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CHAPTER 7 – CHEMICAL SYSTEMS IN EQUILIBRIUM
Dynamic Equilibrium
 Observations of some chemical systems in a closed system
(where no matter is able to escape to the surroundings) will
appear as though no change is occurring, i.e. an equilibrium
has been achieved
• however, this does not mean that there is no
activity at the molecular level
• in fact chemical systems in equilibrium will have a
balance of “reactants moving forward towards
“products” and the “products” moving in the reverse
direction towards ‘reactants”
 this is considered dynamic equilibrium
 Examples of chemical systems in equilibrium:
1. Solubility Equilibrium
• in solubility equilibrium the undissolved solute particles
continuously dissolve into solution, while an equal number of
dissolved solute particles in solution crystallize or precipitate
out of solution
2. Phase Equilibrium
• in phase equilibrium particles in both phases are gaining or
losing kinetic energy, such that they are moving from one
phase to another, while an equal number are moving in the
reverse direction
e.g. H2O(l) → H20(g)
3. Chemical Reaction Equilibrium
 Quantitative reactions are those reactions where all the limiting
reagent is used up in the reaction
e.g. CaCO3 (s) → CaO(s) + CO2(g)
2
where the CO2 (in an open system, like this, escapes so that the
reverse reaction cannot take place)
 If this reaction takes place in a closed container, both reactants
and products can be found
 This is not considered to be a quantitative reaction
 can be explained by considering the reverse reaction:
CaCO3(s) ---- CaO(s) + CO2(g)
 The equilibrium equation is
CaCO3(s) <==>CaO(s) + CO2(g)
 Using this knowledge leads to two types of manufacturing
processes;
1) continuous processing where as in the limestone example - as
limestone is used up, it is replaced with more to keep the reaction
going
2) batch processing where the reaction is done in a closed
container and the products are taken out when the reaction is
complete. i.e. microwave popcorn
N2O4(g) → 2NO2(g
Another example is
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 reactions such as these have led to this generalization for
reversible reactions in closed systems:
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Percent Reaction at Chemical Equilibrium
 Using the reaction
one can see how the reaction proceeds quickly causing the darkpurple colour of the iodine vapour fading and then becoming
constant
 Table 3 shows how the system reaches a state of equilibrium in 2
ways-in terms of percentage reaction and in terms of an
equilibrium constant
 Looking at system 1;
 theoretically, using the mole ratio we would expect a quantity of
2mmol/L of HI(g) but in fact the yield was 1.56mmol/L
 Percent reaction is the yield of product measured at equilibrium
compared with the maximum possible yield of product
If we did the same percent reaction calculations for the other 2
systems we would see this;
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 Table 4 summarizes that the percent reaction is constant at 78%
and to communicate this the percent reaction is written above the
equilibrium arrow
 All chemical reactions are considered to be reversible
 These reactions fall into 3 categories:
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An ICE table can be useful to organize information needed to solve
stoichiometric problems involving equilibrium systems.
ICE stands for Initial, Change and Equilibrium.
Look at the Sample Problems on pages 433-436.
HW Nelson 12: Page 437 PP #’s 6 & 7
Page 437 SR #’s 1-9 odd numbers
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Equilibrium Law in Chemical Reactions
 when considering chemical reactions that reach an equilibrium
point both reactants and products will be in the mixture
 repetitive testing of various chemical reactions that reach
equilibrium while beginning with different initial concentrations of
reactants show that the concentrations of the reactants and
products at equilibrium can be related through the equilibrium law:
Consider the reaction: aA + bB → cC + dD
The Equilibrium Law can be expressed as;
K = [C]c[D]d
[A]a[B]b
where: A, B, C and D are chemical entities in gas or aqueous
phases
a, b, c and d are the coefficients of the balanced
equation
K is the equilibrium constant
 The law of mass action is a relation that states that the value of
the equilibrium constant is constant for a given temperature —
when the concentrations of the reactants and products at
equilibrium are arranged according to the equilibrium law
expression they will be equal to the equilibrium constant

 The equilibrium law describes the behaviour of almost all gaseous
and aqueous chemical equilibria
the concentrations of the chemical entities are expressed in
moles per Litre (mol/L)
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Kinetics Argument for the Equilibrium Law
 Consider the following example to demonstrate how analyzing
rates of reaction can yield the equilibrium law:
N204(g) ↔ 2NO2(g)
• Experimental work has shown the rate law equation for the
forward reaction is:
rf= kf [N204]
while the rate law equation for the reverse reaction is:
r r=
kr [N02]2
• at equilibrium the rate of reaction in the forward direction is
equal to the rate of reaction for the reverse reaction
rf =rr
kr [N204]=kr [N02]2
manipulation of the above equation yields
kf = [N02]2
kr = [N204]
……(1)
the equilibrium law expression can be written as
K = [N02]2 ……….(2)
[N204]
we can set equation (1) equal to equation (2), so that:
K = kf
Kr
 from the unit on rates of reaction we understood that the
magnitude of the rate constant will determine the degree to which
a reaction will proceed in the direction specified
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 Considering that the equilibrium constant is a ratio of the rate
constant for the forward reaction to the rate constant for the
reverse reaction, the magnitude of the equilibrium constant will
allow us to determine whether products or reactants are favoured
in an equilibrium reaction
K
>>1
.
=
<<1
Description
Reaction proceeds to
completion
Products >> Reactants
Products . Reactants
=
Products << Reactants
Example
2O3 (g) ↔ 3O2 (g) K = 2.0 x
1057
CO (g) +H2O(g) ↔CO2 (g) +H2 (g)
at 700º C
K = 5.09
N2 (g) + O2 (g) ↔ 2NO2 (g)
K = 1.0 x 10-25
 Depending on the reaction the equilibrium constant may or may
not vary significantly with temperature
Eg. Production of ammonia from nitrogen and hydrogen gas
@T=25°C
K = 4.26x108
@T=300°C
K = 1.02 x 10-5
@T=400ºC
K = 8.00 x 10-7
Homogeneous and Heterogeneous Equilibria
 Homogeneous equilibrium involves reactants and products in a
single phase, eg. all aqueous or all gas
 Heterogeneous equilibrium involves reactants and products in
more than one phase, eg. liquid and gas
 Consider the following reaction with solki and gas phases:
3Fe(s) + 4H2O(g) ↔ Fe304 (s) + 4H2 (g)
the equilibrium law expression is:
K = [Fe304] [H2]4
[Fe]3 [H2O]4
But, the concentration of the pure solids Fe304 and Fe do NOT
change during the reaction as their densities are fixed; they do
not affect equilibrium
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As such the quantities [Fe304] and [Fe] are constant and could be
incorporated in the equilibrium constant
K [Fe]3
= [Fe3O4] [H2]4
[Fe3O4] [Fe]3 [H2O]4
Or K' = [H2]4
where K’ is the new equilibrium constant
[H2O]4
 Similarly, in heterogeneous reactions that involve pure liquids and
another phase, the concentration of the pure liquid is constant
(fixed density) and can also be incorporated into the equilibrium
constant
 The equilibrium constant reported in reference tables for
heterogeneous equilibria will already incorporate the densities of
pure solids or liquids
 In the case where a heterogeneous reaction involves ions and a
pure solid, pure liquid, or gas, then a net ionic equation is
developed to eliminate spectator ions and show only those ions
altered or involved in the chemical reaction
eg. see page 447 sample problem
Determining Equilibrium Concentrations
 Simple calculations involving the equilibrium constant requires
knowledge of a combination of the concentrations of the reactants
and products and the equilibrium constant
 An example involves the following reaction:
H2 (g) + CO2 (g) ↔ H2O(g) +CO(g)
Where the concentrations are given as;
[H2] = 0.24 mol/L
[H20] = 0.88 mol/L
[C02] = 1.80 mol/L
[CO] = 0.88 mol/L
the equilibrium law expression is:
K = [H2O][ CO]
[H2][ CO2]
K = (0.88)(0.88) = 1.8
(0.24)(1.80)
 Variations of this question may involve:
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a) Determining the concentration of one of the reactants or
products (K given along with other reactant and product
concentrations)
b) Determining K where balanced chemical equation and other
information makes it possible to calculate concentrations
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Qualitative Changes In Equilibrium Systems
 Le Chatelier’s Principle:
“When a chemical system at equilibrium is disturbed by a change
in a property, the system adjusts in a way that opposes the
change”
Le Chatelier’s Principle & Concentration Changes
 this principle predicts if more reactant is added to a system at
equilibrium, then the system will have a change in the equilibrium
called an equilibrium shift
 E.g.
 to improve the yield more HF(g) can be added to the system, the
added reactant “disturbs” the system and the system shifts to the
right using up more of the CCl4 and producing more Freon-12 and
making a new equilibrium state
 in chemical equilibrium shifts, the imposed change in
concentration is usually only partly counteracted and the
concentrations are usually different than the original
concentrations (see Fig. 2 below) – adding CO2
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 by removing a product (decreasing concentration) will also shift an
equilibrium forward, to the right (see Fig. 3)
 An example of the effects of both forward and reverse shifts can
be seen in Fig. 4 (on page 452)
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 other examples of how Le Chatelier’s Principle is used are;
as more NO(g) is removed by reacting it with O2 (g) the system
shifts to the right and more HNO3 is produced
as blood circulates to the lungs, the high concentration of O2 (g)
shifts the equilibrium to the right and the haemoglobin gets
oxygenated
as the oxygenated blood travels through the body the reverse
happens, the equilibrium shifts to the left and more oxygen is
released
Rate Theory & Concentration Changes
 kinetic theory explains equilibrium shifts when a reactant
concentration increases by assuming that when more reactant is
added, more collisions are occurring in the forward direction
 the reverse reaction, in this case, will also occur until a new
equilibrium is reached where the reaction rates are equal again
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 the rates of the new equilibrium will be faster than the original
rates because there are now more particles and therefore more
collisions
 if a substance is removed causing an equilibrium shift, the initial
effect is to decrease one of the rates
 additions or removal of substances in the solid or liquid state DO
NOT change the concentration of that substance
Le Chatelier’s Principle & Temperature Changes
 energy can be treated as though it were a reactant or product
 e.g.
when the system is heated the equilibrium shifts to the right and
more nitrogen dioxide is produced as seen by a darkening of the
reddish-brown colour observed
Rate Theory & Energy Changes
 kinetic theory can explain the shifts that occur when energy in a
system is increased or decreased
 rate theory explains this by assuming that both forward & reverse
rates are slower at lower temperatures, but that the reverse rate
decreases more than the forward rate
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 even though the rates remain equal, the result observes is the
production of more product & more energy
 this causes concentration changes that will increase the reverse
rate and decrease the forward rate until they become equal at the
new lower temperature (see Fig. 7)
 in endothermic reactions, the situation is reversed
Le Chatelier’s Principle & Gas Volume Changes
 Boyle’s Law states that P & V are directly related when T is
constant
 decreasing the volume of a container to half its original value will
double the concentration of every gas in the container
 changing the volume may cause a shift in the equilibrium
 to predict whether this shift may occur we have to look at the total
number of moles of gas reactants and gas products
e.g.
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 if the volume is decreased the overall pressure is increased.
 Le Chatelier’s Principle suggests that the system will react in a way
that resists the change i.e. in a way to reduce the pressure (see
Fig. 9)
 systems with equal numbers of gas molecules on each side of the
equation will not shift after a change in volume
 systems involving only solids or liquids are not affected by
changes in pressure
 substances in these states are virtually incompressible and
reactions involving them cannot counteract pressure changes
Rate Theory & Gas Volume Changes
 kinetic theory explains the effect of decrease of volume by
assuming that both the forward and reverse reaction rates
increase because the concentrations of reactants and products
both increase
 however, the forward rate increases more than the reverse
because there are more particles involved in the forward reaction
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 which means the number of collisions is greater for the forward
direction
 rates are about the same but more product is produced
 the shift causes concentration changes that will increase the
reverse rate and decrease the forward rate until they become
equal again
Changes that do not affect the Position of Equilibrium
Systems
 adding catalysts – decreases the time required to reach the
equilibrium position but does not affect the final position of
equilibrium
 lowers the activation energy for both the forward & reverse
reactions by an equal amount
 equilibrium establishes itself faster but at the same position as if
no catalyst were present
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 adding inert gases – the pressure of a system can be changed by
adding a gas while keeping the volume constant
 if the gas is inert, the equilibrium position of the system will not
change
 presence of the inert gas changes the probability of successful
collisions for both reactants & products equally, which means
there will be no shift in equilibrium
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Quantitative Changes in Equilibrium Systems
 Remember a large value of K means the reaction favours the
formation of products
 A small value of K means the reaction favours the formation of
reactants
The Reaction Quotient, Q
 In a reaction with both reactants and products present we don’t
always know the direction of the reaction
 We can use the concentrations and the equilibrium law expression
to produce a trial value called the reaction quotient or Q
 Q is similar to K BUT Q is not always calculated when the system
is at equilibrium
 The value of Q can mean 3 things:
1. Q =K, and the system is at equilibrium
2. Q>K, system must shift left (toward reactants) to reach
equilibrium, because the product to reactant ratio is too
high
3. Q<K, system must shift to the right (toward products) to
reach equilibrium, because the product to reactant ratio is
too low
Look at the Sample Problems in your textbook, pages 464-480.
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The Solubility Product Constant
Solubility Product
 special case of equilibrium involves cases where excess solute is in
equilibrium with its aqueous solution
 can be done by either preparing a saturated solution with excess
solute remaining or by mixing 2 solutions of salt that results with a
product that precipitates out of solution
 Example:
 for any solute that forms ions in solution, the solubility
equilibrium constant is found by multiplying the concentrations
of the ions in solution raised to the power equal to the
coefficient of each entity in the balanced equation
 this value of K is called the solubility product constant, Ksp
 in the above example, Ksp = [Cu+(aq)][Cl-(aq)]
Ksp = 1.7 x 10-7 at 25°C
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 other examples include:
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Calculating Solubility Using Ksp Values
See Sample Problem on page 484.
Predicting Precipitation
 from the Grade 11 course we used Solubility Tables to predict
whether a precipitate would form or not
 from section 7.5, we used the reaction quotient, Q, to
determine whether the reaction was at equilibrium
 we can use Q to determine whether a precipitate will form
when we mix solutions of metal cations and non-metal anions
 we can calculate Q to see if the ions are present in a high
concentration, in which case we will see a precipitate form or
not
 the reaction quotient, Q, is sometimes called the trial ion
product
 to predict whether a precipitate will form or not we compare
the Ksp to the value of Q
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Example:
Using Trial Ion Product
 Ksp is the value of a system at equilibrium, Q however can be
measured at any time
 this is important in predicting whether or not a precipitate will
form
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The Common Ion Effect
 when an equilibrium exists in a solution involving ions, the
equilibrium can be shifted by dissolving into the solution another
compound that adds a common ion, or reacts with one of the ions
already in solution
example: NaCl = Na+(aq) + Cl-(aq)
add a few drops of HCl and additional crystals of NaCl will
form
 the explanation for this from Le Chatelier’s Principle is that when
we add HCl it releases large numbers of Cl ions into solution which
will cause the reaction to shift to the left….producing more NaCl to
precipitate out
 the lowering of the solubility of an ionic compound (in this case
the NaCl) by adding a common ion is called the common ion
effect
 See Sample Problem on p.491
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SSC
CH
H44U
UC
CH
HE
EM
MIISST
TR
RY
YN
NO
OT
TE
ESS
Spontaneous Reaction
◙
whether a reaction occurs at all- slow or fast.
◙
We’ll look at predicting whether reaction is spontaneous first (then rates of
reactions).
◙
It can have exothermic and endothermic reactions which are spontaneous.
◙
Given the necessary Ea to begin with, reaction occurs without continued
assistance.
◙
As well, see lab on heats of soln; make note that combution or exothermic
reactions tend to be spontaneous while dissolution of NH4NO3 is
endothermic.
Entropy
 Spontaneity depends on ∆H° and change in entropy.
 Entropy is defined as the degree of randomness or disorder.
 Each chemical or physical reaction process has change in entropy.
 “Law of Disorder”: A spontaneous reaction in which an isolated system
always proceeds in a direction of increased entropy.
Eg.
Solid –> Liquid
Liquid –> Gas
∆S>0
∆S>0
0 0 –> 0 0 0
∆S>0
Increasing # of moles
T1 –> T2
(T2 > T1)
Two liquids –>
0
∆S>0
Mixture of two liquids
∆S>
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Solid Solute in –>
Liquid solvent.
soln
∆S>0
 Perfect crystal of pure substance at absolute zero (0 k) • S = 0
 Standard Entropy: Change from 0 k –> 298 k (23° C) and 100 kPa.
 There are values for standard entropy of elements.
S =
J
-----------k · mol
 Standard Entropy Change
o S increases with
temperature.
o Standard S Is change
between 0 and 298 k for
a chemical reaction:
∆ S° = ∑ S° products — ∑ S° reactants
Eg. 4 Fe (s) + 3O 2 (g) –> 2 Fe 2O 3(s)
Gibbs Free Energy
 A thermodynamic quantity that relates enthalpy and entropy.
 Predicts whether a reaction is spontaneous or not.
∆ G = ∆ H — T∆ S°
________________________________________________________
__________
At a standard state:
∆ G° = ∆ H — T∆ S°
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∆ G° < 0,
reaction is spontaneous.
∆ G° > 0,
reaction is not spontaneous.
∆ G° = 0,
reaction is in equilibrium.
Lets look at 4 cases that affect sign of ∆ G°
1.
2 H 2O 2 –> 2 H 2O + O 2 (g)
∆ H° = – 197 KJ
o Exothermic.
o ∆ S° is positive, more moles result.
• Therefore, ∆ G° < 0 at all temperatures; spontaneous.
2.
NH 3 (g) + HCl (g) –> NH 4CL (S)
∆ H° = – 176 KJ
o Exothermic.
o ∆ S° is positive, less moles are produced.
o Occurs at low temperatures.
∆ S° = 94.6 – [ 192.5 + 186.9]
= – 284.8
o At low T –> T∆ S is smaller than ∆ H°;
therefore, it is spontaneous.
o At high T –> T∆ S is larger than ∆ H°;
therefore, it is nonspontaneous.
3.
2 H 2O(g) –>
2 H 2 (g) + O2 (g)
∆ H° = + 484 KJ
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o Endothermic.
o ∆ S° is positive, more moles are produced.
∆ S° = [ 2 (69.9) + 205.1] – 2 (188.8)
= – 32.7
o At low T –> T∆ S is smaller than ∆ H°;
therefore, it is nonspontaneous.
o At high T –> T∆ S is larger than ∆ H°;
therefore, it is spontaneous around 5000°C.
o Reaction is entropy driven.
4.
3 O 2 (g) –> 2 O 3 (g)
∆ H° = + 285 KJ
o Endothermic.
o ∆ S° is positive, less moles are produced.
• Therefore, ∆ G° > 0 is nonspontaneous or spontaneous in
reverse.
Calculating ∆ G
∆ G = ∆ H — T∆ S
 Use example. NH3 (g) + HCL (g) –> NH4CL(S)
∆ H = ∑n∆H°f products — ∑n∆H°f reactants
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Standard Free Energy of Formation
∆ G° reaction = ∑n∆G°f products — ∑n∆G°f reactants
o Note Error Within the text on Pg. 222.
NH3 (g) + HCL (g) –> NH4CL(S)
Predicting Changes in Spontaineity
∆ G° = ∆ H — T∆ S° = 0
T∆ S° = ∆ H°
• Therefore,
T =
∆ H°
-----------∆ S°
o For NH3 (g) + HCL (g) –> NH4CL(S)
∆ H° = –
∆ S° = –
176.0 KJ
0.2948 KJ • K – 1
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• Therefore, T =
– 176.0
-----------– 0.2948
= 618.0 K (345°C)
• Therefore, above 618.0 K, the reaction is nonspontaneous.
o Assumes ∆ S° and ∆ H° do not change substantially from a
standard state.