Download K eq

Chemistry II
Chapter 14
Chemical Equilibrium
Speed of a chemical reaction is determined by kinetics
 How fast a reaction goes
Extent of a chemical reaction is determined by
thermodynamics
 How far a reaction goes
Reaction Dynamics
• forward reaction:
reactants  products
 therefore the [reactant] decreases and the [product] increases
 as [reactant] decreases, the forward reaction rate decreases
• reverse reaction:
products  reactants
 assuming the products are not allowed to escape
 as [product] increases, the reverse reaction rate increases
• processes that proceed in both the forward and reverse direction are said to be
reversible
 reactants ⇌ products
Reaction Dynamics
Rate
Initially,
only
the
Because
As
Once
the equilibrium
forward
the
reactant
reaction
isforward
established,
concentration
proceeds
Eventually,
the
reaction
proceeds
reaction
takes
place.
itin
decreases,
the
makes
forward
products
the
and
forward
reverse
and uses
reaction
reactions
reactants.
the
reverse
direction
as
fast
asslows.
As
the products
at in
thethe
same
accumulate,
rate,direction.
sothe
the
itproceed
proceeds
forward
reverse
concentrations
reaction
ofspeeds
all materials
At
this time
equilibrium
isup.
established.
stay constant.
Rate Forward
Rate Reverse
Time
Concentration 
H2(g) + I2(g) ⇌ 2 HI(g)
Since the
Asreactant
the reaction
proceeds,
the [H2] and
concentrations
[I2]are
decrease
and the
decreasing,
[HI]
increases
the
forward
reaction
And
sincerate
the
slows down
product
concentration is
increasing, the
reverse reaction
rate speeds up
Since the [HI] at
equilibrium is larger than
the [H2] or [I2], we say the
position of equilibrium
favors
products
At
Once
equilibrium,
equilibrium
the is
forward
established,
reaction
the rate is
the
concentrations
same as theno
reverse
longer
reaction
change rate
Time 
Equilibrium Established
Dynamic Equilibrium
• some reactions reach equilibrium only after almost all the
reactant molecules are consumed –
equilibrium favors the products
• other reactions reach equilibrium when only a small percentage
of the reactant molecules are consumed –
equilibrium favors the reactants
An Analogy:
Pwad(left) ⇌ Pwad(right)
Rules:
1. Each student wads up two paper wads.
2. You must start and stop as the timekeeper says.
3. Throw only one paper wad at a time.
4. If a paper wad lands next to you, you must throw it back.
# of paper wads on right
K eq 
# of paper wads on left
Equal Number of Students on Each Side of
the Classroom
Time
(s)
0
Left
Pwad (left)
Right
K eq 
All
0
Pwad (right)
# of paper wads on right
# of paper wads on left
70
5
10
15
Number of Paper Wads
60
50
40
30
20
10
0
0
20
5
10
Time (s)
15
20
Most Students on the Left Side– 2 Students
on the Right Side
Time
(s)
0
Left
Pwad (left) ⇌ Pwad (right)
Right
K eq 
0
All
# of paper wads on right
# of paper wads on left
70
5
10
15
Number of Paper Wads
60
50
40
30
20
10
0
0
20
5
10
Time (s)
15
20
Most Students on the Left Side– 2 Students
on the Right Side
Time
(s)
0
Left
Pwad (left) ⇌ Pwad (right)
Right
K eq 
All
0
# of paper wads on right
# of paper wads on left
70
5
10
15
Number of Paper Wads
60
50
40
30
20
10
0
0
5
10
Time (s)
20
15
20
Common Misconceptions
• Equilibrium means equal amounts of reactant and product.
 No - A reaction can be at equilibrium and have more paper wads
on the product side of the room
• A reaction at equilibrium has stopped.
 No - The paper wads keep flying in both directions even after
equilibrium is achieved
• Equilibrium can only be achieved by starting with reactants.
 No - Equilibrium can also be achieved when starting with all of the
paper wads on the product side of the room.
The Equilibrium Constant
• [reactants] and [products] are not equal at equilibrium, there is a
relationship between them
Law of Mass Action:
aA + bB ⇌ cC + dD,
Keq (or Kc) is called the equilibrium constant
 Units vary from reaction to reaction, unitless in this case

C  D
K eq 
a
b
A  B
c
d
The Equilibrium Constant
so for the reaction:
2 N2O5 ⇌ 4 NO2 + O2

NO2   O 2 
K eq 
2
N 2O5 
4
Units: mol3/L3
The Equilibrium Constant
Significance of Keq
• Keq >> 1, when the reaction reaches equilibrium there will be more
product than reactant molecules
 the position of equilibrium favors products
• Keq << 1, when the reaction reaches equilibrium there will be more
reactant molecules than product molecules
 the position of equilibrium favors reactants
A Large Equilibrium Constant
Remember:
does not tell us about
how fast, only how far
A Small Equilibrium Constant
Relationships between K
and Chemical Equations
• reaction is written backwards, the equilibrium constant is inverted
for the reaction aA + bB ⇌ cC + dD
the equilibrium constant expression is:
for the reaction cC + dD ⇌ aA + bB
the equilibrium constant expression is:

C  D

a
b
A  B

A   B
K backward 
c
d
C  D
c
K forward
d
K backward 
a
1
Kforward
b
Relationships between K
and Chemical Equations
• coefficients of an equation are multiplied by a factor,
K is raised to that factor
for the reaction aA + bB ⇌ cC
equilibrium constant expression is:

C
K original 
Aa  Bb
c
Knew  K
for the reaction 2aA + 2bB ⇌ 2cC
equilibrium constant expression is:
2c

C
K new 
A2 a  B2 b
2
c



C

 
a
b 
 A   B 
n
original
Relationships between K
and Chemical Equations
• when you add equations to get a new equation, Keq for the new equation is the
product of the equilibrium constants of the old equations
for the reactions (1) aA ⇌ bB and
(2) bB ⇌ cC the K expressions are:

B
K1 
Aa
b
for the overall reaction aA ⇌ cC
the K expression is:

C
K2 
b
B
c
Knew  K1  K2
c

C
K new 
a
A
b
c


B
C


a
b
A B
Ex 14.2 – Compute the equilibrium constant at 25°C
for the reaction NH3(g) ⇌ 0.5 N2(g) + 1.5 H2(g)
Given: for N2(g) + 3 H2(g) ⇌ 2 NH3(g), Keq = 3.7 x 108 at 25°C
Find: Keq for NH3(g) ⇌ 0.5N2(g) + 1.5H2(g), at 25°C
Concept Plan:
K
K’
Relationships: Kbackward = 1/Kforward, Knew = Koldn
Solution:
N2(g) + 3 H2(g)  2 NH3(g)
2 NH3(g) ⇌ N2(g) + 3 H2(g)
NH3(g) ⇌ 0.5 N2(g) + 1.5 H2(g)
K1 = 3.7 x 108
1
1
K2 

K1 3.7 108
K '  K 2 
1
2
1




8
 3.7 10 
K '  5.2 10 5
1
2
Equilibrium Constant in Terms of Pressure
Reactions Involving Gases
• the concentration of a gas in a mixture is proportional to its
partial pressure
aA(g) + bB(g) ⇌ cC(g) + dD(g)

C  D
Kc 
a
b
A  B
c
d
PC  PD
Kp  a
b
PA  PB
c
or
d
Kc and Kp
• when calculating Kp, partial pressures are always in atm
• the values of Kp and Kc are not necessarily the same
 because of the difference in units
• the relationship between them is:
Kp  Kc  RT 
n
Δn = c + d – (a + b) or no. moles of
reactants minus no. Moles products
When n = 0, Kp = Kc
Ex 14.3 – Find Kc for the reaction 2 NO(g) + O2(g) ⇌ 2
NO2(g), given Kp = 2.2 x 1012 @ 25°C
Given: Kp = 2.2 x 1012 atm-1
Find: Kc
Concept Plan:
Relationships:
Kp
KP
Kc 
RT n
K p  Kc RT 
n
Solution:
2 NO(g) + O2(g) ⇌ 2 NO2(g)
n = 2  3 = -1
Check:
Kc
Kc 

Kp
RT n
2.2  1012
0.08206
atm L
mol  K
 298 K

1
 5.4  1013
K has units L mol-1
since there are more moles of reactant than product, Kc should be
larger than Kp, and it is
Heterogeneous Equilibria: Reactions
Involving Solids and Liquids
aA(s) + bB(aq) ⇌ cC(l) + dD(aq)

D
Kc 
b
B
d
• Concentrations of solids and liquids doesn’t change
 its amount can change, but the amount of it in solution
doesn’t because it isn’t in solution
• solids and liquids are not included in the Keq expression
Heterogeneous Equilibria

CO2 
Kc 
2
CO
Kp 
PCO 2
2
PCO
The amount of C is
different, amounts of CO
and CO2 remain the
same. C has no effect on
the position of
equilibrium.
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• To find K measure the amounts of reactants and
products in a mixture at equilibrium
• may have different amounts of reactants and
products, but the value of the equilibrium constant
will always be the same
 as long as the temperature is kept constant
Initial and Equilibrium Concentrations for
H2(g) + I2(g) ⇌ 2HI(g) @ 445°C
Equilibrium
Initial
[H2]
0.50
0.0
0.50
1.0
[I2]
0.50
[HI] [H2]
0.0
[I2]
0.11 0.11
[HI]
0.78
Equilibrium
Constant
Keq 
[HI] 2
[H 2 ][I 2 ]
[0.78]2
 50
[0.11][0.11]
0.50 0.055 0.055 0.39
[0.39]2
 50
[0.055][0.055]
0.50 0.50 0.165 0.165 1.17
[1.17]2
 50
[0.165][0.165]
0.0
0.5
0.0
0.53 0.033 0.934
[0.934]2
 50
[0.53][0.033]
37
Calculating Equilibrium Constants from
Measured Equilibrium Concentrations
• Stoichiometry can be used to determine the equilibrium [reactants] and
[products] if you know initial concentrations and one equilibrium concentration
e.g. 2A(aq) + B(aq) ⇌ 4C(aq) with initial concentrations [A] = 1.00 M, [B] = 1.00 M,
and [C] = 0. You then measure the equilibrium concentration of C as [C] = 0.50
M.
initial molarity
[A]
[B]
[C]
1.00
1.00
0
change in concentration -½(0.50) -¼(0.50) +0.50
equilibrium molarity
0.75
0.88
0.50
Calculating Equilibrium
Concentrations
e.g. 2A(aq) + B(aq) ⇌ 4C(aq)
initial molarity
[A]
[B]
[C]
1.00
1.00
0
change in concentration -½(0.50) -¼(0.50) +0.50
equilibrium molarity
0.75
0.88
0.50
Referred to as ICE table (initial, change, equilibrium)
Keq = [C]4/[A]2[B] = 0.13
Ex 14.6  Find the value of Kc for the reaction
2 CH4(g) ⇌ C2H2(g) + 3 H2(g) at 1700°C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035
M
Construct an ICE table
[CH4] [C2H2] [H2]
for the reaction
initial
0.115 0.000 0.000
change
+0.035
for the substance
whose equilibrium
equilibrium
0.035
concentration is
known, calculate the
change in
concentration
Ex 14.6  Find the value of Kc for the reaction
2 CH4(g) ⇌ C2H2(g) + 3 H2(g) at 1700°C if the initial
[CH4] = 0.115 M and the equilibrium [C2H2] = 0.035
M
use the known change
[CH4] [C2H2] [H2]
to determine the
change in the other
initial
0.115 0.000 0.000
materials
-2(0.035) +0.035 +3(0.035)
change
add the change to the
initial concentration to
0.105
equilibrium
0.035
0.045
get the equilibrium
concentration in each
column
use the equilibrium
concentrations to
calculate Kc

C 2 H 2 H 2 
Kc 
2
CH 4 
3

0.0350.105
2 2


0
.
020
mol
L
2
0.045
3
The Reaction Quotient
• reaction mixture not at equilibrium;
how can we determine which
direction it will proceed?
• the answer is to compare the current
concentration ratios to the
equilibrium constant
reaction quotient, Q
for the gas phase reaction
aA + bB ⇌ cC + dD
the reaction quotient is:
c
d

C  D
Qc 
Aa  Bb
PC  PD
Qp 
a
b
PA  PB
c
d
The Reaction Quotient:
Predicting the Direction of Change
• if Q > K, the reaction will proceed fastest in the reverse direction
 Q must decrease, the [products] will decrease and [reactants] will
increase
• if Q < K, the reaction will proceed fastest in the forward direction
 Q must increase, the [products] will increase and [reactants] will
decrease
• if Q = K, the reaction is at equilibrium
 the [products] and [reactants] will not change
• if a reaction mixture contains just reactants, Q = 0, and the reaction
will proceed in the forward direction
• if a reaction mixture contains just products, Q = ∞, and the reaction will
proceed in the reverse direction
Q, K, and the Direction of Reaction
47
Ex 14.7 – For the reaction below, which direction will it
proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm & PICl = 0.355
atm
Given: for I2(g) + Cl2(g) ⇌ 2 ICl(g), Kp = 81.9
Find: direction reaction will proceed
Concept Plan:
PI2, PCl2, PICl
Q
Relationships: If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse
Solution:
I2(g) + Cl2(g) ⇌ 2 ICl(g)
Kp = 81.9

PICl
0.355
Qp 

PI 2  PCl 2 0.114   0.102 
2
2
Qp  10.8
since Q (10.8) < K (81.9), the reaction will proceed to the right
Ex 14.8 If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and Kc = 2.00 @
1000°C, find the [CO2]eq for the reaction given.
Given:
Sort: You’re given the
2 COF2 ⇌ CO2 + CF4
reaction and Kc. You’re also
[COF2]eq = 0.255 M, [CF4]eq = 0.118 M
given the [X]eq of all but one
Find: [CO2]eq
of the chemicals
Concept
Strategize: You can calculate
[CO2]
Plan: K, [COF2], [CF4]
the missing concentration by
using the equilibrium
CO2 CF4 
constant expression
Relationships:
Kc 
COF2 2
Solve: Solve the equilibrium
constant expression for the
unknown quantity by
substituting in the given
amounts
Check: Round to 1 sig fig
and substitute back in
Solution:
Check:
CO2 CF4 
COF2 2
2

COF2 
CO2   K c 
CF4 
2

0.255
 2.00 
 1.10 M
0.118
Kc 
Units & Magnitude OK
Finding Equilibrium Concentrations
When Given the Equilibrium Constant
and Initial Concentrations or Pressures
• first decide which direction the reaction will proceed
 compare Q to K
• define the changes of all materials in terms of x
 use the coefficient from the chemical equation for the coefficient of
x
 the x change is + for materials on the side the reaction is
proceeding toward
 the x change is  for materials on the side the reaction is
proceeding away from
• solve for x
 for 2nd order equations, take square roots of both sides or use the
quadratic formula
 may be able to simplify and approximate answer for very large or
small equilibrium constants
Ex 14.11  For the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) @
25°C, Kp = 81.9. If the initial partial pressures are all
0.100 atm, find the equilibrium concentrations
Construct an ICE
table for the reaction
determine the
direction the reaction
is proceeding
[I2]
0.100
initial
change
equilibrium
[Cl2]
0.100
[ICl]
0.100

PICl
0.100
Qp 

PI2  PCl 2 0.100  0.100
2
2
Qp  1
since Qp(1) < Kp(81.9), the reaction is proceeding forward
Ex 14.11  For the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) @
25°C, Kp = 81.9. If the initial partial pressures are all
0.100 atm, find the equilibrium concentrations
represent the change
in the partial
pressures in terms of
x
sum the columns to
find the equilibrium
concentrations in
terms of x
substitute into the
equilibrium constant
expression and solve
for x
[I2]
0.100
x
[Cl2]
0.100
x
[ICl]
0.100
+2x
initial
change
equilibrium 0.100x 0.100x 0.100+2x
2
PICl
Kp 
PI2  PCl 2
2
2


0.100  2 x 
0.100  2 x 
81.9 

0.100  x   0.100  x  0.100  x 2
Ex 14.11  For the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) @
25°C, Kp = 81.9. If the initial partial pressures are all
0.100 atm, find the equilibrium concentrations
substitute into the
equilibrium constant
expression and solve
for x
[I2]
0.100
x
[Cl2]
0.100
x
[ICl]
0.100
+2x
initial
change
equilibrium 0.100x 0.100x 0.100+2x

0.100  2 x 2 0.100  2 x 
81.9 

2
0.100  x 
0.100  x 
81.9 0.100  x   0.100  2 x 
81.9 0.100   81.9  x   0.100  2 x 
81.9 0.100   0.100  2 x  81.9  x 
81.9 0.100  0.100  2 x  81.9 x
0.805  11.05 x
0.0729  x
Ex 14.11  For the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) @
25°C, Kp = 81.9. If the initial partial pressures are all
0.100 atm, find the equilibrium concentrations
substitute x into the
equilibrium
concentration
definition and solve
[I2]
[Cl2]
0.100 0.100
x
x
-0.0729
-0.0729
[ICl]
0.100
+2x
2(-0.0729)
initial
change
0.027 0.100x
0.027 0.100+2x
0.246
equilibrium 0.100x
0.0729  x
PI2  0.100  x  0.100  0.0729  0.027 atm
PCl 2  0.100  x  0.100  0.0729  0.027 atm
PICl  0.100  2 x  0.100  20.0729  0.246 atm
Disturbing and Re-establishing
Equilibrium
• once a reaction is at equilibrium, the concentrations
•
•
of all the reactants and products remain the same
however if the conditions are changed, the
concentrations of all the chemicals will change until
equilibrium is re-established
the new concentrations will be different, but the
equilibrium constant will be the same
 unless you change the temperature
Le Châtelier’s Principle
• Le Châtelier's Principle:
if a system at equilibrium is disturbed, the position of
equilibrium will shift to minimize the disturbance
 disturbances all involve making the system open
 Concentration, temperature, volume, pressure
An Analogy: Population Changes
The result will be people
moving from Country B into
Country A faster than people
moving from Country A into
Country B. This will continue
until a new equilibrium
between the populations is
When the populations of Country A and Country B
established,
however
the
new
When
an influx ofthe
population
enters
Country
B the
are
in equilibrium,
emigration
rates
between
populations
have different
from
outside
Country
A, itwill
disturbs
the
two somewhere
states are equal
so the
populations
stay constant.
numbers
of people
the B.
old
equilibrium established between
Country
A and than
Country
ones.
The Effect of Concentration
Changes on Equilibrium
• Adding reactant :
aA + bB ⇌ cC + dD
System shifts in a direction to minimize the disturbance
• increases the amount of products until a new equilibrium
is found
 that has the same K

C  D
K eq 
a
b
A  B
c
d
The Effect of Concentration
Changes on Equilibrium
• Removing product:
aA + bB ⇌ cC + dD
System shifts in a direction to minimize the disturbance
• will increase the amounts of products and decrease the
amounts of reactants
 you can use this to drive a reaction to completion!

C  D
K eq 
a
b
A  B
c
d
The Effect of Adding a Gas to a
Gas Phase Reaction at Equilibrium
• adding a gaseous reactant increases its partial pressure,
•
causing the equilibrium to the right
 increasing its partial pressure increases its concentration
 does not increase the partial pressure of the other gases
in the mixture
adding an inert gas to the mixture has no effect on the
position of equilibrium
 does not effect the partial pressures of the gases in the
reaction
Tro, Chemistry: A Molecular Approach
86
The Effect of Concentration
Changes on Equilibrium
Describe the effect of adding more NO2(g) to
the following reaction:
N2O4(g) ⇌ NO2(g)
The Effect of Concentration Changes on
Equilibrium
Q=K
When NO2 is added, some of
it combines to make more
N2O4
Q>K
Q=K
The Effect of Concentration Changes on
Equilibrium
Describe the effect of adding more N2O4(g) to
the following reaction:
N2O4(g) ⇌ NO2(g)
The Effect of Concentration
Changes on Equilibrium
N2O4(g) ⇌ NO2(g)
When N2O4 is added,
reaction shifts to make
more NO2
Summarizing
If a chemical system is in equilibrium:
• Inc. [reactant] (Q < K) reaction shifts to right
• Inc. [product] (Q > K) reaction shifts to left
• Dec. [reactant] (Q > K) reaction shifts to left
• Dec. [product] (Q < K) reaction shifts to right
Effect of Volume Change
on Equilibrium
• decreasing the size of the container increases the concentration of
•
•
•
•
all the gases in the container
 increases their partial pressures
if their partial pressures increase, then the total pressure in the
container will increase
according to Le Châtelier’s Principle, the equilibrium should shift to
remove that pressure
the way the system reduces the pressure is to reduce the number of
gas molecules in the container
when the volume decreases, the equilibrium shifts to the side
with fewer gas molecules
The Effect of Volume Changes on
Equilibrium
Since
When
there
theare
pressure
more gas
is
decreased
molecules
by increasing
on the the
volume,
reactantsthe
side
position
of the of
equilibrium
reaction,shifts
when
toward
the the
side
pressure
with the
is increased
greater number
the
ofposition
molecules
of equilibrium
– the reactant
shifts towardside.
the products.
93
Disturbing Equilibrium: Reducing the
Volume
• for solids, liquids, or solutions, changing the size of the container has no effect
•
•
•
•
on the concentration, therefore no effect on the position of equilibrium
decreasing the container volume will increase the total pressure
 Boyle’s Law
 if the total pressure increases, the partial pressures of all the gases will
increase
 Dalton’s Law of Partial Pressures
decreasing the container volume increases the concentration of all gases
 same number of moles, but different number of liters, resulting in a
different molarity
since the total pressure increases, the position of equilibrium will shift to
decrease the pressure by removing gas molecules
 shift toward the side with fewer gas molecules
at the new equilibrium position, the partial pressures of gaseous reactants and
products will be such that the value of the equilibrium constant is the same
Summarizing
If a chemical system is in equilibrium:
• Dec. volume causes reaction to shift in direction that has fewer moles
of gas particles
• Inc. volume causes reaction to shift in direction that has the greater
number of moles of gas particles
• If equal moles of gas on both sides, no effect
• Adding an inert gas has no effect
The Effect of Temperature Changes on
Equilibrium Position
• exothermic reactions release energy and endothermic
•
reactions absorb energy
if we write Heat as a product in an exothermic reaction or
as a reactant in an endothermic reaction, it will help us use
Le Châtelier’s Principle to predict the effect of temperature
changes
 even though heat is not matter and not written in a
proper equation
The Effect of Temperature Changes
on Equilibrium for Exothermic
Reactions
• for an exothermic reaction, heat is a product
• increasing the temperature is like adding heat
• according to Le Châtelier’s Principle, the equilibrium will
•
shift away from the added heat
adding heat to an exothermic reaction will decrease
the [products] and increase the [reactants]
adding heat to an exothermic reaction will decrease
the value of K
how will decreasing the temperature affect the system?
aA + bB ⇌ cC + dD + Heat
c
d

C  D
Kc 
a
b
A  B
The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
• for an endothermic reaction, heat is a reactant
• increasing the temperature is like adding heat
• according to Le Châtelier’s Principle, the equilibrium will shift away
from the added heat
adding heat to an endothermic reaction will decrease the
[reactants] and increase the [products]
adding heat to an endothermic reaction will increase the value of
K
• how will decreasing the temperature affect the system?
Heat + aA + bB ⇌ cC + dD

C  D
Kc 
a
b
A  B
c
d
The Effect of Temperature
Changes on Equilibrium
Not Changing the Position of
Equilibrium - Catalysts
• catalysts provide an alternative, more efficient mechanism
• works for both forward and reverse reactions
• affects the rate of the forward and reverse reactions by the
•
same factor
therefore catalysts do not affect the position of
equilibrium
Practice - Le Châtelier’s
Principle
2 SO2(g) + O2(g) ⇌ 2 SO3(g)
H° = -198 kJ
How will each of the following changes affect the equilibrium
concentrations of each gas once equilibrium is re-established?








adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding the inert gas helium to the container
adding a catalyst to the mixture
Practice - Le Châtelier’s Principle
2 SO2(g) + O2(g) ⇌ 2 SO3(g)








adding more O2 to the container
condensing and removing SO3
compressing the gases
cooling the container
doubling the volume of the container
warming the mixture
adding helium to the container
adding a catalyst to the mixture
shift to SO3
shift to SO3
shift to SO3
shift to SO3
shift to SO2
shift to SO2
no effect
no effect