
Change amounts of reactants
and/or products
Adding product makes Q > K
Removing reactant makes Q > K
Adding reactant makes Q < K
Removing product makes Q < K
Determine the effect on Q, will tell you the
direction of shift: remember that it is
important to know what Q is equal to.
Change in Pressure # 1
This is usually done by changing volume.
Decreasing the volume
increases the pressure
System shifts in the direction that has the less
moles of gas: less gas = less pressure
Because partial pressures (and
concentrations) change a new equilibrium
must be reached and established.
System tries to minimize the moles of gas.
Change in Pressure # 2
This is usually done by changing volume.
Increasing the volume
decreases the pressure
System shifts in the direction that has more
moles of gas: more gas = more pressure
Because partial pressures (and
concentrations) change a new equilibrium
must be reached and established.
System tries to maximize the moles of gas.
Change in Pressure # 3
By the addition of an inert gas.
Partial pressures of reactants and product
are not changed .
No effect on equilibrium position.
This is a common question on the AP exam
so be aware of it. It should be easy points.
It might phrase it as an inert gas or it might
use one of the Noble Gases:

He
Ne Ar
Kr
Xe
Rn
Change in Temperature
Affects the rates of both the forward and reverse
reactions.
Doesn’t just change the equilibrium position,
changes the equilibrium constant.
The direction & magnitude of the shift depends on
whether the reaction is exothermic or endothermic.
On many questions, it states a temperature that K
is measured in. This is more to make the question
technically correct. Look back on our solved
problems. Temperature was not calculated in.
Change in Temperature :Exothermic
DH < 0 It is a negative number.
The reaction releases heat.
Think of heat as a product since you know
an equation can be written with a heat term:
A + B  C + D + Heat
Raising temperature (adding heat) effectively
increases the amount of heat (product)
Shifts to left.
Remember the steps to solving LeChatelier’s
problems.
Change in Temperature :Endothermic
DH > 0 It is a positive number.
The reaction absorbs heat.
Think of heat as a reactant since you know an
equation can be written with a heat term.
A + B + Heat  C + D
Lowering temperature push toward reactants.
Shifts to left.
Remember the steps to solving LeChatelier’s
problems.
Changes in Temperature: Summary
1. Place the heat term in the reaction.
2. Determine if heat is being added or taken
away from the reaction.




If heat is being added to an exothermic reaction,
the reaction shifts to the left.
If heat is being added to an endothermic
reaction, the reaction shifts to the right.
If heat is being taken away from an exothermic
reaction, the reaction shifts to the right
If heat is being taken away from an endothermic
reaction, the reaction shifts to the left.