Chemical
Equilibrium
AP CHEMISTRY
Chamberlain
1
What is Equilibrium?


Reactions don’t always have a beginning and an end. They
find a ____________________ and have to work through
stress to find it.
Many reactions ____________________________ finish.
o They are __________________________
o Can go ____________________ or
_____________________ without extra energy used

Equilibrium state is reached when a reaction’s forward
progress is…
Dynamic Equilibrium


When a system reaches _______________________ the
forward and reverse reactions continue to occur, but at
_______________________.
After equilibrium the concentrations of reactants and
products remain ________________________.
The Equilibrium Constant Expression:
aA + bB  gG + hH
2
The Equilibrium Constant (Kc)

Constant regardless of the…

Denoted by the symbol ________ and called…

The constant is only good at a given _____________________.
If the temperature is changed, so is the constant.

Concentrations of products appear in the _________________
and concentrations of reactants appear in the
_____________________.

The _____________________ of the concentrations are
identical to the _____________________________________
in the chemical equation.
Quick Check: Write the equilibrium expression for:
N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)
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Example 14.1 If the equilibrium concentrations of COCl2 and Cl2
are the same at 395 °C, find the equilibrium concentration of CO
in the reaction:
CO(g) + Cl2(g) ↔ COCl2(g)
Kc = 1.2 x 103 at 395 °C
Modifying the Chemical Equation

For the reverse reaction, K is the _____________________ of
K for the forward reaction.

When an equation is divided by two, K for the new reaction is
the ________________________ of K for the original reaction.

General Rule: when the coefficients of an equation are
multiplied by a common factor (n) to produce a new equation,

It should be clear that we must…
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
So…
o If we reverse a reaction, to find the new Kc we would…
o If we double a reaction, to find the new Kc we would…
o If we third a reaction, to find the new Kc we would…
Example 14.2 The equilibrium constant for the reaction:
½ H2(g) + ½ I2(g) ↔ HI(g)
(a)
at 718 K is 7.07
What is the value of Kc at 718 K for the reaction
HI(g) ↔ ½ H2(g) + ½ I2(g)
(b) What is the value of Kc at 718 K for the reaction
H2(g) + I2(g)
↔
2 HI(g)
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Try It Out #1
N2(g) + 3H2(g) ↔ 2NH3(g)
For the equation above the Kc = 3.6 x 108. What is the value of Kc
at the same temp for:
NH3 ↔ ½ N2 + 3/2 H2
Try It Out #2
N2(g) + 3H2(g) ↔ 2NH3(g)
For the equation above the Kc = 3.6 x 108. Determine the value for
Kc for the reaction:
1/3 N2 + H2 ↔ 2/3 NH3
Equilibrium Constant Involving Gases, Kp

In reactions involving gases, it is often convenient to measure
partial pressures rather than molarities.

In these cases, a partial pressure equilibrium constant, _____,
is used.
Kp =
6
Where Δn =
Example
2H2S ↔2H2 + S2
Kp = 1.2 x 10-2
Kc =?
Example 14.3 Consider the equilibrium between dinitrogen tetroxide
and nitrogen dioxide:
N2O4(g) ↔2 NO2(g)
Kp = 0.660 at 319 K
(a) What is the value of Kc for this reaction?
(b) What is the value of Kp for the reaction 2 NO2(g) ↔ N2O4(g)?
(c) If the equilibrium partial pressure of NO2(g) is 0.332 atm, what is the
equilibrium partial pressure of N2O4(g)?
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Combining Equilibrium Constant Expressions

Suppose we need: N2O(g) + 3/2 O2(g) ↔2 NO2(g)
Kc(1) = ??
and we’re given:
N2O(g) + ½ O2(g) ↔ 2 NO(g)
2 NO(g) + O2(g) ↔
2 NO2(g)
•
•
Kc(2) = 1.7 x 10–13
Kc(3) = 4.67 x 1013
_________________the given equations gives the desired
equation.
_________________ the given values of K gives the
equilibrium constant for the overall reaction.
Example - #25 p. 607
What is the value of Kp at 298 K for the reaction ½ CH4 (g) + H2O
(g) ↔ ½ CO2 (g) + 2H2 (g), given the following data at 298 K?
CO2 (g) + H2 (g) ↔ CO (g) + H2O (g)
Kp = 9.80 x 10-6
CO (g) + 3 H2 (g) ↔ CH4 (g) + H2O (g)
Kp = 8.00 x 1024
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Try It Out - #27 p. 608
Determine Kc at 298 K for the reaction 2CH4 (g) ↔ C2H2 (g) + 3 H2
(g), given the following data at 298 K.
CH4 (g) + H2O (g) ↔ CO (g) + 3 H2 (g)
Kp = 1.2 x 10-25
2C2H2 (g) + 3O2 (g) ↔ 4CO (g) + 2H2O (g)
Kp = 1.1 x 102
H2 (g) + ½ O2 (g) ↔ H2O (g)
Kp = 1.1 x 1040
Equilibria Involving Pure Liquids and Solids

The equilibrium constant expression does not include terms
for pure ____________________ and __________________
phases because their concentrations do not change in a
reaction.

Although the amounts of pure solid and liquid phases change
during a reaction, these phases remain pure and their
__________________________ do not change.
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
Example: CaCO3 (s) ↔ CaO (s) + CO2 (g)
Try It Out
Write the Kc expression for:

C(s) + H2O (g) ↔ CO(g) + H2(g)

MgCO3(s) ↔ MgO(s) + CO2(g)

2H2(g) + O2(g) ↔ 2H2O(l)

H2S(g) + I2(s) ↔ 2HI(g) + S(s)
Example 14.4
The reaction of steam and coke (a form of carbon)
produces a mixture of carbon monoxide and hydrogen, called watergas. This reaction has long been used to make combustible gases from
coal:
C(s) + H2O(g)
↔
CO(g) + H2(g)
Write the equilibrium constant expression for Kc for this reaction.
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Equilibrium Constants: When do we need them?

A very ________________ value of Kc or Kp signifies that a
reaction goes (essentially) to completion.

A very _______________ value of Kc or Kp signifies that the
forward reaction, as written, occurs only to a slight extent.

An equilibrium constant expression applies only to a reversible
reaction at equilibrium.

Although a reaction may be ____________________________
_____________________, it may be _____________________
_____________________

Thermodynamics tells us

Kinetics tells us
Examples:
2H2 (g) + O2(g) ↔ 2 H2O(g)
Kp = 1.4 x 10 83 @ 298 K
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C(s)+ H2O(g) ↔CO(g) + H2(g)

Kp = 1.6 x 10-21 @ 298 K
So, a reaction is most likely to reach a state of equilibrium in which
significant quantities of both reactants and products are present if
the numerical value of Kc and Kp is neither very large nor very small,
roughly in the range from ______________________
The Reaction Quotient, Q

For nonequilibrium conditions, the expression having the same form
as Kc or Kp is called the…

The reaction quotient is not ____________________ for a reaction,
but is useful for predicting the _____________________ in which a
net change must occur to establish equilibrium.

To determine the net change, we compare the magnitude of Qc to
that of Kc:
o If Qc < Kc
o
If Qc = Kc
o
If Qc > Kc
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Example
CO(g) + 2 H2(g) ↔ CH3OH (g)
If the Kc for this reaction = 14.5, then what direction will the reaction
proceed if the initial concentrations are CO = 0.100 M , H2 = 0.100 M,
CH3OH = 0 M
Try it Out - #49
If a gaseous mixture at 588 K contains 20.0 g each of CO and H 2O
and 25.0 g each of CO2 and H2, in what direction will a net
reaction occur to reach equilibrium? Explain
CO (g) + H2O (g) ↔ CO2 (g) + H2 (g)
Kc = 31.4 at 588 K
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Quick Check
For the reaction 2NO(g) ↔ N2 (g) + O2 (g), Kc = 9.92 at 2273 K.
1. Write the equilibrium expression (Kc) for this reaction.
2. What is Kc for this reaction:
2N2 (g) + 2O2 (g) ↔ 4NO(g)
3. At equilibrium, the concentration of nitrogen and oxygen are
both 1.5 x 10-3 M. What is the concentration of NO?
4. What is Kp for the reaction?
Stresses on Equilibrium

Chemists make a living ___________________ equilibrium.
They want to “knock it out of whack” to maximize a certain
____________________.

Stresses can be applied to shift equilibrium and maximize the
amount of product, as you saw in the POGIL.
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LeChatelier’s Principle

When any change in concentration, temperature, pressure, or
volume is imposed on a system at equilibrium, the system
responds by attaining a new equilibrium condition that
minimizes the impact of the imposed change.
Change in Concentration:

At equilibrium, ________________________

If the concentration of one of the _________________ is
increased, the ____________________ of the reaction
quotient increases.

Q is now ________________ Kc.

This condition is only temporary, however, because the
concentrations of all species must change in such a way so as
to make _____________________ again.

In order to do this, the concentrations of the products
_________________; the equilibrium is shifted to the
______________________.

An increase in concentration…

Remember that this is a change in concentration so changing
the amount of a pure solid or liquid which is already in
equilibrium should not make the reaction consume the
additional solid or liquid.
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
Example: 2HgO (g) ↔2Hg (l) + O2 (g)
∆ H = +43.4 kcal
1. Add HgO, what happens to equilibrium?
2. Add O2, what happens to equilibrium?
3. Add Hg, what happens to equilibrium?
Example 14.7
Water can be removed from an equilibrium mixture in the
reaction of 1-octanol and acetic acid, for example, by using a solid drying agent
that is insoluble in the reaction mixture. Describe how the removal of a small
quantity of water affects the equilibrium.
CH3(CH2)6CH2OH(soln) + CH3COOH(soln)↔ CH3(CH2)6CH2OCOCH3(soln) + H2O(soln)
Change in Pressure

When the external pressure is ________________ (or system
volume is reduced), an equilibrium shifts in the direction
producing the _________________ number of moles of a gas.

When the external pressure is ___________________ (or the
system volume is increased), an equilibrium shifts in the
direction producing the __________________ number of
moles of gas.
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
If there is no change in the number of moles of gas in a
reaction, changes in external pressure (or system volume)
have no effect on an equilibrium.
Example: 2HgO (g) ↔2Hg (l) + O2 (g)
∆ H = +43.4 kcal
1. Add pressure, what happens to equilibrium?
2. Increase volume, what happens to equilibrium?
Example 14.8 An equilibrium mixture of O2(g), SO2(g), and SO3(g) is
transferred from a 1.00-L flask to a 2.00-L flask. In which direction does
a net reaction proceed to restore equilibrium? The balanced equation
for the reaction is
2 SO3(g) ↔ 2 SO2(g) + O2(g)
Change in Temperature

Raising the temperature of an equilibrium mixture shifts
equilibrium in the direction of the _________________
reaction; lowering the temperature shifts equilibrium in the
direction of the ____________________ reaction.
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o Consider heat as though it is a _________________ of
an exothermic reaction or as a _________________ of
an endothermic reaction, and apply Le Chatelier’s
principle
Example 14.9 Is the amount of NO(g) formed from given amounts of
N2(g) and O2(g),
N2(g) + O2(g) ↔
2 NO(g)
ΔH° = +180.5 kJ
greater at high or low temperatures?

A catalyst lowers the activation energy…

Adding a catalyst does not affect an equilibrium state.

A catalyst merely causes equilibrium to be achieved
____________________.
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Determining Values of Equilibrium Constants Experimentally
•
When initial amounts of one or more species, and equilibrium
amounts of one or more species, are given, the amounts of
the remaining species in the equilibrium state and, therefore,
the equilibrium concentrations often can be established.
•
A useful general approach is to tabulate under the chemical
REACTION:
–
the concentrations of substances present INITIALLY
–
CHANGES in these concentrations that occur in
reaching equilibrium
–
the EQUILIBRIUM concentrations.
•
This sort of table is sometimes called a “RICE” table:
Reaction/Initial/Change/Equilibrium.
•
Important note: The “RICE” table may be used with…
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Example Equilibrium involving SO2, O2, and SO3 is important in sulfuric acid
production. When a 0.0200 mol sample of SO 3 is introduced into an evacuated
1.52 L flask @ 900 K, 0.0142 mol SO3 is found to be present at equilibrium.
What is the value of Kp ?
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Example 14.11 In a 10.0-L vessel at 1000 K, 0.250 mol SO2 and 0.200 mol O2
react to form 0.162 mol SO3 at equilibrium. What is Kc, at 1000 K, for the
reaction that is shown here?
2 SO2(g) + O2(g) ↔ 2 SO3(g)
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Try it Out – 14.11 A
A reaction starts with 1.00 mol each of PCl3 and Cl2 in a 1.00 L flask. When
equilibrium is established at 250 degrees C in the reaction PCl3 (g) + Cl2 (g) ↔
PCl5 (g), the amount of PCl5 present is 0.82 mol. What is Kc for this reaction?
Calculating Equilibrium Quantities from Kc and Kp values


When starting with initial reactants and no products and with
the known value of the equilibrium constant, these data are
used to calculate the amount of substances present at
equilibrium.
Typically a RICE table is constructed and the symbol x is used
to identify one of the changes in concentration that occurs in
establishing equilibrium.
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
Then, all the other concentration changes are related to x, the
appropriate terms are substituted into the equilibrium
constant expression, and the equation solved for x.
Example 14.12 Consider the reaction
H2(g) + I2(g) ↔ 2 HI(g)
Kc = 54.3 at 698 K
If we start with 0.500 mol I2(g) and 0.500 mol H2(g) in a 5.25-L vessel at
698 K, how many moles of each gas will be present at equilibrium?
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Example 14.14 Carbon monoxide and chlorine react to form
phosgene, COCl2, which is used in the manufacture of pesticides,
herbicides, and plastics:
COCl2(g) ↔ CO(g) + Cl2(g)
Kc = 1.2 x 103 at 668 K
How much of each substance, in moles, will there be at equilibrium in a
reaction mixture that initially has 0.0100 mol CO, 0.0100 mol Cl2, and
0.100 mol COCl2 in a 10.0-L flask?
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Example 14.15 A sample of phosgene, COCl2(g), is introduced into a
constant-volume vessel at 395 °C and observed to exert an initial
pressure of 0.351 atm. When equilibrium is established for the reaction
CO(g) + Cl2(g) ↔ COCl2(g)
Kp = 22.5
what will be the partial pressure of each gas and the total gas
pressure?
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Try it Out – 14.13 A
Starting with 0.100 mol CO and 0.200 mol Cl2 in a 25.0 L flask, how
many moles of COCl2 will be present at equilibrium in the
following reaction?
CO (g) + Cl2 (g) ↔ COCl2
Kc = 1.2 x 102 at 668 K


For an exothermic reaction; as T increases, Kp decreases
For an endothermic reaction; as T decreases, Kp increases
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Checklist for ICE tables (from chemmybear.com)









Make sure the reversible equation is balanced at the start of the
problem; otherwise, you could end up with wrong amounts in the
table.
The given data should be in amounts, concentrations, partial
pressures, or somehow able to be converted to such. If it is not,
then an ICE table will not help solve the problem.
If the ICE table has the equilibrium in amounts, make sure to
convert equilibrium values to concentrations before plugging in to
solve for Kc.
If the given data is in amounts or concentrations, use the ICE table
to find Kc. If the given data is in partial pressures, use the ICE table
to find Kp. If you desire to convert from one to the other, remember
that
; it is simpler to use the ICE table with
the appropriate givens and convert at the end of the problem.
Enter in known data first, and that should enable you to calculate
the unknown data in the table.
If ever you have a negative value in the Initial or Equilibrium rows,
double check your work! A negative concentration, amount, or
partial pressure is physically impossible. Obviously, the Change row
can contain a value that is a negative number.
Pay attention to the state of each reactant and product. If a
compound is a solid or a liquid, its concentrations are not relevant
to the calculations. Only concentrations of gaseous and aqueous
compounds are used.
In the 'change' row the values will usually be a variable, denoted by
x. It must first be understood which direction the equation is going
to reach equilibrium (from left to right or from right to left). The
value for 'change' in the 'from' direction of the reaction will be the
opposite of x and the 'to' direction will be the positive of x (adding
concentration to one side and take away an equal amount from the
other side).
Know the direction of the reaction. This knowledge will affect the
Change row of the ICE table (for our example, we knew the reaction
would proceed forward, as there was no initial products). Direction
of reaction can be calculated using Q, the reaction quotient. Q is
then compared to a known (read:given) K value.
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
It is easiest to use the same units every time you use an ICE table
(usually molarity is preferred). This will minimize confusion when
calculating the equilibrium constants. ICE tables are usually used for
weak acid or weak base reactions because all of the nature of these
solutions. The amount of acid or base that will dissociate is
unknown (for strong acids and strong bases it can be assumed that
all of the acid or base will dissociate, meaning that the
concentration of the strong acid or base is the same as its
dissociated particles).
*"Partial pressure" may also be substituted for "concentrations" in the
ICE table, if desired (i.e., if the concentrations are not known, Kp instead
of Kc is desired, etc.). "Amount" is also acceptable (the ICE table may be
done in amounts until the equilibrium amounts are found, after which
they will be converted to concentrations). For simplicity, assume that the
word "concentration" can be replaced with "partial pressure" or
"amounts" throughout the non-example portions of this module.
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