Chemical Equilibrium

1
Chemical Equilibrium
Chapter 15
2
Dynamic equilibrium a state when two opposing
processes occur at equal rates.
?
3

• Chemical Equilibrium
• occurs when opposing reactions are proceeding at
  equal rates and the concentrations of products
  and reactants no longer change with time.

Ratef Rater
4
Chemical Equilibrium

• For this to occur, the system must be closed, no
  products or reactants may escape the system.

5
For Reaction A?B

• Forward Reaction
• A?B Rate kfA
• kf forward rate constant
• Reverse reaction
• B?A Rate krB
• kr reverse rate constant

6

• As forward reaction occurs,
• A decreases and forward rate slows.
• B increases and the rate of reverse reaction
  increases.
• Eventually the forward and reverse reactions
  reach the same rate. (Chemical Equilibrium)

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kfA krB

• Once equilibrium is established, the
  concentrations of A and B do not change.

8
Writing equilibrium expressions
Keq products reactants NOTE
at equilibrium
Note eq is the general subscript for an
equilibrium constant. Your textbook uses Keq
throughout. Please look at the equilibrium
constants given on the AP exam. These are the
same as Keq, but are more specific to a type of
reaction (examples gaseous reaction or acid base
reaction).
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Two common ways of describing equilibrium
Kc Concentration of substances in the
reaction are known.
Kp Units of partial pressure are used instead
of concentration.
Kp is used when the reactants and products are
in the gaseous state. Solids and liquids are
left out of equilibrium expressions because their
concentrations do not change during chemical
reactions.
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Law of Mass Action Expresses the relationship
between concentrations of reactants and products
at equilibrium.
Example aA bB ? cC dD
Kc CcDd AaBb
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Kc CcDd AaBb

• The expression above is known as the
  equilibrium-constant expression.
• Kc is the equilibrium constant.
• Subscript c indicates concentration in molarity.

12
Write the equilibrium constant expressions for
the following reaction in the forward direction.
2SO2 (g) O2 (g) ? 2SO3(g)
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Example H2(g) I2(g) ? 2HI(g) The
equilibrium constant for this reaction at 400oC
is 64. What is the equilibrium constant value
for the reverse reaction? Kc fwd
HI2 I2H2 64
x2 x 8
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H2(g) I2(g) ? 2HI(g) Kc rev
H2I2 HI2 Kc
rev 11 x2 Kc rev 1
x2
Reverse
.016
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The Haber Process

• N2(g) 3H2(g) ? 2NH3(g)
• This process for making ammonia is done at high
  temperature and pressure.
• When complete all 3 components are present in the
  closed tank.

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• Regardless of the starting concentrations, at
  equilibrium the relative concentrations of the 3
  gases are the same.

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Equilibrium constant for Haber Process

• The equilibrium-constant depends only on the
  stoichiometry of the reaction, not on its
  mechanism (process by which it occurs).

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• The following equilibrium process has been
  studied at 230ºC
• 2NO(g) O2(g) ?2NO2(g)
• In one experiment the concentration of the
  reacting species at equilibrium are found to be
• NO0.0542M O20.127M NO215.5M.
• Calculate Kc of the reaction at this temperature.

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6.44 x 105Kc has no units
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Kc in terms of pressure

• Example
• aA bB ? cC dD

PA is partial pressure of A in atm, PB is the
partial pressure of B, etc.
21

• Values for Kc and Kp are usually different.
• It is possible to calculate one from the other.

?n is the change in the number of moles of
gas. ?n nproducts - nreactants
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N2O4(g) ?2NO2(g)

• ?n 2-1 1
• For the above reaction, KpKc(RT)

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• For the equilibrium 2SO3(g) ? 2SO2(g) O2(g) at
  temperature 1000K, Kc has a value 4.08 x 10-3.
• Calculate the value for Kp.

Kp 4.08 x 10-3 (.0821 x 1000)1
0.335
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A 31 starting mixture of hydrogen, H2, and
Nitrogen, N2 comes to equilibrium at 500 C. The
mixture at equilibrium is 3.506 NH3, 96.143 N2,
and 0.3506 H2 by volume. The total pressure in
the reaction vessel was 50.0 atm. What is the
value of Kp and Kc for the reaction?
N2 3H2 2NH3
Hint Use Daltons Law of partial pressures to
determine the partial pressure of each component
in the reaction. PA XA PT Since we know we have
100, Partial Pressure ( of substance /100 X
total pressure)
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P NH3 0.03506 x 50.0 atm 1.75 atm P N2
0.96143 x 50.0 atm 48.1 atm P H2
0.003506 x 50.0 atm 0.175 atm
(P NH3)2
(1.75)2
Kp

11.9
(P N2)
(P H2)3
(48.1)
(0.175)3
?n
1 RT
(
)
Kc Kp
-2
(
1 0.0821 x 773
)
4.79 x 104
Kc Kp
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Magnitude of Equilibrium constants

• When Kc is very large the numerator is much
  larger than the denominator.
• The equilibrium lies to the right, or to the
  product side of the equation.

K products reactants
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• When the equilibrium constant is very small the
  equilibrium lies to the left, or the reactant
  side.
• Kgtgt1 Products favored.
• Kltlt1 Reactants favored.

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Direction of equation and K

• Equilibrium can be approached from either
  direction.
• The equilibrium constant for an equation written
  in one direction is the reciprocal of the
  constant for the other direction.

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Direction of equation and K

• N2O4(g) ?2NO2(g)
• For forward Rx

For reverse Rx
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Heterogenous Equilibria

• Substances that are in equilibria and are in
  different phases.
• CaCO3(s) ?CaO(s) CO2(gas)
• Equilibrium constant

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• The concentration of a pure solid or liquid is a
  constant.

• If a pure solid or pure liquid are involved in a
  heterogeneous equilibrium, its concentration is
  not included in the equilibrium-constant
  expression for the Rx.

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• This equation shows that at a given temperature,
  an equilibrium between CaO,CaCO3, CO2 will
  always lead to the same concentration of CO2.
• Pure solids or pure liquids must be present and
  participate for equilibrium to be established.

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• Write the equilibrium expression Kc and Kp for
  the following reaction
• 3Fe(s) 4H2O(g) ?Fe3O4(s) 4H2(g)

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Calculating Equilibrium Constant When all
Equilibrium Concentrations are not known.
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Calculating Equilibrium Constants

• Equilibrium concentrations are often unknown.
• If one equilibrium concentration is known, we can
  use stoichiometry to find the equilibrium
  concentration of the other species.

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Equilibrium Concentrations
Initial concentration
Change in concentration
Equilibrium concentration
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Handout Procedure for calculating equilibrium
concentrations.
H2(g) I2(g) 2HI(g)
H2 I2 2HI
Start
?
Finish
1) First, we design our table so there is a
column for each component in the equation, and
rows for start, ?, and finish.
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H2(g) I2(g) 2HI(g)
H2 I2 2HI
Start 2.0 2.0 0
? 3.50
Finish 3.50

• Next, we fill in what we know.
• Since HI is a product, it is 0 at the start of
  the reaction.
• We also know that at equilibrium, 3.50 mol of HI
  exist.

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H2(g) I2(g) 2HI(g)
H2 I2 2HI
Start 2.0 2.0 0
? 3.50
Finish 3.50
3) We now use stoichiometry to fill in the rest
of the table.
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H2(g) I2(g) 2HI(g)
H2 I2 2HI
Start 2.0 2.0 0
? -1.75 3.50
Finish 3.50
3.50 mol HI 1 mol H2
1.75 mol H2
2 mol HI
It is negative because it is being used up in
the reaction.
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H2(g) I2(g) 2HI(g)
H2 I2 2HI
Start 2.0 2.0 0
? -1.75 -1.75 3.50
Finish 0.25 0.25 3.50
3.50 mol HI 1 mol I2
1.75 mol I2
2 mol HI
Now we can fill in the finish column.
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H2(g) I2(g) 2HI(g)
H2 I2 2HI
Start 2.0 2.0 0
? -1.75 -1.75 3.50
Finish 0.25 0.25 3.50
The quantities in the finished row are used to
calculate Kc.
(HI)2
(3.50)2

Kc
196
(H2)
(I2)
(0.25)
(0.25)
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You try this one! (On back of handout) Nitric
oxide gas, NO, and oxygen gas, O2, react to form
the poisonous gas nitrogen dioxide, NO2, in the
reaction shown below
2NO(g) O2(g) 2NO2(g) 10.0 moles
of NO and 6.00 moles of O2 are placed into an
evacuated 1.00 L vessel, where they begin to
react. At equilibrium, there are 8.80 moles of
NO2 present. Calculate the value of Kc, assuming
that the temperature remains constant throughout
the reaction.
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10.0 moles of NO and 6.00 moles of O2 are placed
into an evacuated 1.00 L vessel, where they begin
to react. At equilibrium, there are 8.80 moles
of NO2 present. Calculate the value of Kc,
assuming that the temperature remains constant
throughout the reaction.
2NO(g) O2(g) 2NO2(g)




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When 3.0 mol of I2 and 4.0 mol of Br2 are placed
in a 2.0 L reactor at 150oC, the following
reaction occurs until equilibrium is
reached I2(g) Br2(g) ? 2IBr(g) Chemical
analysis then shows that the reactor contains 3.2
mol of IBr. What is the value of the equilibrium
constant Kc for the reaction?
Consider M since we have a 2.0 L vessel.
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3.0 mol of I2 and 4.0 mol of Br2 are placed in a
2.0 L reactor at 150oC. Chemical analysis then
shows that the reactor contains 3.2 mol of IBr.
What is the value of the equilibrium constant Kc
for the reaction?
I2(g) Br2(g) ? 2IBr(g)




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If you are dealing with the pressures of gases at
equilibrium, you will be given moles,
temperature, and volume. is the key to
calculating the pressure of each gas at
equilibrium. Stoichiometric equivalents may also
be used.
PV nRT
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Example Handout
2SO3(g) 2SO2(g) O2(g)
Initially, 0.01219 moles of SO3 gas in a 2.0 L
vessel are heated to a temperature of 1000K
where it reaches equilibrium. At equilibrium the
vessel is found to contain 0.00487 moles of SO3
gas. A) Calculate the equilibrium partial
pressures of SO2(g) and O2(g). B) Calculate Keq.
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1) We must first calculate the pressures from
the information given in the problem.
PV nRT P nRT/V P .01219(0.8206)(1000)
/ (2.0) 0.500 atm This is the initial
pressure of SO3 gas.
2SO3(g) 2SO2(g)
O2(g)
2SO3 2SO2 O2
initial .500
?
finish
50
PV nRT P nRT/V P .00487(0.8206)(1000)
/ (2.0) 0.200 atm This is the pressure of
SO3 gas at equilibrium.
2SO3(g) 2SO2(g)
O2(g)
2SO3 2SO2 O2
initial .500
?
finish .200
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2) Next, we must place the information we know
into the table.
2SO3(g) 2SO2(g)
O2(g)
2SO3 2SO2 O2
initial .500
? -.300
finish .200
0
0
52

• Now we use stoichiometric relationships to fill
  in the rest.
• Notice the 22 ratio of SO3 and SO2.
• Notice the 21 ratio of SO3 and O2.

2SO3(g) 2SO2(g)
O2(g)
2SO3 2SO2 O2
initial .500
? -.300 .300 .150
finish .200
0
0
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3) Now we add the columns. The finish row gives
us the partial pressures of each gas at
equilibrium.
2SO3(g) 2SO2(g)
O2(g)
2SO3 2SO2 O2
initial .500
? -.300 .300 .150
finish .200 .300 .150
0
0
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4) Write equilibrium constant expression and
solve for Keq.
2SO3(g) 2SO2(g)
O2(g)
2SO3 2SO2 O2
initial .500
? -.300 .300 .150
finish .200 .300 .150
0
0
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Remember that when gases are involved, we need to
use PVnRT to find the pressures at equilibrium
before we can input the values into the
equilibrium expression. In some cases, a table
wont be needed if the data given is related to
the gases at equilibrium.
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Applications of Equilibrium Constants

• Magnitude of K determines whether reaction will
  proceed.
• Using K we can predict
• Direction of reaction
• Calculate concentrations of reactants and
  products at equilibrium.

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Reaction Quotient (Q)

• This is used when you are given information about
  reactants and products when they have not yet
  reached equilibrium.
• Allows us to determine direction reaction WILL
  go.
• --may be given concentrations before reaction
  begins.
• --may be given concentrations determined
  during a reaction.
• Calculate Q as you do K.
• N2(g) 3H2(g) ? 2NH3(g)
• Compare Q and K as follows.

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Reaction Quotient (Q) and reaction direction

• QK when system is at equilibrium
• QgtK reaction goes to left(reactants)
• QltK reaction goes to right(products)

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Direction of a Reaction
Step 1 Calculate the reaction quotient, Qc
Step 2 Compare Qc to Kc
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case 1 Qc gt Kc
Too much product!
Large numerator.
At equilibrium
case 2 Qc Kc
Too much reactants!
case 3 Qc lt Kc
Large denominator.
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N2(g) 3H2(g) ? 2 NH3(g)

• At the start of a reaction, there are 0.249 mol
  N2, 3.21 x 10-2 mol H2, and 6.42 x 10-4 mol NH3
  in a 3.50 L reaction vessel at 200º C. If the
  equilibrium constant (Kc) for the reaction is
  0.65 at this temperature, decide whether the
  system is at equilibrium. If it is not, predict
  which way the net reaction will proceed.

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Initial Concentrations
63
N2(g) 3H2(g) ? 2 NH2(g)
Kc 0.65
Qc is smaller than Kc. Not at equilibrium too
much reactants. Net Rx goes left to right.
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Handout Problem N2(g) 3H2(g) 2NH3(g)
In the reaction above, the value of Kc at 500C
is 6.0 x 10-2. At some point during the
reaction, the concentrations of each material
were measured. At this point, the concentrations
of each substance were N2 1.0 x 10-5 M, H2
1.5 x 10-3 M, and NH3 1.5 x 10-3 M.
Determine the direction that the reaction was
most likely to proceed when the measurements were
taken.
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Extent of reaction
Kc gtgt 1 (any gt 10)
Forward rxn goes to near completion
Kc ltlt 1 (any lt 0.1)
Reverse rxn goes to near completion
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Calculation of Equilibrium Concentration

• Often only Kc and beginning concentrations are
  known.
• Must be able to solve for equilibrium
  concentrations.

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H2(g) I2(g) ? 2HI(g)
Handout

• A mixture of 0.500 mol H2 and 0.500 mol I2 was
  placed in a 1.00L stainless steel flask at 430ºC.
  Calculate the concentrations of H2, I2, and HI
  at equilibrium. The equilibrium constant Kc for
  the reaction is 54.3 at this temperature.

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H2(g) I2(g) ?2HI(g)
Initial Molarity(M) 0.500 0.500 0.00
Change(M) -x -x 2x
Equlibrium (M) 0.500-x 0.500-x 2x
Start by putting in what we know.
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H2(g) I2(g) ?2HI(g)
H2 I2 2HI
Start (M) 0.500 M 0.500 M 0
? (M) -x -x 2x
Finish (M) 0.500-x 0.500-x 2x
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Must now plug x back into finish row to
calculate actual concentrations.
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At equilibrium theconcentrations are

• H2 (0.500-0.393) 0.107M
• I2 (0.500-0.393) 0.107M
• HI 2(0.393) 0.786M
• Answers can be checked by calculating Kc using
  the equilibrium concentrations.

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H2(g) I2(g) ?2HI(g)
Shouldnt need this on exam.

• For the same reaction and temperature, suppose
  the initial concentrations of HI, H2, and I2 are
  0.0224M, 0.00623M, 0.00414M respectively.
  Calculate the concentrations at equilibrium.

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H2(g) I2(g) ? 2HI(g)
Initial Molarity(M) 0.00623 0.00414 0.0224
Change(M) -x -x 2x
Equlibrium (M) 0.00623-x 0.00414-x 0.02242x
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(No Transcript)
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• This is a quadratic equation of the form
• ax2 bx c 0
• The solution is

a 50.3 b -0.654 c 8.98x10-4
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At equilibrium theconcentrations are

• H2 (0.00623-0.00156) 0.00467M
• I2 (0.00414-0.00156) 0.00258M
• HI (0.02242(0.00156) 0.0255M

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Le Chatelier's Principle

• If a stress is applied to a system in a state of
  equilibrium, the equilibrium will shift so to
  relieve that stress.
• Stress could be a change in
• temperature
• pressure
• concentration of one component.

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? In Reactant or Product Concentration

• Adding a substance to a system at equilibrium
  will shift the reaction by consuming some of the
  added item.
• Removing a substance from a system at equilibrium
  will shift the reaction to form more of the
  removed substance.

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Effects of Volume Pressure

• Reducing the volume of a gas equilibrium system
  (at constant T) will cause a shift in the
  direction that reduces the number of moles of gas
  in the system.
• N2O4 ? 2NO2
• Reducing volume favors which direction?
• Shift to left.

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H2(g) I2(g) ? 2HI(g)

• Changing pressure or volume of this gaseous state
  of equilibrium causes what change?
• No Change.

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Effect of Temperature Change

• Equilibrium constant values (Keq) always change
  with temperature change, but stays the same when
  pressure or concentration is changed.
• Enthalpy of the reaction plays a major roll in
  how heat effects the equilibrium.

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When temperature is increased the reaction is
shifted in the direction that absorbs heat.

• Endothermic ?? (shift to rt)
• Reactants heat ? Products
• Exothermic -?? (shift to left)
• Reactants ? Products heat

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Example The Haber process. ?Hrxn - 48.5
KJ/mol N2(g) 3H2(g) ? 2NH3(g)
Heat
Result
Stress
? more product
?N2
?NH3
? less product
? more product
?P
? less product
?T
hmmm, the Haber process requires high pressure
and high temperature yet, one leads to more
product, the other leads to less productwe will
look at this on a website at the end of the PP.
84
Consider the following equilibrium systems, and
predict the direction of the net reaction in each
case as a result of ? pressure (?volume) on the
system at constant temperature.
2PbS(s) 3 O2(g) ?2PbO(s) 2SO2(g)
Shift to right.
PCl5(g) ?PCl3(g)Cl2(g)
Shift to left.
H2(g) I2(g) ? 2HI(g)
No influence on position of equilibrium.
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Consider the following equilibrium process

• N2F4(g) ?2 NF2(g) ??º 38.5kJ
• Predict the changes in the equilibrium if
• The reacting mixture is heated at constant
  pressure.
• Shift to right.
• NF2 gas is removed at constant T and Pressure.
• Shift to right.

86
Effects of Catalysts

• Catalysts lower the energy barrier between
  reactant and products.
• Ea for both forward and reverse are lowered to
  the same extent.
• Catalysts increase the rate at which equilibrium
  is reached.
• Does NOT change the composition of the
  equilibrium mixture.

87
3O2(g)? 2O3(g) ??º284kJ

• When the above reaction is at equilibrium
• What is the effect of ? pressure
• by ? volume?
• ? O2?
• ? Temperature?
• Adding catalyst?

Shift to right. Shift to right.
Shift to left. No
change, catalysts only help increase
the rate at which equilibrium is reached.