NAME ______________________________
AP NOTES: UNIT 3 (3): EQUILIBRIUM & Le CHATELIER
I-V) Basics to K, the equilibrium constant to ICE Tables to Q
When chemists are in charge of the production of a particular substance, they work to choose the optimal temperature and
pressure that combined with the optimal concentrations yield the greatest amount of product.
Fritz Haber did extensive research on the synthesis of ammonia. He found that the equilibrium concentration of NH3
increased when at low temperature and high pressure…. Yet, the reaction proceeds so slowly at low temperature as to
make the reaction useless. The Zumdahls makes a passionate pitch stating that “…we must study both the
thermodynamics and the kinetics of a reaction before we really understand the factors that control it.” (p. 633). We will get to
thermodynamics shortly … but for now, we can make qualitative predictions as to the effects of changes in concentration,
pressure, and temperature on a system at equilibrium by employing Le Chatelier’s Principle.
Almost too simply stated…. Le Chatelier (luh shot-tell-lē-ā) proposed that when a change is imposed on a system
already at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that
change. This section of the notes is dedicated to the applications of Le Chatelier’s Principle.
VI) Perturbing Chemical Equilibrium & Le Chatelier’s Principle
A) CHEMICAL EQUILIBRIUM: Le Chatelier's Principle: When a *
stress
is applied to a
system already at equilibrium ,the equilibrium shifts due to increased effective collisions,
in a way that relieves ( * consumes
B) Stress: A change in * temperature
) the stress.
, concentration
, pressure
2) shift the equilibrium to the right or favor the forward reaction = * makes “product”
Consider:
2 H2(g) + O2(g)  2 H2O(g) + 483.6 kJ
When the point of equilibrium is shifted to the right then *more H2O is made while H2 & O2 decrease
_____________________________________________________________________
3) shift the equilibrium to the left or favor the reverse reaction = * make more reactants
Consider:
2 H2(g) + O2(g)  2 H2O(g) + 483.6 kJ
When the point of equilibrium is shifted to the left then * more H2 & O2 are made & H2O decreases
_____________________________________________________________________
Comprehension:
When an equilibrium is destroyed by shifting it to the right, then * more product is made, as reactant decreases
When an equilibrium is destroyed by shifting it to the left then * product decreases as reactant is made
246
VII) APPLYING LE CHATELIER: 4 Possible Stresses For A Chemical Reaction Already At Equilibrium
1)
2)
3)
4)
Molar Concentration
Pressure in gaseous systems only (related to a change in [ ] and a change in volume)
Temperature
Catalyst (has no effect upon equilibrium - it alters the rates of both reactions equally)
5) To understand equilibrium you must understand Collision Theory. In essence, the * greater
the number of
effective collisions, the greater the amount of their product
will be made. Once an equilibrium is destroyed, the reaction, will over time, achieve a
new equilibrium…
Stress
Approach to Solving Problems...
Comments
Concentration
Whatever is done to gaseous or
dissolved (aq) reactants in terms
of increasing or decreasing
concentration, also ultimately happens
to the concentration of the products
once the new equilibrium is
established. (and vice versa)
When you add more of 1 reactant …you produce more,
product but you also consume most of the other reactants.
Temperature
An increase in temperature helps
both the exothermic reaction and the
endothermic reaction, BUT, it favors
the endothermic reaction MORE
Process: Determine which reaction (forward or reverse) is
the endothermic reaction and apply the rule.
Pressure
An increase in pressure favors the
reaction which produces the fewer
total number of moles
This stress affects systems with gases.
Adding a catalyst to a system at
equilibrium, has no effect on the
point of equilibrium, Little changes
A catalyst affects the rates of the forward and reverse
reaction equally. It don’t do diddly / squat
to the equilibrium!
Catalyst
When the opposite is done … that is, when you remove
product … you lessen the reactant concentration by shifting
the reaction towards making more product.
Process: Add up and compare the total sum of the
coefficients on each side of the reaction equation
and apply the rule.
We will approach each stress, arithmetically, and as a “cut to the chase” issue. However, the key to
this entire section is to keep asking:
“Where are there the greater number of effective collisions”?
and
“What do those effective collisions make … more reactant or product”?
247
A) Concentration
1) Cut to the Chase: When more of one species is added to a system already at equilibrium,
that will change the number of effective collisions and the reaction will
shift so as to lower the concentration of that component.
When more reactant is added, there is an increase in effective collisions between reactants, and these
collisions make product … Thus the point of equilibrium shifts to the right… making more product and
relieving the “stress” of extra reactant. The reactant concentrations must decrease as this happens
When more product is added, there is an increase in effective collisions between product species, and
these collisions make reactant…. Thus the point of equilibrium shifts to the left … making more
reactant, and relieving the “stress” of extra product.
When product is removed from the system, as it is made (e.g. condensing a gas to a liquid) … we may
infer that the loss of product collisions, means that the reactant collisions are in excess … thus the
reaction shifts to the right, to produce more product….
Whatever you do to the (g) or (aq) concentrations of the reactants, you ultimately do to the product
concentrations (and vice versa), once the new equilibrium is established.
Proof: Given: N2(g) + 3H2(g)  2NH3(g)
with K = 5.96 x 10-2
and equilibrium concentrations of [N2(g)] = 0.399 M, [H2(g) ] = 1.197 M, & [NH3(g)] = 0.202 M
What would happen were an extra 1.000 mol/L N2(g) injected into the reaction vessel?
Well the system is no longer at equilibrium and you want to predict … thus … calculate Q!
Q = (0.202)2
(1.399) (1.197)2
=
1.70 x 10-2
Q is smaller than K thus the reaction will shift to the right and produce more NH3, and
thereby relieving the stress (consuming the extra nitrogen reactant) … Of course, as
NH3 is being produced, the N2 concentration is decreasing BUT so TOO is the [H2]
Thus when adding more N2 reactant …
Where are there the greater number of effective collisions?
Answer: Between reactant species N2 and H2.
And
What do those effective collisions make?
Answer: They make more product … in this case, NH3
248
1) Given the equilibrium:
2SO2(g) + O2(g)  2 SO3(g)
a) When [O2] is increased, at the new equilibrium the [SO3] will *   the [SO2] must be * 
Ask: Where are there greater effective collisions?
Ask: What do those collisions make?
b) When [SO3] is increased at the new equilibrium, the [O2] will *  and the [SO2] must be *
Ask: Where are there greater effective collisions?
Ask: What do those collisions make?
2) Given the equilibrium:
N2(g) + 3H2(g)  2NH3(g)
a) When [N2] is increased, at the new equilibrium, [NH3] will *↑ and the [H2] will * ↓
Ask: Where are there greater effective collisions?
Ask: What do those collisions make?
b) When [NH3] is increased, at the new equilibrium the [H2] and [N2] will * ↑
Ask: Where are there greater effective collisions?
Ask: What do those collisions make?
c) When [NH3] is decreased, by removing ammonia as it is made, at the new equilibrium,
the [H2] and [N2] would * ↓
Ask: Where are there greater effective collisions?
Ask: What do those collisions make?
3) Given the equilibrium: PH3(g) + 3 H2O(ℓ)  H3PO3(aq) + 3 HF(g)
a) When [PH3 ] is increased, at the new equilibrium the [HF] & [H3PO3] will be *↑ and [H2O] *↓
Ask: Where are there greater effective collisions?
Ask: What do those collisions make?
b) When [HF] is increased, at the new equilibrium, [H2O] & [PH3] will be *↑and the [H3PO3] *↓
c) If the [HF] were allowed to decrease, by escaping from the reaction vessel as it were made, then
if a new equilibrium were ever achieved, the [H2O] and [PH3] would *↓
249
For question 4, I am asking about the same issues … but I have blended into this set of questions, other
concepts we are required to keep in mind … .So, GO, GO GADGET THINKER !!!
4) Arsenic can be extracted from its ores by first reacting the ore with oxygen (called, roasting), to form
solid As4O6(s) which is then reduced using carbon to As40
(Zumdahl p. 635)
As4O6(s) + 6 C(s)  As4(g) + 6 CO(g)
Predict the direction of the shift of the equilibrium position in response to each of the following
changes in conditions:
a) Addition of CO will: * shift the point of equilibrium to the left … producing more As4O6 and
carbon, and reducing the concentration of As4 by the time the new equilibrium is reached.
b) Addition of tetraarsenic hexoxide (As4O6) will: * have no effect on the point of equilibrium,
since the amount of a pure solid, cannot affect the equilibrium expression via the Law of
Mass Action.
c) The removal of As4(g) will: * shift the point of equilibrium to the right, causing the
production of even more As4(g) and CO(g), while decreasing or consuming the amounts of the
tetraaarsenic hexoxide and carbon.
d) The removal of C(s) will: * not affect the equilibrium, since it is a solid. The amount of a pure
solid cannot affect the equilibrium expression. It is not included in the Law of Mass Action.
B) Pressure: An increase in pressure favors the reaction which

*produces the fewer number of moles.

*When the number of moles of reactants = the number of moles of
product there is no shift in equilibrium.

* Only systems with gases are affected by pressure.
250
1) There are 3 ways to change the pressure of a reaction system involving gaseous components:
a) *Add or Remove a gaseous reactant or product
b) *Add an inert gas (one which is not involved in the reaction)
c) *Change the volume of the reaction vessel.
2) The issue of adding or removing a gaseous reactant or product has been covered … it is
the effect of changing concentration ….
3) Adding an inert gas (such as Helium or Neon) has absolutely no impact on the point of
equilibrium … since these inert gases have no role in the reaction, they are not part of the
Law of Mass Action … and thus cannot affect K or the point of equilibrium.
4) Ahh!!! … but changing the volume …. that is interesting …. for changing the volume of a
container with gaseous materials, *essentially changes the molar concentrations of the
gaseous species ….
a) Recall: Molarity = mass
volume
INCREASING or decreasing the volume at constant
mass will affect molarity and the # of effective collisions
BUT MORE IMPORTANTLY….
b) Another way of thinking about this is to recognize that when the volume of a
container holding a gaseous system is reduced, the system responds by * reducing its
own volume…. by decreasing the total number of gaseous molecule in the system.
(Zumdahl p. 636)
i)
V = RT (n) and therefore at constant T and P …. V ∝ n
P
directly
proportional
ii) This could increase the rate of dimerization as in: 2 NO2  N2O4(g)
iii) This will also connect up with our work on ENTROPY!
5) Well what about the obverse (the opposite) of the rule for pressure …as it has been proposed?
An increase in pressure (or a reduction in volume) on a gaseous system at equilibrium
will favor that reaction which produces the fewer number of moles
So,
How will a decrease in pressure (or an increase in volume), at constant temperature,
affect a gaseous system at equilibrium?
You know the answer … but the reasoning (the conceptualization) is not so obvious.
251
Ans: A decrease in pressure (or an increase in volume) at constant temperature should
cause the equilibrium to shift in such a way as to favor the reaction which * produces
the greater number of moles.
But WHY?
Okay… Let’s look at the issue:
Given: A decrease in pressure results in the inverse result of increasing volume,
per PV = nRT.
Given: An increase in pressure (or a reduction in volume) on a gaseous system at
equilibrium will favor that reaction which produces the fewer number of moles
Thus, a DECREASE in pressure (or an increase in volume) will favor the reaction
which produces the greater number of moles. In other words, decreasing pressure,
or increasing volume have the same effect on a gaseous system at equilibrium.

In order to compensate for the decreasing pressure and the increasing volume of
the system, the new equilibrium state will favor conditions that INCREASE the
pressure in the system

Pressure (is measured) as species collide with the sides of the container in which
the gas is held. So, the issue deals with the fact, that we must get MORE
collisions at lower pressure (or greater volume). There is only one way to
accomplish that, in a closed system….

The equilibrium will shift so as to favor the reaction which produces MORE
species (as in the greater number of moles …larger “n”).

In theory, the greater of number of moles will account for an increase in collisions
with the wall of the container … thus increasing the pressure.
V = nRT
P
and assuming a closed system….Let us assume, pressure is cut in half (P/2)
thus volume doubles due to their inverse relationship (2V), according to
PV = nRT
2V = nRT The means by which to deal with the change in reducing P is to increase
P/2 n (or T) …but assuming a constant T, the only thing that can change
therefore, to deal with the stress of the changing volume (and thus
decreasing pressure, due to the inverse relationship between the two), is to
increase the number of species (or moles).
Thus, the reaction must shift (at reduced pressure and/or increased volume,
at constant T) so as to favor the reaction which makes the greater number of
moles!
252
1) Practice : Given :
2 NH3(g)  N2(g) + 3H2(g)
Think!
If pressure were increased, which reaction would be favored? *reverse
Analyze:
How many moles does the forward rxn make?
How many moles does the reverse rxn make?
What does that favored reaction make? *NH3
So, the [NH3] would *increase,
while the [N2] and [H2] would *decrease
The equilibrium would have "shifted to the *left when pressure is increased
2) Given : 2 NO2(g)  N2O4(g)
Think!
If pressure were increased, which reaction would be favored? *forward
Analyze:
How many moles does the forward rxn make?
How many moles does the reverse rxn make?
What does that favored reaction make? *N2O4
So, the [NO2] would * decrease,
while the [N2O4] would * increase
The equilibrium would have shifted to the * right under higher pressure
3) Given this system at equilibrium:
ClO(g) + O3(g)  ClO2(g) + O2(g)
Comment upon the effect an increase in pressure would have on the point of equilibrium,
and defend your reasoning.
Assuming the system is at equilibrium, an increase in pressure would * have NO EFFECT
upon shifting the equilibrium, since the number of moles are equivalent for the forward and
the reverse. All molarities will be increased proportionally, and it is recognized that volume
is proportional to the number of moles, but since there is no inequality in the number of
moles (molecules), there is no way for the overall # of molecules to decrease. Thus pressure
has no impact on the equilibrium in this case.
253
4) Given the system at equilibrium: 2 N2(g) + 3 O2(g)  2 N2O3(g)
Comment upon the effect an increase in pressure would have on the point of equilibrium,
and defend your reasoning.
An increase in pressure would shift the equilibrium * to the right, consuming more of the
nitrogen and oxygen (thus decreasing their concentrations) while producing more dinitrogen
trioxide and increasing its concentration… This happens because an increase in pressure
essentially lessens the volume, increasing the molarities and thus increases the number of
effective collisions and since volume is proportional to moles, the reaction responds by
lessening its volume by reducing the number of molecules … which is expressed
by making more of the larger compound N2O3(g) .
_____5) Which of the following systems at equilibrium will "shift to the left" when pressure is
increased ? (a little trick is included here … reflect on WHEN the effects of increased pressure matter)
1)
2 PaI5(s)  2 Pa(s)
2)
2 Al(s) + 3 I2(s) 
3)
2 KrF2(g)+2 H2O(g)  2 Kr(g) + O2(g)+ 4 HF(g)
4)
CaO(s) + SiO2(ℓ) 
+ 5 I2(s)
2 Al I3(s)
CaSiO3(ℓ)
6) Given the system at equilibrium: 2SO2(g) + O2(g)  2 SO3(g)
The equilibrium concentrations at a specific temperature are [SO2] = 0.100 M, [O2] = 0.0500 M
and [SO3] = 2.10 M. Explain the effect the injection of 1.50 moles of helium gas would have
on the point of equilibrium.
*There would be no effect, on the point of equilibrium. Total pressure of the system would be
increased, but the addition of an inert gas not involved in the original reaction has no impact on
the point of equilibrium.
254
C) Temperature: Assuming a system is at equilibrium, an INCREASE IN TEMPERATURE
affects BOTH the exothermic and endothermic reactions BUT the increase favors the
whichever reaction is the * ENDOTHERMIC
MORE.
It should be noted, that
while pressure and concentration can change the point of equilibrium, only a change in
temperature can change the value of the equilibrium constant …
1) Unlike pressure and concentration, a change in temperature disproportionately alters the
rates of the forward and reverse and this results in new equilibrium concentrations not
in the originally expressed ratio of the old equilibrium… Thus, a change in temperature
changes the actual value of Keq … because Keq is dependent upon the rate constants!
Remember … the rate constants k are temperature dependent …thus Keq is
temperature dependent. Essentially, Keq = koverall forward
koverall reverse
2) Le Chatelier’s Principle explains the stress of temperature… Increase temperature, and the
reaction with can “consume” the increase will be favored … Inevitably the reaction which
can “consume” the additional energy is the *endothermic reaction (which may be the forward
or the reverse)
3) When the forward reaction is exothermic … increasing the temperature deCREASES
the value of the equilibrium constant.
4) When the forward reaction is endothermic, increasing the temperature INcreases the
value of the equilibrium constant
First Let’s get the basics out of the way: Given these equilibria, identify the endothermic reaction as
being the forward or reverse.
N2(g) + O2(g) + 182.6 kJ  2 NO(g)
The * forward
rxn. is endothermic
2 H2(g) + O2(g)  2H2O(g) + 483.6 kJ
The * reverse
rxn. is endothermic
H2(g) + I2(g) + 53.0 kJ  2 HI(g)
The * forward
rxn. is endothermic
2CH3OH(l) + O2(g)  2CO2(g) + 4 H2O(l)+1.452 kJ The * reverse
N2(g) + 3H2(g)  2 NH3(g) + 91.8 kJ
The * reverse
rxn. is endothermic
rxn. is endothermic
255
1) Given at equilibrium
2 Al(s) + 3 I2(s)
 2 AlI3(s) + 628.7 kJ
Analysis:
If the temperature were increased which reaction would be favored?
Which reaction is that?
endo or exo
forward or reverse
What is (are) the product(s) of that favored reaction?
Conclusion:
So at the new point of equilibrium:
The [Al] will increase/ decrease / remain the same
The [I2] will increase / decrease / remain the same
The [AlI3] will increase / decrease / remain the same
2) Given the equilibrium:
265.8 kJ + 2 H2O2(l)  2 H2O(l) + O2(g)
Analysis:
If the temperature were increased which reaction would be favored?
Which reaction is that?
endo or exo
forward or reverse
What is (are) the product(s) of that favored reaction?
Conclusion: So at the new point of equilibrium:
The [ H2O2] will increase / decrease / remain the same
The [O2] will increase / decrease / remain the same
The [H2O] will increase / decrease / remain the same
3) Given at equilibrium:
N2(g) + 3H2(g)  2NH3(g) + 91.8 kJ
Analysis:
If the temperature were increased which reaction would be favored?
Which reaction is that?
endo or exo
forward or reverse
What is (are) the product(s) of that favored reaction?
Conclusion: So at the new point of equilibrium:
The [N2] will increase / decrease / remain the same
The [H2] will increase / decrease / remain the same
The [NH3] will increase / decrease / remain the same
256
4) Given at equilibrium :
2 O2(g) + N2(g) + 67.7 kJ  2 NO2(g)
Analysis:
If the temperature were decreased which reaction would be favored?
Which reaction is that?
endo or exo
forward or reverse
What is (are) the product(s) of that favored reaction?
So at the new point of equilibrium:
The [O2] will increase / decrease / remain the same
The [N2] will increase / decrease / remain the same
The [NO2] will increase / decrease / remain the same
D) Catalyst : The addition of a catalyst to a system at equilibrium increases the rates of both the
forward AND reverse reactions, thus * a catalyst has NO EFFECT upon the equilibrium of reaction
It will change the rate of both forward and reverse proportionately.
257
LECHATELIER’S PRINCIPLE SUMMARIZED
Structured Overview of the Effects of the 4 Stresses On Chemical Systems Already At Equilibrium
4 stresses
temperature
catalyst
concentration
pressure
affects endo more
Diddley/Squat
only gases
if  [R] then  [P]
 [R] then  [P]
 [P] then  [R]
The Rule of Berk (Dave) ('02)
Whatever you do to a reactant,
you ultimately do to the product
by the time the new equilibrium
is reached (and vice versa)
an  in press. favors the rxn
which produces the fewer
# of moles
analysis :
determine the endo.
The opposites should be evident ... a decrease in
pressure favors the reaction producing the greater
number of moles and the decrease in temperature
favors the exothermic reaction....
analysis:
Forward rxn produces _____ mol
Reverse rxn produces _____ mol
*TN Consider adding to the worksheets: from CAMS Chemistry Core... reading on p 114 Chlorinating pools and question #2 .... reading 10.5 is not bad
258
PRACTICE 1: LE CHATELIER (JUST THE BASICS)
DIRECTIONS: The following problems deal with issues regarding Le Chatelier's Principle. Complete each as
you have been taught. Think about the number of collisions and what those collisions produce.
___ 1) Given the equilibrium :
N2(g) + 3 H2(g)  2 NH3(g) + 92 kJ
If [H2] were increased, when the reaction reaches its new equilibrium, the [NH3] will have
1) increased
___ 2) Given the equilibrium :
2) decreased
3) remained the same
N2(g) + 3 H2(g) 2 NH3(g) + 92 kJ
If [NH3] were increased, when the reaction reaches its new equilibrium, the [N2] will have
1) increased
___ 3) Given the equilibrium :
2) decreased
3) remained the same
N2(g) + 3 H2(g)  2 NH3(g) + 92 kJ
If the temperature were increased, when the reaction reaches its new equilibrium, the [NH3] will have
1) increased
___ 4) Given the equilibrium :
2) decreased
3) remained the same
N2(g) + 3 H2(g) 2 NH3(g) + 92 kJ
If the pressure were increased, when the reaction reaches its new equilibrium, the [NH3 ] will have
1) increased
2) decreased
3) remained the same
3 O2(g)  2 O3(g)
___5) Given the equilibrium :
Were a catalyst added to the reaction mixture, when the new equilibrium was reached, the [O3 ] would
have
1) increased
2) decreased
3) remained the same
___ 6) Given the equilibrium :
N2(g) + 3 H2(g) 
2 NH3(g) + 92 kJ
When the [NH3(g)] is removed as it is produced, the forward reaction will _________ and
[N2] and [H2] will ________________
1) continue & increase
2) continue & decrease
3) stop & remain the same
259
___ 7) Given the equilibrium:
36 kJ + I2(s) + Cl2(g) 2 ICl (g)
If the temperature of the system were increased, when the new equilibrium is reached the [ICl ] should:
have
1) increased
2) decreased
3) remained the same
____ 8) Given the equilibrium :
36 kJ + I2(s) + Cl2(g)  2 ICl (g)
If [I2] were increased, when the reaction reaches its new equilibrium, the [ICl ] will have
1) increased
2) decreased
3) remained the same
36 kJ + I2(s) + Cl2(g)  2 ICl (g)
___9) Given the equilibrium :
If [ICl] were increased, when the reaction reaches its new equilibrium, the [Cl2 ] will have
1) increased
2) decreased
3) remained the same
___ 10) Given the equilibrium :
36 kJ + I2(s) + Cl2(g)  2 ICl (g)
If a catalyst were added, when the reaction reaches its new equilibrium, the [ICl] would have:
1) increased
2) decreased
___ 11) Given the equilibrium :
3) remained the same
36 kJ + I2(s) + Cl2(g)
 2 ICl (g)
If temperature were decreased, when the reaction reaches its new equilibrium, the [ICl ] will have
1) increased
___ 12) Given the reaction :
2) decreased
36 kJ + I2(s) + Cl2(g)
3) remained the same
 2 ICl (g)
As the ICl is condensed and removed from the reaction vessel, as it is produced:
1) the reaction will stop
2) no new equilibrium will be reached
3) the reverse reaction will begin
260
___ 13) Given the equilibrium :
2 SO2(g) + O2(g)  2 SO3(g) + 188 kJ
If the [SO2] were increased, the [SO3 ] , once the new equilibrium were reached, would have :
1) increased
____ 14) Given the equilibrium :
2) decreased
3) remained the same
2 SO2(g) + O2(g)  2 SO3(g) + 188 kJ
If the temperature on the system is increased, the [O2] should be __________ when the new
equilibrium is reached.
1) greater
___ 15) Given the equilibrium :
2) less
3) the same
2 SO2(g) + O2(g)  2 SO3(g) + 188 kJ
If the temperature on the system is increased, the [SO3] should be __________ when the new
equilibrium is reached.
1) greater
___ 16) Given the equilibrium :
2) less
3) the same
2 SO2(g) + O2(g)  2 SO3(g) + 188 kJ
If the [SO3] were increased, the [O2 ] , once the new equilibrium were reached, would have :
1) increased
___ 17) Given the equilibrium :
2) decreased
3) remained the same
2 SO2(g) + O2(g)  2 SO3(g) + 188 kJ
If the pressure on the system were increased, the [O2] would be __________ when the new equilibrium
was reached.
1) greater
___ 18) Given the equilibrium :
2) less
3) the same
2 SO2(g) + O2(g)  2 SO3(g) + 188 kJ
If a catalyst is added, the [O2] would be __________ when the new equilibrium was reached.
1) greater
___ 19) Given the equilibrium :
2) less
3) the same
2 SO2(g) + O2(g)  2 SO3(g) + 188 kJ
If the [O2] were increased, the [SO2] , (not SO3 ) once the new equilibrium were reached, would have
to be :
1) increased
2) decreased
3) remained the same
261
___ 20) Given the equilibrium :
2 SO2(g) + O2(g)  2 SO3(g) + 188 kJ
If SO3, were removed as fast as it was produced, the [O2] would
1) increased
2) decreased
3) remained the same
___ 21) Given the system at equilibrium : 2 CO(g) + O2(g)  2 CO2(g) + 393 kJ
If the temperature of the system were increased, at the new equilibrium, the [CO2] should have
1) increased
2) decreased
3) remained the same
___ 22) Given the system at equilibrium: 1,822 kJ + 2 HgO(s)  2 Hg(l) + O2(g)
If the temperature were increased, at the new equilibrium, the concentration of Hg would
1) increase
2) decrease
3) remained the same
___ 23) Given the system at equilibrium: 257 kJ + SO2(g) + NO2(g)  SO3(g) + NO(g)
Which of the following will shift the equilibrium to the right ?
1) increase the pressure
2) decrease the pressure
___ 24) Given this system at equilibrium :
3) decrease [SO2]
4) increase the temperature
N2(g)
+ 3 H2(g)  2 NH3(g) + 91.8 kJ
Which of the following would increase the concentration of ammonia ?
1) decrease [N2(g)]
2) increase the temperature
3) increase the pressure
4) decrease [H2(g)]
___ 25) Given this system at equilibrium: H2(g) + Cl2(g)  2 HCl (g) + 770 kJ
Which of the following will favor the forward reaction ?
1) increase the pressure
2) increase [H2(g)]
3) add a catalyst
4) increase the temperature
___ 26) Given the system at equilibrium : 2 CO(g) + O2(g)
An increase in pressure would
1) change the numerical value of the Keq
2) have no effect upon the system

2 CO2(g) + 383 kJ
3) increase the concentration of CO2
4) increase the concentration of O2
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___ 27) Given the reaction at equilibrium:
72 kJ + 2 Ag2O(s)  4 Ag(s) + O2(g)
Which of the following will favor the production of silver ?
1) add more oxygen
2) add catalyst
3) increase the temperature
4) decrease the temperature
___28) Given the reaction at equilibrium:
O2(g) + F2(g)  2 OF(g) + 23 kJ
Which of the following will favor the forward reaction ?
1) increase the temperature
2) increase pressure
3) increase [OF]
4) increase [F2]
For questions 29 – 33, use the following 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
___29) The reaction below is exothermic: 2SO2(g) + O2(g)  2SO3(g)
Le Chatelier's Principle predicts that __________ will result in an increase in the number of moles of
SO3 in the reaction container.
I) increase pressure
II) decrease temperature
III) decrease [O2]
___30) For the endothermic reaction CaCO3(s) CaO(s) + CO2(g), Le Chatelier's principle predicts that
__________ will result in an increase in the number of moles of CO2.
I) increase pressure
II) increase temperature
III) decrease volume
___31) Consider the following reaction at equilibrium: 2NH3(g)  N2(g) + 3H2(g)
Le Chatelier's principle
predicts that the moles of H2 in the reaction container will increase with __________.
I) an decrease in the total pressure (T constant)
II) addition of some N2 to the reaction vessel (V and T constant)
III) some removal of NH3 from the reaction vessel (V and T constant)
___32) Consider the following reaction at equilibrium: C(s) + H2O(g)  CO(g) + H2(g)
Which of the following conditions will increase the partial pressure of CO?
I) increase the volume
II) decrease partial pressure of H2O(g)
III) add more solid carbon
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Your choices for question 33 has been repeated for your convenience.
1)
2)
3)
4)
5)
when only I is correct
when only II is correct
when only I and II are correct
when only II and III are correct
when I, II, and III are each correct
___33) Consider the following reaction at equilibrium.
2CO2(g)  2CO(g) + O2(g) +514 kJ
Le Chatelier's principle predicts that the equilibrium partial pressure of CO (g) can be maximized by
carrying out the reaction __________.
I) by adding helium
II) low pressure
III) low temperature
For 34 – 36 only 1 choice per question is correct.
___34) The effect of a catalyst on equilibrium is to __________.
1) increase the rate of the forward reaction only
2) increase the equilibrium constant so that products are favored
3) increase the rate at which equilibrium is achieved without changing the composition of the
equilibrium mixture
4) shift the equilibrium to the right
___35) In which of the following systems would the number of moles of the substances present at equilibrium
NOT be shifted by a change in the volume of the system at constant temperature?
1) CO(g) + NO(g)  CO2(g) + 1/2 N2(g)
2) NO(g) + O3(g)  NO2(g) + O2(g)
3) N2O4(g)  2 NO2(g)
4) N2(g) + 2 O2(g)  2 NO2(g)
___36) CuO(s) + H2(g)  Cu(s) + H2O(g) + 2.0 kJ
When the substances in the equation above are at equilibrium at pressure P and temperature T, the
equilibrium can be shifted to favor the products by
1) increasing the pressure by means of a moving piston at constant T
2) decreasing the temperature
3) allowing some water vapor to escape at constant P and T
4) adding a catalyst
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Now for something a bit more advanced ….from your friends at the ACS and the College Board:
37) When a sample of NO2 is placed in a container, an equilibrium is rapidly established.
2 NO2(g) ↔ N2O4(g)
The equilibrium mixture is a darker color at high temperatures or at low pressures. Which statement
about the (forward) reaction is true?
1) The reaction is exothermic and NO2 is darker in color than N2O4
2) The reaction is exothermic and N2O4 is darker in color than NO2
3) The reaction is endothermic and NO2 is darker in color than N2O4
4) The reaction is endothermic and N2O4 is darker in color than NO2
38) Consider this reaction at equilibrium:
2 SO2(g) + O2(g) ↔ 2 SO3(g)
ΔH = -198 kJ
Which of these changes would cause an increase in the SO3/SO2 mole ratio?
1)
2)
3)
4)
adding a catalyst
decreasing the temperature
removing O2(g)
decreasing the pressure
39) The solubility of solid silver chromate (Ag2CrO4), which has a Ksp equal to 9.0 x 10-12 at 25°C, is
determined in water and in two different aqueous solutions.
The list of solutions used is: I) pure water
II) 0.1 M AgNO3
III) 0.1 M Na2CrO4
Predict the relative solubility of Ag2CrO4 in the three solutions.
1)
2)
3)
4)
I = II = III
I < II < III
II = III < I
II < III < I
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40) Magnesium oxalate (Mg C2O4) has a Ksp= 8.6 x 10-5 at 25°C.
What will be the result when 100 mL of 0.06 M Mg(NO3)2(aq) is added to 50 mL of 0.06 M Na2C2O4?
Assume the reaction takes place at 25°C.
1)
2)
3)
4)
No precipitate will form, as all products are soluble
A precipitate will form and an excess of Mg2+ ions will remain in solution
A precipitate will form and an excess of C2O42- ions will remain in solution
A precipitate will form but neither ion is present in excess.
41) Solid carbon and gaseous carbon dioxide are placed in a closed vessel and allowed to come to equilibrium
at high temperature according to the reaction equation:
C(s) + CO2(g)  2 CO(g)
When the pressure on the system is increased and the temperature kept constant, what will be the result?
1)
2)
3)
4)
42)
The amount of CO will increase and that of C and CO2 will decrease.
The amount of CO will decrease and that of C and CO2 will increase.
The amount of each of the three substances will increase.
The amount of each of the three substances will decrease.
(Rather Tricky for us, at this point ….but I thought it was worth a try…Try the hints …We’ll see this again….)
A saturated solution of MgF2 contains 1.6 x 10-3 mol/L of dissolved MgF2 at a certain temperature.
What is the Ksp of MgF2 at this temperature?
-6
1) 2.7 x10
2) 3.1 x 10-9
-8
3) 1.6 x 10
4) 6.2 x 10-9
* MgF29(s) + H2O(l)  Mg2+(aq) + 2F1-(aq)
* Ksp = [Mg2+][2F-]2 thus: [1.6 x 10-3][3.2 x10-3]2
Hint 1:* Write out the equation
representing the dissolving of
magnesium fluoride in water ….
Hint 2:* Notice, that twice the
number of mole/L of fluoride ion are
produced relative to magnesium ions.
That is you get:
MgF2 + H2O(l)  Mg2+ + 2F1Hint 3:*Write the Law of Mass
Action ….noting that the 2F-1 also
becomes squared. Don’t include any
solids or liquids, just the magnesium
ion (as “x”) and fluoride as (2x)2
…see where I’m going with this???
ans: 3
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43) The addition of solid Na2SO4 to an aqueous solution in equilibrium with solid BaSO4 (Ksp = 1.5 x 10-9 at 25 °C)
will cause
1)
2)
3)
4)
No change in [Ba2+] in solution
more BaSO4 to dissolve
the precipitation of more BaSO4
an increase in the value of the Ksp of BaSO4
44) Consider the reaction:
AB3(g)  A(g) + 3B(g)
What is the equilibrium constant expression when the initial concentration of AB3 is 0.1 M and the
equilibrium concentration of A is represented by “x”? You can assume the initial concentrations of
A and B are both zero.
1) x – 3x
0.1 – x
2) x – x3
(0.1 – x)2
3) (x)(x3)
(0.1 – 3x)3
4) (x)(3x)3
0.1 – x
ANSWERS:
1) 1 increase a [reactant] and you increase [product]
2) 1 increase a [product] and you increase the [reactant]
3) 2 increase in temp. favors whichever rxn (forward or rev) which is ENDOthermic. In this case, the reverse reaction is the
endothermic, thus the rev. is the favored reaction. This means that ammonia will decompose. Its [ ] will decrease and the [ ] of
the reactants will increase as a result
4) 1 an increase in pressure favors the reaction which produces the fewer number of moles. Since the forward reaction produces 2
moles, and the reverse, 4 moles... the forward reaction must be the favored reaction. Thus, more ammonia is produced.
5) 3 a catalyst has NO affect on the equilibrium. It changes the rates of both reactions equally, thus a null affect is the result.
6) 2 : as you remove product you favor the forward rxn. (This uses up the reactants). Think of it as Nature trying to fill in the hole
you are creating. As you remove product, you don’t allow the reverse rxn to begin. Thus Nature keeps pumping out product–
thus the reactant concentration must decrease ultimately.
7) 1 An increase in temperature favors the ENDOthermic reaction more. The forward reaction, in this case
is the endothermic. Thus, the forward is favored. The forward makes ICl, thus its concentration increases
8) 1
9) 1
10) 3 a catalyst does not affect the equilibrium. The concentrations of each substance will remain the same.
11) 2 Since the forward is the endothermic reaction, a decrease in temperature will slow it down more. ICl
concentrations will therefore drop.
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12) 2 When ICl is removed as it is made, Nature keeps making more. Thus forward reaction will continue. An equilibrium will never
be reached. The reaction CAN NOT REVERSE - because there are no molecules of product to collide with each other, if you
keep removing the product. This is a common technique to prevent a chemical equilibrium.
13) 1
14) 1 An increase in temperature favors the ENDOthermic rxn. The endothermic reaction in this case is the reverse. The reverse rxn
is favored. The reverse reaction makes oxygen.
15) 2 same reasoning as 13). The SO3 is decomposed because the reverse rxn is favored.
16) 1 If you increase the product then you increase the number of collisions between molecules of product. This increase the rate of
the reverse reaction, which in turn makes oxygen.
17) 2 An increase in pressure favors the reaction which PRODUCES the FEWEST number of moles. The forward in this case,
produces only 2 moles, while the reverse produces 3. Ergo, the forward will be favored and the concentration of oxygen will
decrease because it gets used up.
18) 3 A catalyst does not change a system at equilibrium. Both reactions are increased. Thus, as the rate of synthesis increases, the
rate of decomposition increases concomitantly. There is no net change.
19) 2 the forward reaction is favored and SO2 is consumed
20)
21)
22)
23)
2 If the product were removed, no reverse reaction would be initiated. Ergo, it would favor the forward.
2
1
4 Tough one. Choices 1) and 2) play no role since the number of moles in this gaseous system are
equal for both reactions. Increasing the temp. favors the forward and this favors the RIGHT.
24) 3 An increase in pressure will favor the rxn which makes the fewer number of moles, in a gaseous system.
25) 2
26) 3
27) 3 The forward reaction makes Ag. It is endothermic. So, increasing the temperature will help drive the
reaction to the right to make silver.
28) 4 An increase in reactant will help to shift the equilibrium to the right (Favor the forward)
29) 3
30) 2 the forward (making CO2) is endothermic … and an increase in temperature, favors the endothermic … NOTE an increase
in pressure or a decrease in volume (which are essentially the same thing) would result in favoring the reverse reaction, for
only the CO2 would be affected … and it would incur an increase in the number of effective collisions with other CO 2
molecules and cause its decomposition.
31) 4 decreasing the pressure is the inverse of increasing the pressure … So, instead of favoring the reverse reaction, a decrease
would favor the forward, which MAKES H2
32) 1 This is an important question… INcreasing the volume is equivalent to DEcreasing pressure (for gases) … and the moles of
gas between reactants to products is 1:2 (the carbon is a solid and its concentration is not accounted for) … Thus an increase
in volume (or decrease in pressure) shifts the reaction in favor of the greater moles (the reverse of the Le Chatelier rule) and
the partial pressure of CO will increase, by the new point of equilibrium…. Adding more C does not affect the equilibrium.
33) 4 Adding an inert (non-reactive gas) such as helium does not affect the equilibrium.
34) 3
35) 2
36) 3
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37) 1 You are not given a ∆H value (an enthalpy value), so it may seem to be impossible to determine exothermic from endothermic
….so let’s solve the pressure issue first. You will note the balanced reaction equation is your friend, and at low pressure, the
reaction should shift so as to favor the reaction producing the greater number of moles (the obverse of the “normal”
Le Chatelier rule about pressure)
Given the above statement, then NO2(g) must be the darker colored gas … this eliminates choice B and D.
Now, you can infer the thermic nature of the reaction. The equilibrium mixture is darker at high temperatures … thus the
production of NO2 is endothermic - meaning that the forward reaction (about which you are asked) is exothermic. (Phew!)
38) 2 Increasing the SO3/SO2 ratio is just a fancy way of saying ….. “produce more SO 3” … Thus your task is to determine which
change would shift the point of equilibrium to the right and favor the production of more product. Since we are given a
-∆H value we may infer the forward reaction is exothermic.
The endothermic reaction is favored at high temperature, thus
the exothermic would be favored at a lower temperature.
39) 4 This is a corker! It depend upon your grasp of the common ion effect and your understanding that in silver (I) chromate there
is a 2:1 ratio of silver ion to chromate ion.
First the salt is most soluble in pure water, since there is absolutely no common ion
Secondly, the salt will be more soluble in a solution of Na2CrO4 (but less than water) because while there is a common ion
to interfere with the dissolution (the chromate ion), compared to the interference of the silver ion from AgNO3 … You see
the silver chromate provides only “1” chromate to the solution, but it provides 2 silver ions …. Thus the larger concentration
of silver ion in the solution with AgNO3 …would inhibit dissolution of the silver (I) chromate more …making it even less
soluble than in sodium chromate solution.
40) 2 The key here is to recognize that there is twice the volume of magnesium nitrate, as there is sodium oxalate. Thus the oxalate
ion will be consumed completely and half the amount of the magnesium ion would be consumed, leaving magnesium ions
in solution.
41) 2 An increase in pressure will affect the point of equilibrium, so that the pressure stress will be consumed. This means that
the reaction will shift to favor the reaction which produces the fewer number of (gaseous) moles …. Your analysis should
point out that the forward reaction produces 2 moles of gas, while the revers produces only 1 mole of gas. Thus, at the new
equilibrium, the amount of CO will have decreased, and the amount of the carbon and carbon dioxide will have been increased.
42) 3
43) 3 This is the common ion effect in all of its glory … The common ion is the sulfate polyatomic ion (SO 4)244) 4 This can be a bit tricky, and I wanted to give this to you for that reason … You must grasp that the concentration of A = x M
thus the concentration of B (whatever that is) must be (3 times x) … and that concentration is cubed, based upon the
balanced reaction equation. Since AB3 is a gas, and this does go to equilibrium, the [AB 3] changes by its original
concentration minus whatever amount dissociated.
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Push you grasp of it all and integrate Questions 37 – 40 really require a bit’o thought. You may need to pull
from your grasp of kinetics and/or gas laws to complete them.
45) When 2.00 mol of SO2Cl2(g) is placed in a 2.00 L flask at 303 K, 56% of the compound decomposes to
sulfur dioxide gas and chlorine gas (dichlorine).
a) Write the balanced chemical reaction for the decomposition: * SO2Cl(g)
 SO2(g) + Cl2(g)
b) Calculate the Kc for this reaction at this temperature.
Kc = [SO2][Cl2]
[SO2Cl2]
[SO2Cl2] = 2.00 mol = 1.00 M and with 56% consumption, the
2.00 L
change in concentration = 0.56 M
[SO2Cl2] [SO2]
[Cl2]
Initial
1.00
0
0
Change
-0.56
+0.56
0.56
Equilibrium 0.44 M 0.56 M 0.56 M
c) Calculate the Kp for this reaction at 303 K * Kp = Kc(RT)∆n = 0.71(0.08206∙303)1 = 18
d) According to Le Chatelier’s principle, would the percent of SO2Cl2 that decomposes increase,
decrease or stay the same if the mixture were transferred to a 15.00 L.
* The transfer increases volume, which decreases pressure. And a decrease in pressure would
shift the reaction in favor of the products. Thus, a greater percentage of the SO2Cl2 will
decompose.
___46) If Kc = 1 for the equilibrium 2A(g)  B(g) , which describes a relationship between [A] and [B] at
equilibrium? (hint …write the Kc)
1) 2[B] = [A]
2) [B] = [A]2
3) [B] = 2[A]
*ans 2) Based upon: Kc = [B] = 1
[A]2
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47)
Given: CO(g) + Cl2(g) COCl(g) + Cl(g)
Recall that the value of K is temperature dependent, because the rate constants k are temperature dependent
In the above reaction, both the forward and reverse reactions are elementary steps.
At 25°C, the rate constants for the forward and reverse reactions are 1.4 x 10-28 M-1s-1 and 9.3 x 1010 M-1s-1,
respectively.
a) Write the general rate law for the forward reaction * rate = k [CO][Cl2] it is an elementary step,
thus the rate law can be written directly from the chemical equation.
Write the general rate law for the reverse reaction * rate = k [COCl][Cl] it is an elementary step,
thus the rate law can be written directly from the chemical equation.
b) What is the value for the equilibrium constant at 25°C? *K= kforward = 1.4 x 10-28 M-1s-1 = 1.5 x 10-39
kreverse 9.3 x 1010 M-1s-1
c) Are reactants or products more plentiful at equilibrium? Defend your reasoning.
*K is quite small, thus the reactants are much more plentiful that the products at equilibrium,
because the magnitude of K informs us of the success of the forward reaction.
d) Based upon the rate constants, what is the overall order of reaction for the forward reaction?
*Both the forward and reverse are second order, overall, based upon the units of the
rate constants.
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From the 2014 AP Exam Place your answers at the bottom of the page, and on the back. We can look at
graded answers from the good folks at AP Central, later.
48
272
273
274
49) The equilibrium constant Kc for the gas-phase thermal decomposition of cyclopropane to propene equals
1.0 x 105 at 500 K
Hint a1) * Recall that there is a relationship
between pressure and concentration of
gases, because according to PV = nRT,
pressure is directly proportional to moles …
and you were given an equation that relates
the two … What is that equation????
Hint a2) *use: Kp = Kc(RT)∆n
a) What is the value of Kp at 500K?
*Kp = Kc(RT)Δn since Δn = 0, then RT is 1
hence Kp = Kc …Kp = 1.0 x 105
ans: 1.0 x 105
b) What is the equilibrium partial pressure of cyclopropane
at 500K, when the partial pressure of propene is 5.0 atm?
* Kp = (Ppropene)
thus: 1.0 x 105 = (5.0 atm)
(Pcyclopropane)
(x)
Hint b1)* Recall that the expression for Kp
follows that of the Law of Mass Action.
Produts/Reactants
Hint c1: * This is a question about
concentration as a stress… reflect upon Le
Chatelier’s Principle.
Hint d1: This deals with (decreasing)
pressure or increasing volume as a stress.
Recall Le Chatelier’s Principle, and recall
that equilibrium is all about the
stoichiometry of the balanced reaction
equation.
Hint e1: Kc is dependent upon the ratio of
rate constants …so consider the value of Kc
and that ratio….
ans: 5.0 x 10-5 atm
c) Can the equilibrium be shifted, by adding cyclopropane, at constant temperature and pressure?
Explain your thinking (Be sure to have an argument, proof and/or setting)
* Yes, the point of equilibrium can be shifted (argument). Increasing the concentration of
cyclopropane, will cause more (effective) collisions between cyclopropane molecules (relative to the
number of effective collisions between propene molecules) (setting). According to Le Chatelier’s
principle such a stress, resulting in more effective collisions must shift the equilibrium in the direction
forming the product of those collisions. Hence, more propene will be produced, shifting the point
(position) of equilibrium to the right (proof …by quoting a rule/principle as opposed to data).
Keep going for more ….
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d) Can you alter the ratio of the two concentrations, at the point of equilibrium, by adding more
cyclopropane? (Be sure to have an argument, proof and/or setting)
*No, the ratio between the species cannot be altered by adding more cyclopropane (argument). The
ratio is really the value of Kc (setting). The value of Kc is temperature dependent, not concentration
dependent (proof). At the new point (position) of the equilibrium, the ratio (Kc value) will be the same.
e) Can the position of the equilibrium be shifted by increasing the volume of the container, at constant
temperature? Explain your thinking (Be sure to have an argument, proof and/or setting)
* No the position cannot be shifted (argument). Changes in volume have the same effect on the point of
equilibrium, as changes in pressure have, on gaseous systems. And while this is a gaseous system at
equilibrium, according to Le Chatelier’s Principle, such changes can only affect the position of
equilibrium, when there is a difference in the number of moles of gas-phase reactant and gas-phase
product (setting). In this case, based upon the balanced reaction equation, the number of moles of
cyclopropane are equal to the number of moles of propene (proof, citing the balanced equation). Hence
no change in the position of equilibrium will occur.
f) Which has the larger rate constant, the forward reaction or the reverse reaction? Explain your
thinking (Be sure to have an argument, proof and/or setting)
* The forward reaction has the larger rate constant (argument). The temperature dependent Kc = kforward
kreverse
Since the value of Kc = 1.0 x 105, the value of kforward must be larger than that of kreverse (proof and
setting). The magnitude of Kc indicates that the forward reaction proceeds far to the right, before an
equilibrium is established. Since k indicates the fraction of reactant molecules travelling the specific
reaction pathway per time, there must be a greater fraction of reactant molecules (cyclopropame)
experiencing effective collisions (breaking the activation barrier) than propene molecules, in order to
have such a large Kc (proof and setting …again!!!)
276