### PowerPoint

```Disturbing Equilibrium and
Non-equilibrium conditions
Le Châtelier’s Principle, the van’t Hoff
equation, the Reaction Quotient (Q)
Lesson Objectives
Know
-Le Châtelier’s principle outlines how equilibrium is affected by
changes in a system
-Equilibrium can be affected by changes in reactant/product amounts,
changes in volume or pressure, or changes in temperature
-temperature is the only factor that affects the equilibrium constant
Understand -when disturbed from equilibrium, a reaction will shift whichever way
(to products or to reactants) to relieve “stress” in the system
-the reaction quotient is used to determine which way a reaction needs
to shift to reach equilibrium
Do
-analyze changing conditions to determine which way a reaction needs
to shift to regain equilibrium
-solve the van’t Hoff equation for a missing variable, whether the
variable is a K value, DH, or a temperature.
-solve the Q (reaction quotient) equation for non-equilibrium conditions
-use Q to determine which way a reaction will shift to regain equilibrium
Disturbing Equilibrium
LE CHÂTELIER’S PRINCIPLE
Le Chatelier’s Principle

If a system at equilibrium is disturbed, it will
shift to partially counteract the change

Factors that can disturb equilibrium:



Adding or removing reactant or product
(physically or chemically)
Increasing or decreasing volume (which results in
a pressure change for gases)
Increasing or decreasing temperature (adding or
removing heat)
Adding or removing reactant or product


Add reactant or product: reaction shifts to use up
Remove reactant or product: reaction shifts to add
it back in


Substances can be removed by reacting them out –
making precipitate, water, gas, etc.
Only affects equilibrium if the added or removed
substance is:


In the equation (inert or irrelevant substances have no
affect on equilibrium)
Not a solid or liquid (solids and liquids don’t affect
equilibrium because their concentrations don’t change)
Example 1: changing amounts
Iodine chloride decomposes at high
temperatures to iodine and chlorine:
2 ICl(g)  I2(g) + Cl2(g)
If some iodine condenses, which direction
would this reaction shift from equilibrium?
Example 2: changing amounts
Sulfur dioxide and oxygen can combine to
make sulfur trioxide
2 SO2(g) + O2(g)  2SO3(g)
If more oxygen is introduced into this system
while it is at equilibrium, which way will the
reaction shift?
2H2(g) + CO(g)  CH3OH(g)
A.
B.
C.
DH = -90.2 kJ
If a mixture of H2, CO and CH3OH is at
equilibrium, will the concentration of CO
increase, decrease, or remain the same as
equilibrium is reestablished after more H2 is
[CO] will increase
[CO] will decrease
[CO] will stay the same
2H2(g) + CO(g)  CH3OH(g)
A.
B.
C.
DH = -90.2 kJ
If a mixture of H2, CO and CH3OH is at
equilibrium, will the concentration of CO
increase, decrease, or remain the same as
equilibrium is reestablished after CH3OH is
[CO] will increase
[CO] will decrease
[CO] will stay the same
Changing the volume

Increase volume (decrease pressure of gases)


The system will shift to side with more moles of
gas to try to raise pressure
Decrease volume (increase pressure of gases)

The system will shift to the side with fewer moles
of gas to try to lower pressure
Example 3: changing volume
Sulfur dioxide and oxygen can combine to
make sulfur trioxide
2 SO2(g) + O2(g)  2SO3(g)
If the container holding this reaction is
expanded (volume increases), which way
will the reaction shift from equilibrium?
2H2(g) + CO(g)  CH3OH(g)
A.
B.
C.
DH = -90.2 kJ
If a mixture of H2, CO and CH3OH is at
equilibrium, will the concentration of CO
increase, decrease, or remain the same as
equilibrium is reestablished after the volume
of the system is decreased.
[CO] will increase
[CO] will decrease
[CO] will stay the same
Changing the temperature

Increasing temp. is like adding (stressing) heat


Decreasing temp. is like removing heat


System will shift in the endothermic direction
System will shift in the exothermic direction
If heat was shown in the equation, the system
shifts to counteract the addition or removal of
heat (just like reactants or products)


Endothermic: ΔH = positive, written in reactants
Exothermic: ΔH = negative, written in products
Example 4: changing temperatures
Sulfur dioxide and oxygen can combine to
make sulfur trioxide
2 SO2(g) + O2(g)  2SO3(g)
This reaction is highly exothermic, releasing
nearly 200 kJ of energy per mole of product
created. If the temperature of this reaction
was increased, which way would the reaction
shift from equilibrium?
2H2(g) + CO(g)  CH3OH(g)
A.
B.
C.
DH = -90.2 kJ
If a mixture of H2, CO and CH3OH is at
equilibrium, will the concentration of CO
increase, decrease, or remain the same as
equilibrium is reestablished after the
temperature of the system is increased.
[CO] will increase
[CO] will decrease
[CO] will stay the same
2H2(g) + CO(g)  CH3OH(g)
A.
B.
C.
DH = -90.2 kJ
If a mixture of H2, CO and CH3OH is at
equilibrium, will the concentration of CO
increase, decrease, or remain the same as
equilibrium is reestablished after more
[CO] will increase
[CO] will decrease
[CO] will stay the same
Which of the following is NOT true
according to LeChâtelier’s Principle?
A.
B.
C.
D.
E.
A system at equilibrium that has been disturbed
will shift away from the stressor
Removing a product from a system will cause the
system to shift back to reactants
Changing amounts of solids or liquids in a
reaction does not affect a system’s equilibrium
An increase in a gaseous system’s volume will
shift the reaction toward the side with more
moles of gas
A decrease in the temperature of a system at
equilibrium will cause the reaction to shift in the
exothermic direction.
Example 5: Calculating new values
a)
b)
Iodine chloride decomposes at high temperatures:
2 ICl(g)  I2(g) + Cl2(g)
At equilibrium the partial pressures are (in atm)
ICl: 0.43, I2: 0.16 and Cl2: 0.27
Calculate K
If enough iodine condenses to drop its partial
pressure to 0.10 atm, what is the partial pressure of
iodine gas when the equilibrium is reestablished?
CO(g) + Cl2(g) ⇄ COCl2 (g) + heat
Non-equilibrium conditions
THE REACTION QUOTIENT (Q)
Q: the reaction quotient

Q is just like K, except it is for ANY point in
time, not just at equilibrium. You will find Q
(like finding K) and then compare Q to K
Q
=
[products]
[reactants]
COMPARING Q TO K

Q<K
Q =
[products]
[REACTANTS]
reaction must shift to right to produce more products

Q>K
Q = [PRODUCTS]
[reactants]
reaction must shift to left to produce more reactants

Q=K
reaction is AT EQUILIBRIUM
Example 6: Calculating and Using Q
a)
For the reaction:
2NO2 (g)  2NO (g) + O2 (g)
K at a certain temperature is 0.50. Predict the
direction the system will move if one starts with:
PO2 = PNO = PNO2 = 0.10 atm
b)
PNO2 = 0.848 atm, PO2 = 0.0166atm
c)
PNO2 = 0.20atm, PO2 = 0.010atm, PNO = 0.040atm
Effect of Temperature on K
THE VAN’T HOFF EQUATION
Effect of Temperature on K
Of all types of ways to disturb equilibrium,
volume/pressure, changing temperature),
temperature is the only factor to affect the
equilibrium constant.
 Use van’t Hoff equation to find new K , or if
given new K, to find ΔH
Look familiar?
 ln K2 = ΔH 1 – 1
ln k2 = Ea 1 - 1
K1 R T1 T2
k1
R T1
T2
R = 8.31 J/mol·K

Example 7: van’t Hoff equation
For the system
H2(g) + I2(g)  2HI(g)
ΔH = -9.4 kJ and K= 62.5 at 800 K.
What is the equilibrium constant at 333°C?
```