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, big topic which links into other units such as Thermodynamics, Acid/Base Theory, Solution Theory, Bonding, and the behavior of matter, in terms of phase changes. This is an underlying process, for we will connect the basic concepts to the aforementioned studies of matter. 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)

Right now, I’m looking to cover ONLY, Enduring Understandings 6A and 6B & the following Learning Objectives. Okay?

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.

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.

Enduring understanding 6.C: Chemical equilibrium plays an important role in acid-base chemistry and in solubility.

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 equilibrium constant is related to temperature and the difference in Gibbs free energy between reactants and products

Essential knowledge 6.D.1: When the difference in Gibbs free energy between reactants and products (ΔG°) is much larger than the thermal energy (RT), the equilibrium constant is either very small (for ΔG° > 0) or very large (for ΔG° < 0). When ΔG° is comparable to the thermal energy (RT), the equilibrium constant is near 1.

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 a point of equilibrium. A state of equilibrium unto itself is not, a “bad or good” thing. However, in our quest to be able 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.

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

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

· 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 + 40kJ 40kJ + C → B + A

A+B

C

A+B

C

100 100

∆H ∆H

60 60

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

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 changes occur

at the same rate. (ice melts as fast as water freezes or water evaporates at the

same rate as water vapor condenses)

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 reactions occur at the

same rate.

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 ….

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) colliding with N2O4(g)

Important: NO2(g) colliding with NO2(g)

UN-important: N2O4(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

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.

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.

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.

TABLE OF EQUILIBRIUM CONSTANTS

Equilibrium

Constant

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

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)

Kp

The partial pressure of a gas used to describe a gas’s concentration

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.

· It 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.

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

M E M O R I Z E

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….

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) K = [Product]coef [Product]coef

[Reactant]coef [Reactant]coef

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 equal to each species' coefficient from the balanced

reaction equation.

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:

KEq’ = [C]1/2

[A]1/2 [B]

½ A + B ½ C

Now, let [A] = 2 M, [B] = 3 M and [C] = 4 M …

when we solve the two above equilibrium expressions:

KEq’ = [4]1/2 = = 0.472

[2]1/2 [3] 4.24

KEq = [4] = 2 = 0.222

[2][3]2 9

vs.

That makes so much sense

This is one of the reasons why you are always encouraged to balance chemical

reaction equations using whole numbers …. TN: Marshall.edu

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) Practice Writing Equilibrium Expressions….

Write the equilibrium expression for 7a) – f)

Remember: products is/are in the numerator.

Note that the exponents reflect the coefficients of the balanced equation.

a) N2(g) + 3H2(g) 2NH3(g)

* [NH3]2

[N2] [H2]3

Keq =

b) Al2(SO4)3(s) + H2O(l) 2 Al+3(aq) + 3 SO4-2(aq)

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....

Ksp = * [Al+3]2 [SO4-2]3

c) 3 O2(g) 2 O3(g)

* [O3]2

[O2]3

Keq =

Note: This is a nice twist… apply the rules, and think!

d) 2 Fe2O3(s) + 3 C(s) 4 Fe(s) + 3 CO2(g)

Keq = * [CO2]3

Note: This provides a nice twist…apply the rules and think!

e) NH3(g) + HCl(g) NH4Cl(s)

* 1

[NH3][HCl]

Keq =

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 = [Ca2+][Cl-1]2 This is essentially an equilibrium constant expression for a

heterogeneous system. Write the equation which describes the equilibrium expression

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.

7) Simple Calculations for K:

a) Consider the equilibrium: 2 SO2(g) + O2(g) 2 SO3(g) + 791kJ

Equilibrium:

Rate of the Forward Rxn = Rate of the Reverse Rxn

2 SO2(g) + O2(g)

Forward Reaction

Reaction

Rate

Concentrations are measured at any point once equilibrium is reached & then plugged into the "mass action" equation.

2 SO3(g) + 791 kJ

Reverse Reaction

t0 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 = * [SO3]2 Substituion : Keq =* [0.15]2

[SO2]2 [O2] [0.75]2 [0.30]

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] or * 0.0778 = [0.0720 M] [x]

[SO2Cl2] [0.136 M]

*0.010608 = 0.0720x

D) Significance of K

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 or 670,000

[H2]3 [N2] 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 or 1

[N2][O2] 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 : 1 at equilibrium, there is far more reactant, than product

N2(g) + O2g) 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 …

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?

1 10

Where K = 10 and [B] thus: A(g) B(g)

1 [A]

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….

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 1032) 1 x 10-3 3) 1 x 1094) 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 2) 2.5 x 104 3) 1.0 x 10-13 4) 8.9 x 101

___3) Given : A + B C, which Keq value suggests that the favored reaction is the forward reaction ?

1) 8.0 x 10-1 2) 1.5 x 10-7 3) 5.5 x 10-4 4) 6.4 x 101

__4) Which of the following values indicates the least soluble salt at 25 ˚C?

1) Mg(OH)2 Ksp = 8.9 X 10-12 3) Sr(OH)2 Ksp = 3.2 X 10-4

2) Ca(OH)2 Ksp = 1.3 X 10-6 4) Ba(OH)2 Ksp = 5.0 X 10-3

___5) Which equilibrium system will contain the largest concentration of products at 25˚C?

1) AgI(s) ↔ Ag+(aq) + I-(aq) Keq = 8.5 x 10-17

2) CH3COOH(aq) ↔ H+(aq) + CH3COO-(aq) Keq = 1.8 x 10-5

3) Pb2+(aq) + 2Cl-(aq) ↔ PbCl2(s) Keq = 6.3x 104

4) Cu(s) + 2Ag+(aq) ↔ Cu2+(aq) + 2Ag(s) Keq = 2.0 x 1015

___6) Consider the reaction: Al(OH)3(s) Al3+(aq) + 3 OH-1(aq) [hint:: never include (s) and (l)]

Which of the following is the correct equilibrium expression for the reaction?

1) Ksp = [Al3+] 3[OH-1] 3) Ksp = [Al3+] + 3[OH-1]3

2)Ksp = [Al3+] [OH-1]3 4) Ksp = [Al3+] [OH-1]

___7) Given the equilibrium expression: Keq = [A] [B]

[C] [D]

Which pair represents the reactants in the forward reaction?

1) A and B 2) B and D3) C and D 4) A and C

___8) For the system 2A(g) + B(g) 3C(g) , the expression for the equilibrium constant Keq is

1) Keq = [2A] [B] 3) Keq = [A]2 [B]

[3C] [C]3

2) Keq = [3C]4) Keq = [C]3

[2A] [B] [A]2 [B]

___9) Which reaction has the equilibrium expression : Keq = [B] [C]2 ?

[A]3

1) 2 C(g) 3 AB(g) 3) 3 A(g) B(g) + 2 C(g)

2) 2 C(g) 3 A(g) + B(g) 4) 3 AB(g) 2 C(g)

___10 In order for an aqueous solution to be at equilibrium, it must be:

1) dilute2) saturated 3) concentrated4) 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) 310) 2 11) 1

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) when only I is correct

2) when only II is correct

3) when only I and II are correct

4) when only II and III are correct

5) when I, II, and III are each correct

Substance

Ksp

PbCrO4

2.8 x 10-13

PbCO3

1 x 10-14

AgI

8.3 x10-17

_____1) Use the following table of Ksp values.

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

Your choices for question 2 have been repeated for your convenience.

1) when only I is correct

2) when only II is correct

3) when only I and II are correct

4) when only II and III are correct

5) when I, II, and III are each correct

_____2) Use the following table of Ka values.

Compound

Compound + Water

Ka

HCl

HCl + H2O H+1(aq) + Cl-1(aq)

very large … greater than 1

HF

HF+ H2O H+1(aq) + F-1(aq)

3.5 x 10-4

H2S

H2S + H2O H+1(aq) + HS-1(aq)

9.5 x 10-8

An Arrhenius acid is any substance which is the only significant source of H+1 in solution. The

more H+1 produced as the molecules ionize into ions, the stronger the acid. Thus, the larger the

Ka, the greater the amount of H+1 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.

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 ?

1) lead carbonate (Ksp = 7.4 x 10-14)

2) calcium carbonate (Ksp = 8.7 x 10-8)

3) zinc carbonate (Ksp = 1.4 x 10-11)

4) barium carbonate (Ksp = 8.1 x 10-9)

Defend your answer: * It has the largest equilibrium constant. 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 : Keq = [C]3 best represents which of the following reactions ?

[A]2 [B]

1) C(g) A(g) + B(g)3) A(g) + B(g) C(g)

2) 3 C(g) 2 A(g) + B(g)4) 2 A(g) + B(g) 3 C(g)

____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?

1) CdS (Ksp = 3.6 x 10-29)3) FeS (Ksp = 1.3 x 10-17)

2) CoS (Ksp = 3.0 x 10-26)4) HgS (Ksp = 2.4 x 10-52)

____7) Using the choices from #6 .. .which sulfide compound produces the most saturated, yet dilute solution

at 1 atm and 25 °C?

_____ 8) The equilibrium expression : Ksp = [Pb+2] [I-1]2 best represents which of the following reactions ?

1) PbI2(s) Pb(s) + I2(s)

2) PbI2(s) + H2O(l) Pb+2(aq) + 2 I-1(aq)

3) Pb+2(aq) + 2 I-1(aq) PbI2(s) + H2O(l)

4) Pb(s) + I2(s) PbI2(s)

____ 9) Which equilibrium system will contain the LARGEST CONCENTRATION of products at 25 ˚C ?

1) AgI(s) + H2O(l) Ag+1(aq) + I-1(aq) (Ksp = 8.3 x 10-17)

2) CH3COOH(aq) H+1(aq) + CH3COO-1(aq) (Ka = 1.8 x 10-5)

3) Pb+2(aq) + 2 Cl-1(aq) PbCl2(s) + H2O(l) (Kc = 6.3 x 104)

4) Cu(s) + 2 Ag+1(aq) Cu+2(aq) + 2 Ag(s) (Kc = 2.0 x 1015)

Defend your answer: 4 must be correct because has the largest equilibrium constant. 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

for K

_____ 10) Which of the following correctly expresses the Keq for the reaction

Cu(s) + 2 Ag+1(aq) Cu+2(aq) + 2 Ag(s)

1) [Cu+2] [Ag]23) [Ag]2

[Cu] [Ag+1]2 [Cu]

2) [Cu+2]4) [Ag+1]2

[Ag+1]2 [Cu+2]

_____ 11) Which of the following correctly expresses the Keq for the system : CO2(g) CO2(s)

1) 13) [CO2(g)]

[CO2(g)]

2) [CO2(g)]4) [CO2(s)]

[CO2(s)] [CO2(g)]

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

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 where you to reverse the reaction, then the

[A][B]2 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 or 0.0027 or 2.7 x 10-3

3.7 x102

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 Were you to multiply the balanced equation by some

[A][B]2 common factor … then you must account for that by raising

the Law of Mass Action, exponentially by that factor

nA + 2n B 3n C …

[C]3 n

[A][B]2

K’ = [C]3n

[A]n[B]2n

= = Kn

Not Bad!

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

Hint 1: * Notice that the 2nd reaction is the reverse of the first reaction, and the ratio is only ½ of the original.

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

constant for the reaction …. B(g) ½A(g) ?

a) 1 c) 100

b) 10 d) 0.0010

3) Consider the chemical equation and equilibrium constant for the decomposition of ammonia at 25°C

Given: NH3(g) N2(g) + H2(g) K = 1.34 x 10-3

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.

Calculate the equilibrium constant for the following reaction at 25 °C

N2(g) + 3H2(g) 2NH3(g) K’ = * 5.57 x 105

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: A(g) 2B(g)

2B(g) 3C(g)

A(g) 3C(g)

Thus the Koverall = [C]3 …no big problem … Rather, do you see why

[A] manipulation 3 states what it states?

In theory: Koverall = (K1) x (K2)

because: [B]2 x [C]3 = [C]3 (which is the same answer

[A] [B]2 [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:

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

CO2(g) + 3H2(g) CH3OH(g) + H2O(g) K1 =????

CO(g) + H2O(g) CO2(g) + H2(g) K2 = 1.0 x 105

CO(g) + 2H2(g) CH3OH(g) 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 ….

*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”.

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)

KP = (PC)c (PD)d

(PA)a(PB)b

Where P = partial pressures

(This equation is on your tables, and

Think about using either of these two equations

when having

to deal with Partial Pressures of gases and equilibrium

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)

KP = (PSO2)2 PO2

(PSO3)2

Kc = [SO2]2[O2] vs

[SO3]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.

TRY THIS

Hint 1* Consider using

Kp = Kc(RT)∆n , because Kc is given.

Hint 2: *∆n = Prod – Reactants

1) Consider the equilibrium N2(g) + 3H2(g) 2NH3(g) Kc = 9.60 at 300°C.

Calculate Kp for this reaction

* Kp = Kc(RT)∆n

*Kp = (9.60)(0.08206 x 573)-2

*Kp = (9.60)

(0.08206 x 573)2 = 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)

Hint 1:* Kp is given …there is no Kc given so use the parital 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

*Kp = (PNO2)2 thus 0.133 = (PNO2)2

(PN2O4) (2.71)

0.360 = (PNO2)2

thus NO2 or 0.600 atm

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

Hint 1: *Kp and Kc are not necessarily equivalent.

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: 4.50 x 103

*Kc =[CO][H2]2 = (0.24)(1.1)2 = 1.94

[CH3OH] 0.15

*Kp = Kc(RT)∆n

*Kp = (1.94)(0.08026 • 600.15)2 = 4.50 x 103

4) For which of the following reactions is KP = Kc? (more than one choice may be a correct answer)

Defend your reasoning.

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. The reaction in #4 gives

a sum of 0 change in the number of moles ….based upon #moles of product - #moles of reactant and,

is the only one of the examples which meets the requirement for Kp = Kc

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

Equilibrium Concentrations

Equilibrium Constant

[H2] [I2] [HI]

[H2] [I2] [HI]

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.

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

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

Follow this process

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]

Initial

1.00

*0.00

Change

*-0.25

*+0.50

Equilibrium

*0.75

*0.50

Make an ICE table

Solution to b) …. Write the equilibrium expression … plug and chug

Kc = * [B]2 = (0.50)2 = 0.33

[A] (0.75)

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)

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.

(NO2)

atm

(N2O4)

atm

Initial

*0.56

*0.51

Change

*-0.06

*+0.03

Equilibrium

*0.50

*0.54

*Kp = (PN2O4) = 0.54 atm = 2.16 or 2.2

(PNO2)2 (0.50 atm)2

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?

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

*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

4) Determine the value of the equilibrium constant for the reaction: A + 2 B ↔ 2C

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.

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

[A]

[B]

[C]

Initial

1.0

2.0

3.0

Change

+0.80

+1.6

-1.6

Equilibrium

1.8

3.6

1.4

*Kc = [C]2 = (1.4)2 = 0.084

[A] [B]2 (1.8) (3.6)2

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

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 = (0.035)(0105)3 = 0.020

[CH4]2 (0.045)2

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

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 reaction: (Zumdahl p. 624)

PCl5(g) PCl3(g) + Cl2(g)

Hint 1:* Write the equilibrium expression

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

ans: * 8.96 x 10-2

*remove

[PCl5]

[PCl3]

[Cl2]

Initial

8.7 x 10-3

0.298

0.00

Change

-2.00 x 10-3

+2.00 x 10-3

+2.00 x 10-3

Equilibrium

6.7 x 10-3

0.300

2.00 x 10-3

*K = [PCl5(g)] = (6.7 x 10-3)

[PCl3(g)] [Cl2(g)] (0.300)( 2.00 x 10-3)

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)

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

b) Determine the equilibrium constant for the reaction.

iodine

bromine

iodine bromide

Initial

3.0

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

*remove

This one is a bit different … I converted to equilibrium concentrations after working with the mole values … because the overall volume was a 2 L flask …not a 1 L flask as in prior examples

*Kc = [ICl]2 = (1.6)2 = 3.0

[I2][Br2] (0.7)(1.2)

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