The Nature of Chemical Equilibrium


  1. Introduction

    Chemical equilibrium are dynamic.
    Two opposing processes are occurring at equal rates.
    LeChatelier’s Principle guides our explanation of what occurs during equilibrium, both chemical and physical.
    LeChatelier’s Principle - statement

  2. Reversible Reactions

    Forward reaction is read from left to right.
    Reverse reaction is read from right to left.
    Given the proper conditions, every reaction can be reversed.

    Reversible reaction - definition



Refer to figure 18-1 on page 553, reaction is a test tube.

Do the same reaction in a closed container.

At equilibrium we will have a mixture of mercury oxide, mercury and oxygen. The amount of each will remain constant as long as equilibrium is maintained. Both forward and reverse reactions continue to occur.

Chemical equilibrium - definition

As long as the conditions of equilibrium exist, the amount of reactants and products remains the same, since both reactions are occurring at equal rates.

Chemical Equilibrium - definition

Use of two opposing arrows -- never a double headed arrow.

III: Equilibrium, A Dynamic State

Many chemical reactions are reversible if they occur in a closed container. They will reach equilibrium if allowed sufficient time. If one or more of the substances is allowed to escape, no equilibrium can be established.

Which reaction is favored in an equilibrium situation?
If the forward reaction occurs to a large extent before equilibrium is established then we say that the products of the forward reaction are favored and, thus, we will have a higher concentration of products than reactants.

If the forward reaction occurs to a limited extent before equilibrium is established then we say that the reactants are favored or the reverse reaction is favored.


If both reactions occur to approximately the same extent before equilibrium is established then the concentration of reactants and products will be similar.

Equilibrium can be a problem for the chemist or for industrial processes. If equilibrium becomes established the amount of products may be limited. The purpose of an industrial process is to convert as much of the reactants into products.

IV: The Equilibrium Expression

Figure 18.2 page 555

Before the reaction we have all reactants (amount is maximum) and no products (amount is zero). When the reaction begins the amount of reactants decreases while the rate of the reaction decreases. The amount of products increases while the rate of the reverse reaction increases. When the two rates are equal we have equilibrium.

The ratio of products to reactants at a given temperature is a constant according to the following expression where K is the equilibrium constant.

K = [Products] / [Reactants]

Notice that products are in the numerator raised to the power of their coefficients and the reactants are in the denominator raised to the power of their coefficients - NEVER THE POWER OF THE SUBSCRIPT.

The value of K depends only on the temperature. It does not depend on the initial value of the reactants.

A: The Equilibrium Constant
Need to determined the value of K experimentally or find it in the literature.

If K equals one then at equilibrium the concentration of reactants and products are approximately equal.

If the value of K is low then the reverse reaction is favored indicating that the forward reaction proceeds only slightly before the speed of the reverse reaction becomes equal to it.

If the value of K is high the forward reaction is favored.

Only concentrations of substances that can change are included in the expression for K. Pure solids and liquids are not included because their concentrations cannot change.

Equilibrium constant - definition

Chemical Equilibrium expression - definition

B: The H2, I2, HI Equilibrium System

The extent of the reaction is seen by the fading of the color. At some point the color becomes constant indicating equilibrium.

Figure 18-3 page 557

H2(g) + I2(g) = 2 HI(g)
and the equilibrium expression is

K = [HI]2 / [H2][I2]

See values is table 18-1 page 557

If the values for K are not constant for a reaction, then it indicates that the temperature was not constant or that the system did not reach equilibrium.

If you know the concentrations of the reactants and products you can calculate the value for K. If you know the value for all substances except one and you know the value of K you can calculate the concentration of the substance you do not know.

You need a balanced equation to solve any problems dealing with the equilibrium constant expression.

Sample Problem 18-1 page 558

An equilibrium mixture of N2, O2, and NO gases at 1500 K is determined to consist of 6.4 x 10-3 mol/L of N2, 1.7 x 10-3 mol/L of O2, and 1.1 x 10-5 mol/L of NO. What is the equilibrium constant for the system at this temperature if the balanced equation is

N2(g) + O2(g) = 2 NO(g)

Given:

[N
2]=6.4x10-3 mol/L
[O2]=1.7x10-3 mol/L
[NO]=1.1x10-5 mol/L
K=?

K=[NO]2 / [N2][O2]



Homework 18.1


Shifting Equilibrium


Understanding how to shift equilibrium and how factors affect shifts help us convert more reactants into products which is the goal of most chemical reactions.

I: Predicting the Direction of Shift

Le Chatelier’s Principle - statement

physical and chemical - dynamic

pressure, concentration, and temperature

A: Changes in Pressure

only for gases

need to figure total number of moles of gas in reactants and in products - moles and number of particles according to Avogadro’s number

increase pressure favors reaction that produces fewer moles of gas

decrease pressure favors reaction that produces more moles of gas

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


Increase in pressure of a gas in a container increases the concentration of the gas.

B: Changes in Concentration

increase concentration of a reactant or a product increases collisions and therefore increases the rate of the reaction.

A + B = C + D


increase A
shift equilibrium to right
new equilibrium has lower concentration of B

increase B
shift equilibrium to right
new equilibrium has lower concentration of A

increase C
shift equilibrium to left
new equilibrium has lower concentration of D

etc.

Change in concentration does not affect value of K

heterogeneous reaction in one in which the substances are in different physical states

concentrations of pure substances are not changed by adding or removing quantities of these substances because concentration is density dependent and the density of these physical states is constant

can eliminate such substances from the expression for K since their concentration is considered to be 1

CaCO3(s) = CaO(s) + CO2(g)

K= [CO2]

C: Changes in Temperature

Exothermic in one direction - endothermic in the opposite direction

Adding heat favors the endothermic reaction; how does it affect the value of K?

Removing heat favors the exothermic reaction; how does it affect the value of K?

N2(g) + 3 H2(g) = 2 NH3(g) + 92 kJ or
N2(g) + 3 H2(g) = 2 NH3(g) delta H = -92kJ


556 kJ + CaCO3(s) = CaO(s) + CO2(g) or
CaCO3(s) = CaO(s) + CO2(g) delta H = +556 kJ


2 NO2(g) = N2O4(g) + energy


see figure 18-5 page 565

Catalyst affects the rate of both forward and reverse reactions equally and so does not affect the value of K. It only increase the speed of both.

II: Reactions That Go to Completion

do not set up equilibrium

factors are
ions removed from solution because of solubility or ionization
product escapes as a gas
products precipitates

A: Formation of a Gas

H2CO3(aq) ---> H2O(l) + CO2(g)

container must be open to atmosphere

B: Formation of a Precipitate

NaCl(aq) + AgNO3(aq) ---> NaNO3(aq) + AgCl(s)

C: Formation of a Slightly Ionized Product

HCl(aq) + NaOH(aq) ---> NaCl(aq) + H2O(l)

III: Common Ion Effect

Using ions to force a reaction in a particular direction.

one reaction in equilibrium and adding a solution of a substance that has an ion in common with the equilibrium system

HCl(g) + H2O(l) = H3O+(aq) + Cl-(aq)

hydrochloric acid solution is mainly ions

NaCl(s) = Na+(aq) + Cl-1(aq)


add hydrochloric acid solution to a saturated solution of sodium chloride to observe the common ion effect

common ion effect - definition

CH3COOH(aq) + H2O(l) = H3O+1(aq) + CH3COO-1(aq)
acetic acid exists mainly as molecules

NaCH3COO(s) = Na+1(aq) + CH3COO-1(aq)
add sodium acetate to the acetic acid solution

can measure shifts using a pH meter or an indicator

Homework 18.2


III: Equilibria of Acids, Bases, and Salts

A: Ionization Constant of a Weak Acid

CH3COOH + H2O = H3O+1 + CH3COO-1

K = [H3O+1] [CH3COO-1] / [CH3COOH] [H2O]
K[H2O] = [H3O+1] [CH3COO-1] / [CH3COOH]
Ka = [H3O+1] [CH3COO-1] / [CH3COOH]

Acid ionization constant - abbreviation

temperature dependent

value of Ka for weak acid

value of Ka for strong acid

table 18-2 page 570 - Ionization of Acetic Acid

adding sodium acetate, NaCH3COO, to acetic acid:
-effect on equilibrium
- effect on the concentration of the hydronium ion
- effect on the concentration of the other species in the equilibrium expression

weak acid plus salt of that weak acid e.g. acetic acid plus sodium acetate

B: Buffers

Buffered solution - definition

start with weak acid plus the salt of the weak acid; can react with either an acid or a base and not change pH dramatically

to evaluate the effect, start with the equilibrium expression for the weak acid, know what ions you are adding to the buffer, apply Le Chatelier’s principle

CH3COOH + H2O = H3O+1 + CH3COO-1

add an acid to the above solution, what happens?

Also applies to a solution of a weak base and a salt of that weak base.

NH3 + H2O = NH4+1 + OH-1

add ammonium chloride, NH4Cl, to the above solution

blood is buffered between 7.3 and 7.5
buffered aspirin

C: Ionization Constant of Water

self ionization of water
H2O + H2O = H3O+1 + OH-1

Kw = [H3O+1][OH-1]

D: Hydrolysis of Salts

Salts are produced by neutralization reactions.

The strength of the acid and base determines what type of salt is formed i.e. is it one that reacts with water to produce a solution that is not equal to seven or not.

A salt formed from the reaction of a strong base and a weak acid e.g. NaOH + H2CO3 pH > 7

A salt formed from the reaction of a weak base and a strong acid e.g. NH4OH + HCl pH < 7

The ion that comes from the weaker species reacts with water to form a solution whose pH is greater or less than seven.

hydrolysis - definition

1. Anion Hydrolysis

The anion comes from the acid. If the acid is weak it means that it does not lose a proton readily but its anion does accept protons readily.

If HA represents a weak acid then we have the weak acid equilibrium

HA(aq) + H2O(l) = H3O+1 + A-1(aq)

for this reaction
Ka = [H3O+1] [A-1] / [HA]

The anion of this weak acid, A-1(aq) would react with water as

A-1(aq) + H2O(l) = HA(aq) + OH-1(aq)

If the weak acid has a low value of Ka then it indicates a large amount of unionized acid, HA. This indicates that A-1 will have a strong attraction for the hydrogen of water to cause the hydrolysis reaction and the higher the pH of the resulting solution will be and visa versa.

e.g. H2CO3 + H2O = H3O+1 + HCO3-1

K = [H3O+1] [HCO3-1] / [H2CO3] [H2O]
K[H2O] = [H3O+1] [HCO3-1] / [H2CO3] or
Ka = [H3O+1] [HCO3-1] / [H2CO3]

H2CO3 does not lose a proton readily - a weak acid - but its conjugate base does accept protons readily - a strong conjugate base.

figure 18-10 page 572 Indicators in different salt solutions

As a result, the anion, HCO3-1, will react with water to produce H2CO3 + OH-1 or

H2CO3 + H2O = H3O+1 + HCO3-1 and subsequently

HCO3-1 + H2O = H2CO3 + OH-1

Whenever you have the anion of a weak acid present in a solution, it will react with water to form a hydroxide ion.

e.g. sodium carbonate - anion comes from carbonic acid, H2CO3 - therefore, the anion is a strong conjugate base and will react with water to form HCO3-1 + OH-1 as in
CO3-2 + H2O = HCO3-1 + OH-1

2. Cation Hydrolysis

Same situation as above except the salt is produced from a weak base and a strong acid. e.g.

Metal ions in aqueous solution behave as Lewis acids. The positive charge on the metal ion draws electron density from the O-H bond in the water. This increases the bond's polarity making it easier to break.

When the O-H bond breaks, an aqueous proton is released producing an acidic solution.

For a weak base, indicated as B(aq), we have
B(aq) + H2O(l) = BH+1(aq) + OH-1(aq)

and Kb = [BH+1] [OH-1] / [B]

The cation, BH+1 reacts with water to give the following hydrolysis reaction:

BH+1(aq) + H2O(l) = H3O+1(aq) + B(aq)

BH+1 donates a proton to the water which forms a hydronium ion which adds to the acidity of the solution.

Ammonium chloride is an example of such a salt since it comes from the reaction of ammonium hydroxide and hydrochloric acid.

Figure 18-11 page 574 titration curve of weak acid and strong base

3. Hydrolysis in Acid Base Reactions

Hydrolysis explains why the equivalence point of an acid base titration may not be at 7.

Figure 18-11 page 574

Figure 18-12 page 575 titration curve for ammonia and hydrochloric acid.

Salts of weak acids and weak bases can produce solutions that are either acidic, basic, or neutral.

Homework 18-3


4. Solubility Equilibrium

When ionic solids are added to water they dissolve until they reach equilibrium with their dissolved ions. This allows us to write an equilibrium expression for this process and, using balanced chemical equations and solubility data we can determine the concentrations of the ions.

The equilibrium constant can tell us if precipitation occurs when different solutions are mixed.


1. Solubility Product

Saturated solution has the maximum amount of solute dissolved. Depends on temperature.

Saturated and unsaturated v concentrated and dilute

Rules of thumb: soluble if greater than 1 g / 100 g water and insoluble if less than 0.1 g / 100 g water; slightly soluble between these two.

We will be considering solutions whose solute is sparingly soluble e.g. AgCl.

AgCl = Ag+1(aq) + Cl-1(aq)

and K = [Ag+1] [Cl-1] / [AgCl] but because the concentration of a pure solid, such as the [AgCl] in the denominator cannot change we incorporate it into the constant as follows:

K[AgCl] = [Ag+1] [Cl-1] or

Ksp = [Ag+1] [Cl-1]



For the reaction CaF2(s) = Ca+2(aq) + 2F-1(aq)
Ksp = [Ca+2] [F-1]2

We can calculate the value for Ksp from solubility data viz. Appendix Table A-13 on page 901 of your text.

For silver chloride, AgCl, the solubility is 8.9 x 10-5 g of AgCl in 100. g water at 10oC. The molar mass of AgCl is 143.32 g.

AgCl Ksp


Another example is:

For the system:

CaF2(s) = Ca+2(aq) + 2F-1(aq)
the solubility of calcium fluoride is 1.7 x 10-3 g/100. g water at 26oC.

Following the same procedure as we did above gives us a concentration in moles per liter of 2.2 x 10-4 mol/L.

When calcium fluoride dissociates it produces twice as many fluoride ions as calcium ions (1 to 2 molar ratio). The concentration of calcium ions would be 2.2 x 10-4 mol/L and the concentration of fluoride ions would be 2(2.2 x 10-4 mol/L) or 4.4 x 10-4 mol/l.

Note how the concentration of fluoride ions at equilibrium is twice the concentration of the calcium ions.

Then Ksp = [Ca+2] [F-1]2

Ksp = (2.2 x 10-4) (4.4 x 10-4)2
Ksp = 4.3 x 10-11



Table 18-3 page 579 Table of Ksp’s - note only two significant figures

There is a difference between the solubility of a solid and its solubility product constant.

The solubility product constant is an equilibrium constant representing the product of the molar concentrations of its ions in a saturated solution. It has only one value for a given solid at a given temperature.

The solubility of a solid is an equilibrium position that represents the amount of the solid required to form a saturated solution with a specific amount of solvent. It has an infinite number of possible values at a given

temperature and is dependent on other conditions such as the presence of a common ion.

Sample Problem 18-2 page 580

sample prob 18-2


2. Calculating Solubilities

Once you know the Ksp you can determine the solubility.

To find how much barium carbonate, BaCO3, can be dissolved in 1 Liter of water at 25oC. Use table 18-3 to find the Ksp which is 5.1 x 10-9.

BaCO3(s) = Ba+2(aq) + CO3-2(aq)


Ksp = [Ba+2] [CO3-1] = 5.1 x 10-9

barium carbonate dissolves until the product of the molar concentrations of the two ions equals 5.1 x 10-9

The equation above indicates equal amounts of both ions (molar ratio is 1:1).

Thus [Ba+2] = [CO3-2]

Let the concentration of barium ion = x and the concentration of carbonate ion = x since they are equal to each other. Then we have

(x)(x) = 5.1 x 10-9
x=7.14 x 10-5 mol/l which is the solubility of the barium carbonate

Thus the solution concentration is 7.14 x 10-5 M for barium ions and 7.14 x 10-5 for carbonate ions.

Sample Problem 18-3 page 581

Homework 18.4


3. Precipitation Calculations

In the previous section we saw from the equation for barium carbonate that one mole of barium ion and one mole of carbonate ion was produced and we assumed that the concentration of the ions was equal. But in the expression for Ksp it is the product of the concentrations of the two ions that must equal the numeric value of Ksp i.e. they do not have to be equal. The Ksp represents the maximum product of the two ions. If the product is less than the value of Ksp then the solution is unsaturated and no precipitate is evident. If the product is greater than Ksp then a precipitate forms.

If you have a solution of barium chloride and a solution of sodium carbonate and mixed the two solution together a precipitate of barium carbonate can form if the concentration of the barium and the carbonate ions is greater than the Ksp.

The precipitate will continue to form until the concentration of both ions is reduced to the point where their product is equal to the Ksp and we would then have a saturated solution.

Figure 18-13 page 582 Negative ions in presence of certain metal ions.

Sample Problem 18-4 page 583

Homework 18.5

end of notes



Reversible reaction is a chemical reaction in which the products can react to re-form the reactants. back













Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of its products and reactants remain unchanged. back














The equilibrium constant, K, is the ratio of the mathematical product of the concentrations of substances formed at equilibrium to the mathematical product of the concentrations of reacting substances. Each concentration is raised to a power equal to the coefficient of that substance in the chemical equation. back














Chemical equilibrium expression is the equation for K. back














LeChatelier’s Principle: If a system at equilibrium is subjected to a stress, the equilibrium is shifted in the direction that tends to relieve the stress. back














Common ion effect is the phenomenon in which the addition of an ion common to two solutes brings about precipitation or reduced ionization. back














Acid ionization constant is Ka. back














Buffered solution is a solution that can resist changes in pH. back














A hydrolysis reaction is one in which water molecules and dissolved salt ions react. back














Anion hydrolysis is a hydrolysis reaction in which the anion of the salt reacts with the water. back














Cation hydrolysis is a hydrolysis reaction in which the cation of the salt reacts with the water. back