Download Unit 3: 1 Equilibrium and the Constant, K

NAME ___________________________________AP NOTES: UNIT 3 (1): EQUILIBRIUM & THE CONSTANT
Let’s take a moment and reflect on how we’re constructing the coursework. To further our grasp as how
chemists can predict and/or explain chemical phenomena:
First, there was a review of nomenclature, writing empirical formula and dimensional analysis
We began our advanced work, with the underlying process of kinetics (Big Idea 4).
We segued to the study of substances, with a focused look at the gas laws … (Big Idea 2, in part)
We are now moving on to another of the underlying processes … equilibrium (Big Idea 6)
Equilibrium is a big, (I mean, big) topic. It links into other units such as Kinetics, Thermodynamics,
Acid/Base Theory, Solution Theory, Bonding, and the behavior of matter, in terms of phase changes. It is
another underlying process. To control the yield of a reaction chemists need to understand how the position of
equilibrium is affected by conditions such as temperature and pressure. The regulation of chemical
equilibria affects the yield of products in industrial processes and living cells struggle to avoid sinking
into equilibrium Hmm! (Atkins p. 383)
At this point we’ll cover ONLY, Enduring Understandings 6A and 6B & the following Learning Objectives.
BIG IDEA 6: Any bond or intermolecular attraction, that can be formed, can be broken.
These two processes are in dynamic competition sensitive to initial conditions and external
perturbations.
Enduring understanding 6.A:
Chemical equilibrium is a
dynamic, reversible state in
which rates of opposing
processes are equal.
Essential knowledge 6.A.1: In many classes of reactions, it is important to consider both the forward and
reverse reaction.
Essential knowledge 6.A.2: The current state of a system undergoing a reversible reaction can be
characterized by the extent to which reactants have been converted to products. The relative quantities of
reaction components are quantitatively described by the reaction quotient, Q.
Essential knowledge 6.A.3: When a system is at equilibrium, all macroscopic variables, such as
concentrations, partial pressures, and temperature, do not change over time. Equilibrium results from an
equality between the rates of the forward and reverse reactions, at which point Q = K.
Essential knowledge 6.A.4: The magnitude of the equilibrium constant, K, can be used to determine
whether the equilibrium lies toward the reactant side or product side.
Enduring understanding 6.B:
Systems at equilibrium are
responsive to external
perturbations, with the
response leading to a change
in the composition of the
system. understanding 6.C:
Enduring
Chemical equilibrium plays an
important role in acid-base
chemistry and in solubility.
Essential knowledge 6.B.1: Systems at equilibrium respond to disturbances by partially countering the
effect of the disturbance (Le Chatelier’s principle).
Essential knowledge 6.B.2: A disturbance to a system at equilibrium causes Q to differ from K, thereby
taking the system out of the original equilibrium state. The system responds by bringing Q back into
agreement with K, thereby establishing a new equilibrium state.
Essential knowledge 6.C.1: Chemical equilibrium reasoning can be used to describe the proton-transfer
reactions of acid-base chemistry.
Essential knowledge 6.C.2: pH is an important characteristic of aqueous solutions that can be controlled with
buffers. Comparing pH to pKa allows us to determine the protonation state of a molecule with a labile proton.
Essential knowledge 6.C.3: The solubility of a substance can be understood in terms of chemical equilibrium.
Enduring understanding 6.D: The Essential knowledge 6.D.1: When the difference in Gibbs free energy between reactants and products (ΔG°) is
equilibrium constant is related to much larger than the thermal energy (RT), the equilibrium constant is either very small (for ΔG° > 0) or very large
temperature and the difference in (for ΔG° < 0). When ΔG° is comparable to the thermal energy (RT), the equilibrium constant is near 1.
Gibbs free energy between reactants
and products
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Learning objective 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or
environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying
chemical reactions or processes. [See SP 6.2; Essential knowledge 6.A.1]
Learning objective 6.2 The student can, given a manipulation of a chemical reaction or set of reactions (e.g., reversal of reaction or addition
of two reactions), determine the effects of that manipulation on Q or K. [See SP 2.2; Essential knowledge 6.A.2]
Learning objective 6.3 The student can connect kinetics to equilibrium by using reasoning about equilibrium, such as Le Chatelier’s
principle, to infer the relative rates of the forward and reverse reactions. [See SP 7.2; Essential knowledge 6.A.3]
Learning objective 6.4 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K,
use the tendency of Q to approach K to predict and justify the prediction as to whether the reaction will proceed toward products or reactants
as equilibrium is approached. [See SP 2.2, 6.4; Essential knowledge 6.A.3]
Learning objective 6.5 The student can, given data (tabular, graphical, etc.) from which the state of a system at equilibrium can be obtained,
calculate the equilibrium constant, K. [See SP 2.2; Essential knowledge 6.A.3]
Learning objective 6.6 The student can, given a set of initial conditions (concentrations or partial pressures) and the equilibrium constant, K,
use stoichiometric relationships and the law of mass action (Q equals K at equilibrium) to determine qualitatively and/or quantitatively the
conditions at equilibrium for a system involving a single reversible reaction. [See SP 2.2, 6.4; Essential knowledge 6.A.3]
Learning objective 6.7 The student is able, for a reversible reaction that has a large or small K, to determine which chemical species will
have very large versus very small concentrations at equilibrium. [See SP 2.2, 2.3; Essential knowledge 6.A.4]
Learning objective 6.8 The student is able to use Le Chatelier’s principle to predict the direction of the shift resulting from various possible
stresses on a system at chemical equilibrium. [See SP 1.4, 6.4; Essential knowledge 6.B.1]
Learning objective 6.9 The student is able to use Le Chatelier’s principle to design a set of conditions that will optimize a desired outcome,
such as product yield. [See SP 4.2; Essential knowledge 6.B.1]
Learning objective 6.10 The student is able to connect Le Chatelier’s principle to the comparison of Q to K by explaining the effects of the
stress on Q and K. [See SP 1.4, 7.2; Essential knowledge 6.B.2]
As in the other units, we’ll begin with a sweep through of general ideas, basic vocabulary and quick reviews
of work done in Honors Chemistry.
In the unit on Kinetics (Big Idea 4), we studied chemists’ understanding as to how fast a chemical reaction
occurs. In this unit we will study how far a chemical reaction can go. Not all go “to completion”. Some
reactions can experience a reverse reaction. The extent …, the how far of a chemical reaction is determined by
thermodynamics (a later unit). However, we can extract a piece of that unit for our current work, because we
will study an experimentally measurable quantity called the equilibrium constant.
In short, most reactions occur so as to produce products. A reaction with a large equilibrium constant proceeds,
virtually to completion (the virtually complete consumption of reactants to make product). A small value for
the equilibrium constant indicates that nearly all the reactants remain as reactants, producing very small
quantities of product. (Tro p. 648)
Physical changes (e.g. melting-freezing, dissolving-crystallization), and chemical reactions can achieve an
equilibrium. A state of equilibrium unto itself is not, a “bad or good” thing. However, in the quest to
manipulate matter and energy to our own ends, chemists wish to understand the issue, so as to
control/change/affect/avoid/create equilibria, in most chemical systems which achieve equilibrium.
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I) (Dynamic) Equilibrium: In short, an equilibrium exists when the * rate of the forward reaction equals the
rate of the reverse reaction. The concentrations are not (probably) equal …. BUT the concentrations of
all the reactants and products are constant.
A) So, 3 issues define an equilibrium system: Reversible, Dynamic, Constant [reactant]:[product]
B) In unit on kinetics we learned that reaction rates are dependent upon the required activation
energy & rds. You also know that the rates generally increase with increasing concentration of the
reactants (unless, the reaction order is *zero
) …. and the rates decrease with
decreasing concentration of the reactants.
Yet, reflection upon Big Idea 6,
Any bond or intermolecular attraction, that can be formed, can be broken.
These two processes are in dynamic competition sensitive to initial conditions
and external perturbations.
should suggest that as the reactant concentration decreases (and thus the rate), the products due to
their inevitably increasing concentrations, could indeed begin to collide and rupture their newly
formed bonds and engage in a reverse reaction. This activity, OVER TIME, could reconstitute the
reactant concentration to such a point, that the two opposing rates (making product and the
decomposition thereof) could eventually match each other….
http://www.history.com/encyclopedia/fwne/images/ChemicalReactionC3.gif
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II) Most chemical reactions (and virtually all physical changes involved in chemistry) are reversible.
A) We tend to write only 1 equation … but under the correct conditions, a reaction equation, could
symbolize two possible reactions occurring, simultaneously.
1) Before launching into this conversation let’s assume that the ONLY collisions which
can have an effect are summarized below:
reactant colliding with reactant …. to make product
product colliding with product …. to re-constitute reactant
o reactant colliding with product is a non-effective collision & has no effect
on the creation or shifting of the point of equilibrium
2) reactants collide and bond, giving the products (this is called the forward reaction … we
read the FORWARD reaction from left to right)
3) products can react with each other, to reconstitute the reactants (This is called the reverse
reaction and we read the REVERSE reaction equation from right to left … essentially making
the products into reactants.
4) Think of it this way: Every synthesis reaction, can be seen as a decomposition when
read backwards ….
NO(g) + O2(g)  2 NO2(g) + 112.86 kJ can be read in reverse giving a complimentary reaction:
112.86 kJ + 2 NO2(g)  NO(g) + O2(g)
a) Initially students become a bit confused over the change in enthalpy being the same
for both the forward and the reverse reaction … However, when you reflect upon that
the change in enthalpy is the difference between products and reactants, or Hp –Hr
then, it should make a bit more sense.
e.g A + B → C + 40 kJ
100 A+B
60
40 kJ + C → B + A
A+B
100
∆H
∆H
C
60
C
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B) The nature of an equilibrium is interesting, in that whether you begin with only reactants or
with only products, the reaction reaches equilibrium concentration at which the equilibrium
constant is the same, at a specific temperature.
1) No matter what the initial concentrations are, the reaction * always goes in a direction that
ensures that the equilibrium concentrations, when substituted into the equilibrium
expression , give the same equilibrium constant (K)
This idea is summarized very nicely in your B&L text (p. 632)
B) A double-headed arrow () or double arrow ( ) indicate a reaction is a reversible reaction
and that the two opposing reactions are already at equilibrium.
1) Given: 2 SO2(g) + O2(g)  2SO3(g) + 197.78 kJ
a) What visual clue tells you the reaction is at equilibrium? *double/opposing arrows
b) Write the forward reaction: * 2 SO2(g) + O2(g) → 2SO3(g) + 197.78 kJ
c) Write the reverse reaction: * 2SO3(g) + 197.78 kJ → 2 SO2(g) + O2(g)
d) Which reaction (forward or reverse) is exothermic? *forward reaction
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2) Given:
2 H2O(l) + 571 kJ  2 H2(g) + O2(g)
a) What visual clue tells you the reaction is at equilibrium? * double headed arrow
b) Write the forward reaction: * 2 H2O + 571 kJ → 2 H2 + O2
c) Write the reverse reaction: * 2 H2 + O2 → 2 H2O + 571 kJ
d) In terms of energy, (exothermic vs. endothermic) how should you categorize
the reverse reaction? * exothermic
III) There are three forms of equilibrium:
A) phase equilibrium: When the rates of 2 opposing
* physical
at the same rate. (ice melts as fast as water freezes
same rate as water vapor condenses)
changes occur
or water evaporates at the
B) solution equilibrium: When the rate of dissolving occurs at the same rate as * crystallization
A solution equilibrium exists only in saturated solutions, in which a small
amount of solid is present, which is in equilibrium with the dissolved
species.
C) chemical equilibrium: When the rates of 2 opposing * chemical
same rate.
reactions occur at the
Chemmaters Feb 1984
D) Chemical equilibrium is dynamic … there is a good deal of change going on. There could be
a significant amount of bond making and bond breaking going on … It is just that in a dynamic
equilibrium, the net effect of all of this change is a 0 change in the concentrations of reactants and
products.
1) a static equilibrium would be akin to me pushing against a wall, or a tug of war in which
neither party moves forward or back at all ….
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Visualize!
The equilibrium between the decomposition of dinitrogen tetroxide and its synthesis
58 kJ + N2O4(g)

2 NO2(g)
NO2
N2O4
http://www.cartage.org.lb/en/themes/Sciences/Chemistry/Miscellenous/Helpfile/Equilibrium/Equilibrium/eq.gif
1) The following reaction is at equilibrium
N2O4(g)

2 NO2(g)
Being at equilibrium, means: The rate of the forward reaction * equals the rate of the
reverse reaction
AND the [N2O4 ] and [NO2] are * constant
but not necessarily *equal
a) Translation: The product of the forward reaction (nitrogen dioxide) collides and
combines to make N2O4(g) at the same rate as the dinitrogen tetroxide decomposes
into NO2
Recall: Only 2 categories of collisions are ever considered ... The collisions between reactants (which
result in products) & the collisions between products that result in re-constituting the reactants.
For instance: Important: N2O4(g)
Important: NO2(g)
UN-important: N2O4(g)
colliding with N2O4(g)
colliding with NO2(g)
colliding with NO2(g)
2) The following reaction is at equilibrium
N2(g) + 3H2(g)  2NH3(g)
Being at equilibrium, means: The rate of the forward reaction = the rate of the reverse reaction.
OR: The rate of the effective collisions between N2 and H2 = the rate of effective collisions
between molecules of NH3 & NH3
AND the [N2 ] , [H2] and [NH3] are * constant
but not necessarily * equal
a) Another way to look at it is : Ammonia is produced at the same rate as it is
decomposed back into dinitrogen and dihydrogen
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3) The reaction:
2 N2(g) + 3 O2(g)  2 N2O3(g)
is at equilibrium.
This means that the rate of producing * N2O3(g)
equals the rate of
* decomposing N2O3(g) back into N2(g) and O2(g)
a) The concentrations of N2(g), O2(g)
&
N2O3(g) at equilibrium are constant / equal
(circle one)
b) Translation: Dinitrogen trioxide is made at the same rate as it is * decomposed
OR the rate of effective collisions between N2(g) & O2(g) equals the rate of
effective collisions between molecules of N2O3
4) The reaction: 2SO2(g) + O2(g)  2SO3(g) is at equilibrium.
This means that the production of SO3 equals the rate of its decomposition ... OR
the rate of effective collisions between SO2(g) & O2(g) = the rate of effective
collisions between molecules of SO3
(or SO3 and other SO3 molecules)
a) The concentration of each substance is constant / equal
(circle one)
III) D (continued)
2)A chemical equilibrium can be “shifted” (perturbed or stressed, as in Big Idea 6). When
the perturbation (stress) is maintained, the chemical reaction will achieve a new point of
equilibrium, in time. The concentrations of the reactants and products, will be different at
this new point of equilibrium.
a) During this time of perturbation (or stress), *one reaction is inevitably favored or
made to be dominant … and runs at a higher rate.
As the system works away under this stress, in time, under these different conditions a
new equilibrium, at some point will be established. The role of the chemist is to
understand the causative factor(s), consequence(s) and perhaps use it to her/his
advantage.
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b) Were we to assume a reaction at equilibrium, the point of equilibrium can be shifted
(temporarily destroyed, stressed, perturbed) to favor one of the competing reactions
over the other. It will shift in such a way as to reduce the stress, and reach a new
point of equilibrium
Learn this jargon … it is
specific to equilibrium,
and we use it pretty
often…
i) Shift the (point of) equilibrium to the right = favor the forward reaction
and make more product.
Shift the (point of) equilibrium to the left = favor the reverse reaction
and make reactant
e.g.) Given a system at equilibrium:
58 kJ + N2O4(g)  2 NO2(g)
i) at equilibrium the rate of the forward reaction which makes NO2 is equal to
the rate of the reverse reaction which is *the decomposition of NO2 or rather,
the making of dimer, N2O4
ii) upon manipulation, the equilibrium is altered so as to produce more NO2
We would say that the equilibrium was *shifted to the right OR that
we favored the forward reaction.
OR…
iii) upon manipulation, the equilibrium could be altered so as to produce more
N2O4 … We would say that the equilibrium was *shifted to the left, OR that
we favored the reverse reaction.
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III) The Equilibrium Constant (Keq et. al.)
A) There are many different types of equilibrium constants … each has a special designation depending
upon the system it describes. Many of the commonly used symbols are listed in the following table.
Equilibrium
Constant
TABLE OF EQUILIBRIUM CONSTANTS
The Equilibrium Described by the Constant
Kc
A general designation, of c (concentration) in terms of molarity
Keq
A general designation, similar to Kc used for a chemical equilibrium
Kp
The partial pressure of a gas used to describe a gas’s concentration
Ka
The ionization of a(n) (weak) acid in water (a = acid)
Kb
The dissociation of a (weak) base in water (b = base)
Kw
The ionization for water = 1 x 10-14 [H2O(l)  H3O+1(aq) + OH-1(aq) ]
Ksp
The dissociation of an ionic salt in water (sp = solubility product)
B) The equilibrium constant of a reaction is a
 unit-less,
 temperature-dependent,
 ratio,
which has a magnitude (size) indicating the relative amount of product to the amount of
reactant at the point of equilibrium.
The constant suggests the relative success of the forward reaction, before equilibrium is
achieved. It is calculated using the Law of Mass Action.
1) Very generally … the larger the constant, the greater the production of product, before
equilibrium is established.
MEMORIZE
a) Hence a large equilibrium constant (Keq ≫1) favors making product … and moves
towards completion. Such reactions tend to have rather large, negative ∆H values
b) A small equilibrium constant (Keq ≪1) indicates that very little product is produced
by the time a point of equilibrium was reached … and that most of the reactants remain
un-reacted. The forward reaction does not proceed (does not get) very far by the time
an equilibrium system is reached.
c) When K ≈ 1 neither the forward nor the reverse is favored, and the forward reaction
goes “about half-way”…. The molar concentrations should reflect the equilibrium
constant.
d) The time it takes to reach a state of equilibrium is linked to the activation energy
and the reaction rate … It is not linked to the size of the equilibrium constant.
What I want to do now, is to take apart the above definition to show you why it is a ratio,
temperature-dependent, etc. … To do that we need to take a nice long look at The Law of Mass
Action….
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C) Using the Law of Mass Action to Write an Equilibrium Expression (The equation for Kc)
1) The Law of Mass Action is used to calculate the equilibrium constant, using experimentally
obtained molar concentrations.
a) When we write use the Law of Mass Action we are essentially writing out what is
called the equilibrium expression … the equation used to calculate Kc
2) The Law of Mass Action has a highly prescribed format. It includes ONLY aqueous and/or
gaseous components of a reaction. In the event of a heterogeneous equilibrium (in which
multiple phases are present in the system, the Law of Mass Action NEVER, NEVER
includes solids or liquids. Remember … include only gases or aqueous solutions!
a)
coef
coef
K = [Product]
[Product]
coef
coef
[Reactant]
[Reactant]
3) K is a unit-less ratio expressing the molar concentration of the gaseous &/or aqueous
* products
vs the concentration of the gaseous &/or aqueous * reactants
raised by an * exponent
reaction equation.
equal to each species' coefficient from the balanced
4) Note that the equilibrium expression used to calculate the equilibrium constant (K) depends
only upon the reaction stoichiometry!!! (This is quite different from the rate constant of k)
a) For this reason, it is convention to write out the stoichiometry using only whole
numbers. Using fractions can change the equilibrium constant. For example:
KEq = [C]
[A][B]2
A + 2B  C
versus:
½A + B  ½C
KEq’ = [C]1/2
[A]1/2 [B]
Now, let [A] = 2 M, [B] = 3 M and [C] = 4 M …
when we solve the two above equilibrium expressions:
KEq = [4] = 2 = 0.222
[2][3]2
9
That
makes
so much
sense
vs.
KEq’ = [4]1/2 = 2 = 0.472
[2]1/2 [3] 4.24
This is one of the reasons why you are always encouraged to balance chemical
reaction equations using whole numbers ….
TN: Marshall.edu
225
5) In an equilibrium expression, ONLY gaseous and aqueous species are included, for they
are the only species in solution capable of reacting... thus the only factors responsible for
determining the success of the reaction. Solids and liquids are technically unavailable
(un-ionized, or un-dissociated) for reaction, and thus solids and liquids have a constant
concentration … as odd as it is to even think of the term concentration in this light….
TN See Langella equilibrium notes for an example, w/ H2O
6) The equilibrium constant changes with temperature …. Hence the equilibrium constant
for a specific reaction is * temperature dependent (because rate constants, k
are temperature dependent)
a) *endothermic reactions
are more highly favored with an increase in
temperature, than are exothermic reactions …. (more later)
7) So we have here another temperature dependent K constant … Where have we seen this
before? How about k as in the rate constant… Now here is an interesting point for making
connections….
Take Home Message: Equilibrium connects to kinetics because: K = kforward
kreverse
Chemical reactions, whether reversible or non-reversible, follow the chemical kinetics.
Thus we can gain better understanding of the equilibrium constant K by using the concept
of chemical kinetics.
Let kf be the rate of forward reaction and kr be the rate of reverse reaction as shown
below.
kf
aA + bB ↔ cC + dD
kr
Then the forward rate is given by ratef = kf [A]a [B]b and the reverse rate by
rater = kr [C]c [D]d
At equilibrium, the forward rate becomes equal to reverse rate as a result of no net
change in concentrations of either reactants or products. (We have seen this already, in
when dealing with fast steps and having to re-write a fast elementary step.
Thus ratef = rater or kf [A]a [B]b = kr [C]c [D]d
After rearranging the above equation, it takes the following form:
c
d
kf = [C] [D]
a
b
kr
[A] [B]
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Since both kf and kr are constants at a given temperature, their ratio is also a constant that
is equal to the equilibrium constant K.
c
d
(see take home message)
kf = K = [C] [D]
a
b
kr
[A] [B]
The equilibrium constant K is a constant at a particular temperature regardless of
equilibrium concentrations of the species involved because of quotient kf / kr.
https://faculty.ncc.edu/LinkClick.aspx?fileticket=1nktvq0zylA%3D&tabid=1889
8) Practice Writing Equilibrium Expressions….
Write the equilibrium expression for 8a) – f)
a) N2(g) + 3H2(g) 
2NH3(g)
Remember: products is/are in the numerator.
Note that the exponents reflect the coefficients of
the balanced equation.
2
Keq = * [NH3]
[N2] [H2]3
+3
b) Al2(SO4)3(s) + H2O(l)  2 Al
(aq)
+ 3 SO4
-2
(aq)
Ksp = * [Al+3]2 [SO4-2]3
c) 3 O2(g)  2 O3(g)
Keq = * [O3]2
[O2]3
d) 2 Fe2O3(s) + 3 C(s)  4 Fe(s) + 3 CO2(g)
Remember: include only (g) and (aq)
Remember: the product(s) is/are in the numerator.
Note that solids and liquids are EXCLUDED ... thus
the denominator is essentially equal to 1
Note that the exponents reflect the coefficients of
the balanced equation.
Note that this is a "Ksp" because it reflects a
physical change (solubility issue) and not a chemical
reaction....
Note: This is a nice twist… apply the rules, and
think!
Keq = * [CO2]3
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e)
NH3(g)
Note: This provides a nice twist…apply the rules
and think!
+ HCl(g)  NH4Cl(s)
1
Keq = *
[NH3][HCl]
f) 4HCl(g) + O2(g)  H2O(l) + 2Cl2(g)
Keq =* [Cl2]2
[HCl]4 [O2]
g) Given the equilibrium constant expression: K = [CO2]3[H2O]4
Write the balanced reaction equation:
[C3H8] [O2]5
* C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
h) Given the reaction
A(g)  B(g) K = 10.
A reaction mixture initially contains [A] = 1.1M and [B] = 0.0M. Which statement is
true at equilibrium?
and Think: When written correctly, the equilibrium expression, with the substituted concentrations at
equilibrium, must = 10.. Does that give you a hint?
1) The reaction mixture will contain [A] = 1.0M and [B] = 0.1M
2) The reaction mixture will contain [A] = 0.1M and [B] = 1.0M
3) The reaction mixture will contain equal concentrations of A and B
ans* 2
i) Given: Ksp = [Ca ][Cl ] This is essentially an equilibrium constant expression for a
heterogeneous system. Write the equation which describes the equilibrium expression
2+
-1 2
Think! Look at the constant … does it give you any insight?
* CaCl2(s) + H2O(l)  Ca+2(aq) + 2 Cl-(aq)
Why were some substances left out of the equilibrium expression in this
heterogeneous equilibrium system?
* In a heterogeneous equilibrium, substances in the solid and liquid phase are omitted,
due to their constant state or concentration. Only those species in solution or in the gas
phase can affect the equilibrium.
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9) Simple Calculations for K:
a) Consider the equilibrium: 2 SO2(g) + O2(g)  2 SO3(g) + 791kJ
2 SO2(g) + O2(g)
Forward Reaction
Reaction
Rate
Equilibrium:
Rate of the Forward Rxn = Rate of the Reverse Rxn
2 SO3(g) + 791 kJ
Reverse Reaction
t0
Concentrations are measured at any point once
equilibrium is reached & then plugged into the "mass
action" equation.
t1
Time
Using the above reaction equation and the following experimentally determined molar concentrations,
calculate the Keq. At a specific temperature, the equilibrium concentration of [SO2 ] = 0.75 mole/L,
[O2 ] = 0.30 mole/L, and [SO3] = 0.15 mole/L
ans: Keq = 1.33 x 10-1
Keq
2
Substituion : Keq =* [0.15]2
[0.75]2 [0.30]
= * [SO3]
2
[SO2] [O2]
b) Consider the equilibrium:
N2O4(g)  2 NO2(g)
At 100˚C, experimentation shows that 0.10 M = [N2O4(g)] and 0.20 M = [NO2(g)].
Calculate the Keq
ans: Keq = 0.4 or 4 x 10-1
c) At 100°C, Keq = 0.0778 for the following reaction: SO2Cl2(g)  SO2(g) + Cl2(g)
In an equilibrium mixture of the three gases the concentrations of SO2Cl2 and SO2 are
0.136 M and 0.0720 M, respectively. What is [Cl2] in the equilibrium mixture?
ans: [Cl2] = 0.147 M
*Keq = [SO2][Cl2]
[SO2Cl2]
or
* 0.0778 = [0.0720 M] [x]
[0.136 M]
*0.010608 = 0.0720x
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D) Significance of K …a follow up to
1) Recall, back in III) B1: I wrote:
Very generally … the larger the constant, the greater the production of product, before equilibrium is established.
a) Hence a large equilibrium constant (Keq ≫ 1) favors more product … and moves towards completion.
b) A small equilibrium constant (Keq ≪ 1) indicates that very little product was produced by the time a point
of equilibrium was reached … and that most of the reactants remain un-reacted.
c) When K ≈ 1 neither the forward nor the reverse is favored,& the forward reaction goes about half-way
We must now look at what this implies…and what we may infer from the value of K
2) A large equilibrium constant indicates that the numerator (the concentration of the products)
*quite large, indicating that by the time the equilibrium was reached, a good deal of product
existed…
…. and, vice versa ….
3) Consider the following:
3H2(g) + N2g)  2 NH3(g) where K = 6.7 x 105 @ 400°C
Write the equilibrium constant as a fraction…
K = [NH3]2
= 6.7 x 105
3
[H2] [N2]
or
670,000
1
This may be seen, relatively as:
1
:
670,000 ….. product is definitely favored
3H2(g) + N2g)  2 NH3(g)
4) Consider the following: N2(g) + O2(g)  2 NO(g) where K = 4.1 x 10-31 @ 25°C
Write the equilibrium constant as a fraction…
K = [NO]2
= 4.1 x 10-31
[N2][O2]
or
1
41,000,000,000,000,000,000,000,000,000,000
This may be seen, relatively as:
41,000,000,000,000,000,000,000,000,000,000
N2(g) + O2g)
:

1
at equilibrium, there is far more reactant, than product
2 NO(g)
This reaction does not get very far… and thankfully for that … since NO is highly toxic, and that N2 and O2 are the major
constituents of our atmosphere … Imagine the consequences of the two gases reacting easily with each other!!!!! Yet again
here is an example as to the role chemistry can play in explaining the “how” of life on this, our planet …
230
5) Thus: the magnitude of the equilibrium constant K is useful in assessing the status of the
equilibrium:
If K = 1, [C]c [D]d = [A]a [B]b ,
the reaction is in equilibrium
If K >>1, [C]c [D]d >> [A]a [B]b,
the equilibrium lies to the right of the
reaction and favors the products
If K <<1, [C]c [D]d << [A]a [B]b,
the equilibrium lies to the left of the
reaction and favors the reactants
Review the problem in the section practicing the equilibria expression.
Given the reaction
A(g)  B(g) K = 10.
A reaction mixture initially contains [A] = 1.1M and [B] = 0.0M. Which statement is
true at equilibrium?
Were you to apply the “re-write the equilibrium expression as a fraction” …
interpretation, might it help you get the same answer?
Where K = 10
1
and
[B]
[A]
1
10
thus: A(g)  B(g)
1) The reaction mixture will contain [A] = 1.0M and [B] = 0.1M
2) The reaction mixture will contain [A] = 0.1M and [B] = 1.0M
3) The reaction mixture will contain equal concentrations of A and B
Kind’a cool… ain’t it?
Understanding what the concepts
imply can help….
231
BASIC PRACTICE FOR THE MEANING OF K
___1) Given the equilibrium
A + B  C. The greatest concentration of C would be produced if
the equilibrium constant of the reaction were equal to
1) 1 x 10
3
2) 1 x 10
-3
3) 1 x 10
9
4) 1 x 10
-9
___2) Given : A + B  C, The greatest concentration of C would be produced if the equilibrium constant
of the reaction were equal to
1) 5.0 x 10-8
3) 1.0 x 10-13
2) 2.5 x 104
4) 8.9 x 101
___3) Given : A + B  C, which Keq value suggests that the favored reaction is the forward reaction ?
-1
-7
1) 8.0 x 10
-4
2) 1.5 x 10
1
3) 5.5 x 10
4) 6.4 x 10
__4) Which of the following values indicates the least soluble salt at 25 ˚C?
1) Mg(OH)2
Ksp = 8.9 X 10
2) Ca(OH)2
Ksp = 1.3 X 10
-12
-6
-4
3) Sr(OH)2
Ksp = 3.2 X 10
4) Ba(OH)2
Ksp = 5.0 X 10
-3
___5) Which equilibrium system will contain the largest concentration of products at 25˚C?
+
-
-17
1) AgI(s) ↔ Ag (aq) + I (aq)
+
2) CH3COOH(aq) ↔ H (aq) + CH3COO (aq)
3) Pb
2+
Keq = 8.5 x 10
-5
Keq = 1.8 x 10
-
(aq)
+ 2Cl (aq) ↔ PbCl2(s)
+
4) Cu(s) + 2Ag
(aq)
↔ Cu
2+
(aq)
Keq = 6.3x 10
+ 2Ag(s)
Al(OH)3(s) 
___6) Consider the reaction:
4
Keq = 2.0 x 10
Al
3+
(aq) +
15
-1
3 OH
[hint:: never include (s) and (l)]
(aq)
Which of the following is the correct equilibrium expression for the reaction?
3+
-1
1) Ksp = [Al ] 3[OH ]
3+
-1 3
2)Ksp = [Al ] [OH ]
___7) Given the equilibrium expression:
3+
-1 3
3) Ksp = [Al ] + 3[OH ]
3+
-1
4) Ksp = [Al ] [OH ]
Keq =
[A] [B]
[C] [D]
Which pair represents the reactants in the forward reaction?
1) A and B
2) B and D
3) C and D
4) A and C
232
___8) For the system
2A(g) + B(g)  3C(g) , the expression for the equilibrium constant Keq is
2
1) Keq = [2A] [B]
[3C]
3) Keq = [A] [B]
3
[C]
3
2) Keq = [3C]
[2A] [B]
4) Keq = [C]
2
[A] [B]
___9) Which reaction has the equilibrium expression :
1) 2 C(g)  3 AB(g)
2) 2 C(g)  3 A(g) + B(g)
Keq = [B] [C]
3
[A]
2
?
3) 3 A(g)  B(g) + 2 C(g)
4) 3 AB(g)  2 C(g)
___10 In order for an aqueous solution to be at equilibrium, it must be:
1) dilute
2) saturated
3) concentrated
4) supersaturated
___11) Which of the following statements is FALSE?
1) The larger the value for K the longer the period of time it requires the overall reaction to reach a
a point of equilibrium
2) A small value for K indicates that a reaction will reach a point of equilibrium quite quickly in the
reaction, having a significant amount of unreacted reactant, and very little product.
3) The value for K is temperature dependent
4) A large value for K indicates that a reaction will proceed far to the right, favoring the production of
product, prior to the point of equilibrium
Answers: 1) 3
2) 2
3) 4
4) 1
5) 4
6) 2
7) 3
8) 4
9) 3
10) 2
11) 1
233
MORE PRACTICE ABOUT K … This practice deals principally with Ksp … or the solubility product. The
solubility product addresses the extent to which (principally) an ionic compound dissociates into ions. Some
deal with Ka which simply expresses the extent of ionization of a weak acid, in water. But they are really all
predicated upon our work with K, regardless of the sytem, they describe …Okay?
For questions 1 - 2 use the choices:
1)
2)
3)
4)
5)
when only I is correct
when only II is correct
when only I and II are correct
when only II and III are correct
when I, II, and III are each correct
_____1) Use the following table of Ksp values.
Substance
PbCrO4
Ksp
2.8 x 10
PbCO3
1 x 10
AgI
8.3 x10
-13
-14
-17
Assuming the same temperature, volume of water and pressure:
I.
PbCrO4 is more soluble than is PbCO3
II.
PbCrO4 is more soluble than is AgI
III. PbCO3 is more soluble than is AgI
_____2) Use the following table of Ka values.
Compound
HCl
Compound + Water
+1
-1
HCl + H2O  H (aq) + Cl (aq)
HF
HF+ H2O  H
H2S
H2S + H2O  H
+1
(aq)
+1
(aq)
-1
Ka
very large … greater
than 1
+ F
(aq)
3.5 x 10
+ HS
-1
(aq)
9.5 x 10
-4
-8
+1
An Arrhenius acid is any substance which is the only significant source of H in solution. The
+1
more H produced as the molecules ionize into ions, the stronger the acid. Thus, the larger the
+1
Ka, the greater the amount of H produced, when added to water, and the stronger the acid.
Which of the following is (are) accurate assuming the same conditions of temperature and
pressure ?
When dissolved in water:
I) H2S is the strongest acid of the choices.
II) HCl is a stronger acid than is HF.
III) HF is the weakest acid of the choices.
234
3) Note that the equations from question 2's table, for HF and H2S have double headed arrows. The equation
for HCl does not have a double headed arrow. Assuming a typist didn't make an error, explain the
significance.
*The single arrow for HCl suggests that it ionizes in water 100% or rather, the addition of HCl to water proceeds
to completion (the complete ionization of the molecule into its hydrated ions). This means that HCl must be a
strong acid. The other acids (HF and H2S) do NOT ionize completely in water, and this change reaches an
equilibrium, thus the need for the double arrows.
____ 4) Which salt is the MOST soluble in water at 25˚C ?
-14
1) lead carbonate
(Ksp = 7.4 x 10 )
-8
2) calcium carbonate (Ksp = 8.7 x 10 )
-11
3) zinc carbonate
(Ksp = 1.4 x 10 )
-9
4) barium carbonate (Ksp = 8.1 x 10 )
Defend your answer: (Be sure to have an argument, proof and to include any appropriate setting)
-8
* It has the largest equilibrium constant Ksp = 8.7 x 10 . Since the equilibrium constant indicates the
relative amount of reactant to product at equilibrium, the larger the constant indicates the most product that
is made. In this case, the largest constant indicates the greatest ability of the compound to dissolve in water,
or the most soluble compound.
_____5) The expression :
3
Keq = [C]
best represents which of the following reactions ?
2
[A] [B]
1)
C(g)  A(g) + B(g)
3)
2)
3 C(g)  2 A(g) + B(g)
4) 2 A(g)
A(g)
+ B(g)  C(g)
+ B(g)  3 C(g)
235
____6) A saturated solution exists when the maximum amount of solute is dissolved completely, in a specific
volume of water, at a specific temperature and pressure. A saturated solution is an expression of a
physical equilibrium. A saturated solution may be dilute (having the maximum amount of dissolved
species, but that amount is very small, compared to the amount of water) … Or a saturated solution
may be concentrated (having the maximum amount of dissolved species and that amount is relatively
high compared to the amount of water).
Which sulfide compound forms the most concentrated saturated solution at 1 atm and 25 ˚C?
-29
-17
1) CdS (Ksp = 3.6 x 10 )
3) FeS (Ksp = 1.3 x 10 )
-52
4) HgS (Ksp = 2.4 x 10 )
-26
2) CoS (Ksp = 3.0 x 10 )
____7) Using the choices from #6 .. .which sulfide compound produces the most saturated, yet dilute solution
at 1 atm and 25 °C?
+2
-1 2
_____ 8) The equilibrium expression : Ksp = [Pb ] [I ]

1) PbI2(s)
Pb(s) +
+2
3) Pb
-1
(aq)
+ 2I
(aq)
I2(s)

4) Pb(s) +
I2(s)
+2

2) PbI2(s) + H2O(l)
best represents which of the following reactions ?
Pb
-1
(aq)
+ 2I
(aq)
 PbI2(s) + H2O(l)
PbI2(s)
____ 9) Which equilibrium system will contain the LARGEST CONCENTRATION of products at 25 ˚C ?
↔
1) AgI(s) + H2O(l)
2) CH3COOH(aq) ↔ H
+2
3) Pb
(aq)
+ 2 Cl
-1
(aq)
+1
4) Cu(s) + 2 Ag
+1
Ag
+1
-1
(aq)
-17
(aq)
-1
+ CH3COO
(aq)
↔ Cu
(aq)
(Ksp = 8.3 x 10 )
-5
(aq)
↔ PbCl2(s) + H2O(l)
+2
(aq)
+ I
+ 2 Ag(s)
(Ka = 1.8 x 10 )
4
(Kc = 6.3 x 10 )
15
(Kc = 2.0 x 10 )
Defend your answer: (Be sure to have an argument, proof and to include any appropriate setting)
15
* 4 must be correct because has the largest equilibrium constant at 2.0 x 10 . The equilibrium
constant indicates the relative amount of reactant to product, we may infer that the larger the constant
greater the amount of product (or the farther to the right the reaction will procee4). Since the question
asks for the largest concentration of product, the answer must therefore be that with the largest value
236
for K
_____ 10) Which of the following correctly expresses the Keq for the reaction
+1
Cu(s) + 2 Ag
+2
2
1)
[Cu ] [Ag]
+1 2
[Cu] [Ag ]
2)
[Cu ]
+1 2
[Ag ]
+2
+2
(aq)
↔ Cu
(aq)
+ 2 Ag(s)
2
3)
[Ag]
[Cu]
4)
[Ag ]
+2
[Cu ]
+1 2
_____ 11) Which of the following correctly expresses the Keq for the system:
1)
2)
1
[CO2(g)]
[CO2(g)]
[CO2(s)]
3)
[CO2(g)]
4)
[CO2(s)]
[CO2(g)]
CO2(g)  CO2(s)
Answers:
1) 5 The greater the value of Ksp, the more soluble the compound (the more ion or dissolved molecule)
2) 2 It is the only accurate statement. The Ka value is the largest … thus it is the strongest acid, producing the most H+1 in water.
3) given on the page … highlight and check.
4) 2 it is the largest of these very small Ksp values.
5) 4 recall that the equilibrium constant is derived from the equilibrium expression of the Law of Mass Action. The Law of Mass
Action has products (raised to an exponent equivalent to the coefficient of the balanced equation) divided by the reactants …
raised to an exponent …..etc
6) 3
7) 4
8) 2 The Law of mass action has products over reactants and includes ONLY (g) and (aq), never (s) or (l)
9) 4
10) 2
11) 1
237
E) 3 Important Manipulations (pay attention): The relationship between K & the Chemical Equation
When a chemical equation is modified in some way, then the equilibrium constant for the equation
changes because of the modification. The following 3 modifications are relatively common.
1) Were you to reverse the equation, invert the equilibrium constant. (Take the inverse…)
a) e.g. A(g) + 2B(g)  3C(g)
Kforward = [C]3
[A][B]2
where you to reverse the reaction, then the
Kreverse should equal 1 .
Kforward
because Kreverse = [A][B]2
[C]3 …. Piece o’ cake!
b) When Kforward = 3.7 x 102 …. then Kreverse = 1
3.7 x102
or
0.0027 or 2.7 x 10-3
2) If you were to multiply the coefficients, in the equation by a factor, then you should
raise the equilibrium constant exponentially to the same factor.
a) Essentially, here is the rule of thumb:
When the mole ratio is doubled, then square the value of K, when it is tripled,
cube the value of K …. OR when the ratio is cut in half … take the square root.
b) e.g. A(g) + 2B(g)  3C(g)
K = [C]3
[A][B]2
Were you to multiply the balanced equation by some
common factor … then you must account for that by raising
the Law of Mass Action, exponentially by that factor
nA + 2n B  3n C …
K’ = [C]3n
[A]n[B]2n
=
[C]3
[A][B]2
n
= Kn
Not Bad!
238
TRY THIS! (The next 3 problems ultimately combine both of the aforementioned manipulations …but
first try a pretty straight forward one….)
1) The reaction A(g)  2B(g) has an equilibrium constant of K = 0.010. What is the equilibrium
constant for the reverse reaction …. 2B(g)  A(g) ? (hint: use manipulation 1)
ans:100
This is tougher…both manipulation 1 and 2 are required … think … push forward….
2) The reaction A(g)  2B(g) has an equilibrium constant of K = 0.010. What is the equilibrium
constant for the reaction …. B(g)  ½A(g) ?
Hint 1: * Notice that the 2nd reaction
a) 1
b) 10
is the reverse of the first reaction, and
the ratio is only ½ of the original.
c) 100
d) 0.0010
Hint 2: * Since it is a reverse reaction,
take the inverse of K
Hint 3: * Since the values have been
cut by ½ , take the square root of the
inverse of K
ans:*b) or 10
3) Consider the chemical equation and equilibrium constant for the decomposition of ammonia at 25°C
Given: NH3(g)  1⁄2 N2(g) + 3⁄2 H2(g)
K = 1.34 x 10-3
Calculate the equilibrium constant for the following reaction at 25 °C
N2(g) + 3H2(g)  2NH3(g)
K’ = * 5.57 x 105
Hint 1:* Notice that the reaction
is the reverse of the given
reaction AND that the molar ratio
values (while in the same ratio)
have though, been doubled.
Hint 2: * Because it is the reverse
reaction, take the inverse of K…
or 1/1.34 x 10-3
Hint 3: * Since the coefficients
have been doubled … square the
answer to the inverse of K
…when doubled, square, when
cut in half …take the square root.
239
3) If you add two or more individual chemical equations to obtain an overall equation,
then multiply the corresponding equilibrium constants by each other to obtain the
overall equilibrium constant…. (The Rule of Multiple Equilibria)
a) Simplified … Koverall
= (K1) x (K2)
b) consider: A(g)  2B(g)
K1 = [B]2
[A]
and
2B(g)  3C(g)
K2 = [C]3
[B]2
The two equations sum as follows:
Thus the Koverall = [C]3
[A]
In theory:
A(g)  2B(g)
2B(g)  3C(g)
A(g)  3C(g)
…no big problem … Rather, do you see why
manipulation 3 states what it states?
Koverall = (K1) x (K2)
because:
[B]2 x [C]3
[A]
[B]2
=
[C]3 (which is the same answer
[A] as the above addition)
c) For such problems, you may need to run the sum, and use the sum to determine a
Koverall …just as the above problem (and the following one….) requires you to do…
TRY THIS!
1) Predict the equilibrium constant for the first reaction shown here given the equilibrium
constants for the second and third reactions:
CO2(g) + 3H2(g)  CH3OH(g) + H2O(g)
CO(g) + H2O(g) CO2(g) + H2(g)
CO(g) + 2H2(g)  CH3OH(g)
K1 =????
K2 = 1.0 x 105
K3 = 1.4 x 107
* 2CO(g) + 4H2(g)  2CH3OH(g) notice that this is twice
the third equation….
*Koverall = (1.4 x 107)2 or rather 1.96 x 1014
* Koverall = (K1)(K2)(K3) or ….
Hint 1: * Add the reactions, by
cancelling out like species, in order to
get to an overall reaction
Hint 2: *Since the sum is twice that of
the third reaction, square the value
of K3 in order to derive the overall K
Hint 3: *Use Koverall and set it equal to
the product of the other constants,
with K1 being your “x”, or unknown
value. Solve for that “x”
ans: *K1 = 1.40 x 102
*1.96 x 1014 = (x)( 1.0 x 105)( 1.4 x 107)
* K1 = 140 or 1.40 x 102
Check out the Trivedi flash, sections 14.4 – 14.6 for some support refresher…regarding the
effects on equilibrium of “changing the chemical equation”.
240
III) K and its expression in terms of pressure ….
Assignment: Read / Do: Trivedi: 14.10 KC vs KP
A) The role of gases (as seen in Unit 2) is quite important. For gaseous reactions, the partial pressure
of a particular gas is proportional to its molar concentration. Therefore we can express the
equilibrium constant in terms of the partial pressure of the reactants and products. (Tro p. 658)
B) Common expressions of K are Keq is Kc, in which the equilibrium constant is expressed with respect
to molar concentration [ ]. Essentially, there is a relationship between the equilibrium constant in
molar concentration and the equilibrium constant of gases, measured in atmospheres (KP)
1) Ultimately this work will lead us to an equation relating moles of gas and equilibrium
expressions: Given: aA(g) + bB(g) ↔ cC(g) + dD(g)
Think about
using either of
these two
equations
when having
to deal with
Partial Pressures
of gases and
equilibrium
KP = (PC)c (PD)d
(PA)a(PB)b
Where P = partial pressures
(This equation is on your tables, and
it apes the Law of Mass Action)
a) It is linked to a relationship, which IS NOT found on your tables …
Where:
Kp = the value for the partial pressure of a gas
Kc = the equilibrium constant, for molar concentrations
∆n = the difference between the number of moles of gaseous
products and gaseous reactants: Products - Reactants
Kp = Kc(RT)∆n
2) The expression for KP takes the form of the expression for Kc, except that it uses the
partial pressure of each gas in lieu of its concentration.
a) e.g.) Consider the equilibrium: 2SO3(g)  2SO2(g) + O2(g)
Kc = [SO2]2[O2]
[SO3]2
vs
KP = (PSO2)2 PO2
(PSO3)2
b) Partial pressure in atm and molar concentration are NOT necessarily equal!
* So KP ≠ Kc, necessarily. They are equivalent only when ∆n = 0
c) Okay, they aren’t necessarily equivalent ….but, we can relate the two ….
i) When you are given partial pressures in atm you may substitute right into:
KP = (PC)c (PD)d
(PA)a(PB)b
ii) But if given Kc and asked for Kp (or vice versa) you need to convert from
one to the other using Kp = Kc(RT)∆n
Again, when the total number of moles of gas is the same after the reaction as
before the reaction, then ∆n =0 and that means that Kp = Kc Because,
Kp = Kc(RT)0 converts RT to 1 .. AND remember that the expression to find Kp
mimics the equilibrium expression based upon the Law of Mass Action, which is
quite convenient for gaseous systems.
241
TRY THIS
1) Consider the equilibrium N2(g) + 3H2(g)  2NH3(g) Kc = 9.60 at 300°C.
Calculate Kp for this reaction
Hint 1* Consider using
Kp = Kc(RT)∆n , because Kc is given.
Hint 2: *∆n = Prod – Reactants
* Kp = Kc(RT)∆n
*Kp = (9.60)(0.08206 x 573)-2
*Kp = (9.60)
(0.08206 x 573)2
ans: 4.34 x 10-3 atm
=
2) Dinitrogen tetroxide in its liquid state was used as one of the fuels on the lunar lander for the NASA Apollo
missions. In the gas phase, it decomposes to gaseous nitrogen dioxide.
N2O4(g)  2 NO2(g)
Consider an experiment in which gaseous N2O4(g) was placed in a flask and allowed to reach equilibrium
at a temperature where Kp = 0.133. At equilibrium, the pressure of dinitrogen tetroxide was found to
equal 2.71 atm. Calculate the equilibrium pressure of NO2(g).
(Zumdahl p 623)
*Kp = (PNO2)2
(PN2O4)
thus 0.133 = (PNO2)2
(2.71)
*0.360 = (PNO2)2
*thus √0.360 = 𝑃NO2 or 0.600 atm
Hint 1:* Kp is given …there is no Kc given
so use the partial pressure equivalent of
the Law of Mass Action, setting the partial
pressure of nitrogen dioxide (the product)
raised to its coefficient of 2, divided by
the partial pressure of dinitrogen
tetroxide.
Hint 2: *Plug and Chug … you are given
Kp and the partial pressure of dinitrogen
tetroxide.
ans: 0.600 atm
242
3) Slightly More Challenging: At 327°C the equilibrium concentrations are
[CH3OH] = 0.150 M, [CO] = 0.240 M and [H2] = 1.10 M for the reaction: CH3OH(g) ↔ CO(g) + 2 H2(g)
Calculate KP at this temperature
2
*Kc =[CO][H2]
[CH3OH]
=
Hint 1: *Kp and Kc are not necessarily equivalent.
2
(0.240)(1.10)
0.150
= 1.94
∆n
*Kp = Kc(RT)
*Kp = (1.94)(0.08026 • 600.15)2 = 4.50 x 103
Hint 2: *Write the equilibrium expression solving
for Kc ….
Hint 3: *Calculate Kc
Hint 4: * Use the value for Kc to find Kp … did you
convert Celsius to Kelvin???
ans: 4.70 x 103
4) For which of the following reactions is KP = Kc? (more than one choice may be a correct answer)
Defend your reasoning. (Be sure to have an argument, proof and to include any appropriate setting)
1) 2 NH3(g) + CO2(g) ↔ N2CH4O(s) + H2O(g)
2) 2 NBr3(s) ↔ N2(g) + 3Br2(g)
3) 2 KClO3(s) ↔ 2 KCl(s) + 3O2(g)
4) CuO(s) + H2(g) ↔ Cu(𝓁) + H2O(g)
* When ∆n=0 in the equation linking Kp and Kc: Kp =Kc (RT)∆n , then the expression RT is converted
to a value of 1, giving an conversion of Kp = Kc(1) or rather Kp = Kc. (setting) The reaction in #4 gives
a sum of 0 change in the number of moles ….based upon #moles of product - #moles of reactant (proof)
Choice 4 is the only one of the examples which meets the requirement for Kp = Kc (answer)
243
IV)  Using ICE Tables to Calculate K from Measured (experimentally determined) Concentrations
A) For any reaction, the equilibrium concentrations depend on the initial concentrations (hence the
I of ICE) … But the equilibrium constant is always the same at a specific temperature, regardless
of the initial concentrations.
Take a look at the following table and questions… Study the concentrations and the
calculations … what are they telling the scientist in you?
Initial and Equilibrium Concentrations for the Reaction:
H2(g) + I2(g)  2 HI(g) at 445 °C
Initial Concentrations
[H2]
[I2]
[HI]
Equilibrium Concentrations
[H2]
[I2]
[HI]
Equilibrium Constant
Kc = [HI]2
[H2][I2]
1
0.50
0.50
0.0
0.11
0.11
0.78
(0.78)2
= 50
(0.11)(0.11)
2
0.0
0.0
0.50
0.055
0.055
0.39
(0.39)2
= 50
(0.055)(0.055)
3
0.50
0.50
0.50
0.165
0.165
1.17
(1.17)2
= 50
(0.165)(0.165)
4
1.0
0.50
0.0
0.53
0.033
0.934
(0.934)2 = 50
(0.53)(0.033)
5
0.50
1.0
0.0
0.033
0.53
0.934
(0.934)2 = 50
(0.033)(0.53)
1) Study trials 1 and 2 … what may you infer regarding the initial concentrations being all reactant
or all product? * The reaction goes in a direction which establishes an equilibrium, with the
same equilibrium constant.
2) Study trials 4 and 5, must the molar concentrations of the products be equal to each other at
equilibrium? * No …. the product concentrations depend upon the initial concentrations and the
stoichiometric ratio of the balanced reaction equation.
3) The synthesis of hydrogen iodide is an endothermic process, with a ∆Hformation of +26.5 kJ/mol.
Were the temperature of the reaction lowered from 445°C to 100°C, which reaction (the forward or
the reverse) should be favored?
*The endothermic reaction depends upon energy … A lower temperature should hamper the
endothermic reaction. Hence, the exothermic decomposition of HI would be favored.
244
B) ICE table: a decidedly nice process to analyze the changes between initial concentrations,
and the equilibrium concentrations.
ICE stands for: Initial, Change, Equilibrium
[A] [B]
Initial
Change
Equilibrium
Follow this process
1) an ICE table can help you analyze the initial [ ], the change and the final
equilibrium [ ]
a) Use the balanced equation (as a guide) to start an ICE table and use the
problem information, to fill in as much of the table as you can. This
includes the “change” of the reactant or product….
b) Use the change and the stoichiometric ratio from the balanced chemical
equation to complete the table. … Be sure the columns add up to reflect
the correct equilibrium concentrations.
c) Use the balanced reaction equation to write an equilibrium expression and
sub in the appropriate values to calculate K
2) Assume the equilibrium:
A(g)  2 B(g)
A sealed container has been charged with 1.00 M of A, and an initial concentration of
0.00 M of B. At equilibrium, the concentration of A is 0.75 M.
a) Calculate the equilibrium concentrations of A and B
b) Calculate the equilibrium constant (Kc)
Solution to a) … Study the stoichiometry of the balanced equation.
It is 1:2 … That is for every molar change in A, B is TWICE the change
[A]
[B]
1.00
*0.00
Change
*-0.25
*+0.50
Equilibrium
*0.75
*0.50
Make an ICE table
Initial
Solution to b) …. Write the equilibrium expression … plug and chug
Kc = * [B]2
[A]
=
(0.50)2 = 0.33
(0.75)
245
2) NO2, a brown gas, and N2O4, a colorless gas, exist in equilibrium 2NO2(g)  N2O4(g)
Chemistry is a “way”
a) For an ICE Table
Use the balanced
equation (as a
guide) & fill in as
much of the table as
you can.
b) Use the change
and the ratios of the
balanced chemical
equation to
complete the table.
c) Write an
equilibrium
expression to plug
and chug
A closed container at 25°C is charged with NO2 at a partial pressure of 0.56 atm and
with N2O4 at a partial pressure of 0.51 atm. At equilibrium the partial pressure of N2O4
is found to be 0.54 atm.
a) What is the partial pressure of NO2 at equilibrium? (ans: 0.50 atm)
b) What is the value of Kp? (ans: 2.2)
Initial
(NO2)
atm
*0.56
(N2O4)
atm
*0.51
Change
*-0.06
*+0.03
Equilibrium
*0.50
*0.54
*Kp = (PN2O4)
(PNO2)2
=
0.54 atm = 2.16 or 2.2
(0.50 atm)2
Hint 1: * You are given the partial pressure
of the gases, in atmospheres … This means
that you can write the equilibrium
expression by using the Law of Mass
Action, substituting in pressures, for
concentrations.
Hint 2: * This is a perfect time to use an
ICE table … Take a look at the change in
partial pressure and please be sure to
note that there is a stoichiometric effect
here of 2 to 1 …
Hint 3: * Write the equilibrium expression,
using the partial pressure version and be
sure to plug in the equilibrium pressures,
from your ICE table.
3) Consider the following reaction, at equilibrium: CO(g) + 2 H2(g)  CH3OH(g)
A reaction mixture at 780°C initially contains [CO] = 0.500 M and [H2] =1.00 M.
At equilibrium the CO concentration is experimentally determined to be 0.150 M.
What is the value of Kc?
*Kc = [CH3OH]
[CO] [H2]2
*remove
[CO]
[H2]
[CH3OH]
Initial
0.500
1.00
0.00
Change
-0.350
-0.700
+0.350
Equilibrium
0.150
0.300
0.350
Hint 1: * It asks for the equilibrium
constant, so write the equilibrium
expression … hopefully you will recognize
that you require the equilibrium
concentrations … and thus need to figure
those concentration out … How ????
Hint 2: * Create an ICE table
Hint 3: * Follow our process re: ICE tables
… be sure to watch those stoichiometric
ratios … .
ans: 25.9
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4) Determine the value of the equilibrium constant for the reaction: A + 2 B ↔ 2C
when 1.0 mol of A, 2.0 mol of B, and 3.0 mol of C are
placed in 1.0 L vessel and allowed to come to equilibrium,
The final or equilibrium concentration of C = 1.4 mol/L
remove
*
Initial
Change
Equilibrium
[A]
[B]
[C]
1.0
2.0
3.0
+0.80
+1.6
-1.6
1.8
3.6
1.4
*Kc = [C]2
[A] [B]2
=
Hint 1: * Write the equilibrium expression
using the Law of Mass Action.
Hint 2:* Use an ICE Table. Find the
change in moles and final moles at
equilibrium. You should then convert all
mole values to molarities but this problem
is easier since the volume is a 1 L flask. …
Hint 3: * In the ICE table, don’t forget that
the coefficients of the balanced equation
must be used to determine the C the
change in concentration for the species.
(1.4)2
= 0.084
(1.8) (3.6)2
Chemistry is a “way”
5) Given: 2 CH4(g)  C2H2(g) + 3 H2(g) A reaction mixture at
1700 °C initially contains [CH4] = 0.115 M. At equilibrium, the
mixture contains [C2H2] = 0.035 M. What is the value of the
equilibrium constant? ans: 0.020
*remove
[CH4]
[C2H2]
[H2]
Initial
0.115
0.00
0.00
Change
-0.070
+0.035
+0.105
Equilibrium
0.045
0.035
0.105
*Kc = [C2H2] [H2]3
[CH4]2
= (0.035)(0105)3
(0.045)2
a) For an ICE Table:
Use the balanced
equation (as a
guide) & fill in as
much of the table as
you can.
b) Use the change
and the ratios of the
balanced chemical
equation to
complete the table.
c) Write an
equilibrium
expression to plug
and chug
= 0.020
6) Which of the following equations has Kc = KP? (one, or more than one answer may be applicable)
1) PCl5(g)  PCl3(g) + Cl2(g)
2) 2NO(g)  N2(g) + O2(g)
3) CaCO3(s)  CaO(s) + CO2(g)
4) H2O (g) + CO(g)  H2(g) + CO2(g)
5) 2 NOCl(g)  2 NO(g) + Cl2(g)
ans: * 2 and 4
247
7) At a certain temperature, a 1.00 L flask initially contained 0.298 mole of PCl3(g) and 8.70 x 10-3 mole
of PCl5(g). After the system had reached equilibrium, 2.00 x 10-3 mole of Cl2 was found in the flask.
Gaseous phosphorus pentachloride decomposes according to the following reaction. Calculate Kc
(Zumdahl p. 624)
PCl5(g)  PCl3(g) + Cl2(g)

Hint 1:* Write the equilibrium expression
*remove
Initial
Change
Equilibrium
[PCl5]
[PCl3]
[Cl2]
8.7 x 10-3
0.298
0.00
-2.00 x 10-3
+2.00 x 10-3
+2.00 x 10-3
6.7 x 10-3
0.300
2.00 x 10-3
Hint 2: * Convert moles to molarity … Luckily the
volume is 1 L, so the conversion is a direct one …
and run an ICE Table. Think about this … You do
NOT have any chlorine at the start, so it is 0.
The stoichiometric ratios are easy, since
everything is in a 1:1
Hint 3: * Run the Law of Mass Action
*Kc = [PCl3(g)] [Cl2(g)]
[PCl5(g)]
= (0.300)( 2.00 x 10-3)
(6.70 x 10-3)
ans: 8.96 x 10-2
8) 3.0 mol of iodine and 4.0 mol of bromine are placed in a 2.0 L flask at 150°C. The reaction comes to
equilibrium, at which point 3.2 mol of iodine bromide are present.
a) Write the balanced reaction equation for the synthesis of iodine bromide from its elements.
* I2(g) + Br2(g) 2 IBr(g)
b) Determine the equilibrium constant for the reaction.
*remove
iodine
bromine concentrations
iodineafter
bromide
This one is a bit different …
I converted to equilibrium
working
with the mole values … because the overall volume was a 2 L flask …not a 1 L flask
as
in prior examples
Initial
3.0 mol
4.0 mol
0.00
Change
-1.6 mol
-1.6 mol
+3.2 mol
Equilibrium
1.4 mol
2.4 mol
3.2 mol
0.7 M
1.2 M
1.6 M
*Kc = [IBr]2 = (1.6)2
= 3.0
[I2][Br2]
(0.7)(1.2)
Hint 1: * Write the equilibrium expression and
Run an ICE table finding the change in moles.
Remember to take into account the
stoichiometric ratios of the balanced equation.
Hint 2:* Once you have the final mol values at
equilibrium, convert them to molarity values.
Note the volume of the flask is 2 L
Hint 3: * Plug and chug the molarities calculated
from the results of the ICE table into the
equilibrium expression.
ans: *3.0
248